3.1.38 \(\int \frac {-x^3 \log (x)+8 e^{e^x+x} x^3 \log (x)+8 e^{2 e^x+x} x^3 \log (x)+(x^2-x^3+(-2 x^2+2 x^3) \log (x)+e^{e^x} (8 x^2-16 x^2 \log (x))+e^{2 e^x} (4 x^2-8 x^2 \log (x))) \log (\frac {1}{4} (-1-8 e^{e^x}-4 e^{2 e^x}+x))+(2+16 e^{e^x}+8 e^{2 e^x}-2 x) \log ^2(\frac {1}{4} (-1-8 e^{e^x}-4 e^{2 e^x}+x))}{x^5+8 e^{e^x} x^5+4 e^{2 e^x} x^5-x^6+(4 x^3+32 e^{e^x} x^3+16 e^{2 e^x} x^3-4 x^4) \log (\frac {1}{4} (-1-8 e^{e^x}-4 e^{2 e^x}+x))+(4 x+32 e^{e^x} x+16 e^{2 e^x} x-4 x^2) \log ^2(\frac {1}{4} (-1-8 e^{e^x}-4 e^{2 e^x}+x))} \, dx\)

Optimal. Leaf size=33 \[ \frac {\log (x)}{2+\frac {x^2}{\log \left (-\left (1+e^{e^x}\right )^2+\frac {3+x}{4}\right )}} \]

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Rubi [F]  time = 46.34, 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 {-x^3 \log (x)+8 e^{e^x+x} x^3 \log (x)+8 e^{2 e^x+x} x^3 \log (x)+\left (x^2-x^3+\left (-2 x^2+2 x^3\right ) \log (x)+e^{e^x} \left (8 x^2-16 x^2 \log (x)\right )+e^{2 e^x} \left (4 x^2-8 x^2 \log (x)\right )\right ) \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )+\left (2+16 e^{e^x}+8 e^{2 e^x}-2 x\right ) \log ^2\left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )}{x^5+8 e^{e^x} x^5+4 e^{2 e^x} x^5-x^6+\left (4 x^3+32 e^{e^x} x^3+16 e^{2 e^x} x^3-4 x^4\right ) \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )+\left (4 x+32 e^{e^x} x+16 e^{2 e^x} x-4 x^2\right ) \log ^2\left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-(x^3*Log[x]) + 8*E^(E^x + x)*x^3*Log[x] + 8*E^(2*E^x + x)*x^3*Log[x] + (x^2 - x^3 + (-2*x^2 + 2*x^3)*Log
[x] + E^E^x*(8*x^2 - 16*x^2*Log[x]) + E^(2*E^x)*(4*x^2 - 8*x^2*Log[x]))*Log[(-1 - 8*E^E^x - 4*E^(2*E^x) + x)/4
] + (2 + 16*E^E^x + 8*E^(2*E^x) - 2*x)*Log[(-1 - 8*E^E^x - 4*E^(2*E^x) + x)/4]^2)/(x^5 + 8*E^E^x*x^5 + 4*E^(2*
E^x)*x^5 - x^6 + (4*x^3 + 32*E^E^x*x^3 + 16*E^(2*E^x)*x^3 - 4*x^4)*Log[(-1 - 8*E^E^x - 4*E^(2*E^x) + x)/4] + (
4*x + 32*E^E^x*x + 16*E^(2*E^x)*x - 4*x^2)*Log[(-1 - 8*E^E^x - 4*E^(2*E^x) + x)/4]^2),x]

[Out]

