\(\int \frac {-400-800 e^2-400 e^4}{(-16 x+x^2) \log ^2(\frac {16-x}{4 x})} \, dx\) [8473]

Optimal result

Integrand size = 37, antiderivative size = 23 \[ \int \frac {-400-800 e^2-400 e^4}{\left (-16 x+x^2\right ) \log ^2\left (\frac {16-x}{4 x}\right )} \, dx=\frac {25 \left (1+e^2\right )^2}{\log \left (\frac {4-\frac {x}{4}}{x}\right )} \]

[Out]

5/ln((4-1/4*x)/x)*(exp(2)+1)*(5+5*exp(2))

Rubi [F]

\[ \int \frac {-400-800 e^2-400 e^4}{\left (-16 x+x^2\right ) \log ^2\left (\frac {16-x}{4 x}\right )} \, dx=\int \frac {-400-800 e^2-400 e^4}{\left (-16 x+x^2\right ) \log ^2\left (\frac {16-x}{4 x}\right )} \, dx \]

[In]

Int[(-400 - 800*E^2 - 400*E^4)/((-16*x + x^2)*Log[(16 - x)/(4*x)]^2),x]

[Out]

-25*(1 + E^2)^2*Defer[Int][1/((-16 + x)*Log[-1/4 + 4/x]^2), x] + 25*(1 + E^2)^2*Defer[Int][1/(x*Log[-1/4 + 4/x
]^2), x]

Rubi steps \begin{align*} \text {integral}& = -\left (\left (400 \left (1+e^2\right )^2\right ) \int \frac {1}{\left (-16 x+x^2\right ) \log ^2\left (\frac {16-x}{4 x}\right )} \, dx\right ) \\ & = -\left (\left (400 \left (1+e^2\right )^2\right ) \int \frac {1}{(-16+x) x \log ^2\left (\frac {16-x}{4 x}\right )} \, dx\right ) \\ & = -\left (\left (400 \left (1+e^2\right )^2\right ) \int \frac {1}{(-16+x) x \log ^2\left (-\frac {1}{4}+\frac {4}{x}\right )} \, dx\right ) \\ & = -\left (\left (400 \left (1+e^2\right )^2\right ) \int \left (\frac {1}{16 (-16+x) \log ^2\left (-\frac {1}{4}+\frac {4}{x}\right )}-\frac {1}{16 x \log ^2\left (-\frac {1}{4}+\frac {4}{x}\right )}\right ) \, dx\right ) \\ & = -\left (\left (25 \left (1+e^2\right )^2\right ) \int \frac {1}{(-16+x) \log ^2\left (-\frac {1}{4}+\frac {4}{x}\right )} \, dx\right )+\left (25 \left (1+e^2\right )^2\right ) \int \frac {1}{x \log ^2\left (-\frac {1}{4}+\frac {4}{x}\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {-400-800 e^2-400 e^4}{\left (-16 x+x^2\right ) \log ^2\left (\frac {16-x}{4 x}\right )} \, dx=\frac {25 \left (1+e^2\right )^2}{\log \left (-\frac {1}{4}+\frac {4}{x}\right )} \]

[In]

Integrate[(-400 - 800*E^2 - 400*E^4)/((-16*x + x^2)*Log[(16 - x)/(4*x)]^2),x]

[Out]

(25*(1 + E^2)^2)/Log[-1/4 + 4/x]

