3.50.3 \(\int \frac {(64+32 x+12 x^2+x^3) \log (x) \log (\frac {1}{3} e^{\frac {-8-2 x+x^2}{16+4 x}} x)+(32+16 x+2 x^2) \log ^2(\frac {1}{3} e^{\frac {-8-2 x+x^2}{16+4 x}} x)}{16 x+8 x^2+x^3} \, dx\)

Optimal. Leaf size=30 \[ 2 \log (x) \log ^2\left (\frac {1}{3} e^{\frac {1}{4} (-2+x)-\frac {x}{4+x}} x\right ) \]

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Rubi [F]  time = 1.77, 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 {\left (64+32 x+12 x^2+x^3\right ) \log (x) \log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{16+4 x}} x\right )+\left (32+16 x+2 x^2\right ) \log ^2\left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{16+4 x}} x\right )}{16 x+8 x^2+x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((64 + 32*x + 12*x^2 + x^3)*Log[x]*Log[(E^((-8 - 2*x + x^2)/(16 + 4*x))*x)/3] + (32 + 16*x + 2*x^2)*Log[(E
^((-8 - 2*x + x^2)/(16 + 4*x))*x)/3]^2)/(16*x + 8*x^2 + x^3),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (64+32 x+12 x^2+x^3\right ) \log (x) \log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{16+4 x}} x\right )+\left (32+16 x+2 x^2\right ) \log ^2\left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{16+4 x}} x\right )}{x \left (16+8 x+x^2\right )} \, dx\\ &=\int \frac {\left (64+32 x+12 x^2+x^3\right ) \log (x) \log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{16+4 x}} x\right )+\left (32+16 x+2 x^2\right ) \log ^2\left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{16+4 x}} x\right )}{x (4+x)^2} \, dx\\ &=\int \left (\frac {\left (64+32 x+12 x^2+x^3\right ) \log (x) \log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{x (4+x)^2}+\frac {2 \log ^2\left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{x}\right ) \, dx\\ &=2 \int \frac {\log ^2\left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{x} \, dx+\int \frac {\left (64+32 x+12 x^2+x^3\right ) \log (x) \log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{x (4+x)^2} \, dx\\ &=2 \int \frac {\log ^2\left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{x} \, dx+\int \left (\log (x) \log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )+\frac {4 \log (x) \log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{x}-\frac {16 \log (x) \log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{(4+x)^2}\right ) \, dx\\ &=2 \int \frac {\log ^2\left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{x} \, dx+4 \int \frac {\log (x) \log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{x} \, dx-16 \int \frac {\log (x) \log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{(4+x)^2} \, dx+\int \log (x) \log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right ) \, dx\\ &=x \log (x) \log \left (\frac {1}{3} e^{-\frac {8+2 x-x^2}{4 (4+x)}} x\right )+\frac {16 \log (x) \log \left (\frac {1}{3} e^{-\frac {8+2 x-x^2}{4 (4+x)}} x\right )}{4+x}+2 \int \frac {\log ^2\left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{x} \, dx+4 \int \frac {\log (x) \log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{x} \, dx+16 \int \frac {\left (-64-32 x-12 x^2-x^3\right ) \log (x)}{4 x (4+x)^3} \, dx-16 \int \frac {\log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{x (4+x)} \, dx-\int \frac {\left (64+32 x+12 x^2+x^3\right ) \log (x)}{4 (4+x)^2} \, dx-\int \log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right ) \, dx\\ &=-x \log \left (\frac {1}{3} e^{-\frac {8+2 x-x^2}{4 (4+x)}} x\right )+x \log (x) \log \left (\frac {1}{3} e^{-\frac {8+2 x-x^2}{4 (4+x)}} x\right )+\frac {16 \log (x) \log \left (\frac {1}{3} e^{-\frac {8+2 x-x^2}{4 (4+x)}} x\right )}{4+x}-\frac {1}{4} \int \frac {\left (64+32 x+12 x^2+x^3\right ) \log (x)}{(4+x)^2} \, dx+2 \int \frac {\log ^2\left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{x} \, dx+4 \int \frac {\left (-64-32 x-12 x^2-x^3\right ) \log (x)}{x (4+x)^3} \, dx+4 \int \frac {\log (x) \log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{x} \, dx-16 \int \left (\frac {\log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{4 x}-\frac {\log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{4 (4+x)}\right ) \, dx+\int \frac {64+32 x+12 x^2+x^3}{4 (4+x)^2} \, dx\\ &=-x \log \left (\frac {1}{3} e^{-\frac {8+2 x-x^2}{4 (4+x)}} x\right )+x \log (x) \log \left (\frac {1}{3} e^{-\frac {8+2 x-x^2}{4 (4+x)}} x\right )+\frac {16 \log (x) \log \left (\frac {1}{3} e^{-\frac {8+2 x-x^2}{4 (4+x)}} x\right )}{4+x}+\frac {1}{4} \int \frac {64+32 x+12 x^2+x^3}{(4+x)^2} \, dx-\frac {1}{4} \int \left (4 \log (x)+x \log (x)+\frac {64 \log (x)}{(4+x)^2}-\frac {16 \log (x)}{4+x}\right ) \, dx+2 \int \frac {\log ^2\left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{x} \, dx+4 \int \left (-\frac {\log (x)}{x}+\frac {16 \log (x)}{(4+x)^3}\right ) \, dx-4 \int \frac {\log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{x} \, dx+4 \int \frac {\log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{4+x} \, dx+4 \int \frac {\log (x) \log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{x} \, dx\\ &=-x \log \left (\frac {1}{3} e^{-\frac {8+2 x-x^2}{4 (4+x)}} x\right )+x \log (x) \log \left (\frac {1}{3} e^{-\frac {8+2 x-x^2}{4 (4+x)}} x\right )+\frac {16 \log (x) \log \left (\frac {1}{3} e^{-\frac {8+2 x-x^2}{4 (4+x)}} x\right )}{4+x}+\frac {1}{4} \int \left (4+x+\frac {64}{(4+x)^2}-\frac {16}{4+x}\right ) \, dx-\frac {1}{4} \int x \log (x) \, dx+2 \int \frac {\log ^2\left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{x} \, dx-4 \int \frac {\log (x)}{x} \, dx+4 \int \frac {\log (x)}{4+x} \, dx-4 \int \frac {\log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{x} \, dx+4 \int \frac {\log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{4+x} \, dx+4 \int \frac {\log (x) \log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{x} \, dx-16 \int \frac {\log (x)}{(4+x)^2} \, dx+64 \int \frac {\log (x)}{(4+x)^3} \, dx-\int \log (x) \, dx\\ &=2 x+\frac {3 x^2}{16}-\frac {16}{4+x}-x \log (x)-\frac {1}{8} x^2 \log (x)-\frac {32 \log (x)}{(4+x)^2}-\frac {4 x \log (x)}{4+x}+4 \log \left (1+\frac {x}{4}\right ) \log (x)-2 \log ^2(x)-x \log \left (\frac {1}{3} e^{-\frac {8+2 x-x^2}{4 (4+x)}} x\right )+x \log (x) \log \left (\frac {1}{3} e^{-\frac {8+2 x-x^2}{4 (4+x)}} x\right )+\frac {16 \log (x) \log \left (\frac {1}{3} e^{-\frac {8+2 x-x^2}{4 (4+x)}} x\right )}{4+x}-4 \log (4+x)+2 \int \frac {\log ^2\left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{x} \, dx+4 \int \frac {1}{4+x} \, dx-4 \int \frac {\log \left (1+\frac {x}{4}\right )}{x} \, dx-4 \int \frac {\log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{x} \, dx+4 \int \frac {\log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{4+x} \, dx+4 \int \frac {\log (x) \log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{x} \, dx+32 \int \frac {1}{x (4+x)^2} \, dx\\ &=2 x+\frac {3 x^2}{16}-\frac {16}{4+x}-x \log (x)-\frac {1}{8} x^2 \log (x)-\frac {32 \log (x)}{(4+x)^2}-\frac {4 x \log (x)}{4+x}+4 \log \left (1+\frac {x}{4}\right ) \log (x)-2 \log ^2(x)-x \log \left (\frac {1}{3} e^{-\frac {8+2 x-x^2}{4 (4+x)}} x\right )+x \log (x) \log \left (\frac {1}{3} e^{-\frac {8+2 x-x^2}{4 (4+x)}} x\right )+\frac {16 \log (x) \log \left (\frac {1}{3} e^{-\frac {8+2 x-x^2}{4 (4+x)}} x\right )}{4+x}+4 \text {Li}_2\left (-\frac {x}{4}\right )+2 \int \frac {\log ^2\left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{x} \, dx-4 \int \frac {\log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{x} \, dx+4 \int \frac {\log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{4+x} \, dx+4 \int \frac {\log (x) \log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{x} \, dx+32 \int \left (\frac {1}{16 x}-\frac {1}{4 (4+x)^2}-\frac {1}{16 (4+x)}\right ) \, dx\\ &=2 x+\frac {3 x^2}{16}-\frac {8}{4+x}+2 \log (x)-x \log (x)-\frac {1}{8} x^2 \log (x)-\frac {32 \log (x)}{(4+x)^2}-\frac {4 x \log (x)}{4+x}+4 \log \left (1+\frac {x}{4}\right ) \log (x)-2 \log ^2(x)-x \log \left (\frac {1}{3} e^{-\frac {8+2 x-x^2}{4 (4+x)}} x\right )+x \log (x) \log \left (\frac {1}{3} e^{-\frac {8+2 x-x^2}{4 (4+x)}} x\right )+\frac {16 \log (x) \log \left (\frac {1}{3} e^{-\frac {8+2 x-x^2}{4 (4+x)}} x\right )}{4+x}-2 \log (4+x)+4 \text {Li}_2\left (-\frac {x}{4}\right )+2 \int \frac {\log ^2\left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{x} \, dx-4 \int \frac {\log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{x} \, dx+4 \int \frac {\log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{4+x} \, dx+4 \int \frac {\log (x) \log \left (\frac {1}{3} e^{\frac {-8-2 x+x^2}{4 (4+x)}} x\right )}{x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.18, size = 81, normalized size = 2.70 \begin {gather*} \frac {18+\frac {9 x}{2}-\frac {9 x^2}{4}+9 (4+x) \log \left (\frac {1}{3} e^{\frac {1}{4} \left (-6+x+\frac {16}{4+x}\right )} x\right )+(4+x) \log (x) \left (-9+2 \log ^2\left (\frac {1}{3} e^{\frac {1}{4} \left (-6+x+\frac {16}{4+x}\right )} x\right )\right )}{4+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((64 + 32*x + 12*x^2 + x^3)*Log[x]*Log[(E^((-8 - 2*x + x^2)/(16 + 4*x))*x)/3] + (32 + 16*x + 2*x^2)*
Log[(E^((-8 - 2*x + x^2)/(16 + 4*x))*x)/3]^2)/(16*x + 8*x^2 + x^3),x]

