3.59.100 \(\int \frac {e^{\frac {(100 x-40 x^2+4 x^3+e^4 (25 x-10 x^2+x^3)) \log ^2(\frac {e}{3})+(-40 x+8 x^2+e^4 (-10 x+2 x^2)) \log (\frac {e}{3}) \log (x)+(4 x+e^4 x) \log ^2(x)}{\log ^2(\frac {e}{3})}} ((100-80 x+12 x^2+e^4 (25-20 x+3 x^2)) \log ^2(\frac {e}{3})+(8+2 e^4) \log (x)+(4+e^4) \log ^2(x)+\log (\frac {e}{3}) (-40+8 x+e^4 (-10+2 x)+(-40+16 x+e^4 (-10+4 x)) \log (x)))}{\log ^2(\frac {e}{3})} \, dx\)

Optimal. Leaf size=25 \[ e^{\left (4+e^4\right ) x \left (-5+x+\frac {\log (x)}{\log \left (\frac {e}{3}\right )}\right )^2} \]

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Rubi [F]  time = 21.56, 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 {\exp \left (\frac {\left (100 x-40 x^2+4 x^3+e^4 \left (25 x-10 x^2+x^3\right )\right ) \log ^2\left (\frac {e}{3}\right )+\left (-40 x+8 x^2+e^4 \left (-10 x+2 x^2\right )\right ) \log \left (\frac {e}{3}\right ) \log (x)+\left (4 x+e^4 x\right ) \log ^2(x)}{\log ^2\left (\frac {e}{3}\right )}\right ) \left (\left (100-80 x+12 x^2+e^4 \left (25-20 x+3 x^2\right )\right ) \log ^2\left (\frac {e}{3}\right )+\left (8+2 e^4\right ) \log (x)+\left (4+e^4\right ) \log ^2(x)+\log \left (\frac {e}{3}\right ) \left (-40+8 x+e^4 (-10+2 x)+\left (-40+16 x+e^4 (-10+4 x)\right ) \log (x)\right )\right )}{\log ^2\left (\frac {e}{3}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(((100*x - 40*x^2 + 4*x^3 + E^4*(25*x - 10*x^2 + x^3))*Log[E/3]^2 + (-40*x + 8*x^2 + E^4*(-10*x + 2*x^2
))*Log[E/3]*Log[x] + (4*x + E^4*x)*Log[x]^2)/Log[E/3]^2)*((100 - 80*x + 12*x^2 + E^4*(25 - 20*x + 3*x^2))*Log[
E/3]^2 + (8 + 2*E^4)*Log[x] + (4 + E^4)*Log[x]^2 + Log[E/3]*(-40 + 8*x + E^4*(-10 + 2*x) + (-40 + 16*x + E^4*(
-10 + 4*x))*Log[x])))/Log[E/3]^2,x]

[Out]

