3.153 \(\int x \coth (a+2 \log (x)) \, dx\)

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

[Out]

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

________________________________________________________________________________________

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 = {} \[ \int x \coth (a+2 \log (x)) \, dx \]

Verification is Not applicable to the result.

[In]

Int[x*Coth[a + 2*Log[x]],x]

[Out]

Defer[Int][x*Coth[a + 2*Log[x]], x]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.19, size = 26, normalized size = 1.13 \[ (\sinh (a)-\cosh (a)) \tanh ^{-1}\left (x^2 (\sinh (a)+\cosh (a))\right )+\frac {x^2}{2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Coth[a + 2*Log[x]],x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 33, normalized size = 1.43 \[ \frac {1}{2} \, {\left (x^{2} e^{a} - \log \left (x^{2} e^{a} + 1\right ) + \log \left (x^{2} e^{a} - 1\right )\right )} e^{\left (-a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.14, size = 37, normalized size = 1.61 \[ \frac {1}{2} \, x^{2} - \frac {1}{2} \, e^{\left (-a\right )} \log \left (x^{2} e^{a} + 1\right ) + \frac {1}{2} \, e^{\left (-a\right )} \log \left ({\left | x^{2} e^{a} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2 - 1/2*e^(-a)*log(x^2*e^a + 1) + 1/2*e^(-a)*log(abs(x^2*e^a - 1))

________________________________________________________________________________________

maple [A]  time = 0.11, size = 37, normalized size = 1.61 \[ \frac {x^{2}}{2}-\frac {{\mathrm e}^{-a} \ln \left ({\mathrm e}^{a} x^{2}+1\right )}{2}+\frac {{\mathrm e}^{-a} \ln \left ({\mathrm e}^{a} x^{2}-1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.31, size = 36, normalized size = 1.57 \[ \frac {1}{2} \, x^{2} - \frac {1}{2} \, e^{\left (-a\right )} \log \left (x^{2} e^{a} + 1\right ) + \frac {1}{2} \, e^{\left (-a\right )} \log \left (x^{2} e^{a} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.23, size = 25, normalized size = 1.09 \[ \frac {x^2}{2}-\frac {\mathrm {atanh}\left (x^2\,\sqrt {{\mathrm {e}}^{2\,a}}\right )}{\sqrt {{\mathrm {e}}^{2\,a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*coth(a + 2*log(x)),x)

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*coth(a + 2*log(x)), x)

________________________________________________________________________________________