Log[x]/2 + Log[16]*Defer[Int][(x*Log[x])/((1 + 8*E^E^x + 4*E^(2*E^x) - x)*(x^2 + 2*Log[(-1 - 8*E^E^x - 4*E^(2*
E^x) + x)/4])^2), x] - Defer[Int][(x^2*Log[x])/((1 + 8*E^E^x + 4*E^(2*E^x) - x)*(x^2 + 2*Log[(-1 - 8*E^E^x - 4
*E^(2*E^x) + x)/4])^2), x] + 8*Defer[Int][(E^(E^x + x)*x^2*Log[x])/((1 + 8*E^E^x + 4*E^(2*E^x) - x)*(x^2 + 2*L
og[(-1 - 8*E^E^x - 4*E^(2*E^x) + x)/4])^2), x] + 8*Defer[Int][(E^(2*E^x + x)*x^2*Log[x])/((1 + 8*E^E^x + 4*E^(
2*E^x) - x)*(x^2 + 2*Log[(-1 - 8*E^E^x - 4*E^(2*E^x) + x)/4])^2), x] + Defer[Int][(x^3*Log[x])/(x^2 + 2*Log[(-
1 - 8*E^E^x - 4*E^(2*E^x) + x)/4])^2, x] + 2*Defer[Int][(x*Log[x]*Log[(-1 - 8*E^E^x - 4*E^(2*E^x) + x)/4])/((1
+ 8*E^E^x + 4*E^(2*E^x) - x)*(x^2 + 2*Log[(-1 - 8*E^E^x - 4*E^(2*E^x) + x)/4])^2), x] - Defer[Int][x/(x^2 + 2
*Log[(-1 - 8*E^E^x - 4*E^(2*E^x) + x)/4]), x]/2 - Defer[Int][(x*Log[x])/(x^2 + 2*Log[(-1 - 8*E^E^x - 4*E^(2*E^
x) + x)/4]), x] - 2*Defer[Int][(x*Log[x]*Log[-1 - 8*E^E^x - 4*E^(2*E^x) + x])/((1 + 8*E^E^x + 4*E^(2*E^x) - x)
*(x^2 + 2*Log[(-1 - 8*E^E^x - 4*E^(2*E^x) + x)/4])^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {\log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )}{x^3+2 x \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )}+\frac {x \log (x) \left (-x+8 e^{e^x+x} x+8 e^{2 e^x+x} x+\log (16)-2 \left (8 e^{e^x}+4 e^{2 e^x}-x\right ) \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )-2 \log \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2}\right ) \, dx\\ &=\int \frac {\log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )}{x^3+2 x \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )} \, dx+\int \frac {x \log (x) \left (-x+8 e^{e^x+x} x+8 e^{2 e^x+x} x+\log (16)-2 \left (8 e^{e^x}+4 e^{2 e^x}-x\right ) \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )-2 \log \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx\\ &=\int \left (\frac {1}{2 x}-\frac {x}{2 \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )}\right ) \, dx+\int \left (-\frac {x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2}+\frac {8 e^{e^x+x} \left (1+e^{e^x}\right ) x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2}+\frac {x \log (16) \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2}-\frac {2 \left (8 e^{e^x}+4 e^{2 e^x}-x\right ) x \log (x) \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2}-\frac {2 x \log (x) \log \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2}\right ) \, dx\\ &=\frac {\log (x)}{2}-\frac {1}{2} \int \frac {x}{x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )} \, dx-2 \int \frac {\left (8 e^{e^x}+4 e^{2 e^x}-x\right ) x \log (x) \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx-2 \int \frac {x \log (x) \log \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx+8 \int \frac {e^{e^x+x} \left (1+e^{e^x}\right ) x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx+\log (16) \int \frac {x \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx-\int \frac {x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx\\ &=\frac {\log (x)}{2}-\frac {1}{2} \int \frac {x}{x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )} \, dx-2 \int \left (\frac {x \log (x) \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )}{\left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2}-\frac {x \log (x) \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2}\right ) \, dx-2 \int \frac {x \log (x) \log \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx+8 \int \left (\frac {e^{e^x+x} x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2}+\frac {e^{2 