-(((4 + E^4)*(18 - 40*Log[3] + 20*Log[3]^2 + Log[9])*Defer[Int][E^(((4 + E^4)*x*((-5 + x)^2*(-1 + Log[3])^2 +
Log[x]^2))/(-1 + Log[3])^2)*x^(1 - (10*(4 + E^4)*x)/(1 - Log[3]) + (2*(4 + E^4)*x^2)/(1 - Log[3])), x])/(1 - L
og[3])^2) + 3*(4 + E^4)*Defer[Int][E^(((4 + E^4)*x*((-5 + x)^2*(-1 + Log[3])^2 + Log[x]^2))/(-1 + Log[3])^2)*x
^(2 - (10*(4 + E^4)*x)/(1 - Log[3]) + (2*(4 + E^4)*x^2)/(1 - Log[3])), x] + (5*(4 + E^4)*(3 - Log[243])*Defer[
Int][E^(((4 + E^4)*x*((-5 + x)^2*(-1 + Log[3])^2 + Log[x]^2))/(-1 + Log[3])^2)/x^((2*(4 + E^4)*(-5 + x)*x)/(-1
+ Log[3])), x])/(1 - Log[3]) + (4*(4 + E^4)*Defer[Int][E^(((4 + E^4)*x*((-5 + x)^2*(-1 + Log[3])^2 + Log[x]^2
))/(-1 + Log[3])^2)*x^(1 - (10*(4 + E^4)*x)/(1 - Log[3]) + (2*(4 + E^4)*x^2)/(1 - Log[3]))*Log[x], x])/Log[E/3
] - (2*(4 + E^4)*(4 - Log[243])*Defer[Int][(E^(((4 + E^4)*x*((-5 + x)^2*(-1 + Log[3])^2 + Log[x]^2))/(-1 + Log
[3])^2)*Log[x])/x^((2*(4 + E^4)*(-5 + x)*x)/(-1 + Log[3])), x])/(1 - Log[3])^2 + ((4 + E^4)*Defer[Int][(E^(((4
+ E^4)*x*((-5 + x)^2*(-1 + Log[3])^2 + Log[x]^2))/(-1 + Log[3])^2)*Log[x]^2)/x^((2*(4 + E^4)*(-5 + x)*x)/(-1
+ Log[3])), x])/(1 - Log[3])^2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \exp \left (\frac {\left (100 x-40 x^2+4 x^3+e^4 \left (25 x-10 x^2+x^3\right )\right ) \log ^2\left (\frac {e}{3}\right )+\left (-40 x+8 x^2+e^4 \left (-10 x+2 x^2\right )\right ) \log \left (\frac {e}{3}\right ) \log (x)+\left (4 x+e^4 x\right ) \log ^2(x)}{\log ^2\left (\frac {e}{3}\right )}\right ) \left (\left (100-80 x+12 x^2+e^4 \left (25-20 x+3 x^2\right )\right ) \log ^2\left (\frac {e}{3}\right )+\left (8+2 e^4\right ) \log (x)+\left (4+e^4\right ) \log ^2(x)+\log \left (\frac {e}{3}\right ) \left (-40+8 x+e^4 (-10+2 x)+\left (-40+16 x+e^4 (-10+4 x)\right ) \log (x)\right )\right ) \, dx}{\log ^2\left (\frac {e}{3}\right )}\\ &=\frac {\int \exp \left (\frac {\left (4+e^4\right ) x \left ((-5+x)^2 (-1+\log (3))^2+\log ^2(x)\right )}{(-1+\log (3))^2}\right ) \left (4+e^4\right ) x^{-\frac {2 \left (4+e^4\right ) (-5+x) x}{-1+\log (3)}} \left (3 x^2 (-1+\log (3))^2+5 \left (3-8 \log (3)+5 \log ^2(3)\right )-x \left (18-40 \log (3)+20 \log ^2(3)+\log (9)\right )+2 \left (-4+\log (243)+2 x \log \left (\frac {e}{3}\right )\right ) \log (x)+\log ^2(x)\right ) \, dx}{\log ^2\left (\frac {e}{3}\right )}\\ &=\frac {\left (4+e^4\right ) \int \exp \left (\frac {\left (4+e^4\right ) x \left ((-5+x)^2 (-1+\log (3))^2+\log ^2(x)\right )}{(-1+\log (3))^2}\right ) x^{-\frac {2 \left (4+e^4\right ) (-5+x) x}{-1+\log (3)}} \left (3 x^2 (-1+\log (3))^2+5 \left (3-8 \log (3)+5 \log ^2(3)\right )-x \left (18-40 \log (3)+20 \log ^2(3)+\log (9)\right )+2 \left (-4+\log (243)+2 x \log \left (\frac {e}{3}\right )\right ) \log (x)+\log ^2(x)\right ) \, dx}{(1-\log (3))^2}\\ &=\frac {\left (4+e^4\right ) \int \left (5 \exp \left (\frac {\left (4+e^4\right ) x \left ((-5+x)^2 (-1+\log (3))^2+\log ^2(x)\right )}{(-1+\log (3))^2}\right ) x^{-\frac {2 \left (4+e^4\right ) (-5+x) x}{-1+\log (3)}} (3-5 \log (3)) (1-\log (3))+3 \exp \left (\frac {\left (4+e^4\right ) x \left ((-5+x)^2 (-1+\log (3))^2+\log ^2(x)\right )}{(-1+\log (3))^2}\right ) x^{2-\frac {2 \left (4+e^4\right ) (-5+x) x}{-1+\log (3)}} (-1+\log (3))^2-\exp \left (\frac {\left (4+e^4\right ) x \left ((-5+x)^2 (-1+\log (3))^2+\log ^2(x)\right )}{(-1+\log (3))^2}\right ) x^{1-\frac {2 \left (4+e^4\right ) (-5+x) x}{-1+\log (3)}} \left (18-40 \log (3)+20 \log ^2(3)+\log (9)\right )+2 \exp \left (\frac {\left (4+e^4\right ) x \left ((-5+x)^2 (-1+\log (3))^2+\log ^2(x)\right )}{(-1+\log (3))^2}\right ) x^{-\frac {2 \left (4+e^4\right ) (-5+x) x}{-1+\log (3)}} \left (-4+\log (243)+2 x \log \left (\frac {e}{3}\right )\right ) \log (x)+\exp \left (\frac {\left (4+e^4\right ) x \left ((-5+x)^2 (-1+\log (3))^2+\log ^2(x)\right )}{(-1+\log (3))^2}\right ) x^{-\frac {2 \left (4+e^4\right ) (-5+x) x}{-1+\log (3)}} \log ^2(x)\right ) \, dx}{(1-\log (3))^2}\\ &=\left (3 \left (4+e^4\right )\right ) \int \exp \left (\frac {\left (4+e^4\right ) x \left ((-5+x)^2 (-1+\log (3))^2+\log ^2(x)\right )}{(-1+\log (3))^2}\right ) x^{2-\frac {2 \left (4+e^4\right ) (-5+x) x}{-1+\log (3)}} \, dx+\frac {\left (4+e^4\right ) \int \exp \left (\frac {\left (4+e^4\right ) x \left ((-5+x)^2 (-1+\log (3))^2+\log ^2(x)\right )}{(-1+\log (3))^2}\right ) x^{-\frac {2 \left (4+e^4\right ) (-5+x) x}{-1+\log (3)}} \log ^2(x) \, dx}{(1-\log (3))^2}+\frac {\left (2 \left (4+e^4\right )\right ) \int \exp \left (\frac {\left (4+e^4\right ) x \left ((-5+x)^2 (-1+\log (3))^2+\log ^2(x)\right )}{(-1+\log (3))^2}\right ) x^{-\frac {2 \left (4+e^4\right ) (-5+x) x}{-1+\log (3)}} \left (-4+\log (243)+2 x \log \left (\frac {e}{3}\right )\right ) \log (x) \, dx}{(1-\log (3))^2}-\frac {\left (\left (4+e^4\right ) \left (18-40 \log (3)+20 \log ^2(3)+\log (9)\right )\right ) \int \exp \left (\frac {\left (4+e^4\right ) x \left ((-5+x)^2 (-1+\log (3))^2+\log ^2(x)\right )}{(-1+\log (3))^2}\right ) x^{1-\frac {2 \left (4+e^4\right ) (-5+x) x}{-1+\log (3)}} \, dx}{(1-\log (3))^2}+\frac {\left (5 \left (4+e^4\right ) (3-\log (243))\right ) \int \exp \left (\frac {\left (4+e^4\right ) x \left ((-5+x)^2 (-1+\log (3))^2+\log ^2(x)\right )}{(-1+\log (3))^2}\right ) x^{-\frac {2 \left (4+e^4\right ) (-5+x) x}{-1+\log (3)}} \, dx}{1-\log (3)}\\ &=\left (3 \left (4+e^4\right )\right ) \int \exp \left (\frac {\left (4+e^4\right ) x \left ((-5+x)^2 (-1+\log (3))^2+\log ^2(x)\right )}{(-1+\log (3))^2}\right ) x^{2-\frac {10 \left (4+e^4\right ) x}{1-\log (3)}+\frac {2 \left (4+e^4\right ) x^2}{1-\log (3)}} \, dx+\frac {\left (4+e^4\right ) \int \exp \left (\frac {\left (4+e^4\right ) x \left ((-5+x)^2 (-1+\log (3))^2+\log ^2(x)\right )}{(-1+\log (3))^2}\right ) x^{-\frac {2 \left (4+e^4\right ) (-5+x) x}{-1+\log (3)}} \log ^2(x) \, dx}{(1-\log (3))^2}+\frac {\left (2 \left (4+e^4\right )\right ) \int \left (-4 \exp \left (\frac {\left (4+e^4\right ) x \left ((-5+x)^2 (-1+\log (3))^2+\log ^2(x)\right )}{(-1+\log (3))^2}\right ) x^{-\frac {2 \left (4+e^4\right ) (-5+x) x}{-1+\log (3)}} \left (1-\frac {\log (243)}{4}\right ) \log (x)+2 \exp \left (\frac {\left (4+e^4\right ) x \left ((-5+x)^2 (-1+\log (3))^2+\log ^2(x)\right )}{(-1+\log (3))^2}\right ) x^{1-\frac {2 \left (4+e^4\right ) (-5+x) x}{-1+\log (3)}} \log \left (\frac {e}{3}\right ) \log (x)\right ) \, dx}{(1-\log (3))^2}-\frac {\left (\left (4+e^4\right ) \left (18-40 \log (3)+20 \log ^2(3)+\log (9)\right )\right ) \int \exp \left (\frac {\left (4+e^4\right ) x \left ((-5+x)^2 (-1+\log (3))^2+\log ^2(x)\right )}{(-1+\log (3))^2}\right ) x^{1-\frac {10 \left (4+e^4\right ) x}{1-\log (3)}+\frac {2 \left (4+e^4\right ) x^2}{1-\log (3)}} \, dx}{(1-\log (3))^2}+\frac {\left (5 \left (4+e^4\right ) (3-\log (243))\right ) \int \exp \left (\frac {\left (4+e^4\right ) x \left ((-5+x)^2 (-1+\log (3))^2+\log ^2(x)\right )}{(-1+\log (3))^2}\right ) x^{-\frac {2 \left (4+e^4\right ) (-5+x) x}{-1+\log (3)}} \, dx}{1-\log (3)}\\ &=\left (3 \left (4+e^4\right )\right ) \int \exp \left (\frac {\left (4+e^4\right ) x \left ((-5+x)^2 (-1+\log (3))^2+\log ^2(x)\right )}{(-1+\log (3))^2}\right ) x^{2-\frac {10 \left (4+e^4\right ) x}{1-\log (3)}+\frac {2 \left (4+e^4\right ) x^2}{1-\log (3)}} \, dx+\frac {\left (4+e^4\right ) \int \exp \left (\frac {\left (4+e^4\right ) x \left ((-5+x)^2 (-1+\log (3))^2+\log ^2(x)\right )}{(-1+\log (3))^2}\right ) x^{-\frac {2 \left (4+e^4\right ) (-5+x) x}{-1+\log (3)}} \log ^2(x) \, dx}{(1-\log (3))^2}+\frac {\left (4 \left (4+e^4\right )\right ) \int \exp \left (\frac {\left (4+e^4\right ) x \left ((-5+x)^2 (-1+\log (3))^2+\log ^2(x)\right )}{(-1+\log (3))^2}\right ) x^{1-\frac {2 \left (4+e^4\right ) (-5+x) x}{-1+\log (3)}} \log (x) \, dx}{1-\log (3)}-\frac {\left (\left (4+e^4\right ) \left (18-40 \log (3)+20 \log ^2(3)+\log (9)\right )\right ) \int \exp \left (\frac {\left (4+e^4\right ) x \left ((-5+x)^2 (-1+\log (3))^2+\log ^2(x)\right )}{(-1+\log (3))^2}\right ) x^{1-\frac {10 \left (4+e^4\right ) x}{1-\log (3)}+\frac {2 \left (4+e^4\right ) x^2}{1-\log (3)}} \, dx}{(1-\log (3))^2}+\frac {\left (5 \left (4+e^4\right ) (3-\log (243))\right ) \int \exp \left (\frac {\left (4+e^4\right ) x \left ((-5+x)^2 (-1+\log (3))^2+\log ^2(x)\right )}{(-1+\log (3))^2}\right ) x^{-\frac {2 \left (4+e^4\right ) (-5+x) x}{-1+\log (3)}} \, dx}{1-\log (3)}-\frac {\left (2 \left (4+e^4\right ) (4-\log (243))\right ) \int \exp \left (\frac {\left (4+e^4\right ) x \left ((-5+x)^2 (-1+\log (3))^2+\log ^2(x)\right )}{(-1+\log (3))^2}\right ) x^{-\frac {2 \left (4+e^4\right ) (-5+x) x}{-1+\log (3)}} \log (x) \, dx}{(1-\log (3))^2}\\ &=\left (3 \left (4+e^4\right )\right ) \int \exp \left (\frac {\left (4+e^4\right ) x \left ((-5+x)^2 (-1+\log (3))^2+\log ^2(x)\right )}{(-1+\log (3))^2}\right ) x^{2-\frac {10 \left (4+e^4\right ) x}{1-\log (3)}+\frac {2 \left (4+e^4\right ) x^2}{1-\log (3)}} \, dx+\frac {\left (4+e^4\right ) \int \exp \left (\frac {\left (4+e^4\right ) x \left ((-5+x)^2 (-1+\log (3))^2+\log ^2(x)\right )}{(-1+\log (3))^2}\right ) x^{-\frac {2 \left (4+e^4\right ) (-5+x) x}{-1+\log (3)}} \log ^2(x) \, dx}{(1-\log (3))^2}+\frac {\left (4 \left (4+e^4\right )\right ) \int \exp \left (\frac {\left (4+e^4\right ) x \left ((-5+x)^2 (-1+\log (3))^2+\log ^2(x)\right )}{(-1+\log (3))^2}\right ) x^{1-\frac {10 \left (4+e^4\right ) x}{1-\log (3)}+\frac {2 \left (4+e^4\right ) x^2}{1-\log (3)}} \log (x) \, dx}{1-\log (3)}-\frac {\left (\left (4+e^4\right ) \left (18-40 \log (3)+20 \log ^2(3)+\log (9)\right )\right ) \int \exp \left (\frac {\left (4+e^4\right ) x \left ((-5+x)^2 (-1+\log (3))^2+\log ^2(x)\right )}{(-1+\log (3))^2}\right ) x^{1-\frac {10 \left (4+e^4\right ) x}{1-\log (3)}+\frac {2 \left (4+e^4\right ) x^2}{1-\log (3)}} \, dx}{(1-\log (3))^2}+\frac {\left (5 \left (4+e^4\right ) (3-\log (243))\right ) \int \exp \left (\frac {\left (4+e^4\right ) x \left ((-5+x)^2 (-1+\log (3))^2+\log ^2(x)\right )}{(-1+\log (3))^2}\right ) x^{-\frac {2 \left (4+e^4\right ) (-5+x) x}{-1+\log (3)}} \, dx}{1-\log (3)}-\frac {\left (2 \left (4+e^4\right ) (4-\log (243))\right ) \int \exp \left (\frac {\left (4+e^4\right ) x \left ((-5+x)^2 (-1+\log (3))^2+\log ^2(x)\right )}{(-1+\log (3))^2}\right ) x^{-\frac {2 \left (4+e^4\right ) (-5+x) x}{-1+\log (3)}} \log (x) \, dx}{(1-\log (3))^2}\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 3.08, size = 200, normalized size = 8.00 \begin {gather*} \frac {\int e^{\frac {\left (100 x-40 x^2+4 x^3+e^4 \left (25 x-10 x^2+x^3\right )\right ) \log ^2\left (\frac {e}{3}\right )+\left (-40 x+8 x^2+e^4 \left (-10 x+2 x^2\right )\right ) \log \left (\frac {e}{3}\right ) \log (x)+\left (4 x+e^4 x\right ) \log ^2(x)}{\log ^2\left (\frac {e}{3}\right )}} \left (\left (100-80 x+12 x^2+e^4 \left (25-20 x+3 x^2\right )\right ) \log ^2\left (\frac {e}{3}\right )+\left (8+2 e^4\right ) \log (x)+\left (4+e^4\right ) \log ^2(x)+\log \left (\frac {e}{3}\right ) \left (-40+8 x+e^4 (-10+2 x)+\left (-40+16 x+e^4 (-10+4 x)\right ) \log (x)\right )\right ) \, dx}{\log ^2\left (\frac {e}{3}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(((100*x - 40*x^2 + 4*x^3 + E^4*(25*x - 10*x^2 + x^3))*Log[E/3]^2 + (-40*x + 8*x^2 + E^4*(-10*x +
2*x^2))*Log[E/3]*Log[x] + (4*x + E^4*x)*Log[x]^2)/Log[E/3]^2)*((100 - 80*x + 12*x^2 + E^4*(25 - 20*x + 3*x^2)
)*Log[E/3]^2 + (8 + 2*E^4)*Log[x] + (4 + E^4)*Log[x]^2 + Log[E/3]*(-40 + 8*x + E^4*(-10 + 2*x) + (-40 + 16*x +
E^4*(-10 + 4*x))*Log[x])))/Log[E/3]^2,x]