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09

method	result	size
derivativedivides	\(\frac {400 \,{\mathrm e}^{4}+800 \,{\mathrm e}^{2}+400}{16 \ln \left (-\frac {1}{4}+\frac {4}{x}\right )}\)	\(25\)
default	\(-\frac {-400 \,{\mathrm e}^{4}-800 \,{\mathrm e}^{2}-400}{16 \ln \left (-\frac {1}{4}+\frac {4}{x}\right )}\)	\(25\)
parallelrisch	\(-\frac {-400 \,{\mathrm e}^{4}-800 \,{\mathrm e}^{2}-400}{16 \ln \left (-\frac {x -16}{4 x}\right )}\)	\(26\)
norman	\(\frac {25 \,{\mathrm e}^{4}+50 \,{\mathrm e}^{2}+25}{\ln \left (\frac {16-x}{4 x}\right )}\)	\(27\)
risch	\(\frac {25 \,{\mathrm e}^{4}}{\ln \left (\frac {16-x}{4 x}\right )}+\frac {50 \,{\mathrm e}^{2}}{\ln \left (\frac {16-x}{4 x}\right )}+\frac {25}{\ln \left (\frac {16-x}{4 x}\right )}\)	\(51\)

[In]

int((-400*exp(2)^2-800*exp(2)-400)/(x^2-16*x)/ln(1/4*(16-x)/x)^2,x,method=_RETURNVERBOSE)

[Out]

1/16*(400*exp(2)^2+800*exp(2)+400)/ln(-1/4+4/x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {-400-800 e^2-400 e^4}{\left (-16 x+x^2\right ) \log ^2\left (\frac {16-x}{4 x}\right )} \, dx=\frac {25 \, {\left (e^{4} + 2 \, e^{2} + 1\right )}}{\log \left (-\frac {x - 16}{4 \, x}\right )} \]

[In]

integrate((-400*exp(2)^2-800*exp(2)-400)/(x^2-16*x)/log(1/4*(16-x)/x)^2,x, algorithm="fricas")

[Out]

25*(e^4 + 2*e^2 + 1)/log(-1/4*(x - 16)/x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {-400-800 e^2-400 e^4}{\left (-16 x+x^2\right ) \log ^2\left (\frac {16-x}{4 x}\right )} \, dx=\frac {25 + 50 e^{2} + 25 e^{4}}{\log {\left (\frac {4 - \frac {x}{4}}{x} \right )}} \]

[In]

integrate((-400*exp(2)**2-800*exp(2)-400)/(x**2-16*x)/ln(1/4*(16-x)/x)**2,x)

[Out]

(25 + 50*exp(2) + 25*exp(4))/log((4 - x/4)/x)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {-400-800 e^2-400 e^4}{\left (-16 x+x^2\right ) \log ^2\left (\frac {16-x}{4 x}\right )} \, dx=-\frac {25 \, {\left (e^{4} + 2 \, e^{2} + 1\right )}}{2 \, \log \left (2\right ) + \log \left (x\right ) - \log \left (-x + 16\right )} \]

[In]

integrate((-400*exp(2)^2-800*exp(2)-400)/(x^2-16*x)/log(1/4*(16-x)/x)^2,x, algorithm="maxima")

[Out]

-25*(e^4 + 2*e^2 + 1)/(2*log(2) + log(x) - log(-x + 16))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {-400-800 e^2-400 e^4}{\left (-16 x+x^2\right ) \log ^2\left (\frac {16-x}{4 x}\right )} \, dx=\frac {25 \, {\left (e^{4} + 2 \, e^{2} + 1\right )}}{\log \left (-\frac {x - 16}{4 \, x}\right )} \]

[In]

integrate((-400*exp(2)^2-800*exp(2)-400)/(x^2-16*x)/log(1/4*(16-x)/x)^2,x, algorithm="giac")

[Out]

25*(e^4 + 2*e^2 + 1)/log(-1/4*(x - 16)/x)

Mupad [B] (verification not implemented)

Time = 13.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {-400-800 e^2-400 e^4}{\left (-16 x+x^2\right ) \log ^2\left (\frac {16-x}{4 x}\right )} \, dx=\frac {50\,{\mathrm {e}}^2+25\,{\mathrm {e}}^4+25}{\ln \left (-\frac {x-16}{4\,x}\right )} \]

[In]

int((800*exp(2) + 400*exp(4) + 400)/(log(-(x/4 - 4)/x)^2*(16*x - x^2)),x)

[Out]

(50*exp(2) + 25*exp(4) + 25)/log(-(x - 16)/(4*x))