3.96.54 \(\int \frac {4-x+8 x \log (8 e^{-x/4} x)+(-1-2 x^2+2 x^3+2 x \log (x)) \log ^2(8 e^{-x/4} x)}{4 x \log (8 e^{-x/4} x)+(x^3+x \log (x)) \log ^2(8 e^{-x/4} x)} \, dx\)

Optimal. Leaf size=29 \[ 3+2 x-\log \left (x^2+\log (x)+\frac {4}{\log \left (8 e^{-x/4} x\right )}\right ) \]

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Rubi [F]  time = 4.79, 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 {4-x+8 x \log \left (8 e^{-x/4} x\right )+\left (-1-2 x^2+2 x^3+2 x \log (x)\right ) \log ^2\left (8 e^{-x/4} x\right )}{4 x \log \left (8 e^{-x/4} x\right )+\left (x^3+x \log (x)\right ) \log ^2\left (8 e^{-x/4} x\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {-1-2 x^2+2 x^3+2 x \log (x)}{x \left (x^2+\log (x)\right )}+\frac {4-x}{4 x \log \left (8 e^{-x/4} x\right )}+\frac {16+32 x^2-4 x^4+x^5-8 x^2 \log (x)+2 x^3 \log (x)-4 \log ^2(x)+x \log ^2(x)}{4 x \left (x^2+\log (x)\right ) \left (4+x^2 \log \left (8 e^{-x/4} x\right )+\log (x) \log \left (8 e^{-x/4} x\right )\right )}\right ) \, dx\\ &=\frac {1}{4} \int \frac {4-x}{x \log \left (8 e^{-x/4} x\right )} \, dx+\frac {1}{4} \int \frac {16+32 x^2-4 x^4+x^5-8 x^2 \log (x)+2 x^3 \log (x)-4 \log ^2(x)+x \log ^2(x)}{x \left (x^2+\log (x)\right ) \left (4+x^2 \log \left (8 e^{-x/4} x\right )+\log (x) \log \left (8 e^{-x/4} x\right )\right )} \, dx+\int \frac {-1-2 x^2+2 x^3+2 x \log (x)}{x \left (x^2+\log (x)\right )} \, dx\\ &=\log \left (\log \left (8 e^{-x/4} x\right )\right )+\frac {1}{4} \int \frac {16+32 x^2-4 x^4+x^5+2 (-4+x) x^2 \log (x)+(-4+x) \log ^2(x)}{x \left (x^2+\log (x)\right ) \left (4+\left (x^2+\log (x)\right ) \log \left (8 e^{-x/4} x\right )\right )} \, dx+\int \left (2+\frac {-1-2 x^2}{x \left (x^2+\log (x)\right )}\right ) \, dx\\ &=2 x+\log \left (\log \left (8 e^{-x/4} x\right )\right )+\frac {1}{4} \int \left (\frac {16}{x \left (x^2+\log (x)\right ) \left (4+x^2 \log \left (8 e^{-x/4} x\right )+\log (x) \log \left (8 e^{-x/4} x\right )\right )}+\frac {32 x}{\left (x^2+\log (x)\right ) \left (4+x^2 \log \left (8 e^{-x/4} x\right )+\log (x) \log \left (8 e^{-x/4} x\right )\right )}-\frac {4 x^3}{\left (x^2+\log (x)\right ) \left (4+x^2 \log \left (8 e^{-x/4} x\right )+\log (x) \log \left (8 e^{-x/4} x\right )\right )}+\frac {x^4}{\left (x^2+\log (x)\right ) \left (4+x^2 \log \left (8 e^{-x/4} x\right )+\log (x) \log \left (8 e^{-x/4} x\right )\right )}-\frac {8 x \log (x)}{\left (x^2+\log (x)\right ) \left (4+x^2 \log \left (8 e^{-x/4} x\right )+\log (x) \log \left (8 e^{-x/4} x\right )\right )}+\frac {2 x^2 \log (x)}{\left (x^2+\log (x)\right ) \left (4+x^2 \log \left (8 e^{-x/4} x\right )+\log (x) \log \left (8 e^{-x/4} x\right )\right )}+\frac {\log ^2(x)}{\left (x^2+\log (x)\right ) \left (4+x^2 \log \left (8 e^{-x/4} x\right )+\log (x) \log \left (8 e^{-x/4} x\right )\right )}-\frac {4 \log ^2(x)}{x \left (x^2+\log (x)\right ) \left (4+x^2 \log \left (8 e^{-x/4} x\right )+\log (x) \log \left (8 e^{-x/4} x\right )\right )}\right ) \, dx+\int \frac {-1-2 x^2}{x \left (x^2+\log (x)\right )} \, dx\\ &=2 x-\log \left (x^2+\log (x)\right )+\log \left (\log \left (8 e^{-x/4} x\right )\right )+\frac {1}{4} \int \frac {x^4}{\left (x^2+\log (x)\right ) \left (4+x^2 \log \left (8 e^{-x/4} x\right )+\log (x) \log \left (8 e^{-x/4} x\right )\right )} \, dx+\frac {1}{4} \int \frac {\log ^2(x)}{\left (x^2+\log (x)\right ) \left (4+x^2 \log \left (8 e^{-x/4} x\right )+\log (x) \log \left (8 e^{-x/4} x\right )\right )} \, dx+\frac {1}{2} \int \frac {x^2 \log (x)}{\left (x^2+\log (x)\right ) \left (4+x^2 \log \left (8 e^{-x/4} x\right )+\log (x) \log \left (8 e^{-x/4} x\right )\right )} \, dx-2 \int \frac {x \log (x)}{\left (x^2+\log (x)\right ) \left (4+x^2 \log \left (8 e^{-x/4} x\right )+\log (x) \log \left (8 e^{-x/4} x\right )\right )} \, dx+4 \int \frac {1}{x \left (x^2+\log (x)\right ) \left (4+x^2 \log \left (8 e^{-x/4} x\right )+\log (x) \log \left (8 e^{-x/4} x\right )\right )} \, dx+8 \int \frac {x}{\left (x^2+\log (x)\right ) \left (4+x^2 \log \left (8 e^{-x/4} x\right )+\log (x) \log \left (8 e^{-x/4} x\right )\right )} \, dx-\int \frac {x^3}{\left (x^2+\log (x)\right ) \left (4+x^2 \log \left (8 e^{-x/4} x\right )+\log (x) \log \left (8 e^{-x/4} x\right )\right )} \, dx-\int \frac {\log ^2(x)}{x \left (x^2+\log (x)\right ) \left (4+x^2 \log \left (8 e^{-x/4} x\right )+\log (x) \log \left (8 e^{-x/4} x\right )\right )} \, dx\\ &=2 x-\log \left (x^2+\log (x)\right )+\log \left (\log \left (8 e^{-x/4} x\right )\right )+\frac {1}{4} \int \frac {x^4}{\left (x^2+\log (x)\right ) \left (4+\left (x^2+\log (x)\right ) \log \left (8 e^{-x/4} x\right )\right )} \, dx+\frac {1}{4} \int \frac {\log ^2(x)}{\left (x^2+\log (x)\right ) \left (4+\left (x^2+\log (x)\right ) \log \left (8 e^{-x/4} x\right )\right )} \, dx+\frac {1}{2} \int \frac {x^2 \log (x)}{\left (x^2+\log (x)\right ) \left (4+\left (x^2+\log (x)\right ) \log \left (8 e^{-x/4} x\right )\right )} \, dx-2 \int \frac {x \log (x)}{\left (x^2+\log (x)\right ) \left (4+\left (x^2+\log (x)\right ) \log \left (8 e^{-x/4} x\right )\right )} \, dx+4 \int \frac {1}{x \left (x^2+\log (x)\right ) \left (4+\left (x^2+\log (x)\right ) \log \left (8 e^{-x/4} x\right )\right )} \, dx+8 \int \frac {x}{\left (x^2+\log (x)\right ) \left (4+\left (x^2+\log (x)\right ) \log \left (8 e^{-x/4} x\right )\right )} \, dx-\int \frac {x^3}{\left (x^2+\log (x)\right ) \left (4+\left (x^2+\log (x)\right ) \log \left (8 e^{-x/4} x\right )\right )} \, dx-\int \frac {\log ^2(x)}{x \left (x^2+\log (x)\right ) \left (4+\left (x^2+\log (x)\right ) \log \left (8 e^{-x/4} x\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 0.64, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {4-x+8 x \log \left (8 e^{-x/4} x\right )+\left (-1-2 x^2+2 x^3+2 x \log (x)\right ) \log ^2\left (8 e^{-x/4} x\right )}{4 x \log \left (8 e^{-x/4} x\right )+\left (x^3+x \log (x)\right ) \log ^2\left (8 e^{-x/4} x\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.53, size = 56, normalized size = 1.93 \begin {gather*} 2 \, x - \log \left (-x^{3} + 12 \, x^{2} \log \relax (2) + {\left (4 \, x^{2} - x + 12 \, \log \relax (2)\right )} \log \relax (x) + 4 \, \log \relax (x)^{2} + 16\right ) + \log \left (-x + 12 \, \log \relax (2) + 4 \, \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x - log(-x^3 + 12*x^2*log(2) + (4*x^2 - x + 12*log(2))*log(x) + 4*log(x)^2 + 16) + log(-x + 12*log(2) + 4*lo
g(x))

