\(\int \frac {(8 x^2-50 x^3-2 x^4-16 x^5+100 x^6+4 x^7) \log ^2(\frac {1}{x})+(8-50 x-2 x^2) \log (4-25 x-x^2)+\log (\frac {1}{x}) (-50 x-4 x^2+(-8+50 x+2 x^2) \log (4-25 x-x^2))}{(-4 x^2+25 x^3+x^4) \log ^2(\frac {1}{x})} \, dx\) [6249]

Optimal result

Integrand size = 120, antiderivative size = 27 \[ \int \frac {\left (8 x^2-50 x^3-2 x^4-16 x^5+100 x^6+4 x^7\right ) \log ^2\left (\frac {1}{x}\right )+\left (8-50 x-2 x^2\right ) \log \left (4-25 x-x^2\right )+\log \left (\frac {1}{x}\right ) \left (-50 x-4 x^2+\left (-8+50 x+2 x^2\right ) \log \left (4-25 x-x^2\right )\right )}{\left (-4 x^2+25 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=x^4-2 \left (x+\frac {\log (4-x (25+x))}{x \log \left (\frac {1}{x}\right )}\right ) \]

[Out]

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

Rubi [F]

\[ \int \frac {\left (8 x^2-50 x^3-2 x^4-16 x^5+100 x^6+4 x^7\right ) \log ^2\left (\frac {1}{x}\right )+\left (8-50 x-2 x^2\right ) \log \left (4-25 x-x^2\right )+\log \left (\frac {1}{x}\right ) \left (-50 x-4 x^2+\left (-8+50 x+2 x^2\right ) \log \left (4-25 x-x^2\right )\right )}{\left (-4 x^2+25 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=\int \frac {\left (8 x^2-50 x^3-2 x^4-16 x^5+100 x^6+4 x^7\right ) \log ^2\left (\frac {1}{x}\right )+\left (8-50 x-2 x^2\right ) \log \left (4-25 x-x^2\right )+\log \left (\frac {1}{x}\right ) \left (-50 x-4 x^2+\left (-8+50 x+2 x^2\right ) \log \left (4-25 x-x^2\right )\right )}{\left (-4 x^2+25 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx \]

[In]

Int[((8*x^2 - 50*x^3 - 2*x^4 - 16*x^5 + 100*x^6 + 4*x^7)*Log[x^(-1)]^2 + (8 - 50*x - 2*x^2)*Log[4 - 25*x - x^2
] + Log[x^(-1)]*(-50*x - 4*x^2 + (-8 + 50*x + 2*x^2)*Log[4 - 25*x - x^2]))/((-4*x^2 + 25*x^3 + x^4)*Log[x^(-1)
]^2),x]

[Out]

-2*x + x^4 - 2*Log[4 - 25*x - x^2]*LogIntegral[x^(-1)] + 2*Defer[Int][(-25 - 2*x)/(x*(-4 + 25*x + x^2)*Log[x^(
-1)]), x] - 2*Defer[Int][Log[4 - 25*x - x^2]/(x^2*Log[x^(-1)]^2), x] - (100*Defer[Int][LogIntegral[x^(-1)]/(-2
5 + Sqrt[641] - 2*x), x])/Sqrt[641] + (4*(641 - 25*Sqrt[641])*Defer[Int][LogIntegral[x^(-1)]/(25 - Sqrt[641] +
2*x), x])/641 - (100*Defer[Int][LogIntegral[x^(-1)]/(25 + Sqrt[641] + 2*x), x])/Sqrt[641] + (4*(641 + 25*Sqrt
[641])*Defer[Int][LogIntegral[x^(-1)]/(25 + Sqrt[641] + 2*x), x])/641

Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (8 x^2-50 x^3-2 x^4-16 x^5+100 x^6+4 x^7\right ) \log ^2\left (\frac {1}{x}\right )+\left (8-50 x-2 x^2\right ) \log \left (4-25 x-x^2\right )+\log \left (\frac {1}{x}\right ) \left (-50 x-4 x^2+\left (-8+50 x+2 x^2\right ) \log \left (4-25 x-x^2\right )\right )}{x^2 \left (-4+25 x+x^2\right ) \log ^2\left (\frac {1}{x}\right )} \, dx \\ & = \int \left (\frac {2 \left (-25-2 x+4 x \log \left (\frac {1}{x}\right )-25 x^2 \log \left (\frac {1}{x}\right )-x^3 \log \left (\frac {1}{x}\right )-8 x^4 \log \left (\frac {1}{x}\right )+50 x^5 \log \left (\frac {1}{x}\right )+2 x^6 \log \left (\frac {1}{x}\right )\right )}{x \left (-4+25 x+x^2\right ) \log \left (\frac {1}{x}\right )}+\frac {2 \left (-1+\log \left (\frac {1}{x}\right )\right ) \log \left (4-25 x-x^2\right )}{x^2 \log ^2\left (\frac {1}{x}\right )}\right ) \, dx \\ & = 2 \int \frac {-25-2 x+4 x \log \left (\frac {1}{x}\right )-25 x^2 \log \left (\frac {1}{x}\right )-x^3 \log \left (\frac {1}{x}\right )-8 x^4 \log \left (\frac {1}{x}\right )+50 x^5 \log \left (\frac {1}{x}\right )+2 x^6 \log \left (\frac {1}{x}\right )}{x \left (-4+25 x+x^2\right ) \log \left (\frac {1}{x}\right )} \, dx+2 \int \frac {\left (-1+\log \left (\frac {1}{x}\right )\right ) \log \left (4-25 x-x^2\right )}{x^2 \log ^2\left (\frac {1}{x}\right )} \, dx \\ & = 2 \int \frac {25+2 x-x \left (4-25 x-x^2-8 x^3+50 x^4+2 x^5\right ) \log \left (\frac {1}{x}\right )}{x \left (4-25 x-x^2\right ) \log \left (\frac {1}{x}\right )} \, dx+2 \int \left (-\frac {\log \left (4-25 x-x^2\right )}{x^2 \log ^2\left (\frac {1}{x}\right )}+\frac {\log \left (4-25 x-x^2\right )}{x^2 \log \left (\frac {1}{x}\right )}\right ) \, dx \\ & = 2 \int \left (-1+2 x^3+\frac {-25-2 x}{x \left (-4+25 x+x^2\right ) \log \left (\frac {1}{x}\right )}\right ) \, dx-2 \int \frac {\log \left (4-25 x-x^2\right )}{x^2 \log ^2\left (\frac {1}{x}\right )} \, dx+2 \int \frac {\log \left (4-25 x-x^2\right )}{x^2 \log \left (\frac {1}{x}\right )} \, dx \\ & = -2 x+x^4-2 \log \left (4-25 x-x^2\right ) \text {li}\left (\frac {1}{x}\right )+2 \int \frac {-25-2 x}{x \left (-4+25 x+x^2\right ) \log \left (\frac {1}{x}\right )} \, dx-2 \int \frac {\log \left (4-25 x-x^2\right )}{x^2 \log ^2\left (\frac {1}{x}\right )} \, dx+2 \int \frac {(-25-2 x) \text {li}\left (\frac {1}{x}\right )}{4-25 x-x^2} \, dx \\ & = -2 x+x^4-2 \log \left (4-25 x-x^2\right ) \text {li}\left (\frac {1}{x}\right )+2 \int \frac {-25-2 x}{x \left (-4+25 x+x^2\right ) \log \left (\frac {1}{x}\right )} \, dx-2 \int \frac {\log \left (4-25 x-x^2\right )}{x^2 \log ^2\left (\frac {1}{x}\right )} \, dx+2 \int \left (\frac {25 \text {li}\left (\frac {1}{x}\right )}{-4+25 x+x^2}+\frac {2 x \text {li}\left (\frac {1}{x}\right )}{-4+25 x+x^2}\right ) \, dx \\ & = -2 x+x^4-2 \log \left (4-25 x-x^2\right ) \text {li}\left (\frac {1}{x}\right )+2 \int \frac {-25-2 x}{x \left (-4+25 x+x^2\right ) \log \left (\frac {1}{x}\right )} \, dx-2 \int \frac {\log \left (4-25 x-x^2\right )}{x^2 \log ^2\left (\frac {1}{x}\right )} \, dx+4 \int \frac {x \text {li}\left (\frac {1}{x}\right )}{-4+25 x+x^2} \, dx+50 \int \frac {\text {li}\left (\frac {1}{x}\right )}{-4+25 x+x^2} \, dx \\ & = -2 x+x^4-2 \log \left (4-25 x-x^2\right ) \text {li}\left (\frac {1}{x}\right )+2 \int \frac {-25-2 x}{x \left (-4+25 x+x^2\right ) \log \left (\frac {1}{x}\right )} \, dx-2 \int \frac {\log \left (4-25 x-x^2\right )}{x^2 \log ^2\left (\frac {1}{x}\right )} \, dx+4 \int \left (\frac {\left (1-\frac {25}{\sqrt {641}}\right ) \text {li}\left (\frac {1}{x}\right )}{25-\sqrt {641}+2 x}+\frac {\left (1+\frac {25}{\sqrt {641}}\right ) \text {li}\left (\frac {1}{x}\right )}{25+\sqrt {641}+2 x}\right ) \, dx+50 \int \left (-\frac {2 \text {li}\left (\frac {1}{x}\right )}{\sqrt {641} \left (-25+\sqrt {641}-2 x\right )}-\frac {2 \text {li}\left (\frac {1}{x}\right )}{\sqrt {641} \left (25+\sqrt {641}+2 x\right )}\right ) \, dx \\ & = -2 x+x^4-2 \log \left (4-25 x-x^2\right ) \text {li}\left (\frac {1}{x}\right )+2 \int \frac {-25-2 x}{x \left (-4+25 x+x^2\right ) \log \left (\frac {1}{x}\right )} \, dx-2 \int \frac {\log \left (4-25 x-x^2\right )}{x^2 \log ^2\left (\frac {1}{x}\right )} \, dx-\frac {100 \int \frac {\text {li}\left (\frac {1}{x}\right )}{-25+\sqrt {641}-2 x} \, dx}{\sqrt {641}}-\frac {100 \int \frac {\text {li}\left (\frac {1}{x}\right )}{25+\sqrt {641}+2 x} \, dx}{\sqrt {641}}+\frac {1}{641} \left (4 \left (641-25 \sqrt {641}\right )\right ) \int \frac {\text {li}\left (\frac {1}{x}\right )}{25-\sqrt {641}+2 x} \, dx+\frac {1}{641} \left (4 \left (641+25 \sqrt {641}\right )\right ) \int \frac {\text {li}\left (\frac {1}{x}\right )}{25+\sqrt {641}+2 x} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {\left (8 x^2-50 x^3-2 x^4-16 x^5+100 x^6+4 x^7\right ) \log ^2\left (\frac {1}{x}\right )+\left (8-50 x-2 x^2\right ) \log \left (4-25 x-x^2\right )+\log \left (\frac {1}{x}\right ) \left (-50 x-4 x^2+\left (-8+50 x+2 x^2\right ) \log \left (4-25 x-x^2\right )\right )}{\left (-4 x^2+25 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=2 \left (-x+\frac {x^4}{2}-\frac {\log \left (4-25 x-x^2\right )}{x \log \left (\frac {1}{x}\right )}\right ) \]