[Out]

Integrate[E^(((100*x - 40*x^2 + 4*x^3 + E^4*(25*x - 10*x^2 + x^3))*Log[E/3]^2 + (-40*x + 8*x^2 + E^4*(-10*x +
2*x^2))*Log[E/3]*Log[x] + (4*x + E^4*x)*Log[x]^2)/Log[E/3]^2)*((100 - 80*x + 12*x^2 + E^4*(25 - 20*x + 3*x^2))
*Log[E/3]^2 + (8 + 2*E^4)*Log[x] + (4 + E^4)*Log[x]^2 + Log[E/3]*(-40 + 8*x + E^4*(-10 + 2*x) + (-40 + 16*x +
E^4*(-10 + 4*x))*Log[x])), x]/Log[E/3]^2

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fricas [B]  time = 0.82, size = 169, normalized size = 6.76 \begin {gather*} e^{\left (\frac {4 \, x^{3} + {\left (4 \, x^{3} - 40 \, x^{2} + {\left (x^{3} - 10 \, x^{2} + 25 \, x\right )} e^{4} + 100 \, x\right )} \log \relax (3)^{2} + {\left (x e^{4} + 4 \, x\right )} \log \relax (x)^{2} - 40 \, x^{2} + {\left (x^{3} - 10 \, x^{2} + 25 \, x\right )} e^{4} - 2 \, {\left (4 \, x^{3} - 40 \, x^{2} + {\left (x^{3} - 10 \, x^{2} + 25 \, x\right )} e^{4} + 100 \, x\right )} \log \relax (3) + 2 \, {\left (4 \, x^{2} + {\left (x^{2} - 5 \, x\right )} e^{4} - {\left (4 \, x^{2} + {\left (x^{2} - 5 \, x\right )} e^{4} - 20 \, x\right )} \log \relax (3) - 20 \, x\right )} \log \relax (x) + 100 \, x}{\log \relax (3)^{2} - 2 \, \log \relax (3) + 1}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x^2-20*x+25)*exp(4)+12*x^2-80*x+100)*log(1/3*exp(1))^2+(((4*x-10)*exp(4)+16*x-40)*log(x)+(2*x-1
0)*exp(4)+8*x-40)*log(1/3*exp(1))+(4+exp(4))*log(x)^2+(2*exp(4)+8)*log(x))*exp((((x^3-10*x^2+25*x)*exp(4)+4*x^
3-40*x^2+100*x)*log(1/3*exp(1))^2+((2*x^2-10*x)*exp(4)+8*x^2-40*x)*log(x)*log(1/3*exp(1))+(x*exp(4)+4*x)*log(x
)^2)/log(1/3*exp(1))^2)/log(1/3*exp(1))^2,x, algorithm="fricas")

[Out]

e^((4*x^3 + (4*x^3 - 40*x^2 + (x^3 - 10*x^2 + 25*x)*e^4 + 100*x)*log(3)^2 + (x*e^4 + 4*x)*log(x)^2 - 40*x^2 +
(x^3 - 10*x^2 + 25*x)*e^4 - 2*(4*x^3 - 40*x^2 + (x^3 - 10*x^2 + 25*x)*e^4 + 100*x)*log(3) + 2*(4*x^2 + (x^2 -
5*x)*e^4 - (4*x^2 + (x^2 - 5*x)*e^4 - 20*x)*log(3) - 20*x)*log(x) + 100*x)/(log(3)^2 - 2*log(3) + 1))