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giac [B]  time = 0.31, size = 58, normalized size = 2.00 \begin {gather*} 2 \, x - \log \left (-x^{3} + 12 \, x^{2} \log \relax (2) + 4 \, x^{2} \log \relax (x) - x \log \relax (x) + 12 \, \log \relax (2) \log \relax (x) + 4 \, \log \relax (x)^{2} + 16\right ) + \log \left (-x + 12 \, \log \relax (2) + 4 \, \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x - log(-x^3 + 12*x^2*log(2) + 4*x^2*log(x) - x*log(x) + 12*log(2)*log(x) + 4*log(x)^2 + 16) + log(-x + 12*l
og(2) + 4*log(x))

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maple [C]  time = 0.76, size = 342, normalized size = 11.79

method	result	size
risch	\(2 x -\ln \left (\ln \relax (x )+x^{2}\right )+\ln \left (\ln \left ({\mathrm e}^{\frac {x}{4}}\right )+\frac {i \left (\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{-\frac {x}{4}}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-\frac {x}{4}}\right )-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-\frac {x}{4}}\right )^{2}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{-\frac {x}{4}}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-\frac {x}{4}}\right )^{2}+\pi \mathrm {csgn}\left (i x \,{\mathrm e}^{-\frac {x}{4}}\right )^{3}+6 i \ln \relax (2)+2 i \ln \relax (x )\right )}{2}\right )-\ln \left (\ln \left ({\mathrm e}^{\frac {x}{4}}\right )+\frac {i \left (\pi \,x^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{-\frac {x}{4}}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-\frac {x}{4}}\right )-\pi \,x^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-\frac {x}{4}}\right )^{2}-\pi \,x^{2} \mathrm {csgn}\left (i {\mathrm e}^{-\frac {x}{4}}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-\frac {x}{4}}\right )^{2}+\pi \,x^{2} \mathrm {csgn}\left (i x \,{\mathrm e}^{-\frac {x}{4}}\right )^{3}+\ln \relax (x ) \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{-\frac {x}{4}}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-\frac {x}{4}}\right )-\ln \relax (x ) \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-\frac {x}{4}}\right )^{2}-\ln \relax (x ) \pi \,\mathrm {csgn}\left (i {\mathrm e}^{-\frac {x}{4}}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-\frac {x}{4}}\right )^{2}+\ln \relax (x ) \pi \mathrm {csgn}\left (i x \,{\mathrm e}^{-\frac {x}{4}}\right )^{3}+6 i \ln \relax (2) \ln \relax (x )+6 i \ln \relax (2) x^{2}+2 i \ln \relax (x )^{2}+2 i x^{2} \ln \relax (x )+8 i\right )}{2 \ln \relax (x )+2 x^{2}}\right )\)	\(342\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x-ln(ln(x)+x^2)+ln(ln(exp(1/4*x))+1/2*I*(Pi*csgn(I*x)*csgn(I*exp(-1/4*x))*csgn(I*x*exp(-1/4*x))-Pi*csgn(I*x)
*csgn(I*x*exp(-1/4*x))^2-Pi*csgn(I*exp(-1/4*x))*csgn(I*x*exp(-1/4*x))^2+Pi*csgn(I*x*exp(-1/4*x))^3+6*I*ln(2)+2
*I*ln(x)))-ln(ln(exp(1/4*x))+1/2*I*(Pi*x^2*csgn(I*x)*csgn(I*exp(-1/4*x))*csgn(I*x*exp(-1/4*x))-Pi*x^2*csgn(I*x
)*csgn(I*x*exp(-1/4*x))^2-Pi*x^2*csgn(I*exp(-1/4*x))*csgn(I*x*exp(-1/4*x))^2+Pi*x^2*csgn(I*x*exp(-1/4*x))^3+ln
(x)*Pi*csgn(I*x)*csgn(I*exp(-1/4*x))*csgn(I*x*exp(-1/4*x))-ln(x)*Pi*csgn(I*x)*csgn(I*x*exp(-1/4*x))^2-ln(x)*Pi
*csgn(I*exp(-1/4*x))*csgn(I*x*exp(-1/4*x))^2+ln(x)*Pi*csgn(I*x*exp(-1/4*x))^3+6*I*ln(2)*ln(x)+6*I*ln(2)*x^2+2*
I*ln(x)^2+2*I*x^2*ln(x)+8*I)/(ln(x)+x^2))

