∫ e5x+x2 (−512+2560x+1024x2)+e5x+x2 (128x−640x2−256x3) log(1+4 log(e5x+x2+4x _____________x) _____________________________ log (e5x+x2+4x _____________x) )+((−128e5x+x2 x−512x2) log(e5x+x2 +4x x )+(−512e5x+x2 x−2048x2) log2(e5x+x2 +4x x )) log2(1+4 log(e5x+x2+4x _____________x) _____________________________ log (e5x+x2+4x _____________x) )+((32e5x+x2 x2+128x3) log(e5x+x2 +4x x )+(128e5x+x2 x2+512x3) log2(e5x+x2 +4x x )) log3(1+4 log(e5x+x2+4x _____________x) _____________________________ log (e5x+x2+4x _____________x) ) ___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________ ((e5x+x2x+4x2) log(e5x+x2+4x x )+(4e5x+x2x+16x2) log2(e5x+x2+4x x )) log3(1+4 log(e5x+x2+4x _____________x) _____________________________ log (e5x+x2+4x ____________

3.40.60 \(\int \frac {e^{5 x+x^2} (-512+2560 x+1024 x^2)+e^{5 x+x^2} (128 x-640 x^2-256 x^3) \log (\frac {1+4 \log (\frac {e^{5 x+x^2}+4 x}{x})}{\log (\frac {e^{5 x+x^2}+4 x}{x})})+((-128 e^{5 x+x^2} x-512 x^2) \log (\frac {e^{5 x+x^2}+4 x}{x})+(-512 e^{5 x+x^2} x-2048 x^2) \log ^2(\frac {e^{5 x+x^2}+4 x}{x})) \log ^2(\frac {1+4 \log (\frac {e^{5 x+x^2}+4 x}{x})}{\log (\frac {e^{5 x+x^2}+4 x}{x})})+((32 e^{5 x+x^2} x^2+128 x^3) \log (\frac {e^{5 x+x^2}+4 x}{x})+(128 e^{5 x+x^2} x^2+512 x^3) \log ^2(\frac {e^{5 x+x^2}+4 x}{x})) \log ^3(\frac {1+4 \log (\frac {e^{5 x+x^2}+4 x}{x})}{\log (\frac {e^{5 x+x^2}+4 x}{x})})}{((e^{5 x+x^2} x+4 x^2) \log (\frac {e^{5 x+x^2}+4 x}{x})+(4 e^{5 x+x^2} x+16 x^2) \log ^2(\frac {e^{5 x+x^2}+4 x}{x})) \log ^3(\frac {1+4 \log (\frac {e^{5 x+x^2}+4 x}{x})}{\log (\frac {e^{5 x+x^2}+4 x}{x})})} \, dx\)

Optimal. Leaf size=35 \[ 2+16 \left (-x+\frac {4}{\log \left (4+\frac {1}{\log \left (\frac {e^{x (5+x)}+4 x}{x}\right )}\right )}\right )^2 \]

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Rubi [F] time = 37.00, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {e^{5 x+x^2} \left (-512+2560 x+1024 x^2\right )+e^{5 x+x^2} \left (128 x-640 x^2-256 x^3\right ) \log \left (\frac {1+4 \log \left (\frac {e^{5 x+x^2}+4 x}{x}\right )}{\log \left (\frac {e^{5 x+x^2}+4 x}{x}\right )}\right )+\left (\left (-128 e^{5 x+x^2} x-512 x^2\right ) \log \left (\frac {e^{5 x+x^2}+4 x}{x}\right )+\left (-512 e^{5 x+x^2} x-2048 x^2\right ) \log ^2\left (\frac {e^{5 x+x^2}+4 x}{x}\right )\right ) \log ^2\left (\frac {1+4 \log \left (\frac {e^{5 x+x^2}+4 x}{x}\right )}{\log \left (\frac {e^{5 x+x^2}+4 x}{x}\right )}\right )+\left (\left (32 e^{5 x+x^2} x^2+128 x^3\right ) \log \left (\frac {e^{5 x+x^2}+4 x}{x}\right )+\left (128 e^{5 x+x^2} x^2+512 x^3\right ) \log ^2\left (\frac {e^{5 x+x^2}+4 x}{x}\right )\right ) \log ^3\left (\frac {1+4 \log \left (\frac {e^{5 x+x^2}+4 x}{x}\right )}{\log \left (\frac {e^{5 x+x^2}+4 x}{x}\right )}\right )}{\left (\left (e^{5 x+x^2} x+4 x^2\right ) \log \left (\frac {e^{5 x+x^2}+4 x}{x}\right )+\left (4 e^{5 x+x^2} x+16 x^2\right ) \log ^2\left (\frac {e^{5 x+x^2}+4 x}{x}\right )\right ) \log ^3\left (\frac {1+4 \log \left (\frac {e^{5 x+x^2}+4 x}{x}\right )}{\log \left (\frac {e^{5 x+x^2}+4 x}{x}\right )}\right )} \, dx \end {gather*}

Veriﬁcation is not applicable to the result.

[In]

Int[(E^(5*x + x^2)*(-512 + 2560*x + 1024*x^2) + E^(5*x + x^2)*(128*x - 640*x^2 - 256*x^3)*Log[(1 + 4*Log[(E^(5

*x + x^2) + 4*x)/x])/Log[(E^(5*x + x^2) + 4*x)/x]] + ((-128*E^(5*x + x^2)*x - 512*x^2)*Log[(E^(5*x + x^2) + 4*

x)/x] + (-512*E^(5*x + x^2)*x - 2048*x^2)*Log[(E^(5*x + x^2) + 4*x)/x]^2)*Log[(1 + 4*Log[(E^(5*x + x^2) + 4*x)

/x])/Log[(E^(5*x + x^2) + 4*x)/x]]^2 + ((32*E^(5*x + x^2)*x^2 + 128*x^3)*Log[(E^(5*x + x^2) + 4*x)/x] + (128*E