e^x+x} x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2}\right ) \, dx+\log (16) \int \frac {x \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx-\int \frac {x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx\\ &=\frac {\log (x)}{2}-\frac {1}{2} \int \frac {x}{x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )} \, dx-2 \int \frac {x \log (x) \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )}{\left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx+2 \int \frac {x \log (x) \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx-2 \int \frac {x \log (x) \log \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx+8 \int \frac {e^{e^x+x} x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx+8 \int \frac {e^{2 e^x+x} x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx+\log (16) \int \frac {x \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx-\int \frac {x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx\\ &=\frac {\log (x)}{2}-\frac {1}{2} \int \frac {x}{x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )} \, dx+2 \int \frac {x \log (x) \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx-2 \int \left (-\frac {x^3 \log (x)}{2 \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2}+\frac {x \log (x)}{2 \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )}\right ) \, dx-2 \int \frac {x \log (x) \log \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx+8 \int \frac {e^{e^x+x} x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx+8 \int \frac {e^{2 e^x+x} x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx+\log (16) \int \frac {x \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx-\int \frac {x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx\\ &=\frac {\log (x)}{2}-\frac {1}{2} \int \frac {x}{x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )} \, dx+2 \int \frac {x \log (x) \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx-2 \int \frac {x \log (x) \log \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx+8 \int \frac {e^{e^x+x} x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx+8 \int \frac {e^{2 e^x+x} x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx+\log (16) \int \frac {x \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx-\int \frac {x^2 \log (x)}{\left (1+8 e^{e^x}+4 e^{2 e^x}-x\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx+\int \frac {x^3 \log (x)}{\left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )^2} \, dx-\int \frac {x \log (x)}{x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.65, size = 187, normalized size = 5.67 \begin {gather*} \frac {\log (x) \left (2 \left (8 e^{e^x+x}+8 e^{2 e^x+x}+8 e^{e^x} x+4 e^{2 e^x} x-x^2\right ) \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )-(-1+x) \left (\log (16)-2 \log \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )}{2 \left (-1+8 e^{e^x+x}+8 e^{2 e^x+x}+x+8 e^{e^x} x+4 e^{2 e^x} x-x^2\right ) \left (x^2+2 \log \left (\frac {1}{4} \left (-1-8 e^{e^x}-4 e^{2 e^x}+x\right )\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-(x^3*Log[x]) + 8*E^(E^x + x)*x^3*Log[x] + 8*E^(2*E^x + x)*x^3*Log[x] + (x^2 - x^3 + (-2*x^2 + 2*x^
3)*Log[x] + E^E^x*(8*x^2 - 16*x^2*Log[x]) + E^(2*E^x)*(4*x^2 - 8*x^2*Log[x]))*Log[(-1 - 8*E^E^x - 4*E^(2*E^x)
+ x)/4] + (2 + 16*E^E^x + 8*E^(2*E^x) - 2*x)*Log[(-1 - 8*E^E^x - 4*E^(2*E^x) + x)/4]^2)/(x^5 + 8*E^E^x*x^5 + 4
*E^(2*E^x)*x^5 - x^6 + (4*x^3 + 32*E^E^x*x^3 + 16*E^(2*E^x)*x^3 - 4*x^4)*Log[(-1 - 8*E^E^x - 4*E^(2*E^x) + x)/
4] + (4*x + 32*E^E^x*x + 16*E^(2*E^x)*x - 4*x^2)*Log[(-1 - 8*E^E^x - 4*E^(2*E^x) + x)/4]^2),x]

