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

Optimal. Leaf size=40 \[ \frac{x^2}{e^{2 a} x^4+1}-e^{-a} \tan ^{-1}\left (e^a x^2\right )+\frac{x^2}{2} \]

[Out]

x^2/2 + x^2/(1 + E^(2*a)*x^4) - ArcTan[E^a*x^2]/E^a

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Rubi [F]  time = 0.0302817, 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 = {} \[ \int x \tanh ^2(a+2 \log (x)) \, dx \]

Verification is Not applicable to the result.

[In]

Int[x*Tanh[a + 2*Log[x]]^2,x]

[Out]

Defer[Int][x*Tanh[a + 2*Log[x]]^2, x]

Rubi steps

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

Mathematica [A]  time = 0.377913, size = 41, normalized size = 1.02 \[ \frac{x^2}{e^{2 (a+2 \log (x))}+1}-e^{-a} \tan ^{-1}\left (e^a x^2\right )+\frac{x^2}{2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Tanh[a + 2*Log[x]]^2,x]

[Out]

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

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Maple [C]  time = 0.033, size = 57, normalized size = 1.4 \begin{align*}{\frac{{x}^{2}}{2}}+{\frac{{x}^{2}}{1+{{\rm e}^{2\,a}}{x}^{4}}}+{\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))^2,x)

[Out]

1/2*x^2+x^2/(1+exp(2*a)*x^4)+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 = 1.86376, size = 47, normalized size = 1.18 \begin{align*} \frac{1}{2} \, x^{2} - \arctan \left (x^{2} e^{a}\right ) e^{\left (-a\right )} + \frac{x^{2}}{x^{4} e^{\left (2 \, a\right )} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.02991, size = 120, normalized size = 3. \begin{align*} \frac{x^{6} e^{\left (3 \, a\right )} + 3 \, x^{2} e^{a} - 2 \,{\left (x^{4} e^{\left (2 \, a\right )} + 1\right )} \arctan \left (x^{2} e^{a}\right )}{2 \,{\left (x^{4} e^{\left (3 \, a\right )} + e^{a}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \tanh ^{2}{\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))**2,x)

[Out]

Integral(x*tanh(a + 2*log(x))**2, x)

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Giac [A]  time = 1.23874, size = 47, normalized size = 1.18 \begin{align*} \frac{1}{2} \, x^{2} - \arctan \left (x^{2} e^{a}\right ) e^{\left (-a\right )} + \frac{x^{2}}{x^{4} e^{\left (2 \, a\right )} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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