^(5*x + x^2)*x^2 + 512*x^3)*Log[(E^(5*x + x^2) + 4*x)/x]^2)*Log[(1 + 4*Log[(E^(5*x + x^2) + 4*x)/x])/Log[(E^(5

*x + x^2) + 4*x)/x]]^3)/(((E^(5*x + x^2)*x + 4*x^2)*Log[(E^(5*x + x^2) + 4*x)/x] + (4*E^(5*x + x^2)*x + 16*x^2

)*Log[(E^(5*x + x^2) + 4*x)/x]^2)*Log[(1 + 4*Log[(E^(5*x + x^2) + 4*x)/x])/Log[(E^(5*x + x^2) + 4*x)/x]]^3),x]

[Out]

16*x^2 + 2560*Defer[Int][1/(Log[4 + E^(x*(5 + x))/x]*(1 + 4*Log[4 + E^(x*(5 + x))/x])*Log[4 + Log[4 + E^(x*(5

+ x))/x]^(-1)]^3), x] - 512*Defer[Int][1/(x*Log[4 + E^(x*(5 + x))/x]*(1 + 4*Log[4 + E^(x*(5 + x))/x])*Log[4 +

Log[4 + E^(x*(5 + x))/x]^(-1)]^3), x] + 1024*Defer[Int][x/(Log[4 + E^(x*(5 + x))/x]*(1 + 4*Log[4 + E^(x*(5 + x

))/x])*Log[4 + Log[4 + E^(x*(5 + x))/x]^(-1)]^3), x] + 2048*Defer[Int][1/((E^(x*(5 + x)) + 4*x)*Log[4 + E^(x*(

5 + x))/x]*(1 + 4*Log[4 + E^(x*(5 + x))/x])*Log[4 + Log[4 + E^(x*(5 + x))/x]^(-1)]^3), x] - 10240*Defer[Int][x

/((E^(x*(5 + x)) + 4*x)*Log[4 + E^(x*(5 + x))/x]*(1 + 4*Log[4 + E^(x*(5 + x))/x])*Log[4 + Log[4 + E^(x*(5 + x)

)/x]^(-1)]^3), x] - 4096*Defer[Int][x^2/((E^(x*(5 + x)) + 4*x)*Log[4 + E^(x*(5 + x))/x]*(1 + 4*Log[4 + E^(x*(5

+ x))/x])*Log[4 + Log[4 + E^(x*(5 + x))/x]^(-1)]^3), x] + 128*Defer[Int][1/(Log[4 + E^(x*(5 + x))/x]*(1 + 4*L

og[4 + E^(x*(5 + x))/x])*Log[4 + Log[4 + E^(x*(5 + x))/x]^(-1)]^2), x] - 640*Defer[Int][x/(Log[4 + E^(x*(5 + x

))/x]*(1 + 4*Log[4 + E^(x*(5 + x))/x])*Log[4 + Log[4 + E^(x*(5 + x))/x]^(-1)]^2), x] - 256*Defer[Int][x^2/(Log

[4 + E^(x*(5 + x))/x]*(1 + 4*Log[4 + E^(x*(5 + x))/x])*Log[4 + Log[4 + E^(x*(5 + x))/x]^(-1)]^2), x] - 512*Def

er[Int][x/((E^(x*(5 + x)) + 4*x)*Log[4 + E^(x*(5 + x))/x]*(1 + 4*Log[4 + E^(x*(5 + x))/x])*Log[4 + Log[4 + E^(

x*(5 + x))/x]^(-1)]^2), x] + 2560*Defer[Int][x^2/((E^(x*(5 + x)) + 4*x)*Log[4 + E^(x*(5 + x))/x]*(1 + 4*Log[4

+ E^(x*(5 + x))/x])*Log[4 + Log[4 + E^(x*(5 + x))/x]^(-1)]^2), x] + 1024*Defer[Int][x^3/((E^(x*(5 + x)) + 4*x)

*Log[4 + E^(x*(5 + x))/x]*(1 + 4*Log[4 + E^(x*(5 + x))/x])*Log[4 + Log[4 + E^(x*(5 + x))/x]^(-1)]^2), x] - 128

