3.90.24 \(\int \frac {(-1-7 x+8 x^2) \log (\frac {1}{4} (x+8 x^2))+(-x-16 x^2+(1+16 x) \log (x)+(x+8 x^2+(-1-8 x) \log (x)) \log (\frac {1}{4} (x+8 x^2))) \log (-25 x+25 \log (x))+(-2 x^3-16 x^4+(2 x^2+16 x^3) \log (x)) \log ^2(-25 x+25 \log (x))}{(-x^3-8 x^4+(x^2+8 x^3) \log (x)) \log ^2(-25 x+25 \log (x))} \, dx\)

Optimal. Leaf size=32 \[ -3+2 x+\frac {\log \left (\frac {x}{4}+2 x^2\right )}{x \log (25 (-x+\log (x)))} \]

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Rubi [F]  time = 6.46, 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 (-1-7 x+8 x^2\right ) \log \left (\frac {1}{4} \left (x+8 x^2\right )\right )+\left (-x-16 x^2+(1+16 x) \log (x)+\left (x+8 x^2+(-1-8 x) \log (x)\right ) \log \left (\frac {1}{4} \left (x+8 x^2\right )\right )\right ) \log (-25 x+25 \log (x))+\left (-2 x^3-16 x^4+\left (2 x^2+16 x^3\right ) \log (x)\right ) \log ^2(-25 x+25 \log (x))}{\left (-x^3-8 x^4+\left (x^2+8 x^3\right ) \log (x)\right ) \log ^2(-25 x+25 \log (x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-1 - 7*x + 8*x^2)*Log[(x + 8*x^2)/4] + (-x - 16*x^2 + (1 + 16*x)*Log[x] + (x + 8*x^2 + (-1 - 8*x)*Log[x]
)*Log[(x + 8*x^2)/4])*Log[-25*x + 25*Log[x]] + (-2*x^3 - 16*x^4 + (2*x^2 + 16*x^3)*Log[x])*Log[-25*x + 25*Log[
x]]^2)/((-x^3 - 8*x^4 + (x^2 + 8*x^3)*Log[x])*Log[-25*x + 25*Log[x]]^2),x]

[Out]

