3.95.81 \(\int \frac {20 x+4 x^2+e^4 (-60 x^2-144 x^3-24 x^4)+72 x^2 \log (4)+(44 x^2+8 x^3) \log (x)+(10+2 x+e^4 (-30 x-72 x^2-12 x^3)+36 x \log (4)+(22 x+4 x^2) \log (x)) \log (\frac {e^4 (15 x+3 x^2)-9 \log (4)+(-5-x) \log (x)}{3 \log (4)})}{e^4 (-15 x^2-3 x^3)+9 x \log (4)+(5 x+x^2) \log (x)} \, dx\)

Optimal. Leaf size=31 \[ \left (2 x+\log \left (-3+\frac {(5+x) \left (3 e^4 x-\log (x)\right )}{3 \log (4)}\right )\right )^2 \]

________________________________________________________________________________________

Rubi [F]  time = 6.74, 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 {20 x+4 x^2+e^4 \left (-60 x^2-144 x^3-24 x^4\right )+72 x^2 \log (4)+\left (44 x^2+8 x^3\right ) \log (x)+\left (10+2 x+e^4 \left (-30 x-72 x^2-12 x^3\right )+36 x \log (4)+\left (22 x+4 x^2\right ) \log (x)\right ) \log \left (\frac {e^4 \left (15 x+3 x^2\right )-9 \log (4)+(-5-x) \log (x)}{3 \log (4)}\right )}{e^4 \left (-15 x^2-3 x^3\right )+9 x \log (4)+\left (5 x+x^2\right ) \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(20*x + 4*x^2 + E^4*(-60*x^2 - 144*x^3 - 24*x^4) + 72*x^2*Log[4] + (44*x^2 + 8*x^3)*Log[x] + (10 + 2*x + E
^4*(-30*x - 72*x^2 - 12*x^3) + 36*x*Log[4] + (22*x + 4*x^2)*Log[x])*Log[(E^4*(15*x + 3*x^2) - 9*Log[4] + (-5 -
x)*Log[x])/(3*Log[4])])/(E^4*(-15*x^2 - 3*x^3) + 9*x*Log[4] + (5*x + x^2)*Log[x]),x]

[Out]