*Defer[Int][Log[4 + Log[4 + E^(x*(5 + x))/x]^(-1)]^(-1), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {32 \left (4-x \log \left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )\right ) \left (4 e^{x (5+x)} \left (-1+5 x+2 x^2\right )-x \left (e^{x (5+x)}+4 x\right ) \log \left (4+\frac {e^{x (5+x)}}{x}\right ) \log ^2\left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )-4 x \left (e^{x (5+x)}+4 x\right ) \log ^2\left (4+\frac {e^{x (5+x)}}{x}\right ) \log ^2\left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )\right )}{x \left (e^{x (5+x)}+4 x\right ) \log \left (4+\frac {e^{x (5+x)}}{x}\right ) \left (1+4 \log \left (4+\frac {e^{x (5+x)}}{x}\right )\right ) \log ^3\left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )} \, dx\\ &=32 \int \frac {\left (4-x \log \left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )\right ) \left (4 e^{x (5+x)} \left (-1+5 x+2 x^2\right )-x \left (e^{x (5+x)}+4 x\right ) \log \left (4+\frac {e^{x (5+x)}}{x}\right ) \log ^2\left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )-4 x \left (e^{x (5+x)}+4 x\right ) \log ^2\left (4+\frac {e^{x (5+x)}}{x}\right ) \log ^2\left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )\right )}{x \left (e^{x (5+x)}+4 x\right ) \log \left (4+\frac {e^{x (5+x)}}{x}\right ) \left (1+4 \log \left (4+\frac {e^{x (5+x)}}{x}\right )\right ) \log ^3\left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )} \, dx\\ &=32 \int \left (\frac {16 \left (-1+5 x+2 x^2\right ) \left (-4+x \log \left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )\right )}{\left (e^{x (5+x)}+4 x\right ) \log \left (4+\frac {e^{x (5+x)}}{x}\right ) \left (1+4 \log \left (4+\frac {e^{x (5+x)}}{x}\right )\right ) \log ^3\left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )}-\frac {\left (-4+x \log \left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )\right ) \left (-4+20 x+8 x^2-x \log \left (4+\frac {e^{x (5+x)}}{x}\right ) \log ^2\left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )-4 x \log ^2\left (4+\frac {e^{x (5+x)}}{x}\right ) \log ^2\left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )\right )}{x \log \left (4+\frac {e^{x (5+x)}}{x}\right ) \left (1+4 \log \left (4+\frac {e^{x (5+x)}}{x}\right )\right ) \log ^3\left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )}\right ) \, dx\\ &=-\left (32 \int \frac {\left (-4+x \log \left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )\right ) \left (-4+20 x+8 x^2-x \log \left (4+\frac {e^{x (5+x)}}{x}\right ) \log ^2\left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )-4 x \log ^2\left (4+\frac {e^{x (5+x)}}{x}\right ) \log ^2\left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )\right )}{x \log \left (4+\frac {e^{x (5+x)}}{x}\right ) \left (1+4 \log \left (4+\frac {e^{x (5+x)}}{x}\right )\right ) \log ^3\left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )} \, dx\right )+512 \int \frac {\left (-1+5 x+2 x^2\right ) \left (-4+x \log \left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )\right )}{\left (e^{x (5+x)}+4 x\right ) \log \left (4+\frac {e^{x (5+x)}}{x}\right ) \left (1+4 \log \left (4+\frac {e^{x (5+x)}}{x}\right )\right ) \log ^3\left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )} \, dx\\ &=-\left (32 \int \left (-x-\frac {16 \left (-1+5 x+2 x^2\right )}{x \log \left (4+\frac {e^{x (5+x)}}{x}\right ) \left (1+4 \log \left (4+\frac {e^{x (5+x)}}{x}\right )\right ) \log ^3\left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )}+\frac {4 \left (-1+5 x+2 x^2\right )}{\log \left (4+\frac {e^{x (5+x)}}{x}\right ) \left (1+4 \log \left (4+\frac {e^{x (5+x)}}{x}\right )\right ) \log ^2\left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )}+\frac {4}{\log \left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )}\right ) \, dx\right )+512 \int \left (-\frac {-4+x \log \left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )}{\left (e^{x (5+x)}+4 x\right ) \log \left (4+\frac {e^{x (5+x)}}{x}\right ) \left (1+4 \log \left (4+\frac {e^{x (5+x)}}{x}\right )\right ) \log ^3\left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )}+\frac {5 x \left (-4+x \log \left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )\right )}{\left (e^{x (5+x)}+4 x\right ) \log \left (4+\frac {e^{x (5+x)}}{x}\right ) \left (1+4 \log \left (4+\frac {e^{x (5+x)}}{x}\right )\right ) \log ^3\left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )}+\frac {2 x^2 \left (-4+x \log \left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )\right )}{\left (e^{x (5+x)}+4 x\right ) \log \left (4+\frac {e^{x (5+x)}}{x}\right ) \left (1+4 \log \left (4+\frac {e^{x (5+x)}}{x}\right )\right ) \log ^3\left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )}\right ) \, dx\\ &=16 x^2-128 \int \frac {-1+5 x+2 x^2}{\log \left (4+\frac {e^{x (5+x)}}{x}\right ) \left (1+4 \log \left (4+\frac {e^{x (5+x)}}{x}\right )\right ) \log ^2\left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )} \, dx-128 \int \frac {1}{\log \left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )} \, dx+512 \int \frac {-1+5 x+2 x^2}{x \log \left (4+\frac {e^{x (5+x)}}{x}\right ) \left (1+4 \log \left (4+\frac {e^{x (5+x)}}{x}\right )\right ) \log ^3\left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )} \, dx-512 \int \frac {-4+x \log \left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )}{\left (e^{x (5+x)}+4 x\right ) \log \left (4+\frac {e^{x (5+x)}}{x}\right ) \left (1+4 \log \left (4+\frac {e^{x (5+x)}}{x}\right )\right ) \log ^3\left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )} \, dx+1024 \int \frac {x^2 \left (-4+x \log \left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )\right )}{\left (e^{x (5+x)}+4 x\right ) \log \left (4+\frac {e^{x (5+x)}}{x}\right ) \left (1+4 \log \left (4+\frac {e^{x (5+x)}}{x}\right )\right ) \log ^3\left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )} \, dx+2560 \int \frac {x \left (-4+x \log \left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )\right )}{\left (e^{x (5+x)}+4 x\right ) \log \left (4+\frac {e^{x (5+x)}}{x}\right ) \left (1+4 \log \left (4+\frac {e^{x (5+x)}}{x}\right )\right ) \log ^3\left (4+\frac {1}{\log \left (4+\frac {e^{x (5+x)}}{x}\right )}\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A] time = 0.23, size = 61, normalized size = 1.74 \begin {gather*} 32 \left (\frac {x^2}{2}+\frac {8}{\log ^2\left (4+\frac {1}{\log \left (4+\frac {e^{5 x+x^2}}{x}\right )}\right )}-\frac {4 x}{\log \left (4+\frac {1}{\log \left (4+\frac {e^{5 x+x^2}}{x}\right )}\right )}\right ) \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Integrate[(E^(5*x + x^2)*(-512 + 2560*x + 1024*x^2) + E^(5*x + x^2)*(128*x - 640*x^2 - 256*x^3)*Log[(1 + 4*Log

[(E^(5*x + x^2) + 4*x)/x])/Log[(E^(5*x + x^2) + 4*x)/x]] + ((-128*E^(5*x + x^2)*x - 512*x^2)*Log[(E^(5*x + x^2

) + 4*x)/x] + (-512*E^(5*x + x^2)*x - 2048*x^2)*Log[(E^(5*x + x^2) + 4*x)/x]^2)*Log[(1 + 4*Log[(E^(5*x + x^2)

+ 4*x)/x])/Log[(E^(5*x + x^2) + 4*x)/x]]^2 + ((32*E^(5*x + x^2)*x^2 + 128*x^3)*Log[(E^(5*x + x^2) + 4*x)/x] +

