\(\int \frac {-4 \log (2)+7 \log (2) \log (x)-22 x \log ^2(x)+(4 \log (2) \log (x)-8 x \log ^2(x)) \log (\frac {-8 x^2 \log (2)+16 x^3 \log (x)}{\log (x)})}{-\log (2) \log (x)+2 x \log ^2(x)} \, dx\) [8799]

Optimal result

Integrand size = 71, antiderivative size = 23 \[ \int \frac {-4 \log (2)+7 \log (2) \log (x)-22 x \log ^2(x)+\left (4 \log (2) \log (x)-8 x \log ^2(x)\right ) \log \left (\frac {-8 x^2 \log (2)+16 x^3 \log (x)}{\log (x)}\right )}{-\log (2) \log (x)+2 x \log ^2(x)} \, dx=x-4 x \log \left (8 x^2 \left (2 x-\frac {\log (2)}{\log (x)}\right )\right ) \]

[Out]

x-4*ln(4*x^2*(4*x-2*ln(2)/ln(x)))*x

Rubi [F]

\[ \int \frac {-4 \log (2)+7 \log (2) \log (x)-22 x \log ^2(x)+\left (4 \log (2) \log (x)-8 x \log ^2(x)\right ) \log \left (\frac {-8 x^2 \log (2)+16 x^3 \log (x)}{\log (x)}\right )}{-\log (2) \log (x)+2 x \log ^2(x)} \, dx=\int \frac {-4 \log (2)+7 \log (2) \log (x)-22 x \log ^2(x)+\left (4 \log (2) \log (x)-8 x \log ^2(x)\right ) \log \left (\frac {-8 x^2 \log (2)+16 x^3 \log (x)}{\log (x)}\right )}{-\log (2) \log (x)+2 x \log ^2(x)} \, dx \]

[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{align*} \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) \operatorname {LogIntegral}(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 \operatorname {LogIntegral}(x)+\frac {\log (16) \operatorname {LogIntegral}(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 \operatorname {LogIntegral}(x)+\frac {\log (16) \operatorname {LogIntegral}(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{align*}

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {-4 \log (2)+7 \log (2) \log (x)-22 x \log ^2(x)+\left (4 \log (2) \log (x)-8 x \log ^2(x)\right ) \log \left (\frac {-8 x^2 \log (2)+16 x^3 \log (x)}{\log (x)}\right )}{-\log (2) \log (x)+2 x \log ^2(x)} \, dx=x-4 x \log \left (8 x^2 \left (2 x-\frac {\log (2)}{\log (x)}\right )\right ) \]

[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])]

Maple [A] (verified)

Time = 2.88 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17

method	result	size
norman	\(x -4 x \ln \left (\frac {16 x^{3} \ln \left (x \right )-8 x^{2} \ln \left (2\right )}{\ln \left (x \right )}\right )\)	\(27\)
parallelrisch	\(x -4 x \ln \left (\frac {16 x^{3} \ln \left (x \right )-8 x^{2} \ln \left (2\right )}{\ln \left (x \right )}\right )\)	\(27\)
default	\(-12 x \ln \left (2\right )+x -4 \ln \left (\frac {x^{2} \left (2 x \ln \left (x \right )-\ln \left (2\right )\right )}{\ln \left (x \right )}\right ) x\)	\(30\)
risch	\(-4 x \ln \left (-2 x \ln \left (x \right )+\ln \left (2\right )\right )+4 x \ln \left (\ln \left (x \right )\right )-2 i \pi x \operatorname {csgn}\left (\frac {i x^{2} \left (-2 x \ln \left (x \right )+\ln \left (2\right )\right )}{\ln \left (x \right )}\right )^{3}+2 i \pi x \operatorname {csgn}\left (i x^{2}\right )^{3}+2 i \pi x \,\operatorname {csgn}\left (i \left (-2 x \ln \left (x \right )+\ln \left (2\right )\right )\right ) \operatorname {csgn}\left (\frac {i \left (-2 x \ln \left (x \right )+\ln \left (2\right )\right )}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right )+4 i \pi x \operatorname {csgn}\left (\frac {i x^{2} \left (-2 x \ln \left (x \right )+\ln \left (2\right )\right )}{\ln \left (x \right )}\right )^{2}-4 i \pi x -2 i \pi x \operatorname {csgn}\left (\frac {i \left (-2 x \ln \left (x \right )+\ln \left (2\right )\right )}{\ln \left (x \right )}\right )^{2} \operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right )+2 i \pi x \,\operatorname {csgn}\left (\frac {i \left (-2 x \ln \left (x \right )+\ln \left (2\right )\right )}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i x^{2} \left (-2 x \ln \left (x \right )+\ln \left (2\right )\right )}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (i x^{2}\right )+2 i \pi x \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+2 i \pi x \operatorname {csgn}\left (\frac {i \left (-2 x \ln \left (x \right )+\ln \left (2\right )\right )}{\ln \left (x \right )}\right )^{3}-4 i \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-2 i \pi x \operatorname {csgn}\left (\frac {i x^{2} \left (-2 x \ln \left (x \right )+\ln \left (2\right )\right )}{\ln \left (x \right )}\right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 i \pi x \,\operatorname {csgn}\left (\frac {i \left (-2 x \ln \left (x \right )+\ln \left (2\right )\right )}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i x^{2} \left (-2 x \ln \left (x \right )+\ln \left (2\right )\right )}{\ln \left (x \right )}\right )^{2}-2 i \pi x \,\operatorname {csgn}\left (i \left (-2 x \ln \left (x \right )+\ln \left (2\right )\right )\right ) \operatorname {csgn}\left (\frac {i \left (-2 x \ln \left (x \right )+\ln \left (2\right )\right )}{\ln \left (x \right )}\right )^{2}-12 x \ln \left (2\right )-8 x \ln \left (x \right )+x\)	\(392\)