[Out]

(Log[x]*(2*(8*E^(E^x + x) + 8*E^(2*E^x + x) + 8*E^E^x*x + 4*E^(2*E^x)*x - x^2)*Log[(-1 - 8*E^E^x - 4*E^(2*E^x)
+ x)/4] - (-1 + x)*(Log[16] - 2*Log[-1 - 8*E^E^x - 4*E^(2*E^x) + x])))/(2*(-1 + 8*E^(E^x + x) + 8*E^(2*E^x +
x) + x + 8*E^E^x*x + 4*E^(2*E^x)*x - x^2)*(x^2 + 2*Log[(-1 - 8*E^E^x - 4*E^(2*E^x) + x)/4]))

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fricas [B]  time = 0.76, size = 83, normalized size = 2.52 \begin {gather*} \frac {\log \left (\frac {1}{4} \, {\left ({\left (x - 1\right )} e^{\left (2 \, x\right )} - 4 \, e^{\left (2 \, x + 2 \, e^{x}\right )} - 8 \, e^{\left (2 \, x + e^{x}\right )}\right )} e^{\left (-2 \, x\right )}\right ) \log \relax (x)}{x^{2} + 2 \, \log \left (\frac {1}{4} \, {\left ({\left (x - 1\right )} e^{\left (2 \, x\right )} - 4 \, e^{\left (2 \, x + 2 \, e^{x}\right )} - 8 \, e^{\left (2 \, x + e^{x}\right )}\right )} e^{\left (-2 \, x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*exp(exp(x))^2+16*exp(exp(x))-2*x+2)*log(-exp(exp(x))^2-2*exp(exp(x))+1/4*x-1/4)^2+((-8*x^2*log(x
)+4*x^2)*exp(exp(x))^2+(-16*x^2*log(x)+8*x^2)*exp(exp(x))+(2*x^3-2*x^2)*log(x)-x^3+x^2)*log(-exp(exp(x))^2-2*e
xp(exp(x))+1/4*x-1/4)+8*x^3*exp(x)*log(x)*exp(exp(x))^2+8*x^3*exp(x)*log(x)*exp(exp(x))-x^3*log(x))/((16*x*exp
(exp(x))^2+32*x*exp(exp(x))-4*x^2+4*x)*log(-exp(exp(x))^2-2*exp(exp(x))+1/4*x-1/4)^2+(16*x^3*exp(exp(x))^2+32*
x^3*exp(exp(x))-4*x^4+4*x^3)*log(-exp(exp(x))^2-2*exp(exp(x))+1/4*x-1/4)+4*x^5*exp(exp(x))^2+8*x^5*exp(exp(x))
-x^6+x^5),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 3.95, size = 57, normalized size = 1.73 \begin {gather*} -\frac {2 \, \log \relax (2) \log \relax (x) - \log \left (x - 4 \, e^{\left (2 \, e^{x}\right )} - 8 \, e^{\left (e^{x}\right )} - 1\right ) \log \relax (x)}{x^{2} - 4 \, \log \relax (2) + 2 \, \log \left (x - 4 \, e^{\left (2 \, e^{x}\right )} - 8 \, e^{\left (e^{x}\right )} - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*exp(exp(x))^2+16*exp(exp(x))-2*x+2)*log(-exp(exp(x))^2-2*exp(exp(x))+1/4*x-1/4)^2+((-8*x^2*log(x
)+4*x^2)*exp(exp(x))^2+(-16*x^2*log(x)+8*x^2)*exp(exp(x))+(2*x^3-2*x^2)*log(x)-x^3+x^2)*log(-exp(exp(x))^2-2*e
xp(exp(x))+1/4*x-1/4)+8*x^3*exp(x)*log(x)*exp(exp(x))^2+8*x^3*exp(x)*log(x)*exp(exp(x))-x^3*log(x))/((16*x*exp
(exp(x))^2+32*x*exp(exp(x))-4*x^2+4*x)*log(-exp(exp(x))^2-2*exp(exp(x))+1/4*x-1/4)^2+(16*x^3*exp(exp(x))^2+32*
x^3*exp(exp(x))-4*x^4+4*x^3)*log(-exp(exp(x))^2-2*exp(exp(x))+1/4*x-1/4)+4*x^5*exp(exp(x))^2+8*x^5*exp(exp(x))
-x^6+x^5),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.17, size = 39, normalized size = 1.18