[Out]

(18 + (9*x)/2 - (9*x^2)/4 + 9*(4 + x)*Log[(E^((-6 + x + 16/(4 + x))/4)*x)/3] + (4 + x)*Log[x]*(-9 + 2*Log[(E^(
(-6 + x + 16/(4 + x))/4)*x)/3]^2))/(4 + x)

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fricas [B]  time = 0.76, size = 110, normalized size = 3.67 \begin {gather*} \frac {16 \, {\left (x^{2} + 8 \, x + 16\right )} \log \relax (x)^{3} + 8 \, {\left (x^{3} + 2 \, x^{2} - 4 \, {\left (x^{2} + 8 \, x + 16\right )} \log \relax (3) - 16 \, x - 32\right )} \log \relax (x)^{2} + {\left (x^{4} - 4 \, x^{3} + 16 \, {\left (x^{2} + 8 \, x + 16\right )} \log \relax (3)^{2} - 12 \, x^{2} - 8 \, {\left (x^{3} + 2 \, x^{2} - 16 \, x - 32\right )} \log \relax (3) + 32 \, x + 64\right )} \log \relax (x)}{8 \, {\left (x^{2} + 8 \, x + 16\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+16*x+32)*log(1/3*x*exp((x^2-2*x-8)/(4*x+16)))^2+(x^3+12*x^2+32*x+64)*log(x)*log(1/3*x*exp((x
^2-2*x-8)/(4*x+16))))/(x^3+8*x^2+16*x),x, algorithm="fricas")

[Out]