[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))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {-4 \log (2)+7 \log (2) \log (x)-22 x \log ^2(x)+\left (4 \log (2) \log (x)-8 x \log ^2(x)\right ) \log \left (\frac {-8 x^2 \log (2)+16 x^3 \log (x)}{\log (x)}\right )}{-\log (2) \log (x)+2 x \log ^2(x)} \, dx=-4 \, x \log \left (\frac {8 \, {\left (2 \, x^{3} \log \left (x\right ) - x^{2} \log \left (2\right )\right )}}{\log \left (x\right )}\right ) + x \]

[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

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {-4 \log (2)+7 \log (2) \log (x)-22 x \log ^2(x)+\left (4 \log (2) \log (x)-8 x \log ^2(x)\right ) \log \left (\frac {-8 x^2 \log (2)+16 x^3 \log (x)}{\log (x)}\right )}{-\log (2) \log (x)+2 x \log ^2(x)} \, dx=- 4 x \log {\left (\frac {16 x^{3} \log {\left (x \right )} - 8 x^{2} \log {\left (2 \right )}}{\log {\left (x \right )}} \right )} + x \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.52 \[ \int \frac {-4 \log (2)+7 \log (2) \log (x)-22 x \log ^2(x)+\left (4 \log (2) \log (x)-8 x \log ^2(x)\right ) \log \left (\frac {-8 x^2 \log (2)+16 x^3 \log (x)}{\log (x)}\right )}{-\log (2) \log (x)+2 x \log ^2(x)} \, dx=-x {\left (12 \, \log \left (2\right ) - 1\right )} - 4 \, x \log \left (2 \, x \log \left (x\right ) - \log \left (2\right )\right ) - 8 \, x \log \left (x\right ) + 4 \, x \log \left (\log \left (x\right )\right ) \]

[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))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.52 \[ \int \frac {-4 \log (2)+7 \log (2) \log (x)-22 x \log ^2(x)+\left (4 \log (2) \log (x)-8 x \log ^2(x)\right ) \log \left (\frac {-8 x^2 \log (2)+16 x^3 \log (x)}{\log (x)}\right )}{-\log (2) \log (x)+2 x \log ^2(x)} \, dx=-x {\left (12 \, \log \left (2\right ) - 1\right )} - 4 \, x \log \left (2 \, x \log \left (x\right ) - \log \left (2\right )\right ) - 8 \, x \log \left (x\right ) + 4 \, x \log \left (\log \left (x\right )\right ) \]

[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))

Mupad [F(-1)]

Timed out. \[ \int \frac {-4 \log (2)+7 \log (2) \log (x)-22 x \log ^2(x)+\left (4 \log (2) \log (x)-8 x \log ^2(x)\right ) \log \left (\frac {-8 x^2 \log (2)+16 x^3 \log (x)}{\log (x)}\right )}{-\log (2) \log (x)+2 x \log ^2(x)} \, dx=\int -\frac {4\,\ln \left (2\right )+22\,x\,{\ln \left (x\right )}^2-7\,\ln \left (2\right )\,\ln \left (x\right )+\ln \left (\frac {16\,x^3\,\ln \left (x\right )-8\,x^2\,\ln \left (2\right )}{\ln \left (x\right )}\right )\,\left (8\,x\,{\ln \left (x\right )}^2-4\,\ln \left (2\right )\,\ln \left (x\right )\right )}{2\,x\,{\ln \left (x\right )}^2-\ln \left (2\right )\,\ln \left (x\right )} \,d x \]

[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)