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maxima [B]  time = 0.50, size = 53, normalized size = 1.83 \begin {gather*} 2 \, x - \log \left (-\frac {1}{4} \, x^{3} + 3 \, x^{2} \log \relax (2) + \frac {1}{4} \, {\left (4 \, x^{2} - x + 12 \, \log \relax (2)\right )} \log \relax (x) + \log \relax (x)^{2} + 4\right ) + \log \left (-\frac {1}{4} \, x + 3 \, \log \relax (2) + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x - log(-1/4*x^3 + 3*x^2*log(2) + 1/4*(4*x^2 - x + 12*log(2))*log(x) + log(x)^2 + 4) + log(-1/4*x + 3*log(2)
+ log(x))

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mupad [B]  time = 14.00, size = 491, normalized size = 16.93 \begin {gather*} 2\,x-\ln \left (\frac {16\,x+4\,x\,{\ln \relax (x)}^2-17\,x^2\,\ln \relax (x)+4\,x^3\,\ln \relax (x)-16\,{\ln \relax (x)}^2-48\,x^2\,\ln \relax (2)+12\,x^3\,\ln \relax (2)-48\,\ln \relax (2)\,\ln \relax (x)+4\,x\,\ln \relax (x)+4\,x^3-x^4+12\,x\,\ln \relax (2)\,\ln \relax (x)-64}{x}\right )+\ln \left (x\,\left (x-4\right )\right )-\ln \left (9\,x^3\,{\ln \relax (2)}^2-36\,x^2\,{\ln \relax (2)}^2-8\,x+3\,x\,\ln \relax (2)+\frac {9\,x\,{\ln \relax (2)}^2}{2}+\frac {45\,x^2\,\ln \relax (2)}{4}+3\,x^3\,\ln \relax (2)+\frac {45\,x^4\,\ln \relax (2)}{2}-6\,x^5\,\ln \relax (2)-18\,{\ln \relax (2)}^2+\frac {515\,x^2}{8}-\frac {287\,x^3}{32}+30\,x^4-\frac {23\,x^5}{16}-\frac {7\,x^6}{2}+x^7+32\right )+\ln \left (\frac {1}{x^2}\right )+\ln \left (32\,x-384\,\ln \relax (2)-128\,\ln \relax (x)-\frac {261\,x^2\,{\ln \relax (2)}^2}{2}+432\,x^2\,{\ln \relax (2)}^3-72\,x^3\,{\ln \relax (2)}^2-108\,x^3\,{\ln \relax (2)}^3-261\,x^4\,{\ln \relax (2)}^2+72\,x^5\,{\ln \relax (2)}^2+72\,{\ln \relax (2)}^2\,\ln \relax (x)-\frac {515\,x^2\,\ln \relax (x)}{2}+\frac {287\,x^3\,\ln \relax (x)}{8}-120\,x^4\,\ln \relax (x)+\frac {23\,x^5\,\ln \relax (x)}{4}+14\,x^6\,\ln \relax (x)-4\,x^7\,\ln \relax (x)+96\,x\,\ln \relax (2)-54\,x\,{\ln \relax (2)}^2-\frac {1539\,x^2\,\ln \relax (2)}{2}-54\,x\,{\ln \relax (2)}^3+\frac {951\,x^3\,\ln \relax (2)}{8}-357\,x^4\,\ln \relax (2)+\frac {159\,x^5\,\ln \relax (2)}{4}+36\,x^6\,\ln \relax (2)-12\,x^7\,\ln \relax (2)+32\,x\,\ln \relax (x)+216\,{\ln \relax (2)}^3-8\,x^2+\frac {515\,x^3}{8}-\frac {287\,x^4}{32}+30\,x^5-\frac {23\,x^6}{16}-\frac {7\,x^7}{2}+x^8-12\,x\,\ln \relax (2)\,\ln \relax (x)-18\,x\,{\ln \relax (2)}^2\,\ln \relax (x)-45\,x^2\,\ln \relax (2)\,\ln \relax (x)-12\,x^3\,\ln \relax (2)\,\ln \relax (x)-90\,x^4\,\ln \relax (2)\,\ln \relax (x)+24\,x^5\,\ln \relax (2)\,\ln \relax (x)+144\,x^2\,{\ln \relax (2)}^2\,\ln \relax (x)-36\,x^3\,{\ln \relax (2)}^2\,\ln \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x*log(8*x*exp(-x/4)) - x + log(8*x*exp(-x/4))^2*(2*x*log(x) - 2*x^2 + 2*x^3 - 1) + 4)/(log(8*x*exp(-x/4
))^2*(x*log(x) + x^3) + 4*x*log(8*x*exp(-x/4))),x)