1/8*(16*(x^2 + 8*x + 16)*log(x)^3 + 8*(x^3 + 2*x^2 - 4*(x^2 + 8*x + 16)*log(3) - 16*x - 32)*log(x)^2 + (x^4 -
4*x^3 + 16*(x^2 + 8*x + 16)*log(3)^2 - 12*x^2 - 8*(x^3 + 2*x^2 - 16*x - 32)*log(3) + 32*x + 64)*log(x))/(x^2 +
8*x + 16)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, {\left (x^{2} + 8 \, x + 16\right )} \log \left (\frac {1}{3} \, x e^{\left (\frac {x^{2} - 2 \, x - 8}{4 \, {\left (x + 4\right )}}\right )}\right )^{2} + {\left (x^{3} + 12 \, x^{2} + 32 \, x + 64\right )} \log \left (\frac {1}{3} \, x e^{\left (\frac {x^{2} - 2 \, x - 8}{4 \, {\left (x + 4\right )}}\right )}\right ) \log \relax (x)}{x^{3} + 8 \, x^{2} + 16 \, x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+16*x+32)*log(1/3*x*exp((x^2-2*x-8)/(4*x+16)))^2+(x^3+12*x^2+32*x+64)*log(x)*log(1/3*x*exp((x
^2-2*x-8)/(4*x+16))))/(x^3+8*x^2+16*x),x, algorithm="giac")

[Out]

integrate((2*(x^2 + 8*x + 16)*log(1/3*x*e^(1/4*(x^2 - 2*x - 8)/(x + 4)))^2 + (x^3 + 12*x^2 + 32*x + 64)*log(1/
3*x*e^(1/4*(x^2 - 2*x - 8)/(x + 4)))*log(x))/(x^3 + 8*x^2 + 16*x), x)

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maple [C]  time = 0.42, size = 944, normalized size = 31.47

method	result	size
risch	\(-4 \ln \relax (3) \ln \relax (x )^{2}+2 \ln \relax (x )^{3}+2 \ln \relax (3)^{2} \ln \relax (x )-\frac {\pi ^{2} \ln \relax (x ) \mathrm {csgn}\left (i {\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right )^{2} \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right )^{4}}{2}+\pi ^{2} \ln \relax (x ) \mathrm {csgn}\left (i {\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right )^{5}-\frac {\pi ^{2} \ln \relax (x ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right )^{6}}{2}-2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right ) \ln \relax (x )^{2}-\frac {\pi ^{2} \ln \relax (x ) \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right )^{4}}{2}+\pi ^{2} \ln \relax (x ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right )^{5}+\left (-2 i \ln \relax (x ) \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right )+2 i \ln \relax (x ) \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right )^{2}+2 i \ln \relax (x ) \pi \,\mathrm {csgn}\left (i {\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right )^{2}-2 i \ln \relax (x ) \pi \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right )^{3}-4 \ln \relax (3) \ln \relax (x )+4 \ln \relax (x )^{2}\right ) \ln \left ({\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right )+2 \ln \relax (x ) \ln \left ({\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right )^{2}-2 i \pi \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right )^{3} \ln \relax (x )^{2}-2 i \pi \ln \relax (x ) \ln \relax (3) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right )^{2}+2 i \pi \ln \relax (x ) \ln \relax (3) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right )^{3}-\frac {\pi ^{2} \ln \relax (x ) \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right )^{2} \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right )^{2}}{2}+\pi ^{2} \ln \relax (x ) \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right )^{3}+\pi ^{2} \ln \relax (x ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right )^{2} \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right )^{3}-2 \pi ^{2} \ln \relax (x ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right )^{4}+2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right )^{2} \ln \relax (x )^{2}+2 i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right )^{2} \ln \relax (x )^{2}-2 i \pi \ln \relax (x ) \ln \relax (3) \mathrm {csgn}\left (i {\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right )^{2}+2 i \pi \ln \relax (x ) \ln \relax (3) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {\left (x -4\right ) \left (2+x \right )}{4 x +16}}\right )\)	\(944\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^2+16*x+32)*ln(1/3*x*exp((x^2-2*x-8)/(4*x+16)))^2+(x^3+12*x^2+32*x+64)*ln(x)*ln(1/3*x*exp((x^2-2*x-8)
/(4*x+16))))/(x^3+8*x^2+16*x),x,method=_RETURNVERBOSE)

[Out]