4*x + 4*x^2 - 20*Log[5 + x] - 4*(5 - 9*Log[4])*Defer[Int][(15*E^4*x + 3*E^4*x^2 - 9*Log[4] - 5*Log[x] - x*Log[
x])^(-1), x] - 4*(1 - 15*E^4)*Defer[Int][x/(15*E^4*x + 3*E^4*x^2 - 9*Log[4] - 5*Log[x] - x*Log[x]), x] + 12*E^
4*Defer[Int][x^2/(15*E^4*x + 3*E^4*x^2 - 9*Log[4] - 5*Log[x] - x*Log[x]), x] - 180*Log[4]*Defer[Int][1/((5 + x
)*(15*E^4*x + 3*E^4*x^2 - 9*Log[4] - 5*Log[x] - x*Log[x])), x] - 2*(1 - 15*E^4 + 18*Log[4])*Defer[Int][Log[(3*
E^4*x*(5 + x) - 9*Log[4] - (5 + x)*Log[x])/Log[64]]/(15*E^4*x + 3*E^4*x^2 - 9*Log[4] - 5*Log[x] - x*Log[x]), x
] - 10*Defer[Int][Log[(3*E^4*x*(5 + x) - 9*Log[4] - (5 + x)*Log[x])/Log[64]]/(x*(15*E^4*x + 3*E^4*x^2 - 9*Log[
4] - 5*Log[x] - x*Log[x])), x] + 72*E^4*Defer[Int][(x*Log[(3*E^4*x*(5 + x) - 9*Log[4] - (5 + x)*Log[x])/Log[64
]])/(15*E^4*x + 3*E^4*x^2 - 9*Log[4] - 5*Log[x] - x*Log[x]), x] + 12*E^4*Defer[Int][(x^2*Log[(3*E^4*x*(5 + x)
- 9*Log[4] - (5 + x)*Log[x])/Log[64]])/(15*E^4*x + 3*E^4*x^2 - 9*Log[4] - 5*Log[x] - x*Log[x]), x] - 4*Defer[I
nt][(x*Log[x]*Log[(3*E^4*x*(5 + x) - 9*Log[4] - (5 + x)*Log[x])/Log[64]])/(15*E^4*x + 3*E^4*x^2 - 9*Log[4] - 5
*Log[x] - x*Log[x]), x] + 22*Defer[Int][(Log[x]*Log[(3*E^4*x*(5 + x) - 9*Log[4] - (5 + x)*Log[x])/Log[64]])/(-
15*E^4*x - 3*E^4*x^2 + 9*Log[4] + 5*Log[x] + x*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {20 x+e^4 \left (-60 x^2-144 x^3-24 x^4\right )+x^2 (4+72 \log (4))+\left (44 x^2+8 x^3\right ) \log (x)+\left (10+2 x+e^4 \left (-30 x-72 x^2-12 x^3\right )+36 x \log (4)+\left (22 x+4 x^2\right ) \log (x)\right ) \log \left (\frac {e^4 \left (15 x+3 x^2\right )-9 \log (4)+(-5-x) \log (x)}{3 \log (4)}\right )}{e^4 \left (-15 x^2-3 x^3\right )+9 x \log (4)+\left (5 x+x^2\right ) \log (x)} \, dx\\ &=\int \frac {2 \left (5-36 e^4 x^2-6 e^4 x^3-x \left (-1+15 e^4-18 \log (4)\right )+x (11+2 x) \log (x)\right ) \left (-2 x-\log \left (\frac {3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)}{\log (64)}\right )\right )}{x \left (3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)\right )} \, dx\\ &=2 \int \frac {\left (5-36 e^4 x^2-6 e^4 x^3-x \left (-1+15 e^4-18 \log (4)\right )+x (11+2 x) \log (x)\right ) \left (-2 x-\log \left (\frac {3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)}{\log (64)}\right )\right )}{x \left (3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)\right )} \, dx\\ &=2 \int \left (\frac {2 \left (-5+36 e^4 x^2+6 e^4 x^3-x \left (1-15 e^4+18 \log (4)\right )-11 x \log (x)-2 x^2 \log (x)\right )}{15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)}+\frac {\left (-5+36 e^4 x^2+6 e^4 x^3-x \left (1-15 e^4+18 \log (4)\right )-11 x \log (x)-2 x^2 \log (x)\right ) \log \left (\frac {3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)}{\log (64)}\right )}{x \left (15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)\right )}\right ) \, dx\\ &=2 \int \frac {\left (-5+36 e^4 x^2+6 e^4 x^3-x \left (1-15 e^4+18 \log (4)\right )-11 x \log (x)-2 x^2 \log (x)\right ) \log \left (\frac {3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)}{\log (64)}\right )}{x \left (15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)\right )} \, dx+4 \int \frac {-5+36 e^4 x^2+6 e^4 x^3-x \left (1-15 e^4+18 \log (4)\right )-11 x \log (x)-2 x^2 \log (x)}{15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)} \, dx\\ &=2 \int \left (-\frac {5 \log \left (\frac {3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)}{\log (64)}\right )}{x \left (15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)\right )}+\frac {36 e^4 x \log \left (\frac {3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)}{\log (64)}\right )}{15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)}+\frac {6 e^4 x^2 \log \left (\frac {3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)}{\log (64)}\right )}{15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)}+\frac {\left (-1+15 e^4-18 \log (4)\right ) \log \left (\frac {3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)}{\log (64)}\right )}{15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)}-\frac {2 x \log (x) \log \left (\frac {3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)}{\log (64)}\right )}{15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)}+\frac {11 \log (x) \log \left (\frac {3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)}{\log (64)}\right )}{-15 e^4 x-3 e^4 x^2+9 \log (4)+5 \log (x)+x \log (x)}\right ) \, dx+4 \int \left (\frac {x (11+2 x)}{5+x}+\frac {-25-\left (1-30 e^4\right ) x^2+3 e^4 x^3-x \left (10-75 e^4-9 \log (4)\right )}{(5+x) \left (15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)\right )}\right ) \, dx\\ &=4 \int \frac {x (11+2 x)}{5+x} \, dx+4 \int \frac {-25-\left (1-30 e^4\right ) x^2+3 e^4 x^3-x \left (10-75 e^4-9 \log (4)\right )}{(5+x) \left (15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)\right )} \, dx-4 \int \frac {x \log (x) \log \left (\frac {3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)}{\log (64)}\right )}{15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)} \, dx-10 \int \frac {\log \left (\frac {3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)}{\log (64)}\right )}{x \left (15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)\right )} \, dx+22 \int \frac {\log (x) \log \left (\frac {3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)}{\log (64)}\right )}{-15 e^4 x-3 e^4 x^2+9 \log (4)+5 \log (x)+x \log (x)} \, dx+\left (12 e^4\right ) \int \frac {x^2 \log \left (\frac {3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)}{\log (64)}\right )}{15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)} \, dx+\left (72 e^4\right ) \int \frac {x \log \left (\frac {3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)}{\log (64)}\right )}{15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)} \, dx-\left (2 \left (1-15 e^4+18 \log (4)\right )\right ) \int \frac {\log \left (\frac {3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)}{\log (64)}\right )}{15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)} \, dx\\ &=4 \int \left (1+2 x-\frac {5}{5+x}\right ) \, dx+4 \int \left (\frac {\left (-1+15 e^4\right ) x}{15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)}+\frac {3 e^4 x^2}{15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)}-\frac {5 \left (1-\frac {9 \log (4)}{5}\right )}{15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)}-\frac {45 \log (4)}{(5+x) \left (15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)\right )}\right ) \, dx-4 \int \frac {x \log (x) \log \left (\frac {3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)}{\log (64)}\right )}{15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)} \, dx-10 \int \frac {\log \left (\frac {3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)}{\log (64)}\right )}{x \left (15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)\right )} \, dx+22 \int \frac {\log (x) \log \left (\frac {3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)}{\log (64)}\right )}{-15 e^4 x-3 e^4 x^2+9 \log (4)+5 \log (x)+x \log (x)} \, dx+\left (12 e^4\right ) \int \frac {x^2 \log \left (\frac {3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)}{\log (64)}\right )}{15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)} \, dx+\left (72 e^4\right ) \int \frac {x \log \left (\frac {3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)}{\log (64)}\right )}{15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)} \, dx-\left (2 \left (1-15 e^4+18 \log (4)\right )\right ) \int \frac {\log \left (\frac {3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)}{\log (64)}\right )}{15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)} \, dx\\ &=4 x+4 x^2-20 \log (5+x)-4 \int \frac {x \log (x) \log \left (\frac {3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)}{\log (64)}\right )}{15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)} \, dx-10 \int \frac {\log \left (\frac {3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)}{\log (64)}\right )}{x \left (15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)\right )} \, dx+22 \int \frac {\log (x) \log \left (\frac {3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)}{\log (64)}\right )}{-15 e^4 x-3 e^4 x^2+9 \log (4)+5 \log (x)+x \log (x)} \, dx+\left (12 e^4\right ) \int \frac {x^2}{15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)} \, dx+\left (12 e^4\right ) \int \frac {x^2 \log \left (\frac {3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)}{\log (64)}\right )}{15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)} \, dx+\left (72 e^4\right ) \int \frac {x \log \left (\frac {3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)}{\log (64)}\right )}{15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)} \, dx-\left (4 \left (1-15 e^4\right )\right ) \int \frac {x}{15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)} \, dx-(4 (5-9 \log (4))) \int \frac {1}{15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)} \, dx-(180 \log (4)) \int \frac {1}{(5+x) \left (15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)\right )} \, dx-\left (2 \left (1-15 e^4+18 \log (4)\right )\right ) \int \frac {\log \left (\frac {3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)}{\log (64)}\right )}{15 e^4 x+3 e^4 x^2-9 \log (4)-5 \log (x)-x \log (x)} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [B]  time = 0.08, size = 71, normalized size = 2.29 \begin {gather*} -2 \left (-2 x^2-2 x \log \left (\frac {3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)}{\log (64)}\right )-\frac {1}{2} \log ^2\left (\frac {3 e^4 x (5+x)-9 \log (4)-(5+x) \log (x)}{\log (64)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(20*x + 4*x^2 + E^4*(-60*x^2 - 144*x^3 - 24*x^4) + 72*x^2*Log[4] + (44*x^2 + 8*x^3)*Log[x] + (10 + 2
*x + E^4*(-30*x - 72*x^2 - 12*x^3) + 36*x*Log[4] + (22*x + 4*x^2)*Log[x])*Log[(E^4*(15*x + 3*x^2) - 9*Log[4] +
(-5 - x)*Log[x])/(3*Log[4])])/(E^4*(-15*x^2 - 3*x^3) + 9*x*Log[4] + (5*x + x^2)*Log[x]),x]

