3.159 \(\int \frac {\tanh ^2(a+2 \log (x))}{x^3} \, dx\)

Optimal. Leaf size=59 \[ -e^a \tan ^{-1}\left (e^a x^2\right )-\frac {3 e^{2 a} x^2}{2 \left (e^{2 a} x^4+1\right )}-\frac {1}{2 x^2 \left (e^{2 a} x^4+1\right )} \]

[Out]

-1/2/x^2/(1+exp(2*a)*x^4)-3/2*exp(2*a)*x^2/(1+exp(2*a)*x^4)-exp(a)*arctan(exp(a)*x^2)

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Rubi [F]  time = 0.05, 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 = {} \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^3} \, dx \]

Verification is Not applicable to the result.

[In]

Int[Tanh[a + 2*Log[x]]^2/x^3,x]

[Out]

Defer[Int][Tanh[a + 2*Log[x]]^2/x^3, x]

Rubi steps

\begin {align*} \int \frac {\tanh ^2(a+2 \log (x))}{x^3} \, dx &=\int \frac {\tanh ^2(a+2 \log (x))}{x^3} \, dx\\ \end {align*}

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Mathematica [A]  time = 0.40, size = 40, normalized size = 0.68 \[ \frac {-\frac {2}{e^{-2 (a+2 \log (x))}+1}-1}{2 x^2}+e^a \tan ^{-1}\left (\frac {e^{-a}}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[a + 2*Log[x]]^2/x^3,x]

[Out]

(-1 - 2/(1 + E^(-2*(a + 2*Log[x]))))/(2*x^2) + E^a*ArcTan[1/(E^a*x^2)]

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fricas [A]  time = 0.62, size = 51, normalized size = 0.86 \[ -\frac {3 \, x^{4} e^{\left (2 \, a\right )} + 2 \, {\left (x^{6} e^{\left (3 \, a\right )} + x^{2} e^{a}\right )} \arctan \left (x^{2} e^{a}\right ) + 1}{2 \, {\left (x^{6} e^{\left (2 \, a\right )} + x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(a+2*log(x))^2/x^3,x, algorithm="fricas")

[Out]

-1/2*(3*x^4*e^(2*a) + 2*(x^6*e^(3*a) + x^2*e^a)*arctan(x^2*e^a) + 1)/(x^6*e^(2*a) + x^2)

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giac [A]  time = 0.12, size = 39, normalized size = 0.66 \[ -\arctan \left (x^{2} e^{a}\right ) e^{a} - \frac {3 \, x^{4} e^{\left (2 \, a\right )} + 1}{2 \, {\left (x^{6} e^{\left (2 \, a\right )} + x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(a+2*log(x))^2/x^3,x, algorithm="giac")

[Out]

-arctan(x^2*e^a)*e^a - 1/2*(3*x^4*e^(2*a) + 1)/(x^6*e^(2*a) + x^2)

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maple [C]  time = 0.09, size = 66, normalized size = 1.12 \[ \frac {-\frac {3 \,{\mathrm e}^{2 a} x^{4}}{2}-\frac {1}{2}}{x^{2} \left (1+{\mathrm e}^{2 a} x^{4}\right )}+\frac {\left (\munderset {\textit {\_R} =\RootOf \left ({\mathrm e}^{2 a}+\textit {\_Z}^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (-4 \,{\mathrm e}^{2 a}-5 \textit {\_R}^{2}\right ) x^{2}-\textit {\_R} \right )\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(a+2*ln(x))^2/x^3,x)

[Out]

(-3/2*exp(2*a)*x^4-1/2)/x^2/(1+exp(2*a)*x^4)+1/2*sum(_R*ln((-4*exp(2*a)-5*_R^2)*x^2-_R),_R=RootOf(exp(2*a)+_Z^
2))

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maxima [A]  time = 0.44, size = 37, normalized size = 0.63 \[ \arctan \left (\frac {e^{\left (-a\right )}}{x^{2}}\right ) e^{a} - \frac {1}{2 \, x^{2}} - \frac {e^{\left (2 \, a\right )}}{x^{2} {\left (\frac {1}{x^{4}} + e^{\left (2 \, a\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(a+2*log(x))^2/x^3,x, algorithm="maxima")

[Out]

arctan(e^(-a)/x^2)*e^a - 1/2/x^2 - e^(2*a)/(x^2*(1/x^4 + e^(2*a)))

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mupad [B]  time = 1.06, size = 47, normalized size = 0.80 \[ -\mathrm {atan}\left (x^2\,\sqrt {{\mathrm {e}}^{2\,a}}\right )\,\sqrt {{\mathrm {e}}^{2\,a}}-\frac {\frac {3\,{\mathrm {e}}^{2\,a}\,x^4}{2}+\frac {1}{2}}{{\mathrm {e}}^{2\,a}\,x^6+x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(a + 2*log(x))^2/x^3,x)

[Out]

- atan(x^2*exp(2*a)^(1/2))*exp(2*a)^(1/2) - ((3*x^4*exp(2*a))/2 + 1/2)/(x^6*exp(2*a) + x^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh ^{2}{\left (a + 2 \log {\relax (x )} \right )}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(a+2*ln(x))**2/x**3,x)

[Out]

Integral(tanh(a + 2*log(x))**2/x**3, x)

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