3.88.99 \(\int \frac {-4 \log (2)+7 \log (2) \log (x)-22 x \log ^2(x)+(4 \log (2) \log (x)-8 x \log ^2(x)) \log (\frac {-8 x^2 \log (2)+16 x^3 \log (x)}{\log (x)})}{-\log (2) \log (x)+2 x \log ^2(x)} \, dx\)

Optimal. Leaf size=23 \[ x-4 x \log \left (8 x^2 \left (2 x-\frac {\log (2)}{\log (x)}\right )\right ) \]

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Rubi [F]  time = 0.81, 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 \log (2)+7 \log (2) \log (x)-22 x \log ^2(x)+\left (4 \log (2) \log (x)-8 x \log ^2(x)\right ) \log \left (\frac {-8 x^2 \log (2)+16 x^3 \log (x)}{\log (x)}\right )}{-\log (2) \log (x)+2 x \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-4*Log[2] + 7*Log[2]*Log[x] - 22*x*Log[x]^2 + (4*Log[2]*Log[x] - 8*x*Log[x]^2)*Log[(-8*x^2*Log[2] + 16*x^
3*Log[x])/Log[x]])/(-(Log[2]*Log[x]) + 2*x*Log[x]^2),x]

[Out]

x - 4*x*Log[8*x^2*(2*x - Log[2]/Log[x])] - 4*LogIntegral[x] + (Log[16]*LogIntegral[x])/Log[2] - 4*Log[2]*Defer
[Int][(Log[2] - 2*x*Log[x])^(-1), x] + 8*Defer[Int][x/(-Log[2] + 2*x*Log[x]), x] + Log[2]*Log[16]*Defer[Int][(
Log[2]^2 - x*Log[4]*Log[x])^(-1), x] - Log[256]*Defer[Int][x/(-Log[2]^2 + x*Log[4]*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 \log (2)-7 \log (2) \log (x)+22 x \log ^2(x)-\left (4 \log (2) \log (x)-8 x \log ^2(x)\right ) \log \left (\frac {-8 x^2 \log (2)+16 x^3 \log (x)}{\log (x)}\right )}{\log (x) (\log (2)-2 x \log (x))} \, dx\\ &=\int \left (\frac {\log (16)-\log (128) \log (x)+22 x \log ^2(x)}{\log (x) (\log (2)-2 x \log (x))}-4 \log \left (8 x^2 \left (2 x-\frac {\log (2)}{\log (x)}\right )\right )\right ) \, dx\\ &=-\left (4 \int \log \left (8 x^2 \left (2 x-\frac {\log (2)}{\log (x)}\right )\right ) \, dx\right )+\int \frac {\log (16)-\log (128) \log (x)+22 x \log ^2(x)}{\log (x) (\log (2)-2 x \log (x))} \, dx\\ &=-4 x \log \left (8 x^2 \left (2 x-\frac {\log (2)}{\log (x)}\right )\right )+4 \int \frac {-\log (2)+\log (4) \log (x)-6 x \log ^2(x)}{\log (x) (\log (2)-2 x \log (x))} \, dx+\int \left (-11+\frac {\log (16)}{\log (2) \log (x)}+\frac {\log (2) \log (16)+x \log (256)}{\log ^2(2)-x \log (4) \log (x)}\right ) \, dx\\ &=-11 x-4 x \log \left (8 x^2 \left (2 x-\frac {\log (2)}{\log (x)}\right )\right )+4 \int \left (3-\frac {1}{\log (x)}+\frac {2 x+\log (2)}{-\log (2)+2 x \log (x)}\right ) \, dx+\frac {\log (16) \int \frac {1}{\log (x)} \, dx}{\log (2)}+\int \frac {\log (2) \log (16)+x \log (256)}{\log ^2(2)-x \log (4) \log (x)} \, dx\\ &=x-4 x \log \left (8 x^2 \left (2 x-\frac {\log (2)}{\log (x)}\right )\right )+\frac {\log (16) \text {li}(x)}{\log (2)}-4 \int \frac {1}{\log (x)} \, dx+4 \int \frac {2 x+\log (2)}{-\log (2)+2 x \log (x)} \, dx+\int \left (\frac {\log (2) \log (16)}{\log ^2(2)-x \log (4) \log (x)}-\frac {x \log (256)}{-\log ^2(2)+x \log (4) \log (x)}\right ) \, dx\\ &=x-4 x \log \left (8 x^2 \left (2 x-\frac {\log (2)}{\log (x)}\right )\right )-4 \text {li}(x)+\frac {\log (16) \text {li}(x)}{\log (2)}+4 \int \left (-\frac {\log (2)}{\log (2)-2 x \log (x)}+\frac {2 x}{-\log (2)+2 x \log (x)}\right ) \, dx+(\log (2) \log (16)) \int \frac {1}{\log ^2(2)-x \log (4) \log (x)} \, dx-\log (256) \int \frac {x}{-\log ^2(2)+x \log (4) \log (x)} \, dx\\ &=x-4 x \log \left (8 x^2 \left (2 x-\frac {\log (2)}{\log (x)}\right )\right )-4 \text {li}(x)+\frac {\log (16) \text {li}(x)}{\log (2)}+8 \int \frac {x}{-\log (2)+2 x \log (x)} \, dx-(4 \log (2)) \int \frac {1}{\log (2)-2 x \log (x)} \, dx+(\log (2) \log (16)) \int \frac {1}{\log ^2(2)-x \log (4) \log (x)} \, dx-\log (256) \int \frac {x}{-\log ^2(2)+x \log (4) \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 23, normalized size = 1.00 \begin {gather*} x-4 x \log \left (8 x^2 \left (2 x-\frac {\log (2)}{\log (x)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*Log[2] + 7*Log[2]*Log[x] - 22*x*Log[x]^2 + (4*Log[2]*Log[x] - 8*x*Log[x]^2)*Log[(-8*x^2*Log[2] +
16*x^3*Log[x])/Log[x]])/(-(Log[2]*Log[x]) + 2*x*Log[x]^2),x]

