\(\int \frac {-160 x+68 x^2-8 x^3-4 x \log (7)+(-72 x+16 x^2) \log (7) \log (\frac {4}{x})-8 x \log ^2(7) \log ^2(\frac {4}{x})}{16-8 x+x^2+(8-2 x) \log (7) \log (\frac {4}{x})+\log ^2(7) \log ^2(\frac {4}{x})} \, dx\) [6054]

Optimal result

Integrand size = 90, antiderivative size = 27 \[ \int \frac {-160 x+68 x^2-8 x^3-4 x \log (7)+\left (-72 x+16 x^2\right ) \log (7) \log \left (\frac {4}{x}\right )-8 x \log ^2(7) \log ^2\left (\frac {4}{x}\right )}{16-8 x+x^2+(8-2 x) \log (7) \log \left (\frac {4}{x}\right )+\log ^2(7) \log ^2\left (\frac {4}{x}\right )} \, dx=5-x^2 \left (4+\frac {4}{4-x+\log (7) \log \left (\frac {4}{x}\right )}\right ) \]

[Out]

5-x^2*(4+4/(4+ln(7)*ln(4/x)-x))

Rubi [F]

\[ \int \frac {-160 x+68 x^2-8 x^3-4 x \log (7)+\left (-72 x+16 x^2\right ) \log (7) \log \left (\frac {4}{x}\right )-8 x \log ^2(7) \log ^2\left (\frac {4}{x}\right )}{16-8 x+x^2+(8-2 x) \log (7) \log \left (\frac {4}{x}\right )+\log ^2(7) \log ^2\left (\frac {4}{x}\right )} \, dx=\int \frac {-160 x+68 x^2-8 x^3-4 x \log (7)+\left (-72 x+16 x^2\right ) \log (7) \log \left (\frac {4}{x}\right )-8 x \log ^2(7) \log ^2\left (\frac {4}{x}\right )}{16-8 x+x^2+(8-2 x) \log (7) \log \left (\frac {4}{x}\right )+\log ^2(7) \log ^2\left (\frac {4}{x}\right )} \, dx \]

[In]

Int[(-160*x + 68*x^2 - 8*x^3 - 4*x*Log[7] + (-72*x + 16*x^2)*Log[7]*Log[4/x] - 8*x*Log[7]^2*Log[4/x]^2)/(16 -
8*x + x^2 + (8 - 2*x)*Log[7]*Log[4/x] + Log[7]^2*Log[4/x]^2),x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {68 x^2-8 x^3+x (-160-4 \log (7))+\left (-72 x+16 x^2\right ) \log (7) \log \left (\frac {4}{x}\right )-8 x \log ^2(7) \log ^2\left (\frac {4}{x}\right )}{16-8 x+x^2+(8-2 x) \log (7) \log \left (\frac {4}{x}\right )+\log ^2(7) \log ^2\left (\frac {4}{x}\right )} \, dx \\ & = \int \frac {4 x \left (17 x-2 x^2-40 \left (1+\frac {\log (7)}{40}\right )-2 (9-2 x) \log (7) \log \left (\frac {4}{x}\right )-2 \log ^2(7) \log ^2\left (\frac {4}{x}\right )\right )}{\left (4-x+\log (7) \log \left (\frac {4}{x}\right )\right )^2} \, dx \\ & = 4 \int \frac {x \left (17 x-2 x^2-40 \left (1+\frac {\log (7)}{40}\right )-2 (9-2 x) \log (7) \log \left (\frac {4}{x}\right )-2 \log ^2(7) \log ^2\left (\frac {4}{x}\right )\right )}{\left (4-x+\log (7) \log \left (\frac {4}{x}\right )\right )^2} \, dx \\ & = 4 \int \left (-2 x-\frac {x (x+\log (7))}{\left (-4+x-\log (7) \log \left (\frac {4}{x}\right )\right )^2}+\frac {2 x}{-4+x-\log (7) \log \left (\frac {4}{x}\right )}\right ) \, dx \\ & = -4 x^2-4 \int \frac {x (x+\log (7))}{\left (-4+x-\log (7) \log \left (\frac {4}{x}\right )\right )^2} \, dx+8 \int \frac {x}{-4+x-\log (7) \log \left (\frac {4}{x}\right )} \, dx \\ & = -4 x^2-4 \int \left (\frac {x^2}{\left (-4+x-\log (7) \log \left (\frac {4}{x}\right )\right )^2}+\frac {x \log (7)}{\left (-4+x-\log (7) \log \left (\frac {4}{x}\right )\right )^2}\right ) \, dx+8 \int \frac {x}{-4+x-\log (7) \log \left (\frac {4}{x}\right )} \, dx \\ & = -4 x^2-4 \int \frac {x^2}{\left (-4+x-\log (7) \log \left (\frac {4}{x}\right )\right )^2} \, dx+8 \int \frac {x}{-4+x-\log (7) \log \left (\frac {4}{x}\right )} \, dx-(4 \log (7)) \int \frac {x}{\left (-4+x-\log (7) \log \left (\frac {4}{x}\right )\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {-160 x+68 x^2-8 x^3-4 x \log (7)+\left (-72 x+16 x^2\right ) \log (7) \log \left (\frac {4}{x}\right )-8 x \log ^2(7) \log ^2\left (\frac {4}{x}\right )}{16-8 x+x^2+(8-2 x) \log (7) \log \left (\frac {4}{x}\right )+\log ^2(7) \log ^2\left (\frac {4}{x}\right )} \, dx=-4 \left (x^2+\frac {x^2}{4-x+\log (7) \log \left (\frac {4}{x}\right )}\right ) \]