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giac [B]  time = 6.34, size = 118, normalized size = 4.72 \begin {gather*} e^{\left (x^{3} e^{4} + 4 \, x^{3} - 10 \, x^{2} e^{4} + \frac {2 \, x^{2} e^{4} \log \relax (x)}{\log \left (\frac {1}{3} \, e\right )} - 40 \, x^{2} + 25 \, x e^{4} + \frac {x e^{4} \log \relax (x)^{2}}{\log \left (\frac {1}{3} \, e\right )^{2}} + \frac {8 \, x^{2} \log \relax (x)}{\log \left (\frac {1}{3} \, e\right )} - \frac {10 \, x e^{4} \log \relax (x)}{\log \left (\frac {1}{3} \, e\right )} + 100 \, x + \frac {4 \, x \log \relax (x)^{2}}{\log \left (\frac {1}{3} \, e\right )^{2}} - \frac {40 \, x \log \relax (x)}{\log \left (\frac {1}{3} \, e\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x^2-20*x+25)*exp(4)+12*x^2-80*x+100)*log(1/3*exp(1))^2+(((4*x-10)*exp(4)+16*x-40)*log(x)+(2*x-1
0)*exp(4)+8*x-40)*log(1/3*exp(1))+(4+exp(4))*log(x)^2+(2*exp(4)+8)*log(x))*exp((((x^3-10*x^2+25*x)*exp(4)+4*x^
3-40*x^2+100*x)*log(1/3*exp(1))^2+((2*x^2-10*x)*exp(4)+8*x^2-40*x)*log(x)*log(1/3*exp(1))+(x*exp(4)+4*x)*log(x
)^2)/log(1/3*exp(1))^2)/log(1/3*exp(1))^2,x, algorithm="giac")

[Out]

e^(x^3*e^4 + 4*x^3 - 10*x^2*e^4 + 2*x^2*e^4*log(x)/log(1/3*e) - 40*x^2 + 25*x*e^4 + x*e^4*log(x)^2/log(1/3*e)^
2 + 8*x^2*log(x)/log(1/3*e) - 10*x*e^4*log(x)/log(1/3*e) + 100*x + 4*x*log(x)^2/log(1/3*e)^2 - 40*x*log(x)/log
(1/3*e))

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maple [B]  time = 0.44, size = 90, normalized size = 3.60

method	result	size
default	\({\mathrm e}^{\frac {\left (\left (x^{3}-10 x^{2}+25 x \right ) {\mathrm e}^{4}+4 x^{3}-40 x^{2}+100 x \right ) \ln \left (\frac {{\mathrm e}}{3}\right )^{2}+\left (\left (2 x^{2}-10 x \right ) {\mathrm e}^{4}+8 x^{2}-40 x \right ) \ln \relax (x ) \ln \left (\frac {{\mathrm e}}{3}\right )+\left (x \,{\mathrm e}^{4}+4 x \right ) \ln \relax (x )^{2}}{\ln \left (\frac {{\mathrm e}}{3}\right )^{2}}}\)	\(90\)
norman	\({\mathrm e}^{\frac {\left (\left (x^{3}-10 x^{2}+25 x \right ) {\mathrm e}^{4}+4 x^{3}-40 x^{2}+100 x \right ) \ln \left (\frac {{\mathrm e}}{3}\right )^{2}+\left (\left (2 x^{2}-10 x \right ) {\mathrm e}^{4}+8 x^{2}-40 x \right ) \ln \relax (x ) \ln \left (\frac {{\mathrm e}}{3}\right )+\left (x \,{\mathrm e}^{4}+4 x \right ) \ln \relax (x )^{2}}{\ln \left (\frac {{\mathrm e}}{3}\right )^{2}}}\)	\(90\)
risch	\(\frac {{\mathrm e}^{\frac {x \left (x \ln \relax (3)-\ln \relax (x )-5 \ln \relax (3)-x +5\right )^{2} \left (4+{\mathrm e}^{4}\right )}{\left (\ln \relax (3)-1\right )^{2}}} \ln \relax (3)^{2}}{\left (1-\ln \relax (3)\right )^{2}}-\frac {2 \,{\mathrm e}^{\frac {x \left (x \ln \relax (3)-\ln \relax (x )-5 \ln \relax (3)-x +5\right )^{2} \left (4+{\mathrm e}^{4}\right )}{\left (\ln \relax (3)-1\right )^{2}}} \ln \relax (3)}{\left (1-\ln \relax (3)\right )^{2}}+\frac {{\mathrm e}^{\frac {x \left (x \ln \relax (3)-\ln \relax (x )-5 \ln \relax (3)-x +5\right )^{2} \left (4+{\mathrm e}^{4}\right )}{\left (\ln \relax (3)-1\right )^{2}}}}{\left (1-\ln \relax (3)\right )^{2}}\)	\(132\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((3*x^2-20*x+25)*exp(4)+12*x^2-80*x+100)*ln(1/3*exp(1))^2+(((4*x-10)*exp(4)+16*x-40)*ln(x)+(2*x-10)*exp(4
)+8*x-40)*ln(1/3*exp(1))+(4+exp(4))*ln(x)^2+(2*exp(4)+8)*ln(x))*exp((((x^3-10*x^2+25*x)*exp(4)+4*x^3-40*x^2+10
0*x)*ln(1/3*exp(1))^2+((2*x^2-10*x)*exp(4)+8*x^2-40*x)*ln(x)*ln(1/3*exp(1))+(x*exp(4)+4*x)*ln(x)^2)/ln(1/3*exp
(1))^2)/ln(1/3*exp(1))^2,x,method=_RETURNVERBOSE)

