3.148 \(\int x \tanh (a+2 \log (x)) \, dx\)

Optimal. Leaf size=23 \[ \frac{x^2}{2}-e^{-a} \tan ^{-1}\left (e^a x^2\right ) \]

[Out]

x^2/2 - ArcTan[E^a*x^2]/E^a

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Rubi [F]  time = 0.0160746, 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 = {} \[ \int x \tanh (a+2 \log (x)) \, dx \]

Verification is Not applicable to the result.

[In]

Int[x*Tanh[a + 2*Log[x]],x]

[Out]

Defer[Int][x*Tanh[a + 2*Log[x]], x]

Rubi steps

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

Mathematica [A]  time = 0.186758, size = 35, normalized size = 1.52 \[ -\cosh (a) \tan ^{-1}\left (x^2 (\sinh (a)+\cosh (a))\right )+\sinh (a) \tan ^{-1}\left (x^2 (\sinh (a)+\cosh (a))\right )+\frac{x^2}{2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Tanh[a + 2*Log[x]],x]

[Out]

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

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Maple [C]  time = 0.035, size = 41, normalized size = 1.8 \begin{align*}{\frac{{x}^{2}}{2}}+{\frac{i}{2}}{{\rm e}^{-a}}\ln \left ({{\rm e}^{a}}{x}^{2}-i \right ) -{\frac{i}{2}}{{\rm e}^{-a}}\ln \left ({{\rm e}^{a}}{x}^{2}+i \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*tanh(a+2*ln(x)),x)

[Out]

1/2*x^2+1/2*I*exp(-a)*ln(exp(a)*x^2-I)-1/2*I*exp(-a)*ln(exp(a)*x^2+I)

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Maxima [A]  time = 1.91379, size = 26, normalized size = 1.13 \begin{align*} \frac{1}{2} \, x^{2} - \arctan \left (x^{2} e^{a}\right ) e^{\left (-a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2 - arctan(x^2*e^a)*e^(-a)

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Fricas [A]  time = 1.83915, size = 57, normalized size = 2.48 \begin{align*} \frac{1}{2} \,{\left (x^{2} e^{a} - 2 \, \arctan \left (x^{2} e^{a}\right )\right )} e^{\left (-a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*tanh(a + 2*log(x)), x)

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Giac [A]  time = 1.23499, size = 26, normalized size = 1.13 \begin{align*} \frac{1}{2} \, x^{2} - \arctan \left (x^{2} e^{a}\right ) e^{\left (-a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2 - arctan(x^2*e^a)*e^(-a)