3.195 \(\int (e x)^m \coth ^2(d (a+b \log (c x^n))) \, dx\)

Optimal. Leaf size=168 \[ -\frac {2 (e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{2 b d n};\frac {m+1}{2 b d n}+1;e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b d e n}+\frac {(e x)^{m+1} \left (e^{2 a d} \left (c x^n\right )^{2 b d}+1\right )}{b d e n \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}+\frac {(e x)^{m+1} (b d n+m+1)}{b d e (m+1) n} \]

[Out]

(b*d*n+m+1)*(e*x)^(1+m)/b/d/e/(1+m)/n+(e*x)^(1+m)*(1+exp(2*a*d)*(c*x^n)^(2*b*d))/b/d/e/n/(1-exp(2*a*d)*(c*x^n)
^(2*b*d))-2*(e*x)^(1+m)*hypergeom([1, 1/2*(1+m)/b/d/n],[1+1/2*(1+m)/b/d/n],exp(2*a*d)*(c*x^n)^(2*b*d))/b/d/e/n

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Rubi [F]  time = 0.08, 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 = {} \[ \int (e x)^m \coth ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx \]

Verification is Not applicable to the result.

[In]

Int[(e*x)^m*Coth[d*(a + b*Log[c*x^n])]^2,x]

[Out]

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

Rubi steps

\begin {align*} \int (e x)^m \coth ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx &=\int (e x)^m \coth ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx\\ \end {align*}

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Mathematica [A]  time = 14.88, size = 312, normalized size = 1.86 \[ (e x)^m \left (\frac {x}{m+1}-\frac {x^{-2 m} \exp \left (-\frac {(2 m+1) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{b n}\right ) \left ((m+1) x^{2 b d n+2 m+1} \exp \left (\frac {(2 b d n+2 m+1) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{b n}\right ) \, _2F_1\left (1,\frac {m+2 b d n+1}{2 b d n};\frac {m+4 b d n+1}{2 b d n};e^{2 d \left (a+b \log \left (c x^n\right )\right )}\right )+(2 b d n+m+1) e^{\frac {(2 m+1) \left (a+b \log \left (c x^n\right )\right )}{b n}} \, _2F_1\left (1,\frac {m+1}{2 b d n};\frac {m+1}{2 b d n}+1;e^{2 d \left (a+b \log \left (c x^n\right )\right )}\right )+(2 b d n+m+1) e^{\frac {(2 m+1) \left (a+b \log \left (c x^n\right )\right )}{b n}} \coth \left (d \left (a+b \log \left (c x^n\right )\right )\right )\right )}{b d n (2 b d n+m+1)}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*x)^m*(x/(1 + m) - (E^(((1 + 2*m)*(a + b*Log[c*x^n]))/(b*n))*(1 + m + 2*b*d*n)*Coth[d*(a + b*Log[c*x^n])] +
E^(((1 + 2*m)*(a + b*Log[c*x^n]))/(b*n))*(1 + m + 2*b*d*n)*Hypergeometric2F1[1, (1 + m)/(2*b*d*n), 1 + (1 + m)
/(2*b*d*n), E^(2*d*(a + b*Log[c*x^n]))] + E^(((1 + 2*m + 2*b*d*n)*(a - b*n*Log[x] + b*Log[c*x^n]))/(b*n))*(1 +
 m)*x^(1 + 2*m + 2*b*d*n)*Hypergeometric2F1[1, (1 + m + 2*b*d*n)/(2*b*d*n), (1 + m + 4*b*d*n)/(2*b*d*n), E^(2*
d*(a + b*Log[c*x^n]))])/(b*d*E^(((1 + 2*m)*(a - b*n*Log[x] + b*Log[c*x^n]))/(b*n))*n*(1 + m + 2*b*d*n)*x^(2*m)
))

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (e x\right )^{m} \coth \left (b d \log \left (c x^{n}\right ) + a d\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((e*x)^m*coth(b*d*log(c*x^n) + a*d)^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e x\right )^{m} \coth \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x)^m*coth((b*log(c*x^n) + a)*d)^2, x)

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maple [F]  time = 1.56, size = 0, normalized size = 0.00 \[ \int \left (e x \right )^{m} \left (\coth ^{2}\left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m*coth(d*(a+b*ln(c*x^n)))^2,x)

[Out]

int((e*x)^m*coth(d*(a+b*ln(c*x^n)))^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -e^{m} {\left (m + 1\right )} \int \frac {x^{m}}{b c^{b d} d n e^{\left (b d \log \left (x^{n}\right ) + a d\right )} + b d n}\,{d x} + e^{m} {\left (m + 1\right )} \int \frac {x^{m}}{b c^{b d} d n e^{\left (b d \log \left (x^{n}\right ) + a d\right )} - b d n}\,{d x} + \frac {b c^{2 \, b d} d e^{m} n x e^{\left (2 \, b d \log \left (x^{n}\right ) + 2 \, a d + m \log \relax (x)\right )} - {\left (b d e^{m} n + 2 \, e^{m} {\left (m + 1\right )}\right )} x x^{m}}{{\left (m n + n\right )} b c^{2 \, b d} d e^{\left (2 \, b d \log \left (x^{n}\right ) + 2 \, a d\right )} - {\left (m n + n\right )} b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-e^m*(m + 1)*integrate(x^m/(b*c^(b*d)*d*n*e^(b*d*log(x^n) + a*d) + b*d*n), x) + e^m*(m + 1)*integrate(x^m/(b*c
^(b*d)*d*n*e^(b*d*log(x^n) + a*d) - b*d*n), x) + (b*c^(2*b*d)*d*e^m*n*x*e^(2*b*d*log(x^n) + 2*a*d + m*log(x))
- (b*d*e^m*n + 2*e^m*(m + 1))*x*x^m)/((m*n + n)*b*c^(2*b*d)*d*e^(2*b*d*log(x^n) + 2*a*d) - (m*n + n)*b*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {coth}\left (d\,\left (a+b\,\ln \left (c\,x^n\right )\right )\right )}^2\,{\left (e\,x\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(d*(a + b*log(c*x^n)))^2*(e*x)^m,x)

[Out]

int(coth(d*(a + b*log(c*x^n)))^2*(e*x)^m, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e x\right )^{m} \coth ^{2}{\left (a d + b d \log {\left (c x^{n} \right )} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**m*coth(d*(a+b*ln(c*x**n)))**2,x)

[Out]

Integral((e*x)**m*coth(a*d + b*d*log(c*x**n))**2, x)

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