(128*E^(5*x + x^2)*x^2 + 512*x^3)*Log[(E^(5*x + x^2) + 4*x)/x]^2)*Log[(1 + 4*Log[(E^(5*x + x^2) + 4*x)/x])/Log

[(E^(5*x + x^2) + 4*x)/x]]^3)/(((E^(5*x + x^2)*x + 4*x^2)*Log[(E^(5*x + x^2) + 4*x)/x] + (4*E^(5*x + x^2)*x +

16*x^2)*Log[(E^(5*x + x^2) + 4*x)/x]^2)*Log[(1 + 4*Log[(E^(5*x + x^2) + 4*x)/x])/Log[(E^(5*x + x^2) + 4*x)/x]]

^3),x]

[Out]

32*(x^2/2 + 8/Log[4 + Log[4 + E^(5*x + x^2)/x]^(-1)]^2 - (4*x)/Log[4 + Log[4 + E^(5*x + x^2)/x]^(-1)])

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fricas [B] time = 0.68, size = 141, normalized size = 4.03 \begin {gather*} \frac {16 \, {\left (x^{2} \log \left (\frac {4 \, \log \left (\frac {4 \, x + e^{\left (x^{2} + 5 \, x\right )}}{x}\right ) + 1}{\log \left (\frac {4 \, x + e^{\left (x^{2} + 5 \, x\right )}}{x}\right )}\right )^{2} - 8 \, x \log \left (\frac {4 \, \log \left (\frac {4 \, x + e^{\left (x^{2} + 5 \, x\right )}}{x}\right ) + 1}{\log \left (\frac {4 \, x + e^{\left (x^{2} + 5 \, x\right )}}{x}\right )}\right ) + 16\right )}}{\log \left (\frac {4 \, \log \left (\frac {4 \, x + e^{\left (x^{2} + 5 \, x\right )}}{x}\right ) + 1}{\log \left (\frac {4 \, x + e^{\left (x^{2} + 5 \, x\right )}}{x}\right )}\right )^{2}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((((128*x^2*exp(x^2+5*x)+512*x^3)*log((exp(x^2+5*x)+4*x)/x)^2+(32*x^2*exp(x^2+5*x)+128*x^3)*log((exp(

x^2+5*x)+4*x)/x))*log((4*log((exp(x^2+5*x)+4*x)/x)+1)/log((exp(x^2+5*x)+4*x)/x))^3+((-512*x*exp(x^2+5*x)-2048*

x^2)*log((exp(x^2+5*x)+4*x)/x)^2+(-128*x*exp(x^2+5*x)-512*x^2)*log((exp(x^2+5*x)+4*x)/x))*log((4*log((exp(x^2+

5*x)+4*x)/x)+1)/log((exp(x^2+5*x)+4*x)/x))^2+(-256*x^3-640*x^2+128*x)*exp(x^2+5*x)*log((4*log((exp(x^2+5*x)+4*

x)/x)+1)/log((exp(x^2+5*x)+4*x)/x))+(1024*x^2+2560*x-512)*exp(x^2+5*x))/((4*x*exp(x^2+5*x)+16*x^2)*log((exp(x^

2+5*x)+4*x)/x)^2+(x*exp(x^2+5*x)+4*x^2)*log((exp(x^2+5*x)+4*x)/x))/log((4*log((exp(x^2+5*x)+4*x)/x)+1)/log((ex

p(x^2+5*x)+4*x)/x))^3,x, algorithm="fricas")

[Out]

16*(x^2*log((4*log((4*x + e^(x^2 + 5*x))/x) + 1)/log((4*x + e^(x^2 + 5*x))/x))^2 - 8*x*log((4*log((4*x + e^(x^

2 + 5*x))/x) + 1)/log((4*x + e^(x^2 + 5*x))/x)) + 16)/log((4*log((4*x + e^(x^2 + 5*x))/x) + 1)/log((4*x + e^(x

^2 + 5*x))/x))^2

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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((((128*x^2*exp(x^2+5*x)+512*x^3)*log((exp(x^2+5*x)+4*x)/x)^2+(32*x^2*exp(x^2+5*x)+128*x^3)*log((exp(

x^2+5*x)+4*x)/x))*log((4*log((exp(x^2+5*x)+4*x)/x)+1)/log((exp(x^2+5*x)+4*x)/x))^3+((-512*x*exp(x^2+5*x)-2048*

x^2)*log((exp(x^2+5*x)+4*x)/x)^2+(-128*x*exp(x^2+5*x)-512*x^2)*log((exp(x^2+5*x)+4*x)/x))*log((4*log((exp(x^2+

5*x)+4*x)/x)+1)/log((exp(x^2+5*x)+4*x)/x))^2+(-256*x^3-640*x^2+128*x)*exp(x^2+5*x)*log((4*log((exp(x^2+5*x)+4*

x)/x)+1)/log((exp(x^2+5*x)+4*x)/x))+(1024*x^2+2560*x-512)*exp(x^2+5*x))/((4*x*exp(x^2+5*x)+16*x^2)*log((exp(x^

2+5*x)+4*x)/x)^2+(x*exp(x^2+5*x)+4*x^2)*log((exp(x^2+5*x)+4*x)/x))/log((4*log((exp(x^2+5*x)+4*x)/x)+1)/log((ex

p(x^2+5*x)+4*x)/x))^3,x, algorithm="giac")

[Out]

Timed out

________________________________________________________________________________________

maple [C] time = 13.95, size = 4425, normalized size = 126.43


method	result	size

risch	\(\text {Expression too large to display}\)	\(4425\)

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((((128*x^2*exp(x^2+5*x)+512*x^3)*ln((exp(x^2+5*x)+4*x)/x)^2+(32*x^2*exp(x^2+5*x)+128*x^3)*ln((exp(x^2+5*x)

+4*x)/x))*ln((4*ln((exp(x^2+5*x)+4*x)/x)+1)/ln((exp(x^2+5*x)+4*x)/x))^3+((-512*x*exp(x^2+5*x)-2048*x^2)*ln((ex

p(x^2+5*x)+4*x)/x)^2+(-128*x*exp(x^2+5*x)-512*x^2)*ln((exp(x^2+5*x)+4*x)/x))*ln((4*ln((exp(x^2+5*x)+4*x)/x)+1)