[Out]

x - 4*x*Log[8*x^2*(2*x - Log[2]/Log[x])]

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fricas [A]  time = 0.92, size = 27, normalized size = 1.17 \begin {gather*} -4 \, x \log \left (\frac {8 \, {\left (2 \, x^{3} \log \relax (x) - x^{2} \log \relax (2)\right )}}{\log \relax (x)}\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x*log(x)^2+4*log(2)*log(x))*log((16*x^3*log(x)-8*x^2*log(2))/log(x))-22*x*log(x)^2+7*log(2)*log
(x)-4*log(2))/(2*x*log(x)^2-log(2)*log(x)),x, algorithm="fricas")

[Out]

-4*x*log(8*(2*x^3*log(x) - x^2*log(2))/log(x)) + x

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giac [A]  time = 0.15, size = 35, normalized size = 1.52 \begin {gather*} -x {\left (12 \, \log \relax (2) - 1\right )} - 4 \, x \log \left (2 \, x \log \relax (x) - \log \relax (2)\right ) - 8 \, x \log \relax (x) + 4 \, x \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x*log(x)^2+4*log(2)*log(x))*log((16*x^3*log(x)-8*x^2*log(2))/log(x))-22*x*log(x)^2+7*log(2)*log
(x)-4*log(2))/(2*x*log(x)^2-log(2)*log(x)),x, algorithm="giac")

[Out]

-x*(12*log(2) - 1) - 4*x*log(2*x*log(x) - log(2)) - 8*x*log(x) + 4*x*log(log(x))

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maple [A]  time = 0.27, size = 27, normalized size = 1.17