2*x + Defer[Int][Log[(x*(1 + 8*x))/4]/(x^2*(x - Log[x])*Log[25*(-x + Log[x])]^2), x] - Defer[Int][Log[(x*(1 +
8*x))/4]/(x*(x - Log[x])*Log[25*(-x + Log[x])]^2), x] + Defer[Int][1/(x^2*Log[25*(-x + Log[x])]), x] + 8*Defer
[Int][1/(x*Log[25*(-x + Log[x])]), x] - 64*Defer[Int][1/((1 + 8*x)*Log[25*(-x + Log[x])]), x] - Defer[Int][Log
[(x*(1 + 8*x))/4]/(x^2*Log[25*(-x + Log[x])]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\frac {\log (25 (-x+\log (x))) \left (1+16 x+2 x^2 (1+8 x) \log (25 (-x+\log (x)))\right )}{1+8 x}-\frac {\log \left (\frac {1}{4} x (1+8 x)\right ) (-1+x+(x-\log (x)) \log (25 (-x+\log (x))))}{x-\log (x)}}{x^2 \log ^2(25 (-x+\log (x)))} \, dx\\ &=\int \left (2-\frac {(-1+x) \log \left (\frac {1}{4} x (1+8 x)\right )}{x^2 (x-\log (x)) \log ^2(25 (-x+\log (x)))}+\frac {1+16 x-\log \left (\frac {1}{4} x (1+8 x)\right )-8 x \log \left (\frac {1}{4} x (1+8 x)\right )}{x^2 (1+8 x) \log (25 (-x+\log (x)))}\right ) \, dx\\ &=2 x-\int \frac {(-1+x) \log \left (\frac {1}{4} x (1+8 x)\right )}{x^2 (x-\log (x)) \log ^2(25 (-x+\log (x)))} \, dx+\int \frac {1+16 x-\log \left (\frac {1}{4} x (1+8 x)\right )-8 x \log \left (\frac {1}{4} x (1+8 x)\right )}{x^2 (1+8 x) \log (25 (-x+\log (x)))} \, dx\\ &=2 x-\int \left (-\frac {\log \left (\frac {1}{4} x (1+8 x)\right )}{x^2 (x-\log (x)) \log ^2(25 (-x+\log (x)))}+\frac {\log \left (\frac {1}{4} x (1+8 x)\right )}{x (x-\log (x)) \log ^2(25 (-x+\log (x)))}\right ) \, dx+\int \frac {1+16 x-(1+8 x) \log \left (\frac {1}{4} x (1+8 x)\right )}{x^2 (1+8 x) \log (25 (-x+\log (x)))} \, dx\\ &=2 x+\int \left (\frac {1+16 x-\log \left (\frac {1}{4} x (1+8 x)\right )-8 x \log \left (\frac {1}{4} x (1+8 x)\right )}{x^2 \log (25 (-x+\log (x)))}+\frac {8 \left (-1-16 x+\log \left (\frac {1}{4} x (1+8 x)\right )+8 x \log \left (\frac {1}{4} x (1+8 x)\right )\right )}{x \log (25 (-x+\log (x)))}-\frac {64 \left (-1-16 x+\log \left (\frac {1}{4} x (1+8 x)\right )+8 x \log \left (\frac {1}{4} x (1+8 x)\right )\right )}{(1+8 x) \log (25 (-x+\log (x)))}\right ) \, dx+\int \frac {\log \left (\frac {1}{4} x (1+8 x)\right )}{x^2 (x-\log (x)) \log ^2(25 (-x+\log (x)))} \, dx-\int \frac {\log \left (\frac {1}{4} x (1+8 x)\right )}{x (x-\log (x)) \log ^2(25 (-x+\log (x)))} \, dx\\ &=2 x+8 \int \frac {-1-16 x+\log \left (\frac {1}{4} x (1+8 x)\right )+8 x \log \left (\frac {1}{4} x (1+8 x)\right )}{x \log (25 (-x+\log (x)))} \, dx-64 \int \frac {-1-16 x+\log \left (\frac {1}{4} x (1+8 x)\right )+8 x \log \left (\frac {1}{4} x (1+8 x)\right )}{(1+8 x) \log (25 (-x+\log (x)))} \, dx+\int \frac {\log \left (\frac {1}{4} x (1+8 x)\right )}{x^2 (x-\log (x)) \log ^2(25 (-x+\log (x)))} \, dx-\int \frac {\log \left (\frac {1}{4} x (1+8 x)\right )}{x (x-\log (x)) \log ^2(25 (-x+\log (x)))} \, dx+\int \frac {1+16 x-\log \left (\frac {1}{4} x (1+8 x)\right )-8 x \log \left (\frac {1}{4} x (1+8 x)\right )}{x^2 \log (25 (-x+\log (x)))} \, dx\\ &=2 x+8 \int \frac {-1-16 x+(1+8 x) \log \left (\frac {1}{4} x (1+8 x)\right )}{x \log (25 (-x+\log (x)))} \, dx-64 \int \frac {-1-16 x+(1+8 x) \log \left (\frac {1}{4} x (1+8 x)\right )}{(1+8 x) \log (25 (-x+\log (x)))} \, dx+\int \frac {\log \left (\frac {1}{4} x (1+8 x)\right )}{x^2 (x-\log (x)) \log ^2(25 (-x+\log (x)))} \, dx-\int \frac {\log \left (\frac {1}{4} x (1+8 x)\right )}{x (x-\log (x)) \log ^2(25 (-x+\log (x)))} \, dx+\int \frac {1+16 x-(1+8 x) \log \left (\frac {1}{4} x (1+8 x)\right )}{x^2 \log (25 (-x+\log (x)))} \, dx\\ &=2 x+8 \int \left (-\frac {16}{\log (25 (-x+\log (x)))}-\frac {1}{x \log (25 (-x+\log (x)))}+\frac {8 \log \left (\frac {1}{4} x (1+8 x)\right )}{\log (25 (-x+\log (x)))}+\frac {\log \left (\frac {1}{4} x (1+8 x)\right )}{x \log (25 (-x+\log (x)))}\right ) \, dx-64 \int \left (-\frac {1}{(1+8 x) \log (25 (-x+\log (x)))}-\frac {16 x}{(1+8 x) \log (25 (-x+\log (x)))}+\frac {\log \left (\frac {1}{4} x (1+8 x)\right )}{(1+8 x) \log (25 (-x+\log (x)))}+\frac {8 x \log \left (\frac {1}{4} x (1+8 x)\right )}{(1+8 x) \log (25 (-x+\log (x)))}\right ) \, dx+\int \left (\frac {1}{x^2 \log (25 (-x+\log (x)))}+\frac {16}{x \log (25 (-x+\log (x)))}-\frac {\log \left (\frac {1}{4} x (1+8 x)\right )}{x^2 \log (25 (-x+\log (x)))}-\frac {8 \log \left (\frac {1}{4} x (1+8 x)\right )}{x \log (25 (-x+\log (x)))}\right ) \, dx+\int \frac {\log \left (\frac {1}{4} x (1+8 x)\right )}{x^2 (x-\log (x)) \log ^2(25 (-x+\log (x)))} \, dx-\int \frac {\log \left (\frac {1}{4} x (1+8 x)\right )}{x (x-\log (x)) \log ^2(25 (-x+\log (x)))} \, dx\\ &=2 x-8 \int \frac {1}{x \log (25 (-x+\log (x)))} \, dx+16 \int \frac {1}{x \log (25 (-x+\log (x)))} \, dx+64 \int \frac {1}{(1+8 x) \log (25 (-x+\log (x)))} \, dx+64 \int \frac {\log \left (\frac {1}{4} x (1+8 x)\right )}{\log (25 (-x+\log (x)))} \, dx-64 \int \frac {\log \left (\frac {1}{4} x (1+8 x)\right )}{(1+8 x) \log (25 (-x+\log (x)))} \, dx-128 \int \frac {1}{\log (25 (-x+\log (x)))} \, dx-512 \int \frac {x \log \left (\frac {1}{4} x (1+8 x)\right )}{(1+8 x) \log (25 (-x+\log (x)))} \, dx+1024 \int \frac {x}{(1+8 x) \log (25 (-x+\log (x)))} \, dx+\int \frac {\log \left (\frac {1}{4} x (1+8 x)\right )}{x^2 (x-\log (x)) \log ^2(25 (-x+\log (x)))} \, dx-\int \frac {\log \left (\frac {1}{4} x (1+8 x)\right )}{x (x-\log (x)) \log ^2(25 (-x+\log (x)))} \, dx+\int \frac {1}{x^2 \log (25 (-x+\log (x)))} \, dx-\int \frac {\log \left (\frac {1}{4} x (1+8 x)\right )}{x^2 \log (25 (-x+\log (x)))} \, dx\\ &=2 x-8 \int \frac {1}{x \log (25 (-x+\log (x)))} \, dx+16 \int \frac {1}{x \log (25 (-x+\log (x)))} \, dx+64 \int \frac {1}{(1+8 x) \log (25 (-x+\log (x)))} \, dx+64 \int \frac {\log \left (\frac {1}{4} x (1+8 x)\right )}{\log (25 (-x+\log (x)))} \, dx-64 \int \frac {\log \left (\frac {1}{4} x (1+8 x)\right )}{(1+8 x) \log (25 (-x+\log (x)))} \, dx-128 \int \frac {1}{\log (25 (-x+\log (x)))} \, dx-512 \int \left (\frac {\log \left (\frac {1}{4} x (1+8 x)\right )}{8 \log (25 (-x+\log (x)))}-\frac {\log \left (\frac {1}{4} x (1+8 x)\right )}{8 (1+8 x) \log (25 (-x+\log (x)))}\right ) \, dx+1024 \int \left (\frac {1}{8 \log (25 (-x+\log (x)))}-\frac {1}{8 (1+8 x) \log (25 (-x+\log (x)))}\right ) \, dx+\int \frac {\log \left (\frac {1}{4} x (1+8 x)\right )}{x^2 (x-\log (x)) \log ^2(25 (-x+\log (x)))} \, dx-\int \frac {\log \left (\frac {1}{4} x (1+8 x)\right )}{x (x-\log (x)) \log ^2(25 (-x+\log (x)))} \, dx+\int \frac {1}{x^2 \log (25 (-x+\log (x)))} \, dx-\int \frac {\log \left (\frac {1}{4} x (1+8 x)\right )}{x^2 \log (25 (-x+\log (x)))} \, dx\\ &=2 x-8 \int \frac {1}{x \log (25 (-x+\log (x)))} \, dx+16 \int \frac {1}{x \log (25 (-x+\log (x)))} \, dx+64 \int \frac {1}{(1+8 x) \log (25 (-x+\log (x)))} \, dx-128 \int \frac {1}{(1+8 x) \log (25 (-x+\log (x)))} \, dx+\int \frac {\log \left (\frac {1}{4} x (1+8 x)\right )}{x^2 (x-\log (x)) \log ^2(25 (-x+\log (x)))} \, dx-\int \frac {\log \left (\frac {1}{4} x (1+8 x)\right )}{x (x-\log (x)) \log ^2(25 (-x+\log (x)))} \, dx+\int \frac {1}{x^2 \log (25 (-x+\log (x)))} \, dx-\int \frac {\log \left (\frac {1}{4} x (1+8 x)\right )}{x^2 \log (25 (-x+\log (x)))} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 30, normalized size = 0.94 \begin {gather*} 2 x+\frac {\log \left (\frac {1}{4} x (1+8 x)\right )}{x \log (25 (-x+\log (x)))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-1 - 7*x + 8*x^2)*Log[(x + 8*x^2)/4] + (-x - 16*x^2 + (1 + 16*x)*Log[x] + (x + 8*x^2 + (-1 - 8*x)*
Log[x])*Log[(x + 8*x^2)/4])*Log[-25*x + 25*Log[x]] + (-2*x^3 - 16*x^4 + (2*x^2 + 16*x^3)*Log[x])*Log[-25*x + 2
5*Log[x]]^2)/((-x^3 - 8*x^4 + (x^2 + 8*x^3)*Log[x])*Log[-25*x + 25*Log[x]]^2),x]