/ln((exp(x^2+5*x)+4*x)/x))^2+(-256*x^3-640*x^2+128*x)*exp(x^2+5*x)*ln((4*ln((exp(x^2+5*x)+4*x)/x)+1)/ln((exp(x

^2+5*x)+4*x)/x))+(1024*x^2+2560*x-512)*exp(x^2+5*x))/((4*x*exp(x^2+5*x)+16*x^2)*ln((exp(x^2+5*x)+4*x)/x)^2+(x*

exp(x^2+5*x)+4*x^2)*ln((exp(x^2+5*x)+4*x)/x))/ln((4*ln((exp(x^2+5*x)+4*x)/x)+1)/ln((exp(x^2+5*x)+4*x)/x))^3,x,

method=_RETURNVERBOSE)

[Out]

16*x^2-256*I*(-4*I+Pi*x*csgn(I/(2*I*ln(1/4*exp((5+x)*x)+x)+Pi*csgn(I/x*(1/4*exp((5+x)*x)+x))^3+Pi*csgn(I/x)*cs

gn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))+4*I*ln(2)-Pi*csgn(I/x)*csgn(I/x*(1/4*exp((5+x)*x)+x)

)^2-Pi*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-2*I*ln(x)))*csgn(I*(1/2*I+2*I*ln(1/4*exp(

(5+x)*x)+x)+Pi*csgn(I/x*(1/4*exp((5+x)*x)+x))^3+Pi*csgn(I/x)*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5

+x)*x)+x))+4*I*ln(2)-Pi*csgn(I/x)*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-Pi*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1

/4*exp((5+x)*x)+x))^2-2*I*ln(x)))*csgn(I/(2*I*ln(1/4*exp((5+x)*x)+x)+Pi*csgn(I/x*(1/4*exp((5+x)*x)+x))^3+Pi*cs

gn(I/x)*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))+4*I*ln(2)-Pi*csgn(I/x)*csgn(I/x*(1/4*exp((

5+x)*x)+x))^2-Pi*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-2*I*ln(x))*(1/2*I+2*I*ln(1/4*ex

p((5+x)*x)+x)+Pi*csgn(I/x*(1/4*exp((5+x)*x)+x))^3+Pi*csgn(I/x)*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp(

(5+x)*x)+x))+4*I*ln(2)-Pi*csgn(I/x)*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-Pi*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*

(1/4*exp((5+x)*x)+x))^2-2*I*ln(x)))-Pi*x*csgn(I/(2*I*ln(1/4*exp((5+x)*x)+x)+Pi*csgn(I/x*(1/4*exp((5+x)*x)+x))^

3+Pi*csgn(I/x)*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))+4*I*ln(2)-Pi*csgn(I/x)*csgn(I/x*(1/

4*exp((5+x)*x)+x))^2-Pi*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-2*I*ln(x)))*csgn(I/(2*I*

ln(1/4*exp((5+x)*x)+x)+Pi*csgn(I/x*(1/4*exp((5+x)*x)+x))^3+Pi*csgn(I/x)*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*

(1/4*exp((5+x)*x)+x))+4*I*ln(2)-Pi*csgn(I/x)*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-Pi*csgn(I*(1/4*exp((5+x)*x)+x))*

csgn(I/x*(1/4*exp((5+x)*x)+x))^2-2*I*ln(x))*(1/2*I+2*I*ln(1/4*exp((5+x)*x)+x)+Pi*csgn(I/x*(1/4*exp((5+x)*x)+x)

)^3+Pi*csgn(I/x)*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))+4*I*ln(2)-Pi*csgn(I/x)*csgn(I/x*(

1/4*exp((5+x)*x)+x))^2-Pi*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-2*I*ln(x)))^2-Pi*x*csg

n(I*(1/2*I+2*I*ln(1/4*exp((5+x)*x)+x)+Pi*csgn(I/x*(1/4*exp((5+x)*x)+x))^3+Pi*csgn(I/x)*csgn(I*(1/4*exp((5+x)*x

)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))+4*I*ln(2)-Pi*csgn(I/x)*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-Pi*csgn(I*(1/4*ex

p((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-2*I*ln(x)))*csgn(I/(2*I*ln(1/4*exp((5+x)*x)+x)+Pi*csgn(I/x*(1/

4*exp((5+x)*x)+x))^3+Pi*csgn(I/x)*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))+4*I*ln(2)-Pi*csg

n(I/x)*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-Pi*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-2*I*l

n(x))*(1/2*I+2*I*ln(1/4*exp((5+x)*x)+x)+Pi*csgn(I/x*(1/4*exp((5+x)*x)+x))^3+Pi*csgn(I/x)*csgn(I*(1/4*exp((5+x)

*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))+4*I*ln(2)-Pi*csgn(I/x)*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-Pi*csgn(I*(1/4*

exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-2*I*ln(x)))^2+Pi*x*csgn(I/(2*I*ln(1/4*exp((5+x)*x)+x)+Pi*csg

n(I/x*(1/4*exp((5+x)*x)+x))^3+Pi*csgn(I/x)*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))+4*I*ln(

2)-Pi*csgn(I/x)*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-Pi*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x)

)^2-2*I*ln(x))*(1/2*I+2*I*ln(1/4*exp((5+x)*x)+x)+Pi*csgn(I/x*(1/4*exp((5+x)*x)+x))^3+Pi*csgn(I/x)*csgn(I*(1/4*

exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))+4*I*ln(2)-Pi*csgn(I/x)*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-Pi*csg

n(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-2*I*ln(x)))^3+4*I*ln(2)*x-2*I*x*ln(2*I*ln(1/4*exp((

5+x)*x)+x)+Pi*csgn(I/x*(1/4*exp((5+x)*x)+x))^3+Pi*csgn(I/x)*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+

x)*x)+x))+4*I*ln(2)-Pi*csgn(I/x)*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-Pi*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/