[Out]

-2*(-2*x^2 - 2*x*Log[(3*E^4*x*(5 + x) - 9*Log[4] - (5 + x)*Log[x])/Log[64]] - Log[(3*E^4*x*(5 + x) - 9*Log[4]
- (5 + x)*Log[x])/Log[64]]^2/2)

________________________________________________________________________________________

fricas [B]  time = 0.65, size = 71, normalized size = 2.29 \begin {gather*} 4 \, x^{2} + 4 \, x \log \left (\frac {3 \, {\left (x^{2} + 5 \, x\right )} e^{4} - {\left (x + 5\right )} \log \relax (x) - 18 \, \log \relax (2)}{6 \, \log \relax (2)}\right ) + \log \left (\frac {3 \, {\left (x^{2} + 5 \, x\right )} e^{4} - {\left (x + 5\right )} \log \relax (x) - 18 \, \log \relax (2)}{6 \, \log \relax (2)}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x^2+22*x)*log(x)+72*x*log(2)+(-12*x^3-72*x^2-30*x)*exp(4)+2*x+10)*log(1/6*((-x-5)*log(x)-18*log
(2)+(3*x^2+15*x)*exp(4))/log(2))+(8*x^3+44*x^2)*log(x)+144*x^2*log(2)+(-24*x^4-144*x^3-60*x^2)*exp(4)+4*x^2+20
*x)/((x^2+5*x)*log(x)+18*x*log(2)+(-3*x^3-15*x^2)*exp(4)),x, algorithm="fricas")

[Out]

4*x^2 + 4*x*log(1/6*(3*(x^2 + 5*x)*e^4 - (x + 5)*log(x) - 18*log(2))/log(2)) + log(1/6*(3*(x^2 + 5*x)*e^4 - (x
+ 5)*log(x) - 18*log(2))/log(2))^2

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {2 \, {\left (72 \, x^{2} \log \relax (2) + 2 \, x^{2} - 6 \, {\left (2 \, x^{4} + 12 \, x^{3} + 5 \, x^{2}\right )} e^{4} + 2 \, {\left (2 \, x^{3} + 11 \, x^{2}\right )} \log \relax (x) - {\left (3 \, {\left (2 \, x^{3} + 12 \, x^{2} + 5 \, x\right )} e^{4} - 36 \, x \log \relax (2) - {\left (2 \, x^{2} + 11 \, x\right )} \log \relax (x) - x - 5\right )} \log \left (\frac {3 \, {\left (x^{2} + 5 \, x\right )} e^{4} - {\left (x + 5\right )} \log \relax (x) - 18 \, \log \relax (2)}{6 \, \log \relax (2)}\right ) + 10 \, x\right )}}{3 \, {\left (x^{3} + 5 \, x^{2}\right )} e^{4} - 18 \, x \log \relax (2) - {\left (x^{2} + 5 \, x\right )} \log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x^2+22*x)*log(x)+72*x*log(2)+(-12*x^3-72*x^2-30*x)*exp(4)+2*x+10)*log(1/6*((-x-5)*log(x)-18*log
(2)+(3*x^2+15*x)*exp(4))/log(2))+(8*x^3+44*x^2)*log(x)+144*x^2*log(2)+(-24*x^4-144*x^3-60*x^2)*exp(4)+4*x^2+20
*x)/((x^2+5*x)*log(x)+18*x*log(2)+(-3*x^3-15*x^2)*exp(4)),x, algorithm="giac")