-4*ln(3)*ln(x)^2+2*ln(x)^3+2*ln(3)^2*ln(x)-1/2*Pi^2*ln(x)*csgn(I*exp(1/4*(x-4)*(2+x)/(4+x)))^2*csgn(I*x*exp(1/
4*(x-4)*(2+x)/(4+x)))^4+Pi^2*ln(x)*csgn(I*exp(1/4*(x-4)*(2+x)/(4+x)))*csgn(I*x*exp(1/4*(x-4)*(2+x)/(4+x)))^5-1
/2*Pi^2*ln(x)*csgn(I*x*exp(1/4*(x-4)*(2+x)/(4+x)))^6-2*I*Pi*csgn(I*x)*csgn(I*exp(1/4*(x-4)*(2+x)/(4+x)))*csgn(
I*x*exp(1/4*(x-4)*(2+x)/(4+x)))*ln(x)^2-1/2*Pi^2*ln(x)*csgn(I*x)^2*csgn(I*x*exp(1/4*(x-4)*(2+x)/(4+x)))^4+Pi^2
*ln(x)*csgn(I*x)*csgn(I*x*exp(1/4*(x-4)*(2+x)/(4+x)))^5+(-2*I*ln(x)*Pi*csgn(I*x)*csgn(I*exp(1/4*(x-4)*(2+x)/(4
+x)))*csgn(I*x*exp(1/4*(x-4)*(2+x)/(4+x)))+2*I*ln(x)*Pi*csgn(I*x)*csgn(I*x*exp(1/4*(x-4)*(2+x)/(4+x)))^2+2*I*l
n(x)*Pi*csgn(I*exp(1/4*(x-4)*(2+x)/(4+x)))*csgn(I*x*exp(1/4*(x-4)*(2+x)/(4+x)))^2-2*I*ln(x)*Pi*csgn(I*x*exp(1/
4*(x-4)*(2+x)/(4+x)))^3-4*ln(3)*ln(x)+4*ln(x)^2)*ln(exp(1/4*(x-4)*(2+x)/(4+x)))+2*ln(x)*ln(exp(1/4*(x-4)*(2+x)
/(4+x)))^2-2*I*Pi*csgn(I*x*exp(1/4*(x-4)*(2+x)/(4+x)))^3*ln(x)^2-2*I*Pi*ln(x)*ln(3)*csgn(I*x)*csgn(I*x*exp(1/4
*(x-4)*(2+x)/(4+x)))^2+2*I*Pi*ln(x)*ln(3)*csgn(I*x*exp(1/4*(x-4)*(2+x)/(4+x)))^3-1/2*Pi^2*ln(x)*csgn(I*x)^2*cs
gn(I*exp(1/4*(x-4)*(2+x)/(4+x)))^2*csgn(I*x*exp(1/4*(x-4)*(2+x)/(4+x)))^2+Pi^2*ln(x)*csgn(I*x)^2*csgn(I*exp(1/
4*(x-4)*(2+x)/(4+x)))*csgn(I*x*exp(1/4*(x-4)*(2+x)/(4+x)))^3+Pi^2*ln(x)*csgn(I*x)*csgn(I*exp(1/4*(x-4)*(2+x)/(
4+x)))^2*csgn(I*x*exp(1/4*(x-4)*(2+x)/(4+x)))^3-2*Pi^2*ln(x)*csgn(I*x)*csgn(I*exp(1/4*(x-4)*(2+x)/(4+x)))*csgn
(I*x*exp(1/4*(x-4)*(2+x)/(4+x)))^4+2*I*Pi*csgn(I*x)*csgn(I*x*exp(1/4*(x-4)*(2+x)/(4+x)))^2*ln(x)^2+2*I*Pi*csgn
(I*exp(1/4*(x-4)*(2+x)/(4+x)))*csgn(I*x*exp(1/4*(x-4)*(2+x)/(4+x)))^2*ln(x)^2-2*I*Pi*ln(x)*ln(3)*csgn(I*exp(1/
4*(x-4)*(2+x)/(4+x)))*csgn(I*x*exp(1/4*(x-4)*(2+x)/(4+x)))^2+2*I*Pi*ln(x)*ln(3)*csgn(I*x)*csgn(I*exp(1/4*(x-4)
*(2+x)/(4+x)))*csgn(I*x*exp(1/4*(x-4)*(2+x)/(4+x)))

