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

Optimal. Leaf size=23 \[ \frac {25 \left (1+e^2\right )^2}{\log \left (\frac {4-\frac {x}{4}}{x}\right )} \]

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Rubi [F]  time = 0.16, 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 {-400-800 e^2-400 e^4}{\left (-16 x+x^2\right ) \log ^2\left (\frac {16-x}{4 x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 23, normalized size = 1.00 \begin {gather*} \frac {25 \left (1+e^2\right )^2}{\log \left (\frac {4-\frac {x}{4}}{x}\right )} \end {gather*}

Antiderivative was successfully verified.

[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[(4 - x/4)/x]

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fricas [A]  time = 0.61, size = 21, normalized size = 0.91 \begin {gather*} \frac {25 \, {\left (e^{4} + 2 \, e^{2} + 1\right )}}{\log \left (-\frac {x - 16}{4 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.12, size = 21, normalized size = 0.91 \begin {gather*} \frac {25 \, {\left (e^{4} + 2 \, e^{2} + 1\right )}}{\log \left (-\frac {x - 16}{4 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.36, size = 25, normalized size = 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	\(-\frac {-400 \,{\mathrm e}^{4}-800 \,{\mathrm e}^{2}-400}{16 \ln \left (-\frac {1}{4}+\frac {4}{x}\right )}\)	\(25\)
norman	\(\frac {25 \,{\mathrm e}^{4}+50 \,{\mathrm e}^{2}+25}{\ln \left (\frac {16-x}{4 x}\right )}\)	\(27\)
risch	\(\frac {25 \,{\mathrm e}^{4}}{\ln \left (\frac {16-x}{4 x}\right )}+\frac {50 \,{\mathrm e}^{2}}{\ln \left (\frac {16-x}{4 x}\right )}+\frac {25}{\ln \left (\frac {16-x}{4 x}\right )}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.52, size = 27, normalized size = 1.17 \begin {gather*} -\frac {25 \, {\left (e^{4} + 2 \, e^{2} + 1\right )}}{2 \, \log \relax (2) + \log \relax (x) - \log \left (-x + 16\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [B]  time = 5.26, size = 22, normalized size = 0.96 \begin {gather*} \frac {50\,{\mathrm {e}}^2+25\,{\mathrm {e}}^4+25}{\ln \left (-\frac {x-16}{4\,x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.12, size = 19, normalized size = 0.83 \begin {gather*} \frac {25 + 50 e^{2} + 25 e^{4}}{\log {\left (\frac {4 - \frac {x}{4}}{x} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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