[Out]

exp((((x^3-10*x^2+25*x)*exp(4)+4*x^3-40*x^2+100*x)*ln(1/3*exp(1))^2+((2*x^2-10*x)*exp(4)+8*x^2-40*x)*ln(x)*ln(
1/3*exp(1))+(x*exp(4)+4*x)*ln(x)^2)/ln(1/3*exp(1))^2)

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maxima [B]  time = 0.78, size = 142, normalized size = 5.68 \begin {gather*} \frac {{\left (\log \relax (3)^{2} - 2 \, \log \relax (3) + 1\right )} e^{\left (x^{3} e^{4} + 4 \, x^{3} - 10 \, x^{2} e^{4} - \frac {2 \, x^{2} e^{4} \log \relax (x)}{\log \relax (3) - 1} + \frac {x e^{4} \log \relax (x)^{2}}{\log \relax (3)^{2} - 2 \, \log \relax (3) + 1} - 40 \, x^{2} + 25 \, x e^{4} - \frac {8 \, x^{2} \log \relax (x)}{\log \relax (3) - 1} + \frac {10 \, x e^{4} \log \relax (x)}{\log \relax (3) - 1} + \frac {4 \, x \log \relax (x)^{2}}{\log \relax (3)^{2} - 2 \, \log \relax (3) + 1} + 100 \, x + \frac {40 \, x \log \relax (x)}{\log \relax (3) - 1}\right )}}{\log \left (\frac {1}{3} \, e\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x^2-20*x+25)*exp(4)+12*x^2-80*x+100)*log(1/3*exp(1))^2+(((4*x-10)*exp(4)+16*x-40)*log(x)+(2*x-1
0)*exp(4)+8*x-40)*log(1/3*exp(1))+(4+exp(4))*log(x)^2+(2*exp(4)+8)*log(x))*exp((((x^3-10*x^2+25*x)*exp(4)+4*x^
3-40*x^2+100*x)*log(1/3*exp(1))^2+((2*x^2-10*x)*exp(4)+8*x^2-40*x)*log(x)*log(1/3*exp(1))+(x*exp(4)+4*x)*log(x
)^2)/log(1/3*exp(1))^2)/log(1/3*exp(1))^2,x, algorithm="maxima")

[Out]

(log(3)^2 - 2*log(3) + 1)*e^(x^3*e^4 + 4*x^3 - 10*x^2*e^4 - 2*x^2*e^4*log(x)/(log(3) - 1) + x*e^4*log(x)^2/(lo
g(3)^2 - 2*log(3) + 1) - 40*x^2 + 25*x*e^4 - 8*x^2*log(x)/(log(3) - 1) + 10*x*e^4*log(x)/(log(3) - 1) + 4*x*lo
g(x)^2/(log(3)^2 - 2*log(3) + 1) + 100*x + 40*x*log(x)/(log(3) - 1))/log(1/3*e)^2

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mupad [B]  time = 8.25, size = 362, normalized size = 14.48 \begin {gather*} {\left (\frac {1}{9}\right )}^{\frac {100\,x+25\,x\,{\mathrm {e}}^4-10\,x^2\,{\mathrm {e}}^4+x^3\,{\mathrm {e}}^4-40\,x^2+4\,x^3}{{\ln \relax (3)}^2-2\,\ln \relax (3)+1}}\,x^{\frac {2\,\left (20\,x+5\,x\,{\mathrm {e}}^4-x^2\,{\mathrm {e}}^4-4\,x^2\right )}{\ln \relax (3)-1}}\,{\mathrm {e}}^{\frac {x^3\,{\mathrm {e}}^4\,{\ln \relax (3)}^2}{{\ln \relax (3)}^2-2\,\ln \relax (3)+1}}\,{\mathrm {e}}^{-\frac {10\,x^2\,{\mathrm {e}}^4\,{\ln \relax (3)}^2}{{\ln \relax (3)}^2-2\,\ln \relax (3)+1}}\,{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^4\,{\ln \relax (x)}^2}{{\ln \relax (3)}^2-2\,\ln \relax (3)+1}}\,{\mathrm {e}}^{\frac {x^3\,{\mathrm {e}}^4}{{\ln \relax (3)}^2-2\,\ln \relax (3)+1}}\,{\mathrm {e}}^{-\frac {10\,x^2\,{\mathrm {e}}^4}{{\ln \relax (3)}^2-2\,\ln \relax (3)+1}}\,{\mathrm {e}}^{\frac {100\,x\,{\ln \relax (3)}^2}{{\ln \relax (3)}^2-2\,\ln \relax (3)+1}}\,{\mathrm {e}}^{\frac {100\,x}{{\ln \relax (3)}^2-2\,\ln \relax (3)+1}}\,{\mathrm {e}}^{\frac {4\,x^3\,{\ln \relax (3)}^2}{{\ln \relax (3)}^2-2\,\ln \relax (3)+1}}\,{\mathrm {e}}^{-\frac {40\,x^2\,{\ln \relax (3)}^2}{{\ln \relax (3)}^2-2\,\ln \relax (3)+1}}\,{\mathrm {e}}^{\frac {25\,x\,{\mathrm {e}}^4\,{\ln \relax (3)}^2}{{\ln \relax (3)}^2-2\,\ln \relax (3)+1}}\,{\mathrm {e}}^{\frac {4\,x\,{\ln \relax (x)}^2}{{\ln \relax (3)}^2-2\,\ln \relax (3)+1}}\,{\mathrm {e}}^{\frac {4\,x^3}{{\ln \relax (3)}^2-2\,\ln \relax (3)+1}}\,{\mathrm {e}}^{-\frac {40\,x^2}{{\ln \relax (3)}^2-2\,\ln \relax (3)+1}}\,{\mathrm {e}}^{\frac {25\,x\,{\mathrm {e}}^4}{{\ln \relax (3)}^2-2\,\ln \relax (3)+1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((log(x)^2*(4*x + x*exp(4)) + log(exp(1)/3)^2*(100*x + exp(4)*(25*x - 10*x^2 + x^3) - 40*x^2 + 4*x^3)
- log(exp(1)/3)*log(x)*(40*x + exp(4)*(10*x - 2*x^2) - 8*x^2))/log(exp(1)/3)^2)*(log(exp(1)/3)*(8*x + log(x)*(
16*x + exp(4)*(4*x - 10) - 40) + exp(4)*(2*x - 10) - 40) + log(x)*(2*exp(4) + 8) + log(x)^2*(exp(4) + 4) + log
(exp(1)/3)^2*(exp(4)*(3*x^2 - 20*x + 25) - 80*x + 12*x^2 + 100)))/log(exp(1)/3)^2,x)