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maxima [B]  time = 0.53, size = 123, normalized size = 4.10 \begin {gather*} \frac {16 \, {\left (x^{2} + 8 \, x + 16\right )} \log \relax (x)^{3} + 8 \, {\left (x^{3} - 2 \, x^{2} {\left (2 \, \log \relax (3) - 1\right )} - 16 \, x {\left (2 \, \log \relax (3) + 1\right )} - 64 \, \log \relax (3) - 32\right )} \log \relax (x)^{2} + {\left (x^{4} - 4 \, x^{3} {\left (2 \, \log \relax (3) + 1\right )} + 4 \, {\left (4 \, \log \relax (3)^{2} - 4 \, \log \relax (3) - 3\right )} x^{2} + 32 \, {\left (4 \, \log \relax (3)^{2} + 4 \, \log \relax (3) + 1\right )} x + 256 \, \log \relax (3)^{2} + 256 \, \log \relax (3) + 64\right )} \log \relax (x)}{8 \, {\left (x^{2} + 8 \, x + 16\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+16*x+32)*log(1/3*x*exp((x^2-2*x-8)/(4*x+16)))^2+(x^3+12*x^2+32*x+64)*log(x)*log(1/3*x*exp((x
^2-2*x-8)/(4*x+16))))/(x^3+8*x^2+16*x),x, algorithm="maxima")

[Out]

1/8*(16*(x^2 + 8*x + 16)*log(x)^3 + 8*(x^3 - 2*x^2*(2*log(3) - 1) - 16*x*(2*log(3) + 1) - 64*log(3) - 32)*log(
x)^2 + (x^4 - 4*x^3*(2*log(3) + 1) + 4*(4*log(3)^2 - 4*log(3) - 3)*x^2 + 32*(4*log(3)^2 + 4*log(3) + 1)*x + 25
6*log(3)^2 + 256*log(3) + 64)*log(x))/(x^2 + 8*x + 16)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {\left (2\,x^2+16\,x+32\right )\,{\ln \left (\frac {x\,{\mathrm {e}}^{-\frac {-x^2+2\,x+8}{4\,x+16}}}{3}\right )}^2+\ln \relax (x)\,\left (x^3+12\,x^2+32\,x+64\right )\,\ln \left (\frac {x\,{\mathrm {e}}^{-\frac {-x^2+2\,x+8}{4\,x+16}}}{3}\right )}{x^3+8\,x^2+16\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log((x*exp(-(2*x - x^2 + 8)/(4*x + 16)))/3)^2*(16*x + 2*x^2 + 32) + log((x*exp(-(2*x - x^2 + 8)/(4*x + 16
)))/3)*log(x)*(32*x + 12*x^2 + x^3 + 64))/(16*x + 8*x^2 + x^3),x)

[Out]

int((log((x*exp(-(2*x - x^2 + 8)/(4*x + 16)))/3)^2*(16*x + 2*x^2 + 32) + log((x*exp(-(2*x - x^2 + 8)/(4*x + 16
)))/3)*log(x)*(32*x + 12*x^2 + x^3 + 64))/(16*x + 8*x^2 + x^3), x)

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sympy [B]  time = 1.11, size = 114, normalized size = 3.80 \begin {gather*} 2 \log {\relax (x )}^{3} + \left (2 \log {\relax (3 )}^{2} + 6 \log {\relax (3 )} + \frac {17}{2}\right ) \log {\relax (x )} + \frac {\left (x^{4} - 8 x^{3} \log {\relax (3 )} - 4 x^{3} - 80 x^{2} - 64 x^{2} \log {\relax (3 )} - 512 x - 256 x \log {\relax (3 )} - 1024 - 512 \log {\relax (3 )}\right ) \log {\relax (x )}}{8 x^{2} + 64 x + 128} + \frac {\left (x^{2} - 4 x \log {\relax (3 )} - 2 x - 16 \log {\relax (3 )} - 8\right ) \log {\relax (x )}^{2}}{x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**2+16*x+32)*ln(1/3*x*exp((x**2-2*x-8)/(4*x+16)))**2+(x**3+12*x**2+32*x+64)*ln(x)*ln(1/3*x*exp(
(x**2-2*x-8)/(4*x+16))))/(x**3+8*x**2+16*x),x)

[Out]

2*log(x)**3 + (2*log(3)**2 + 6*log(3) + 17/2)*log(x) + (x**4 - 8*x**3*log(3) - 4*x**3 - 80*x**2 - 64*x**2*log(
3) - 512*x - 256*x*log(3) - 1024 - 512*log(3))*log(x)/(8*x**2 + 64*x + 128) + (x**2 - 4*x*log(3) - 2*x - 16*lo
g(3) - 8)*log(x)**2/(x + 4)

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