3.166 \(\int (e x)^m \coth ^2(a+2 \log (x)) \, dx\)

Optimal. Leaf size=79 \[ -\frac{(e x)^{m+1} \, _2F_1\left (1,\frac{m+1}{4};\frac{m+5}{4};e^{2 a} x^4\right )}{e}+\frac{(e x)^{m+1}}{e \left (1-e^{2 a} x^4\right )}+\frac{(e x)^{m+1}}{e (m+1)} \]

[Out]

(e*x)^(1 + m)/(e*(1 + m)) + (e*x)^(1 + m)/(e*(1 - E^(2*a)*x^4)) - ((e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/
4, (5 + m)/4, E^(2*a)*x^4])/e

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Rubi [F]  time = 0.0679443, 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 = {} \[ \int (e x)^m \coth ^2(a+2 \log (x)) \, dx \]

Verification is Not applicable to the result.

[In]

Int[(e*x)^m*Coth[a + 2*Log[x]]^2,x]

[Out]

Defer[Int][(e*x)^m*Coth[a + 2*Log[x]]^2, x]

Rubi steps

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

Mathematica [B]  time = 0.369473, size = 165, normalized size = 2.09 \[ \frac{x (e x)^m \left (\frac{x^4 (\sinh (a)+\cosh (a)) \left ((m+5) x^4 (\sinh (a)+\cosh (a)) \, _2F_1\left (2,\frac{m+9}{4};\frac{m+13}{4};x^4 (\cosh (2 a)+\sinh (2 a))\right )+2 (m+9) (\cosh (a)-\sinh (a)) \, _2F_1\left (2,\frac{m+5}{4};\frac{m+9}{4};x^4 (\cosh (2 a)+\sinh (2 a))\right )\right )}{(m+5) (m+9)}+\frac{(\cosh (2 a)-\sinh (2 a)) \, _2F_1\left (2,\frac{m+1}{4};\frac{m+5}{4};x^4 (\cosh (2 a)+\sinh (2 a))\right )}{m+1}\right )}{(\cosh (a)-\sinh (a))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*x)^m*Coth[a + 2*Log[x]]^2,x]

[Out]

(x*(e*x)^m*((x^4*(Cosh[a] + Sinh[a])*(2*(9 + m)*Hypergeometric2F1[2, (5 + m)/4, (9 + m)/4, x^4*(Cosh[2*a] + Si
nh[2*a])]*(Cosh[a] - Sinh[a]) + (5 + m)*x^4*Hypergeometric2F1[2, (9 + m)/4, (13 + m)/4, x^4*(Cosh[2*a] + Sinh[
2*a])]*(Cosh[a] + Sinh[a])))/((5 + m)*(9 + m)) + (Hypergeometric2F1[2, (1 + m)/4, (5 + m)/4, x^4*(Cosh[2*a] +
Sinh[2*a])]*(Cosh[2*a] - Sinh[2*a]))/(1 + m)))/(Cosh[a] - Sinh[a])^2

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Maple [F]  time = 0.027, size = 0, normalized size = 0. \begin{align*} \int \left ( ex \right ) ^{m} \left ({\rm coth} \left (a+2\,\ln \left ( x \right ) \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x)^m*coth(a + 2*log(x))^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (e x\right )^{m} \coth \left (a + 2 \, \log \left (x\right )\right )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((e*x)^m*coth(a + 2*log(x))^2, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x)^m*coth(a + 2*log(x))^2, x)