3.163 \(\int \frac{\coth ^2(a+2 \log (x))}{x^2} \, dx\)

Optimal. Leaf size=86 \[ \frac{2 e^{2 a} x^3}{1-e^{2 a} x^4}-\frac{1}{x \left (1-e^{2 a} x^4\right )}-\frac{1}{2} e^{a/2} \tan ^{-1}\left (e^{a/2} x\right )+\frac{1}{2} e^{a/2} \tanh ^{-1}\left (e^{a/2} x\right ) \]

[Out]

-(1/(x*(1 - E^(2*a)*x^4))) + (2*E^(2*a)*x^3)/(1 - E^(2*a)*x^4) - (E^(a/2)*ArcTan[E^(a/2)*x])/2 + (E^(a/2)*ArcT
anh[E^(a/2)*x])/2

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Rubi [F]  time = 0.0448182, 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 = {} \[ \int \frac{\coth ^2(a+2 \log (x))}{x^2} \, dx \]

Verification is Not applicable to the result.

[In]

Int[Coth[a + 2*Log[x]]^2/x^2,x]

[Out]

Defer[Int][Coth[a + 2*Log[x]]^2/x^2, x]

Rubi steps

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

Mathematica [C]  time = 3.12673, size = 153, normalized size = 1.78 \[ \frac{16}{231} e^{2 a} x^3 \left (e^{2 a} x^4+1\right )^2 \text{HypergeometricPFQ}\left (\left \{\frac{3}{4},2,2,2\right \},\left \{1,1,\frac{15}{4}\right \},e^{2 a} x^4\right )+\frac{e^{-2 a} \left (\left (-e^{8 a} x^{16}-56 e^{6 a} x^{12}+362 e^{4 a} x^8+632 e^{2 a} x^4+343\right ) \, _2F_1\left (\frac{3}{4},1;\frac{7}{4};e^{2 a} x^4\right )+3 e^{6 a} x^{12}-241 e^{4 a} x^8-1163 e^{2 a} x^4-343\right )}{384 x^5} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Coth[a + 2*Log[x]]^2/x^2,x]

[Out]

(-343 - 1163*E^(2*a)*x^4 - 241*E^(4*a)*x^8 + 3*E^(6*a)*x^12 + (343 + 632*E^(2*a)*x^4 + 362*E^(4*a)*x^8 - 56*E^
(6*a)*x^12 - E^(8*a)*x^16)*Hypergeometric2F1[3/4, 1, 7/4, E^(2*a)*x^4])/(384*E^(2*a)*x^5) + (16*E^(2*a)*x^3*(1
+ E^(2*a)*x^4)^2*HypergeometricPFQ[{3/4, 2, 2, 2}, {1, 1, 15/4}, E^(2*a)*x^4])/231

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Maple [C]  time = 0.033, size = 101, normalized size = 1.2 \begin{align*}{\frac{-2\,{{\rm e}^{2\,a}}{x}^{4}+1}{x \left ({{\rm e}^{2\,a}}{x}^{4}-1 \right ) }}+{\frac{\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{2}+{{\rm e}^{a}} \right ) }{\it \_R}\,\ln \left ( \left ( -5\,{{\it \_R}}^{4}+4\,{{\rm e}^{2\,a}} \right ) x-{{\it \_R}}^{3} \right ) }{4}}+{\frac{\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{2}-{{\rm e}^{a}} \right ) }{\it \_R}\,\ln \left ( \left ( -5\,{{\it \_R}}^{4}+4\,{{\rm e}^{2\,a}} \right ) x-{{\it \_R}}^{3} \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-2*exp(2*a)*x^4+1)/x/(exp(2*a)*x^4-1)+1/4*sum(_R*ln((-5*_R^4+4*exp(2*a))*x-_R^3),_R=RootOf(_Z^2+exp(a)))+1/4*
sum(_R*ln((-5*_R^4+4*exp(2*a))*x-_R^3),_R=RootOf(_Z^2-exp(a)))

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Maxima [A]  time = 1.56413, size = 93, normalized size = 1.08 \begin{align*} \frac{1}{2} \, \arctan \left (\frac{e^{\left (-\frac{1}{2} \, a\right )}}{x}\right ) e^{\left (\frac{1}{2} \, a\right )} - \frac{1}{4} \, e^{\left (\frac{1}{2} \, a\right )} \log \left (\frac{\frac{1}{x} - e^{\left (\frac{1}{2} \, a\right )}}{\frac{1}{x} + e^{\left (\frac{1}{2} \, a\right )}}\right ) - \frac{1}{x} + \frac{e^{\left (2 \, a\right )}}{x{\left (\frac{1}{x^{4}} - e^{\left (2 \, a\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.60401, size = 236, normalized size = 2.74 \begin{align*} -\frac{8 \, x^{4} e^{\left (2 \, a\right )} + 2 \,{\left (x^{5} e^{\left (2 \, a\right )} - x\right )} \arctan \left (x e^{\left (\frac{1}{2} \, a\right )}\right ) e^{\left (\frac{1}{2} \, a\right )} -{\left (x^{5} e^{\left (2 \, a\right )} - x\right )} e^{\left (\frac{1}{2} \, a\right )} \log \left (\frac{x^{2} e^{a} + 2 \, x e^{\left (\frac{1}{2} \, a\right )} + 1}{x^{2} e^{a} - 1}\right ) - 4}{4 \,{\left (x^{5} e^{\left (2 \, a\right )} - x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(8*x^4*e^(2*a) + 2*(x^5*e^(2*a) - x)*arctan(x*e^(1/2*a))*e^(1/2*a) - (x^5*e^(2*a) - x)*e^(1/2*a)*log((x^2
*e^a + 2*x*e^(1/2*a) + 1)/(x^2*e^a - 1)) - 4)/(x^5*e^(2*a) - x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(coth(a + 2*log(x))**2/x**2, x)

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Giac [A]  time = 1.12148, size = 104, normalized size = 1.21 \begin{align*} -\frac{1}{2} \, \arctan \left (x e^{\left (\frac{1}{2} \, a\right )}\right ) e^{\left (\frac{1}{2} \, a\right )} - \frac{1}{4} \, e^{\left (\frac{1}{2} \, a\right )} \log \left (\frac{{\left | 2 \, x e^{a} - 2 \, e^{\left (\frac{1}{2} \, a\right )} \right |}}{{\left | 2 \, x e^{a} + 2 \, e^{\left (\frac{1}{2} \, a\right )} \right |}}\right ) - \frac{2 \, x^{4} e^{\left (2 \, a\right )} - 1}{x^{5} e^{\left (2 \, a\right )} - x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*arctan(x*e^(1/2*a))*e^(1/2*a) - 1/4*e^(1/2*a)*log(abs(2*x*e^a - 2*e^(1/2*a))/abs(2*x*e^a + 2*e^(1/2*a)))
- (2*x^4*e^(2*a) - 1)/(x^5*e^(2*a) - x)