[Out]

integrate(-2*(72*x^2*log(2) + 2*x^2 - 6*(2*x^4 + 12*x^3 + 5*x^2)*e^4 + 2*(2*x^3 + 11*x^2)*log(x) - (3*(2*x^3 +
12*x^2 + 5*x)*e^4 - 36*x*log(2) - (2*x^2 + 11*x)*log(x) - x - 5)*log(1/6*(3*(x^2 + 5*x)*e^4 - (x + 5)*log(x)
- 18*log(2))/log(2)) + 10*x)/(3*(x^3 + 5*x^2)*e^4 - 18*x*log(2) - (x^2 + 5*x)*log(x)), x)

________________________________________________________________________________________

maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (4 x^{2}+22 x \right ) \ln \relax (x )+72 x \ln \relax (2)+\left (-12 x^{3}-72 x^{2}-30 x \right ) {\mathrm e}^{4}+2 x +10\right ) \ln \left (\frac {\left (-x -5\right ) \ln \relax (x )-18 \ln \relax (2)+\left (3 x^{2}+15 x \right ) {\mathrm e}^{4}}{6 \ln \relax (2)}\right )+\left (8 x^{3}+44 x^{2}\right ) \ln \relax (x )+144 x^{2} \ln \relax (2)+\left (-24 x^{4}-144 x^{3}-60 x^{2}\right ) {\mathrm e}^{4}+4 x^{2}+20 x}{\left (x^{2}+5 x \right ) \ln \relax (x )+18 x \ln \relax (2)+\left (-3 x^{3}-15 x^{2}\right ) {\mathrm e}^{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((4*x^2+22*x)*ln(x)+72*x*ln(2)+(-12*x^3-72*x^2-30*x)*exp(4)+2*x+10)*ln(1/6*((-x-5)*ln(x)-18*ln(2)+(3*x^2+
15*x)*exp(4))/ln(2))+(8*x^3+44*x^2)*ln(x)+144*x^2*ln(2)+(-24*x^4-144*x^3-60*x^2)*exp(4)+4*x^2+20*x)/((x^2+5*x)
*ln(x)+18*x*ln(2)+(-3*x^3-15*x^2)*exp(4)),x)

[Out]

int((((4*x^2+22*x)*ln(x)+72*x*ln(2)+(-12*x^3-72*x^2-30*x)*exp(4)+2*x+10)*ln(1/6*((-x-5)*ln(x)-18*ln(2)+(3*x^2+
15*x)*exp(4))/ln(2))+(8*x^3+44*x^2)*ln(x)+144*x^2*ln(2)+(-24*x^4-144*x^3-60*x^2)*exp(4)+4*x^2+20*x)/((x^2+5*x)
*ln(x)+18*x*ln(2)+(-3*x^3-15*x^2)*exp(4)),x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -2 \, \int \frac {72 \, x^{2} \log \relax (2) + 2 \, x^{2} - 6 \, {\left (2 \, x^{4} + 12 \, x^{3} + 5 \, x^{2}\right )} e^{4} + 2 \, {\left (2 \, x^{3} + 11 \, x^{2}\right )} \log \relax (x) - {\left (3 \, {\left (2 \, x^{3} + 12 \, x^{2} + 5 \, x\right )} e^{4} - 36 \, x \log \relax (2) - {\left (2 \, x^{2} + 11 \, x\right )} \log \relax (x) - x - 5\right )} \log \left (\frac {3 \, {\left (x^{2} + 5 \, x\right )} e^{4} - {\left (x + 5\right )} \log \relax (x) - 18 \, \log \relax (2)}{6 \, \log \relax (2)}\right ) + 10 \, x}{3 \, {\left (x^{3} + 5 \, x^{2}\right )} e^{4} - 18 \, x \log \relax (2) - {\left (x^{2} + 5 \, x\right )} \log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x^2+22*x)*log(x)+72*x*log(2)+(-12*x^3-72*x^2-30*x)*exp(4)+2*x+10)*log(1/6*((-x-5)*log(x)-18*log
(2)+(3*x^2+15*x)*exp(4))/log(2))+(8*x^3+44*x^2)*log(x)+144*x^2*log(2)+(-24*x^4-144*x^3-60*x^2)*exp(4)+4*x^2+20
*x)/((x^2+5*x)*log(x)+18*x*log(2)+(-3*x^3-15*x^2)*exp(4)),x, algorithm="maxima")