4*exp((5+x)*x)+x))^2-2*I*ln(x))+2*I*x*ln(1/2*I+2*I*ln(1/4*exp((5+x)*x)+x)+Pi*csgn(I/x*(1/4*exp((5+x)*x)+x))^3+

Pi*csgn(I/x)*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))+4*I*ln(2)-Pi*csgn(I/x)*csgn(I/x*(1/4*

exp((5+x)*x)+x))^2-Pi*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-2*I*ln(x)))/(Pi*csgn(I/(2*

I*ln(1/4*exp((5+x)*x)+x)+Pi*csgn(I/x*(1/4*exp((5+x)*x)+x))^3+Pi*csgn(I/x)*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/

x*(1/4*exp((5+x)*x)+x))+4*I*ln(2)-Pi*csgn(I/x)*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-Pi*csgn(I*(1/4*exp((5+x)*x)+x)

)*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-2*I*ln(x)))*csgn(I*(1/2*I+2*I*ln(1/4*exp((5+x)*x)+x)+Pi*csgn(I/x*(1/4*exp((

5+x)*x)+x))^3+Pi*csgn(I/x)*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))+4*I*ln(2)-Pi*csgn(I/x)*

csgn(I/x*(1/4*exp((5+x)*x)+x))^2-Pi*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-2*I*ln(x)))*

csgn(I/(2*I*ln(1/4*exp((5+x)*x)+x)+Pi*csgn(I/x*(1/4*exp((5+x)*x)+x))^3+Pi*csgn(I/x)*csgn(I*(1/4*exp((5+x)*x)+x

))*csgn(I/x*(1/4*exp((5+x)*x)+x))+4*I*ln(2)-Pi*csgn(I/x)*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-Pi*csgn(I*(1/4*exp((

5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-2*I*ln(x))*(1/2*I+2*I*ln(1/4*exp((5+x)*x)+x)+Pi*csgn(I/x*(1/4*exp

((5+x)*x)+x))^3+Pi*csgn(I/x)*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))+4*I*ln(2)-Pi*csgn(I/x

)*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-Pi*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-2*I*ln(x))

)-Pi*csgn(I/(2*I*ln(1/4*exp((5+x)*x)+x)+Pi*csgn(I/x*(1/4*exp((5+x)*x)+x))^3+Pi*csgn(I/x)*csgn(I*(1/4*exp((5+x)

*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))+4*I*ln(2)-Pi*csgn(I/x)*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-Pi*csgn(I*(1/4*

exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-2*I*ln(x)))*csgn(I/(2*I*ln(1/4*exp((5+x)*x)+x)+Pi*csgn(I/x*(

1/4*exp((5+x)*x)+x))^3+Pi*csgn(I/x)*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))+4*I*ln(2)-Pi*c

sgn(I/x)*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-Pi*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-2*I

*ln(x))*(1/2*I+2*I*ln(1/4*exp((5+x)*x)+x)+Pi*csgn(I/x*(1/4*exp((5+x)*x)+x))^3+Pi*csgn(I/x)*csgn(I*(1/4*exp((5+

x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))+4*I*ln(2)-Pi*csgn(I/x)*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-Pi*csgn(I*(1/

4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-2*I*ln(x)))^2-Pi*csgn(I*(1/2*I+2*I*ln(1/4*exp((5+x)*x)+x)+

Pi*csgn(I/x*(1/4*exp((5+x)*x)+x))^3+Pi*csgn(I/x)*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))+4

*I*ln(2)-Pi*csgn(I/x)*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-Pi*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)

*x)+x))^2-2*I*ln(x)))*csgn(I/(2*I*ln(1/4*exp((5+x)*x)+x)+Pi*csgn(I/x*(1/4*exp((5+x)*x)+x))^3+Pi*csgn(I/x)*csgn

(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))+4*I*ln(2)-Pi*csgn(I/x)*csgn(I/x*(1/4*exp((5+x)*x)+x))^

2-Pi*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-2*I*ln(x))*(1/2*I+2*I*ln(1/4*exp((5+x)*x)+x

)+Pi*csgn(I/x*(1/4*exp((5+x)*x)+x))^3+Pi*csgn(I/x)*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))

+4*I*ln(2)-Pi*csgn(I/x)*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-Pi*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+

x)*x)+x))^2-2*I*ln(x)))^2+Pi*csgn(I/(2*I*ln(1/4*exp((5+x)*x)+x)+Pi*csgn(I/x*(1/4*exp((5+x)*x)+x))^3+Pi*csgn(I/

x)*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))+4*I*ln(2)-Pi*csgn(I/x)*csgn(I/x*(1/4*exp((5+x)*

x)+x))^2-Pi*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-2*I*ln(x))*(1/2*I+2*I*ln(1/4*exp((5+

x)*x)+x)+Pi*csgn(I/x*(1/4*exp((5+x)*x)+x))^3+Pi*csgn(I/x)*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)

*x)+x))+4*I*ln(2)-Pi*csgn(I/x)*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-Pi*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*

exp((5+x)*x)+x))^2-2*I*ln(x)))^3+4*I*ln(2)-2*I*ln(2*I*ln(1/4*exp((5+x)*x)+x)+Pi*csgn(I/x*(1/4*exp((5+x)*x)+x))

^3+Pi*csgn(I/x)*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))+4*I*ln(2)-Pi*csgn(I/x)*csgn(I/x*(1

/4*exp((5+x)*x)+x))^2-Pi*csgn(I*(1/4*exp((5+x)*x)+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-2*I*ln(x))+2*I*ln(1/2*I

+2*I*ln(1/4*exp((5+x)*x)+x)+Pi*csgn(I/x*(1/4*exp((5+x)*x)+x))^3+Pi*csgn(I/x)*csgn(I*(1/4*exp((5+x)*x)+x))*csgn

(I/x*(1/4*exp((5+x)*x)+x))+4*I*ln(2)-Pi*csgn(I/x)*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-Pi*csgn(I*(1/4*exp((5+x)*x)