[Out]

2*x + Log[(x*(1 + 8*x))/4]/(x*Log[25*(-x + Log[x])])

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fricas [A]  time = 0.45, size = 40, normalized size = 1.25 \begin {gather*} \frac {2 \, x^{2} \log \left (-25 \, x + 25 \, \log \relax (x)\right ) + \log \left (2 \, x^{2} + \frac {1}{4} \, x\right )}{x \log \left (-25 \, x + 25 \, \log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^3+2*x^2)*log(x)-16*x^4-2*x^3)*log(25*log(x)-25*x)^2+(((-8*x-1)*log(x)+8*x^2+x)*log(2*x^2+1/4
*x)+(16*x+1)*log(x)-16*x^2-x)*log(25*log(x)-25*x)+(8*x^2-7*x-1)*log(2*x^2+1/4*x))/((8*x^3+x^2)*log(x)-8*x^4-x^
3)/log(25*log(x)-25*x)^2,x, algorithm="fricas")

[Out]

(2*x^2*log(-25*x + 25*log(x)) + log(2*x^2 + 1/4*x))/(x*log(-25*x + 25*log(x)))

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giac [A]  time = 0.31, size = 37, normalized size = 1.16 \begin {gather*} 2 \, x - \frac {2 \, \log \relax (2) - \log \left (8 \, x + 1\right ) - \log \relax (x)}{x \log \left (-25 \, x + 25 \, \log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^3+2*x^2)*log(x)-16*x^4-2*x^3)*log(25*log(x)-25*x)^2+(((-8*x-1)*log(x)+8*x^2+x)*log(2*x^2+1/4
*x)+(16*x+1)*log(x)-16*x^2-x)*log(25*log(x)-25*x)+(8*x^2-7*x-1)*log(2*x^2+1/4*x))/((8*x^3+x^2)*log(x)-8*x^4-x^
3)/log(25*log(x)-25*x)^2,x, algorithm="giac")

[Out]

2*x - (2*log(2) - log(8*x + 1) - log(x))/(x*log(-25*x + 25*log(x)))

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maple [C]  time = 0.21, size = 114, normalized size = 3.56

method	result	size
risch	\(2 x -\frac {i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (x +\frac {1}{8}\right )\right ) \mathrm {csgn}\left (i x \left (x +\frac {1}{8}\right )\right )-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left (x +\frac {1}{8}\right )\right )^{2}-i \pi \,\mathrm {csgn}\left (i \left (x +\frac {1}{8}\right )\right ) \mathrm {csgn}\left (i x \left (x +\frac {1}{8}\right )\right )^{2}+i \pi \mathrm {csgn}\left (i x \left (x +\frac {1}{8}\right )\right )^{3}+4 \ln \relax (2)-2 \ln \relax (x )-2 \ln \left (x +\frac {1}{8}\right )}{2 x \ln \left (25 \ln \relax (x )-25 x \right )}\)	\(114\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((16*x^3+2*x^2)*ln(x)-16*x^4-2*x^3)*ln(25*ln(x)-25*x)^2+(((-8*x-1)*ln(x)+8*x^2+x)*ln(2*x^2+1/4*x)+(16*x+1
)*ln(x)-16*x^2-x)*ln(25*ln(x)-25*x)+(8*x^2-7*x-1)*ln(2*x^2+1/4*x))/((8*x^3+x^2)*ln(x)-8*x^4-x^3)/ln(25*ln(x)-2
5*x)^2,x,method=_RETURNVERBOSE)

[Out]

2*x-1/2*(I*Pi*csgn(I*x)*csgn(I*(x+1/8))*csgn(I*x*(x+1/8))-I*Pi*csgn(I*x)*csgn(I*x*(x+1/8))^2-I*Pi*csgn(I*(x+1/
8))*csgn(I*x*(x+1/8))^2+I*Pi*csgn(I*x*(x+1/8))^3+4*ln(2)-2*ln(x)-2*ln(x+1/8))/x/ln(25*ln(x)-25*x)