method	result	size
risch	\(\frac {\ln \relax (x )}{2}-\frac {x^{2} \ln \relax (x )}{2 \left (x^{2}+2 \ln \left (-{\mathrm e}^{2 \,{\mathrm e}^{x}}-2 \,{\mathrm e}^{{\mathrm e}^{x}}+\frac {x}{4}-\frac {1}{4}\right )\right )}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*exp(exp(x))^2+16*exp(exp(x))-2*x+2)*ln(-exp(exp(x))^2-2*exp(exp(x))+1/4*x-1/4)^2+((-8*x^2*ln(x)+4*x^2)
*exp(exp(x))^2+(-16*x^2*ln(x)+8*x^2)*exp(exp(x))+(2*x^3-2*x^2)*ln(x)-x^3+x^2)*ln(-exp(exp(x))^2-2*exp(exp(x))+
1/4*x-1/4)+8*x^3*exp(x)*ln(x)*exp(exp(x))^2+8*x^3*exp(x)*ln(x)*exp(exp(x))-x^3*ln(x))/((16*x*exp(exp(x))^2+32*
x*exp(exp(x))-4*x^2+4*x)*ln(-exp(exp(x))^2-2*exp(exp(x))+1/4*x-1/4)^2+(16*x^3*exp(exp(x))^2+32*x^3*exp(exp(x))
-4*x^4+4*x^3)*ln(-exp(exp(x))^2-2*exp(exp(x))+1/4*x-1/4)+4*x^5*exp(exp(x))^2+8*x^5*exp(exp(x))-x^6+x^5),x,meth
od=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 1.44, size = 40, normalized size = 1.21 \begin {gather*} -\frac {x^{2} \log \relax (x)}{2 \, {\left (x^{2} - 4 \, \log \relax (2) + 2 \, \log \left (x - 4 \, e^{\left (2 \, e^{x}\right )} - 8 \, e^{\left (e^{x}\right )} - 1\right )\right )}} + \frac {1}{2} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*exp(exp(x))^2+16*exp(exp(x))-2*x+2)*log(-exp(exp(x))^2-2*exp(exp(x))+1/4*x-1/4)^2+((-8*x^2*log(x
)+4*x^2)*exp(exp(x))^2+(-16*x^2*log(x)+8*x^2)*exp(exp(x))+(2*x^3-2*x^2)*log(x)-x^3+x^2)*log(-exp(exp(x))^2-2*e
xp(exp(x))+1/4*x-1/4)+8*x^3*exp(x)*log(x)*exp(exp(x))^2+8*x^3*exp(x)*log(x)*exp(exp(x))-x^3*log(x))/((16*x*exp
(exp(x))^2+32*x*exp(exp(x))-4*x^2+4*x)*log(-exp(exp(x))^2-2*exp(exp(x))+1/4*x-1/4)^2+(16*x^3*exp(exp(x))^2+32*
x^3*exp(exp(x))-4*x^4+4*x^3)*log(-exp(exp(x))^2-2*exp(exp(x))+1/4*x-1/4)+4*x^5*exp(exp(x))^2+8*x^5*exp(exp(x))
-x^6+x^5),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\ln \left (\frac {x}{4}-2\,{\mathrm {e}}^{{\mathrm {e}}^x}-{\mathrm {e}}^{2\,{\mathrm {e}}^x}-\frac {1}{4}\right )}^2\,\left (16\,{\mathrm {e}}^{{\mathrm {e}}^x}-2\,x+8\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}+2\right )-\ln \left (\frac {x}{4}-2\,{\mathrm {e}}^{{\mathrm {e}}^x}-{\mathrm {e}}^{2\,{\mathrm {e}}^x}-\frac {1}{4}\right )\,\left (\ln \relax (x)\,\left (2\,x^2-2\,x^3\right )+{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (16\,x^2\,\ln \relax (x)-8\,x^2\right )+{\mathrm {e}}^{2\,{\mathrm {e}}^x}\,\left (8\,x^2\,\ln \relax (x)-4\,x^2\right )-x^2+x^3\right )-x^3\,\ln \relax (x)+8\,x^3\,{\mathrm {e}}^{x+{\mathrm {e}}^x}\,\ln \relax (x)+8\,x^3\,{\mathrm {e}}^{x+2\,{\mathrm {e}}^x}\,\ln \relax (x)}{{\ln \left (\frac {x}{4}-2\,{\mathrm {e}}^{{\mathrm {e}}^x}-{\mathrm {e}}^{2\,{\mathrm {e}}^x}-\frac {1}{4}\right )}^2\,\left (4\,x+32\,x\,{\mathrm {e}}^{{\mathrm {e}}^x}+16\,x\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}-4\,x^2\right )+8\,x^5\,{\mathrm {e}}^{{\mathrm {e}}^x}+\ln \left (\frac {x}{4}-2\,{\mathrm {e}}^{{\mathrm {e}}^x}-{\mathrm {e}}^{2\,{\mathrm {e}}^x}-\frac {1}{4}\right )\,\left (32\,x^3\,{\mathrm {e}}^{{\mathrm {e}}^x}+4\,x^3-4\,x^4+16\,x^3\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}\right )+x^5-x^6+4\,x^5\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x/4 - 2*exp(exp(x)) - exp(2*exp(x)) - 1/4)^2*(16*exp(exp(x)) - 2*x + 8*exp(2*exp(x)) + 2) - log(x/4 -
2*exp(exp(x)) - exp(2*exp(x)) - 1/4)*(log(x)*(2*x^2 - 2*x^3) + exp(exp(x))*(16*x^2*log(x) - 8*x^2) + exp(2*ex
p(x))*(8*x^2*log(x) - 4*x^2) - x^2 + x^3) - x^3*log(x) + 8*x^3*exp(exp(x))*exp(x)*log(x) + 8*x^3*exp(2*exp(x))
*exp(x)*log(x))/(log(x/4 - 2*exp(exp(x)) - exp(2*exp(x)) - 1/4)^2*(4*x + 32*x*exp(exp(x)) + 16*x*exp(2*exp(x))
- 4*x^2) + 8*x^5*exp(exp(x)) + log(x/4 - 2*exp(exp(x)) - exp(2*exp(x)) - 1/4)*(32*x^3*exp(exp(x)) + 4*x^3 - 4
*x^4 + 16*x^3*exp(2*exp(x))) + x^5 - x^6 + 4*x^5*exp(2*exp(x))),x)