+x))*csgn(I/x*(1/4*exp((5+x)*x)+x))^2-2*I*ln(x)))^2

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maxima [B] time = 0.63, size = 221, normalized size = 6.31 \begin {gather*} \frac {16 \, {\left (x^{2} \log \left (4 \, \log \left (4 \, x + e^{\left (x^{2} + 5 \, x\right )}\right ) - 4 \, \log \relax (x) + 1\right )^{2} + x^{2} \log \left (\log \left (4 \, x + e^{\left (x^{2} + 5 \, x\right )}\right ) - \log \relax (x)\right )^{2} - 2 \, {\left (x^{2} \log \left (\log \left (4 \, x + e^{\left (x^{2} + 5 \, x\right )}\right ) - \log \relax (x)\right ) + 4 \, x\right )} \log \left (4 \, \log \left (4 \, x + e^{\left (x^{2} + 5 \, x\right )}\right ) - 4 \, \log \relax (x) + 1\right ) + 8 \, x \log \left (\log \left (4 \, x + e^{\left (x^{2} + 5 \, x\right )}\right ) - \log \relax (x)\right ) + 16\right )}}{\log \left (4 \, \log \left (4 \, x + e^{\left (x^{2} + 5 \, x\right )}\right ) - 4 \, \log \relax (x) + 1\right )^{2} - 2 \, \log \left (4 \, \log \left (4 \, x + e^{\left (x^{2} + 5 \, x\right )}\right ) - 4 \, \log \relax (x) + 1\right ) \log \left (\log \left (4 \, x + e^{\left (x^{2} + 5 \, x\right )}\right ) - \log \relax (x)\right ) + \log \left (\log \left (4 \, x + e^{\left (x^{2} + 5 \, x\right )}\right ) - \log \relax (x)\right )^{2}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((((128*x^2*exp(x^2+5*x)+512*x^3)*log((exp(x^2+5*x)+4*x)/x)^2+(32*x^2*exp(x^2+5*x)+128*x^3)*log((exp(

x^2+5*x)+4*x)/x))*log((4*log((exp(x^2+5*x)+4*x)/x)+1)/log((exp(x^2+5*x)+4*x)/x))^3+((-512*x*exp(x^2+5*x)-2048*

x^2)*log((exp(x^2+5*x)+4*x)/x)^2+(-128*x*exp(x^2+5*x)-512*x^2)*log((exp(x^2+5*x)+4*x)/x))*log((4*log((exp(x^2+

5*x)+4*x)/x)+1)/log((exp(x^2+5*x)+4*x)/x))^2+(-256*x^3-640*x^2+128*x)*exp(x^2+5*x)*log((4*log((exp(x^2+5*x)+4*

x)/x)+1)/log((exp(x^2+5*x)+4*x)/x))+(1024*x^2+2560*x-512)*exp(x^2+5*x))/((4*x*exp(x^2+5*x)+16*x^2)*log((exp(x^

2+5*x)+4*x)/x)^2+(x*exp(x^2+5*x)+4*x^2)*log((exp(x^2+5*x)+4*x)/x))/log((4*log((exp(x^2+5*x)+4*x)/x)+1)/log((ex

p(x^2+5*x)+4*x)/x))^3,x, algorithm="maxima")

[Out]

16*(x^2*log(4*log(4*x + e^(x^2 + 5*x)) - 4*log(x) + 1)^2 + x^2*log(log(4*x + e^(x^2 + 5*x)) - log(x))^2 - 2*(x

^2*log(log(4*x + e^(x^2 + 5*x)) - log(x)) + 4*x)*log(4*log(4*x + e^(x^2 + 5*x)) - 4*log(x) + 1) + 8*x*log(log(

4*x + e^(x^2 + 5*x)) - log(x)) + 16)/(log(4*log(4*x + e^(x^2 + 5*x)) - 4*log(x) + 1)^2 - 2*log(4*log(4*x + e^(

x^2 + 5*x)) - 4*log(x) + 1)*log(log(4*x + e^(x^2 + 5*x)) - log(x)) + log(log(4*x + e^(x^2 + 5*x)) - log(x))^2)

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mupad [F] time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {\left (\left (-128\,x^2\,{\mathrm {e}}^{x^2+5\,x}-512\,x^3\right )\,{\ln \left (\frac {4\,x+{\mathrm {e}}^{x^2+5\,x}}{x}\right )}^2+\left (-32\,x^2\,{\mathrm {e}}^{x^2+5\,x}-128\,x^3\right )\,\ln \left (\frac {4\,x+{\mathrm {e}}^{x^2+5\,x}}{x}\right )\right )\,{\ln \left (\frac {4\,\ln \left (\frac {4\,x+{\mathrm {e}}^{x^2+5\,x}}{x}\right )+1}{\ln \left (\frac {4\,x+{\mathrm {e}}^{x^2+5\,x}}{x}\right )}\right )}^3+\left (\left (512\,x\,{\mathrm {e}}^{x^2+5\,x}+2048\,x^2\right )\,{\ln \left (\frac {4\,x+{\mathrm {e}}^{x^2+5\,x}}{x}\right )}^2+\left (128\,x\,{\mathrm {e}}^{x^2+5\,x}+512\,x^2\right )\,\ln \left (\frac {4\,x+{\mathrm {e}}^{x^2+5\,x}}{x}\right )\right )\,{\ln \left (\frac {4\,\ln \left (\frac {4\,x+{\mathrm {e}}^{x^2+5\,x}}{x}\right )+1}{\ln \left (\frac {4\,x+{\mathrm {e}}^{x^2+5\,x}}{x}\right )}\right )}^2+{\mathrm {e}}^{x^2+5\,x}\,\left (256\,x^3+640\,x^2-128\,x\right )\,\ln \left (\frac {4\,\ln \left (\frac {4\,x+{\mathrm {e}}^{x^2+5\,x}}{x}\right )+1}{\ln \left (\frac {4\,x+{\mathrm {e}}^{x^2+5\,x}}{x}\right )}\right )-{\mathrm {e}}^{x^2+5\,x}\,\left (1024\,x^2+2560\,x-512\right )}{{\ln \left (\frac {4\,\ln \left (\frac {4\,x+{\mathrm {e}}^{x^2+5\,x}}{x}\right )+1}{\ln \left (\frac {4\,x+{\mathrm {e}}^{x^2+5\,x}}{x}\right )}\right )}^3\,\left (\left (4\,x\,{\mathrm {e}}^{x^2+5\,x}+16\,x^2\right )\,{\ln \left (\frac {4\,x+{\mathrm {e}}^{x^2+5\,x}}{x}\right )}^2+\left (x\,{\mathrm {e}}^{x^2+5\,x}+4\,x^2\right )\,\ln \left (\frac {4\,x+{\mathrm {e}}^{x^2+5\,x}}{x}\right )\right )} \,d x \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(-(log((4*log((4*x + exp(5*x + x^2))/x) + 1)/log((4*x + exp(5*x + x^2))/x))^2*(log((4*x + exp(5*x + x^2))/x

