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

Optimal. Leaf size=20 \[ e^a \tan ^{-1}\left (e^a x^2\right )+\frac{1}{2 x^2} \]

[Out]

1/(2*x^2) + E^a*ArcTan[E^a*x^2]

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Rubi [F]  time = 0.0244611, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int \frac{\tanh (a+2 \log (x))}{x^3} \, dx \]

Verification is Not applicable to the result.

[In]

Int[Tanh[a + 2*Log[x]]/x^3,x]

[Out]

Defer[Int][Tanh[a + 2*Log[x]]/x^3, x]

Rubi steps

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

Mathematica [A]  time = 0.153145, size = 40, normalized size = 2. \[ \cosh (a) \left (-\tan ^{-1}\left (\frac{\cosh (a)-\sinh (a)}{x^2}\right )\right )-\sinh (a) \tan ^{-1}\left (\frac{\cosh (a)-\sinh (a)}{x^2}\right )+\frac{1}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[a + 2*Log[x]]/x^3,x]

[Out]

1/(2*x^2) - ArcTan[(Cosh[a] - Sinh[a])/x^2]*Cosh[a] - ArcTan[(Cosh[a] - Sinh[a])/x^2]*Sinh[a]

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Maple [C]  time = 0.027, size = 44, normalized size = 2.2 \begin{align*}{\frac{1}{2\,{x}^{2}}}+{\frac{\sum _{{\it \_R}={\it RootOf} \left ({{\rm e}^{2\,a}}+{{\it \_Z}}^{2} \right ) }{\it \_R}\,\ln \left ( \left ( 4\,{{\rm e}^{2\,a}}+5\,{{\it \_R}}^{2} \right ){x}^{2}-{\it \_R} \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/x^2+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 = 1.6208, size = 26, normalized size = 1.3 \begin{align*} -\arctan \left (\frac{e^{\left (-a\right )}}{x^{2}}\right ) e^{a} + \frac{1}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arctan(e^(-a)/x^2)*e^a + 1/2/x^2

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Fricas [A]  time = 2.09518, size = 55, normalized size = 2.75 \begin{align*} \frac{2 \, x^{2} \arctan \left (x^{2} e^{a}\right ) e^{a} + 1}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh{\left (a + 2 \log{\left (x \right )} \right )}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(tanh(a + 2*log(x))/x**3, x)

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Giac [A]  time = 1.20641, size = 22, normalized size = 1.1 \begin{align*} \arctan \left (x^{2} e^{a}\right ) e^{a} + \frac{1}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(x^2*e^a)*e^a + 1/2/x^2