[In]

Integrate[((8*x^2 - 50*x^3 - 2*x^4 - 16*x^5 + 100*x^6 + 4*x^7)*Log[x^(-1)]^2 + (8 - 50*x - 2*x^2)*Log[4 - 25*x
- x^2] + Log[x^(-1)]*(-50*x - 4*x^2 + (-8 + 50*x + 2*x^2)*Log[4 - 25*x - x^2]))/((-4*x^2 + 25*x^3 + x^4)*Log[
x^(-1)]^2),x]

[Out]

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

Maple [A] (verified)

Time = 23.58 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04

method	result	size
risch	\(\frac {2 \ln \left (-x^{2}-25 x +4\right )}{x \ln \left (x \right )}+x^{4}-2 x\)	\(28\)
parallelrisch	\(\frac {50 x^{5} \ln \left (\frac {1}{x}\right )-100 x^{2} \ln \left (\frac {1}{x}\right )+1234 x \ln \left (\frac {1}{x}\right )-100 \ln \left (-x^{2}-25 x +4\right )}{50 x \ln \left (\frac {1}{x}\right )}\)	\(51\)

[In]

int(((4*x^7+100*x^6-16*x^5-2*x^4-50*x^3+8*x^2)*ln(1/x)^2+((2*x^2+50*x-8)*ln(-x^2-25*x+4)-4*x^2-50*x)*ln(1/x)+(
-2*x^2-50*x+8)*ln(-x^2-25*x+4))/(x^4+25*x^3-4*x^2)/ln(1/x)^2,x,method=_RETURNVERBOSE)

[Out]

2/x/ln(x)*ln(-x^2-25*x+4)+x^4-2*x

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {\left (8 x^2-50 x^3-2 x^4-16 x^5+100 x^6+4 x^7\right ) \log ^2\left (\frac {1}{x}\right )+\left (8-50 x-2 x^2\right ) \log \left (4-25 x-x^2\right )+\log \left (\frac {1}{x}\right ) \left (-50 x-4 x^2+\left (-8+50 x+2 x^2\right ) \log \left (4-25 x-x^2\right )\right )}{\left (-4 x^2+25 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=\frac {{\left (x^{5} - 2 \, x^{2}\right )} \log \left (\frac {1}{x}\right ) - 2 \, \log \left (-x^{2} - 25 \, x + 4\right )}{x \log \left (\frac {1}{x}\right )} \]

[In]

integrate(((4*x^7+100*x^6-16*x^5-2*x^4-50*x^3+8*x^2)*log(1/x)^2+((2*x^2+50*x-8)*log(-x^2-25*x+4)-4*x^2-50*x)*l
og(1/x)+(-2*x^2-50*x+8)*log(-x^2-25*x+4))/(x^4+25*x^3-4*x^2)/log(1/x)^2,x, algorithm="fricas")