method	result	size
norman	\(x -4 x \ln \left (\frac {16 x^{3} \ln \relax (x )-8 x^{2} \ln \relax (2)}{\ln \relax (x )}\right )\)	\(27\)
risch	\(-4 x \ln \left (-2 x \ln \relax (x )+\ln \relax (2)\right )+4 x \ln \left (\ln \relax (x )\right )+2 i \pi x \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+2 i \pi x \mathrm {csgn}\left (i x^{2}\right )^{3}+2 i \pi x \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (\frac {i \left (-2 x \ln \relax (x )+\ln \relax (2)\right )}{\ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i x^{2} \left (-2 x \ln \relax (x )+\ln \relax (2)\right )}{\ln \relax (x )}\right )-4 i \pi x -2 i \pi x \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (\frac {i x^{2} \left (-2 x \ln \relax (x )+\ln \relax (2)\right )}{\ln \relax (x )}\right )^{2}+4 i \pi x \mathrm {csgn}\left (\frac {i x^{2} \left (-2 x \ln \relax (x )+\ln \relax (2)\right )}{\ln \relax (x )}\right )^{2}-2 i \pi x \mathrm {csgn}\left (\frac {i \left (-2 x \ln \relax (x )+\ln \relax (2)\right )}{\ln \relax (x )}\right )^{2} \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right )-4 i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+2 i \pi x \mathrm {csgn}\left (\frac {i \left (-2 x \ln \relax (x )+\ln \relax (2)\right )}{\ln \relax (x )}\right )^{3}-2 i \pi x \mathrm {csgn}\left (\frac {i x^{2} \left (-2 x \ln \relax (x )+\ln \relax (2)\right )}{\ln \relax (x )}\right )^{3}-2 i \pi x \,\mathrm {csgn}\left (\frac {i \left (-2 x \ln \relax (x )+\ln \relax (2)\right )}{\ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i x^{2} \left (-2 x \ln \relax (x )+\ln \relax (2)\right )}{\ln \relax (x )}\right )^{2}-2 i \pi x \,\mathrm {csgn}\left (i \left (-2 x \ln \relax (x )+\ln \relax (2)\right )\right ) \mathrm {csgn}\left (\frac {i \left (-2 x \ln \relax (x )+\ln \relax (2)\right )}{\ln \relax (x )}\right )^{2}+2 i \pi x \,\mathrm {csgn}\left (i \left (-2 x \ln \relax (x )+\ln \relax (2)\right )\right ) \mathrm {csgn}\left (\frac {i \left (-2 x \ln \relax (x )+\ln \relax (2)\right )}{\ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right )-12 x \ln \relax (2)-8 x \ln \relax (x )+x\)	\(392\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-8*x*ln(x)^2+4*ln(2)*ln(x))*ln((16*x^3*ln(x)-8*x^2*ln(2))/ln(x))-22*x*ln(x)^2+7*ln(2)*ln(x)-4*ln(2))/(2*
x*ln(x)^2-ln(2)*ln(x)),x,method=_RETURNVERBOSE)

[Out]

x-4*x*ln((16*x^3*ln(x)-8*x^2*ln(2))/ln(x))

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maxima [A]  time = 0.48, size = 35, normalized size = 1.52 \begin {gather*} -x {\left (12 \, \log \relax (2) - 1\right )} - 4 \, x \log \left (2 \, x \log \relax (x) - \log \relax (2)\right ) - 8 \, x \log \relax (x) + 4 \, x \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x*log(x)^2+4*log(2)*log(x))*log((16*x^3*log(x)-8*x^2*log(2))/log(x))-22*x*log(x)^2+7*log(2)*log
(x)-4*log(2))/(2*x*log(x)^2-log(2)*log(x)),x, algorithm="maxima")

[Out]

-x*(12*log(2) - 1) - 4*x*log(2*x*log(x) - log(2)) - 8*x*log(x) + 4*x*log(log(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {4\,\ln \relax (2)+22\,x\,{\ln \relax (x)}^2-7\,\ln \relax (2)\,\ln \relax (x)+\ln \left (\frac {16\,x^3\,\ln \relax (x)-8\,x^2\,\ln \relax (2)}{\ln \relax (x)}\right )\,\left (8\,x\,{\ln \relax (x)}^2-4\,\ln \relax (2)\,\ln \relax (x)\right )}{2\,x\,{\ln \relax (x)}^2-\ln \relax (2)\,\ln \relax (x)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*log(2) + 22*x*log(x)^2 - 7*log(2)*log(x) + log((16*x^3*log(x) - 8*x^2*log(2))/log(x))*(8*x*log(x)^2 -
4*log(2)*log(x)))/(2*x*log(x)^2 - log(2)*log(x)),x)

[Out]

int(-(4*log(2) + 22*x*log(x)^2 - 7*log(2)*log(x) + log((16*x^3*log(x) - 8*x^2*log(2))/log(x))*(8*x*log(x)^2 -
4*log(2)*log(x)))/(2*x*log(x)^2 - log(2)*log(x)), x)

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sympy [A]  time = 0.43, size = 26, normalized size = 1.13 \begin {gather*} - 4 x \log {\left (\frac {16 x^{3} \log {\relax (x )} - 8 x^{2} \log {\relax (2 )}}{\log {\relax (x )}} \right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x*ln(x)**2+4*ln(2)*ln(x))*ln((16*x**3*ln(x)-8*x**2*ln(2))/ln(x))-22*x*ln(x)**2+7*ln(2)*ln(x)-4*
ln(2))/(2*x*ln(x)**2-ln(2)*ln(x)),x)

[Out]

-4*x*log((16*x**3*log(x) - 8*x**2*log(2))/log(x)) + x

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