[Out]

(1/9)^((100*x + 25*x*exp(4) - 10*x^2*exp(4) + x^3*exp(4) - 40*x^2 + 4*x^3)/(log(3)^2 - 2*log(3) + 1))*x^((2*(2
0*x + 5*x*exp(4) - x^2*exp(4) - 4*x^2))/(log(3) - 1))*exp((x^3*exp(4)*log(3)^2)/(log(3)^2 - 2*log(3) + 1))*exp
(-(10*x^2*exp(4)*log(3)^2)/(log(3)^2 - 2*log(3) + 1))*exp((x*exp(4)*log(x)^2)/(log(3)^2 - 2*log(3) + 1))*exp((
x^3*exp(4))/(log(3)^2 - 2*log(3) + 1))*exp(-(10*x^2*exp(4))/(log(3)^2 - 2*log(3) + 1))*exp((100*x*log(3)^2)/(l
og(3)^2 - 2*log(3) + 1))*exp((100*x)/(log(3)^2 - 2*log(3) + 1))*exp((4*x^3*log(3)^2)/(log(3)^2 - 2*log(3) + 1)
)*exp(-(40*x^2*log(3)^2)/(log(3)^2 - 2*log(3) + 1))*exp((25*x*exp(4)*log(3)^2)/(log(3)^2 - 2*log(3) + 1))*exp(
(4*x*log(x)^2)/(log(3)^2 - 2*log(3) + 1))*exp((4*x^3)/(log(3)^2 - 2*log(3) + 1))*exp(-(40*x^2)/(log(3)^2 - 2*l
og(3) + 1))*exp((25*x*exp(4))/(log(3)^2 - 2*log(3) + 1))

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sympy [B]  time = 1.92, size = 90, normalized size = 3.60 \begin {gather*} e^{\frac {\left (4 x + x e^{4}\right ) \log {\relax (x )}^{2} + \left (8 x^{2} - 40 x + \left (2 x^{2} - 10 x\right ) e^{4}\right ) \log {\left (\frac {e}{3} \right )} \log {\relax (x )} + \left (4 x^{3} - 40 x^{2} + 100 x + \left (x^{3} - 10 x^{2} + 25 x\right ) e^{4}\right ) \log {\left (\frac {e}{3} \right )}^{2}}{\log {\left (\frac {e}{3} \right )}^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x**2-20*x+25)*exp(4)+12*x**2-80*x+100)*ln(1/3*exp(1))**2+(((4*x-10)*exp(4)+16*x-40)*ln(x)+(2*x-
10)*exp(4)+8*x-40)*ln(1/3*exp(1))+(4+exp(4))*ln(x)**2+(2*exp(4)+8)*ln(x))*exp((((x**3-10*x**2+25*x)*exp(4)+4*x
**3-40*x**2+100*x)*ln(1/3*exp(1))**2+((2*x**2-10*x)*exp(4)+8*x**2-40*x)*ln(x)*ln(1/3*exp(1))+(x*exp(4)+4*x)*ln
(x)**2)/ln(1/3*exp(1))**2)/ln(1/3*exp(1))**2,x)

[Out]

exp(((4*x + x*exp(4))*log(x)**2 + (8*x**2 - 40*x + (2*x**2 - 10*x)*exp(4))*log(E/3)*log(x) + (4*x**3 - 40*x**2
+ 100*x + (x**3 - 10*x**2 + 25*x)*exp(4))*log(E/3)**2)/log(E/3)**2)

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