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maxima [A]  time = 0.51, size = 50, normalized size = 1.56 \begin {gather*} \frac {4 \, x^{2} \log \relax (5) + 2 \, x^{2} \log \left (-x + \log \relax (x)\right ) - 2 \, \log \relax (2) + \log \left (8 \, x + 1\right ) + \log \relax (x)}{2 \, x \log \relax (5) + x \log \left (-x + \log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^3+2*x^2)*log(x)-16*x^4-2*x^3)*log(25*log(x)-25*x)^2+(((-8*x-1)*log(x)+8*x^2+x)*log(2*x^2+1/4
*x)+(16*x+1)*log(x)-16*x^2-x)*log(25*log(x)-25*x)+(8*x^2-7*x-1)*log(2*x^2+1/4*x))/((8*x^3+x^2)*log(x)-8*x^4-x^
3)/log(25*log(x)-25*x)^2,x, algorithm="maxima")

[Out]

(4*x^2*log(5) + 2*x^2*log(-x + log(x)) - 2*log(2) + log(8*x + 1) + log(x))/(2*x*log(5) + x*log(-x + log(x)))

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mupad [B]  time = 8.10, size = 183, normalized size = 5.72 \begin {gather*} 2\,x+\frac {\frac {\ln \left (2\,x^2+\frac {x}{4}\right )}{x}-\frac {\ln \left (25\,\ln \relax (x)-25\,x\right )\,\left (x-\ln \relax (x)\right )\,\left (16\,x-\ln \left (2\,x^2+\frac {x}{4}\right )-8\,x\,\ln \left (2\,x^2+\frac {x}{4}\right )+1\right )}{x\,\left (8\,x+1\right )\,\left (x-1\right )}}{\ln \left (25\,\ln \relax (x)-25\,x\right )}+\ln \left (2\,x^2+\frac {x}{4}\right )\,\left (\frac {1}{x-x^2}+\frac {x-1}{x-x^2}-\frac {\ln \relax (x)}{x-x^2}\right )-\frac {2\,x+\frac {1}{8}}{-x^2+\frac {7\,x}{8}+\frac {1}{8}}+\frac {\ln \relax (x)\,\left (2\,x+\frac {1}{8}\right )}{-x^3+\frac {7\,x^2}{8}+\frac {x}{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(25*log(x) - 25*x)^2*(2*x^3 - log(x)*(2*x^2 + 16*x^3) + 16*x^4) + log(25*log(x) - 25*x)*(x - log(x)*(1
6*x + 1) - log(x/4 + 2*x^2)*(x - log(x)*(8*x + 1) + 8*x^2) + 16*x^2) + log(x/4 + 2*x^2)*(7*x - 8*x^2 + 1))/(lo
g(25*log(x) - 25*x)^2*(x^3 - log(x)*(x^2 + 8*x^3) + 8*x^4)),x)

[Out]

2*x + (log(x/4 + 2*x^2)/x - (log(25*log(x) - 25*x)*(x - log(x))*(16*x - log(x/4 + 2*x^2) - 8*x*log(x/4 + 2*x^2
) + 1))/(x*(8*x + 1)*(x - 1)))/log(25*log(x) - 25*x) + log(x/4 + 2*x^2)*(1/(x - x^2) + (x - 1)/(x - x^2) - log
(x)/(x - x^2)) - (2*x + 1/8)/((7*x)/8 - x^2 + 1/8) + (log(x)*(2*x + 1/8))/(x/8 + (7*x^2)/8 - x^3)

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sympy [A]  time = 0.50, size = 24, normalized size = 0.75 \begin {gather*} 2 x + \frac {\log {\left (2 x^{2} + \frac {x}{4} \right )}}{x \log {\left (- 25 x + 25 \log {\relax (x )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x**3+2*x**2)*ln(x)-16*x**4-2*x**3)*ln(25*ln(x)-25*x)**2+(((-8*x-1)*ln(x)+8*x**2+x)*ln(2*x**2+1
/4*x)+(16*x+1)*ln(x)-16*x**2-x)*ln(25*ln(x)-25*x)+(8*x**2-7*x-1)*ln(2*x**2+1/4*x))/((8*x**3+x**2)*ln(x)-8*x**4
-x**3)/ln(25*ln(x)-25*x)**2,x)

[Out]

2*x + log(2*x**2 + x/4)/(x*log(-25*x + 25*log(x)))

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