[Out]

int((log(x/4 - 2*exp(exp(x)) - exp(2*exp(x)) - 1/4)^2*(16*exp(exp(x)) - 2*x + 8*exp(2*exp(x)) + 2) - log(x/4 -
2*exp(exp(x)) - exp(2*exp(x)) - 1/4)*(log(x)*(2*x^2 - 2*x^3) + exp(exp(x))*(16*x^2*log(x) - 8*x^2) + exp(2*ex
p(x))*(8*x^2*log(x) - 4*x^2) - x^2 + x^3) - x^3*log(x) + 8*x^3*exp(x + exp(x))*log(x) + 8*x^3*exp(x + 2*exp(x)
)*log(x))/(log(x/4 - 2*exp(exp(x)) - exp(2*exp(x)) - 1/4)^2*(4*x + 32*x*exp(exp(x)) + 16*x*exp(2*exp(x)) - 4*x
^2) + 8*x^5*exp(exp(x)) + log(x/4 - 2*exp(exp(x)) - exp(2*exp(x)) - 1/4)*(32*x^3*exp(exp(x)) + 4*x^3 - 4*x^4 +
16*x^3*exp(2*exp(x))) + x^5 - x^6 + 4*x^5*exp(2*exp(x))), x)

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sympy [A]  time = 2.48, size = 39, normalized size = 1.18 \begin {gather*} - \frac {x^{2} \log {\relax (x )}}{2 x^{2} + 4 \log {\left (\frac {x}{4} - e^{2 e^{x}} - 2 e^{e^{x}} - \frac {1}{4} \right )}} + \frac {\log {\relax (x )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*exp(exp(x))**2+16*exp(exp(x))-2*x+2)*ln(-exp(exp(x))**2-2*exp(exp(x))+1/4*x-1/4)**2+((-8*x**2*ln
(x)+4*x**2)*exp(exp(x))**2+(-16*x**2*ln(x)+8*x**2)*exp(exp(x))+(2*x**3-2*x**2)*ln(x)-x**3+x**2)*ln(-exp(exp(x)
)**2-2*exp(exp(x))+1/4*x-1/4)+8*x**3*exp(x)*ln(x)*exp(exp(x))**2+8*x**3*exp(x)*ln(x)*exp(exp(x))-x**3*ln(x))/(
(16*x*exp(exp(x))**2+32*x*exp(exp(x))-4*x**2+4*x)*ln(-exp(exp(x))**2-2*exp(exp(x))+1/4*x-1/4)**2+(16*x**3*exp(
exp(x))**2+32*x**3*exp(exp(x))-4*x**4+4*x**3)*ln(-exp(exp(x))**2-2*exp(exp(x))+1/4*x-1/4)+4*x**5*exp(exp(x))**
2+8*x**5*exp(exp(x))-x**6+x**5),x)

[Out]

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

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