[Out]

((x^5 - 2*x^2)*log(1/x) - 2*log(-x^2 - 25*x + 4))/(x*log(1/x))

Sympy [F(-2)]

Exception generated. \[ \int \frac {\left (8 x^2-50 x^3-2 x^4-16 x^5+100 x^6+4 x^7\right ) \log ^2\left (\frac {1}{x}\right )+\left (8-50 x-2 x^2\right ) \log \left (4-25 x-x^2\right )+\log \left (\frac {1}{x}\right ) \left (-50 x-4 x^2+\left (-8+50 x+2 x^2\right ) \log \left (4-25 x-x^2\right )\right )}{\left (-4 x^2+25 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(((4*x**7+100*x**6-16*x**5-2*x**4-50*x**3+8*x**2)*ln(1/x)**2+((2*x**2+50*x-8)*ln(-x**2-25*x+4)-4*x**2
-50*x)*ln(1/x)+(-2*x**2-50*x+8)*ln(-x**2-25*x+4))/(x**4+25*x**3-4*x**2)/ln(1/x)**2,x)

[Out]

Exception raised: TypeError >> '>' not supported between instances of 'Poly' and 'int'

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {\left (8 x^2-50 x^3-2 x^4-16 x^5+100 x^6+4 x^7\right ) \log ^2\left (\frac {1}{x}\right )+\left (8-50 x-2 x^2\right ) \log \left (4-25 x-x^2\right )+\log \left (\frac {1}{x}\right ) \left (-50 x-4 x^2+\left (-8+50 x+2 x^2\right ) \log \left (4-25 x-x^2\right )\right )}{\left (-4 x^2+25 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=\frac {{\left (x^{5} - 2 \, x^{2}\right )} \log \left (x\right ) + 2 \, \log \left (-x^{2} - 25 \, x + 4\right )}{x \log \left (x\right )} \]

[In]

integrate(((4*x^7+100*x^6-16*x^5-2*x^4-50*x^3+8*x^2)*log(1/x)^2+((2*x^2+50*x-8)*log(-x^2-25*x+4)-4*x^2-50*x)*l
og(1/x)+(-2*x^2-50*x+8)*log(-x^2-25*x+4))/(x^4+25*x^3-4*x^2)/log(1/x)^2,x, algorithm="maxima")

[Out]

((x^5 - 2*x^2)*log(x) + 2*log(-x^2 - 25*x + 4))/(x*log(x))

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {\left (8 x^2-50 x^3-2 x^4-16 x^5+100 x^6+4 x^7\right ) \log ^2\left (\frac {1}{x}\right )+\left (8-50 x-2 x^2\right ) \log \left (4-25 x-x^2\right )+\log \left (\frac {1}{x}\right ) \left (-50 x-4 x^2+\left (-8+50 x+2 x^2\right ) \log \left (4-25 x-x^2\right )\right )}{\left (-4 x^2+25 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=x^{4} - 2 \, x + \frac {2 \, \log \left (-x^{2} - 25 \, x + 4\right )}{x \log \left (x\right )} \]

[In]

integrate(((4*x^7+100*x^6-16*x^5-2*x^4-50*x^3+8*x^2)*log(1/x)^2+((2*x^2+50*x-8)*log(-x^2-25*x+4)-4*x^2-50*x)*l
og(1/x)+(-2*x^2-50*x+8)*log(-x^2-25*x+4))/(x^4+25*x^3-4*x^2)/log(1/x)^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 11.37 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {\left (8 x^2-50 x^3-2 x^4-16 x^5+100 x^6+4 x^7\right ) \log ^2\left (\frac {1}{x}\right )+\left (8-50 x-2 x^2\right ) \log \left (4-25 x-x^2\right )+\log \left (\frac {1}{x}\right ) \left (-50 x-4 x^2+\left (-8+50 x+2 x^2\right ) \log \left (4-25 x-x^2\right )\right )}{\left (-4 x^2+25 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=x^4-2\,x-\frac {2\,\ln \left (-x^2-25\,x+4\right )}{x\,\ln \left (\frac {1}{x}\right )} \]

[In]

int(-(log(4 - x^2 - 25*x)*(50*x + 2*x^2 - 8) - log(1/x)^2*(8*x^2 - 50*x^3 - 2*x^4 - 16*x^5 + 100*x^6 + 4*x^7)
+ log(1/x)*(50*x - log(4 - x^2 - 25*x)*(50*x + 2*x^2 - 8) + 4*x^2))/(log(1/x)^2*(25*x^3 - 4*x^2 + x^4)),x)

[Out]

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