)*(128*x*exp(5*x + x^2) + 512*x^2) + log((4*x + exp(5*x + x^2))/x)^2*(512*x*exp(5*x + x^2) + 2048*x^2)) - exp(

5*x + x^2)*(2560*x + 1024*x^2 - 512) - log((4*log((4*x + exp(5*x + x^2))/x) + 1)/log((4*x + exp(5*x + x^2))/x)

)^3*(log((4*x + exp(5*x + x^2))/x)^2*(128*x^2*exp(5*x + x^2) + 512*x^3) + log((4*x + exp(5*x + x^2))/x)*(32*x^

2*exp(5*x + x^2) + 128*x^3)) + exp(5*x + x^2)*log((4*log((4*x + exp(5*x + x^2))/x) + 1)/log((4*x + exp(5*x + x

^2))/x))*(640*x^2 - 128*x + 256*x^3))/(log((4*log((4*x + exp(5*x + x^2))/x) + 1)/log((4*x + exp(5*x + x^2))/x)

)^3*(log((4*x + exp(5*x + x^2))/x)*(x*exp(5*x + x^2) + 4*x^2) + log((4*x + exp(5*x + x^2))/x)^2*(4*x*exp(5*x +

x^2) + 16*x^2))),x)

[Out]

int(-(log((4*log((4*x + exp(5*x + x^2))/x) + 1)/log((4*x + exp(5*x + x^2))/x))^2*(log((4*x + exp(5*x + x^2))/x

)*(128*x*exp(5*x + x^2) + 512*x^2) + log((4*x + exp(5*x + x^2))/x)^2*(512*x*exp(5*x + x^2) + 2048*x^2)) - exp(

5*x + x^2)*(2560*x + 1024*x^2 - 512) - log((4*log((4*x + exp(5*x + x^2))/x) + 1)/log((4*x + exp(5*x + x^2))/x)

)^3*(log((4*x + exp(5*x + x^2))/x)^2*(128*x^2*exp(5*x + x^2) + 512*x^3) + log((4*x + exp(5*x + x^2))/x)*(32*x^

2*exp(5*x + x^2) + 128*x^3)) + exp(5*x + x^2)*log((4*log((4*x + exp(5*x + x^2))/x) + 1)/log((4*x + exp(5*x + x

^2))/x))*(640*x^2 - 128*x + 256*x^3))/(log((4*log((4*x + exp(5*x + x^2))/x) + 1)/log((4*x + exp(5*x + x^2))/x)

)^3*(log((4*x + exp(5*x + x^2))/x)*(x*exp(5*x + x^2) + 4*x^2) + log((4*x + exp(5*x + x^2))/x)^2*(4*x*exp(5*x +

x^2) + 16*x^2))), x)

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sympy [B] time = 42.86, size = 82, normalized size = 2.34 \begin {gather*} 16 x^{2} + \frac {- 128 x \log {\left (\frac {4 \log {\left (\frac {4 x + e^{x^{2} + 5 x}}{x} \right )} + 1}{\log {\left (\frac {4 x + e^{x^{2} + 5 x}}{x} \right )}} \right )} + 256}{\log {\left (\frac {4 \log {\left (\frac {4 x + e^{x^{2} + 5 x}}{x} \right )} + 1}{\log {\left (\frac {4 x + e^{x^{2} + 5 x}}{x} \right )}} \right )}^{2}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((((128*x**2*exp(x**2+5*x)+512*x**3)*ln((exp(x**2+5*x)+4*x)/x)**2+(32*x**2*exp(x**2+5*x)+128*x**3)*ln

((exp(x**2+5*x)+4*x)/x))*ln((4*ln((exp(x**2+5*x)+4*x)/x)+1)/ln((exp(x**2+5*x)+4*x)/x))**3+((-512*x*exp(x**2+5*

x)-2048*x**2)*ln((exp(x**2+5*x)+4*x)/x)**2+(-128*x*exp(x**2+5*x)-512*x**2)*ln((exp(x**2+5*x)+4*x)/x))*ln((4*ln

((exp(x**2+5*x)+4*x)/x)+1)/ln((exp(x**2+5*x)+4*x)/x))**2+(-256*x**3-640*x**2+128*x)*exp(x**2+5*x)*ln((4*ln((ex

p(x**2+5*x)+4*x)/x)+1)/ln((exp(x**2+5*x)+4*x)/x))+(1024*x**2+2560*x-512)*exp(x**2+5*x))/((4*x*exp(x**2+5*x)+16

*x**2)*ln((exp(x**2+5*x)+4*x)/x)**2+(x*exp(x**2+5*x)+4*x**2)*ln((exp(x**2+5*x)+4*x)/x))/ln((4*ln((exp(x**2+5*x

)+4*x)/x)+1)/ln((exp(x**2+5*x)+4*x)/x))**3,x)

[Out]

16*x**2 + (-128*x*log((4*log((4*x + exp(x**2 + 5*x))/x) + 1)/log((4*x + exp(x**2 + 5*x))/x)) + 256)/log((4*log

((4*x + exp(x**2 + 5*x))/x) + 1)/log((4*x + exp(x**2 + 5*x))/x))**2

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