[In]

Integrate[(-160*x + 68*x^2 - 8*x^3 - 4*x*Log[7] + (-72*x + 16*x^2)*Log[7]*Log[4/x] - 8*x*Log[7]^2*Log[4/x]^2)/
(16 - 8*x + x^2 + (8 - 2*x)*Log[7]*Log[4/x] + Log[7]^2*Log[4/x]^2),x]

[Out]

-4*(x^2 + x^2/(4 - x + Log[7]*Log[4/x]))

Maple [A] (verified)

Time = 1.65 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04

method	result	size
risch	\(-4 x^{2}-\frac {4 x^{2}}{4+\ln \left (7\right ) \ln \left (\frac {4}{x}\right )-x}\)	\(28\)
norman	\(\frac {-20 x^{2}+4 x^{3}-4 \ln \left (7\right ) \ln \left (\frac {4}{x}\right ) x^{2}}{4+\ln \left (7\right ) \ln \left (\frac {4}{x}\right )-x}\)	\(42\)
parallelrisch	\(\frac {-20 x^{2}+4 x^{3}-4 \ln \left (7\right ) \ln \left (\frac {4}{x}\right ) x^{2}}{4+\ln \left (7\right ) \ln \left (\frac {4}{x}\right )-x}\)	\(42\)
derivativedivides	\(\frac {4 \left (4-\frac {4 \ln \left (7\right ) \ln \left (\frac {4}{x}\right )}{x}-\frac {20}{x}\right ) x^{2}}{\frac {4 \ln \left (7\right ) \ln \left (\frac {4}{x}\right )}{x}+\frac {16}{x}-4}\)	\(48\)
default	\(\frac {4 \left (4-\frac {4 \ln \left (7\right ) \ln \left (\frac {4}{x}\right )}{x}-\frac {20}{x}\right ) x^{2}}{\frac {4 \ln \left (7\right ) \ln \left (\frac {4}{x}\right )}{x}+\frac {16}{x}-4}\)	\(48\)

[In]

int((-8*x*ln(7)^2*ln(4/x)^2+(16*x^2-72*x)*ln(7)*ln(4/x)-4*x*ln(7)-8*x^3+68*x^2-160*x)/(ln(7)^2*ln(4/x)^2+(-2*x
+8)*ln(7)*ln(4/x)+x^2-8*x+16),x,method=_RETURNVERBOSE)

[Out]

-4*x^2-4*x^2/(4+ln(7)*ln(4/x)-x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.52 \[ \int \frac {-160 x+68 x^2-8 x^3-4 x \log (7)+\left (-72 x+16 x^2\right ) \log (7) \log \left (\frac {4}{x}\right )-8 x \log ^2(7) \log ^2\left (\frac {4}{x}\right )}{16-8 x+x^2+(8-2 x) \log (7) \log \left (\frac {4}{x}\right )+\log ^2(7) \log ^2\left (\frac {4}{x}\right )} \, dx=-\frac {4 \, {\left (x^{2} \log \left (7\right ) \log \left (\frac {4}{x}\right ) - x^{3} + 5 \, x^{2}\right )}}{\log \left (7\right ) \log \left (\frac {4}{x}\right ) - x + 4} \]

[In]

integrate((-8*x*log(7)^2*log(4/x)^2+(16*x^2-72*x)*log(7)*log(4/x)-4*x*log(7)-8*x^3+68*x^2-160*x)/(log(7)^2*log
(4/x)^2+(-2*x+8)*log(7)*log(4/x)+x^2-8*x+16),x, algorithm="fricas")

[Out]

-4*(x^2*log(7)*log(4/x) - x^3 + 5*x^2)/(log(7)*log(4/x) - x + 4)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {-160 x+68 x^2-8 x^3-4 x \log (7)+\left (-72 x+16 x^2\right ) \log (7) \log \left (\frac {4}{x}\right )-8 x \log ^2(7) \log ^2\left (\frac {4}{x}\right )}{16-8 x+x^2+(8-2 x) \log (7) \log \left (\frac {4}{x}\right )+\log ^2(7) \log ^2\left (\frac {4}{x}\right )} \, dx=- 4 x^{2} - \frac {4 x^{2}}{- x + \log {\left (7 \right )} \log {\left (\frac {4}{x} \right )} + 4} \]

