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]

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

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Rubi [F]  time = 0.02, 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 = {} \[ \int x \tanh (a+2 \log (x)) \, dx \]

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*}

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Mathematica [A]  time = 0.19, 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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fricas [A]  time = 0.51, size = 22, normalized size = 0.96 \[ \frac {1}{2} \, {\left (x^{2} e^{a} - 2 \, \arctan \left (x^{2} e^{a}\right )\right )} e^{\left (-a\right )} \]

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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giac [A]  time = 0.14, size = 19, normalized size = 0.83 \[ \frac {1}{2} \, x^{2} - \arctan \left (x^{2} e^{a}\right ) e^{\left (-a\right )} \]

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)

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maple [C]  time = 0.11, size = 41, normalized size = 1.78 \[ \frac {x^{2}}{2}+\frac {i {\mathrm e}^{-a} \ln \left ({\mathrm e}^{a} x^{2}-i\right )}{2}-\frac {i {\mathrm e}^{-a} \ln \left ({\mathrm e}^{a} x^{2}+i\right )}{2} \]

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 = 0.43, size = 19, normalized size = 0.83 \[ \frac {1}{2} \, x^{2} - \arctan \left (x^{2} e^{a}\right ) e^{\left (-a\right )} \]

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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mupad [B]  time = 1.07, size = 25, normalized size = 1.09 \[ \frac {x^2}{2}-\frac {\mathrm {atan}\left (x^2\,\sqrt {{\mathrm {e}}^{2\,a}}\right )}{\sqrt {{\mathrm {e}}^{2\,a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*tanh(a + 2*log(x)),x)

[Out]

x^2/2 - atan(x^2*exp(2*a)^(1/2))/exp(2*a)^(1/2)

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

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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