[Out]

-2*integrate((72*x^2*log(2) + 2*x^2 - 6*(2*x^4 + 12*x^3 + 5*x^2)*e^4 + 2*(2*x^3 + 11*x^2)*log(x) - (3*(2*x^3 +
12*x^2 + 5*x)*e^4 - 36*x*log(2) - (2*x^2 + 11*x)*log(x) - x - 5)*log(1/6*(3*(x^2 + 5*x)*e^4 - (x + 5)*log(x)
- 18*log(2))/log(2)) + 10*x)/(3*(x^3 + 5*x^2)*e^4 - 18*x*log(2) - (x^2 + 5*x)*log(x)), x)

________________________________________________________________________________________

mupad [B]  time = 6.66, size = 38, normalized size = 1.23 \begin {gather*} {\left (2\,x+\ln \left (-\frac {3\,\ln \relax (2)+\frac {\ln \relax (x)\,\left (x+5\right )}{6}-\frac {{\mathrm {e}}^4\,\left (3\,x^2+15\,x\right )}{6}}{\ln \relax (2)}\right )\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((20*x + log(-(3*log(2) + (log(x)*(x + 5))/6 - (exp(4)*(15*x + 3*x^2))/6)/log(2))*(2*x + 72*x*log(2) - exp(
4)*(30*x + 72*x^2 + 12*x^3) + log(x)*(22*x + 4*x^2) + 10) + log(x)*(44*x^2 + 8*x^3) + 144*x^2*log(2) - exp(4)*
(60*x^2 + 144*x^3 + 24*x^4) + 4*x^2)/(18*x*log(2) + log(x)*(5*x + x^2) - exp(4)*(15*x^2 + 3*x^3)),x)

[Out]

(2*x + log(-(3*log(2) + (log(x)*(x + 5))/6 - (exp(4)*(15*x + 3*x^2))/6)/log(2)))^2

________________________________________________________________________________________

sympy [B]  time = 0.76, size = 76, normalized size = 2.45 \begin {gather*} 4 x^{2} + 4 x \log {\left (\frac {\frac {\left (- x - 5\right ) \log {\relax (x )}}{6} + \frac {\left (3 x^{2} + 15 x\right ) e^{4}}{6} - 3 \log {\relax (2 )}}{\log {\relax (2 )}} \right )} + \log {\left (\frac {\frac {\left (- x - 5\right ) \log {\relax (x )}}{6} + \frac {\left (3 x^{2} + 15 x\right ) e^{4}}{6} - 3 \log {\relax (2 )}}{\log {\relax (2 )}} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x**2+22*x)*ln(x)+72*x*ln(2)+(-12*x**3-72*x**2-30*x)*exp(4)+2*x+10)*ln(1/6*((-x-5)*ln(x)-18*ln(2
)+(3*x**2+15*x)*exp(4))/ln(2))+(8*x**3+44*x**2)*ln(x)+144*x**2*ln(2)+(-24*x**4-144*x**3-60*x**2)*exp(4)+4*x**2
+20*x)/((x**2+5*x)*ln(x)+18*x*ln(2)+(-3*x**3-15*x**2)*exp(4)),x)

[Out]

4*x**2 + 4*x*log(((-x - 5)*log(x)/6 + (3*x**2 + 15*x)*exp(4)/6 - 3*log(2))/log(2)) + log(((-x - 5)*log(x)/6 +
(3*x**2 + 15*x)*exp(4)/6 - 3*log(2))/log(2))**2

________________________________________________________________________________________