[In]

integrate((-8*x*ln(7)**2*ln(4/x)**2+(16*x**2-72*x)*ln(7)*ln(4/x)-4*x*ln(7)-8*x**3+68*x**2-160*x)/(ln(7)**2*ln(
4/x)**2+(-2*x+8)*ln(7)*ln(4/x)+x**2-8*x+16),x)

[Out]

-4*x**2 - 4*x**2/(-x + log(7)*log(4/x) + 4)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.70 \[ \int \frac {-160 x+68 x^2-8 x^3-4 x \log (7)+\left (-72 x+16 x^2\right ) \log (7) \log \left (\frac {4}{x}\right )-8 x \log ^2(7) \log ^2\left (\frac {4}{x}\right )}{16-8 x+x^2+(8-2 x) \log (7) \log \left (\frac {4}{x}\right )+\log ^2(7) \log ^2\left (\frac {4}{x}\right )} \, dx=\frac {4 \, {\left (x^{2} \log \left (7\right ) \log \left (x\right ) - {\left (2 \, \log \left (7\right ) \log \left (2\right ) + 5\right )} x^{2} + x^{3}\right )}}{2 \, \log \left (7\right ) \log \left (2\right ) - \log \left (7\right ) \log \left (x\right ) - x + 4} \]

[In]

integrate((-8*x*log(7)^2*log(4/x)^2+(16*x^2-72*x)*log(7)*log(4/x)-4*x*log(7)-8*x^3+68*x^2-160*x)/(log(7)^2*log
(4/x)^2+(-2*x+8)*log(7)*log(4/x)+x^2-8*x+16),x, algorithm="maxima")

[Out]

4*(x^2*log(7)*log(x) - (2*log(7)*log(2) + 5)*x^2 + x^3)/(2*log(7)*log(2) - log(7)*log(x) - x + 4)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {-160 x+68 x^2-8 x^3-4 x \log (7)+\left (-72 x+16 x^2\right ) \log (7) \log \left (\frac {4}{x}\right )-8 x \log ^2(7) \log ^2\left (\frac {4}{x}\right )}{16-8 x+x^2+(8-2 x) \log (7) \log \left (\frac {4}{x}\right )+\log ^2(7) \log ^2\left (\frac {4}{x}\right )} \, dx=-4 \, x^{2} - \frac {4}{\frac {\log \left (7\right ) \log \left (\frac {4}{x}\right )}{x^{2}} - \frac {1}{x} + \frac {4}{x^{2}}} \]

[In]

integrate((-8*x*log(7)^2*log(4/x)^2+(16*x^2-72*x)*log(7)*log(4/x)-4*x*log(7)-8*x^3+68*x^2-160*x)/(log(7)^2*log
(4/x)^2+(-2*x+8)*log(7)*log(4/x)+x^2-8*x+16),x, algorithm="giac")

[Out]

-4*x^2 - 4/(log(7)*log(4/x)/x^2 - 1/x + 4/x^2)

Mupad [B] (verification not implemented)

Time = 11.78 (sec) , antiderivative size = 85, normalized size of antiderivative = 3.15 \[ \int \frac {-160 x+68 x^2-8 x^3-4 x \log (7)+\left (-72 x+16 x^2\right ) \log (7) \log \left (\frac {4}{x}\right )-8 x \log ^2(7) \log ^2\left (\frac {4}{x}\right )}{16-8 x+x^2+(8-2 x) \log (7) \log \left (\frac {4}{x}\right )+\log ^2(7) \log ^2\left (\frac {4}{x}\right )} \, dx=8\,x-\frac {\frac {8\,x^2\,\ln \left (\frac {4}{x}\right )}{x+\ln \left (7\right )}+\frac {4\,x\,\left (8\,x+x\,\ln \left (7\right )-x^2\right )}{\ln \left (7\right )\,\left (x+\ln \left (7\right )\right )}}{\ln \left (\frac {4}{x}\right )-\frac {x-4}{\ln \left (7\right )}}+\frac {8\,{\ln \left (7\right )}^2}{x+\ln \left (7\right )}-4\,x^2 \]

[In]

int(-(160*x + 4*x*log(7) - 68*x^2 + 8*x^3 + 8*x*log(7)^2*log(4/x)^2 + log(7)*log(4/x)*(72*x - 16*x^2))/(log(7)
^2*log(4/x)^2 - 8*x + x^2 - log(7)*log(4/x)*(2*x - 8) + 16),x)

[Out]

8*x - ((8*x^2*log(4/x))/(x + log(7)) + (4*x*(8*x + x*log(7) - x^2))/(log(7)*(x + log(7))))/(log(4/x) - (x - 4)
/log(7)) + (8*log(7)^2)/(x + log(7)) - 4*x^2