[Out]

2*x - log((16*x + 4*x*log(x)^2 - 17*x^2*log(x) + 4*x^3*log(x) - 16*log(x)^2 - 48*x^2*log(2) + 12*x^3*log(2) -
48*log(2)*log(x) + 4*x*log(x) + 4*x^3 - x^4 + 12*x*log(2)*log(x) - 64)/x) + log(x*(x - 4)) - log(9*x^3*log(2)^
2 - 36*x^2*log(2)^2 - 8*x + 3*x*log(2) + (9*x*log(2)^2)/2 + (45*x^2*log(2))/4 + 3*x^3*log(2) + (45*x^4*log(2))
/2 - 6*x^5*log(2) - 18*log(2)^2 + (515*x^2)/8 - (287*x^3)/32 + 30*x^4 - (23*x^5)/16 - (7*x^6)/2 + x^7 + 32) +
log(1/x^2) + log(32*x - 384*log(2) - 128*log(x) - (261*x^2*log(2)^2)/2 + 432*x^2*log(2)^3 - 72*x^3*log(2)^2 -
108*x^3*log(2)^3 - 261*x^4*log(2)^2 + 72*x^5*log(2)^2 + 72*log(2)^2*log(x) - (515*x^2*log(x))/2 + (287*x^3*log
(x))/8 - 120*x^4*log(x) + (23*x^5*log(x))/4 + 14*x^6*log(x) - 4*x^7*log(x) + 96*x*log(2) - 54*x*log(2)^2 - (15
39*x^2*log(2))/2 - 54*x*log(2)^3 + (951*x^3*log(2))/8 - 357*x^4*log(2) + (159*x^5*log(2))/4 + 36*x^6*log(2) -
12*x^7*log(2) + 32*x*log(x) + 216*log(2)^3 - 8*x^2 + (515*x^3)/8 - (287*x^4)/32 + 30*x^5 - (23*x^6)/16 - (7*x^
7)/2 + x^8 - 12*x*log(2)*log(x) - 18*x*log(2)^2*log(x) - 45*x^2*log(2)*log(x) - 12*x^3*log(2)*log(x) - 90*x^4*
log(2)*log(x) + 24*x^5*log(2)*log(x) + 144*x^2*log(2)^2*log(x) - 36*x^3*log(2)^2*log(x))

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sympy [B]  time = 1.29, size = 53, normalized size = 1.83 \begin {gather*} 2 x + \log {\left (- \frac {x}{4} + \log {\relax (x )} + 3 \log {\relax (2 )} \right )} - \log {\left (- \frac {x^{3}}{4} + 3 x^{2} \log {\relax (2 )} + \left (x^{2} - \frac {x}{4} + 3 \log {\relax (2 )}\right ) \log {\relax (x )} + \log {\relax (x )}^{2} + 4 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x + log(-x/4 + log(x) + 3*log(2)) - log(-x**3/4 + 3*x**2*log(2) + (x**2 - x/4 + 3*log(2))*log(x) + log(x)**2
+ 4)

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