\(\int \text {csch}^3(a+b \log (c x^n)) \, dx\) [158]

Optimal result

Integrand size = 13, antiderivative size = 69 \[ \int \text {csch}^3\left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {8 e^{3 a} x \left (c x^n\right )^{3 b} \operatorname {Hypergeometric2F1}\left (3,\frac {3 b+\frac {1}{n}}{2 b},\frac {1}{2} \left (5+\frac {1}{b n}\right ),e^{2 a} \left (c x^n\right )^{2 b}\right )}{1+3 b n} \]

[Out]

-8*exp(3*a)*x*(c*x^n)^(3*b)*hypergeom([3, 1/2*(3*b+1/n)/b],[5/2+1/2/b/n],exp(2*a)*(c*x^n)^(2*b))/(3*b*n+1)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {5665, 5667, 269, 371} \[ \int \text {csch}^3\left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {8 e^{3 a} x \left (c x^n\right )^{3 b} \operatorname {Hypergeometric2F1}\left (3,\frac {3 b+\frac {1}{n}}{2 b},\frac {1}{2} \left (5+\frac {1}{b n}\right ),e^{2 a} \left (c x^n\right )^{2 b}\right )}{3 b n+1} \]

[In]

Int[Csch[a + b*Log[c*x^n]]^3,x]

[Out]

(-8*E^(3*a)*x*(c*x^n)^(3*b)*Hypergeometric2F1[3, (3*b + n^(-1))/(2*b), (5 + 1/(b*n))/2, E^(2*a)*(c*x^n)^(2*b)]
)/(1 + 3*b*n)

Rule 269

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 5665

Int[Csch[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[x/(n*(c*x^n)^(1/n)), Subst[Int[
x^(1/n - 1)*Csch[d*(a + b*Log[x])]^p, x], x, c*x^n], x] /; FreeQ[{a, b, c, d, n, p}, x] && (NeQ[c, 1] || NeQ[n
, 1])

Rule 5667

Int[Csch[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Dist[2^p/E^(a*d*p), Int[(e*x)^m
*(1/(x^(b*d*p)*(1 - 1/(E^(2*a*d)*x^(2*b*d)))^p)), x], x] /; FreeQ[{a, b, d, e, m}, x] && IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (x \left (c x^n\right )^{-1/n}\right ) \text {Subst}\left (\int x^{-1+\frac {1}{n}} \text {csch}^3(a+b \log (x)) \, dx,x,c x^n\right )}{n} \\ & = \frac {\left (8 e^{-3 a} x \left (c x^n\right )^{-1/n}\right ) \text {Subst}\left (\int \frac {x^{-1-3 b+\frac {1}{n}}}{\left (1-e^{-2 a} x^{-2 b}\right )^3} \, dx,x,c x^n\right )}{n} \\ & = \frac {\left (8 e^{-3 a} x \left (c x^n\right )^{-1/n}\right ) \text {Subst}\left (\int \frac {x^{-1+3 b+\frac {1}{n}}}{\left (-e^{-2 a}+x^{2 b}\right )^3} \, dx,x,c x^n\right )}{n} \\ & = -\frac {8 e^{3 a} x \left (c x^n\right )^{3 b} \operatorname {Hypergeometric2F1}\left (3,\frac {3 b+\frac {1}{n}}{2 b},\frac {1}{2} \left (5+\frac {1}{b n}\right ),e^{2 a} \left (c x^n\right )^{2 b}\right )}{1+3 b n} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.01 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.46 \[ \int \text {csch}^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {-4 x \left (1+b n \coth \left (a+b \log \left (c x^n\right )\right )\right ) \text {csch}\left (a+b \log \left (c x^n\right )\right )+8 e^a (-1+b n) x \left (c x^n\right )^b \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (1+\frac {1}{b n}\right ),\frac {1}{2} \left (3+\frac {1}{b n}\right ),e^{2 \left (a+b \log \left (c x^n\right )\right )}\right )}{8 b^2 n^2} \]

[In]

Integrate[Csch[a + b*Log[c*x^n]]^3,x]

[Out]

(-4*x*(1 + b*n*Coth[a + b*Log[c*x^n]])*Csch[a + b*Log[c*x^n]] + 8*E^a*(-1 + b*n)*x*(c*x^n)^b*Hypergeometric2F1
[1, (1 + 1/(b*n))/2, (3 + 1/(b*n))/2, E^(2*(a + b*Log[c*x^n]))])/(8*b^2*n^2)

Maple [F]

\[\int {\operatorname {csch}\left (a +b \ln \left (c \,x^{n}\right )\right )}^{3}d x\]

[In]

int(csch(a+b*ln(c*x^n))^3,x)

[Out]

int(csch(a+b*ln(c*x^n))^3,x)

Fricas [F]

\[ \int \text {csch}^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\int { \operatorname {csch}\left (b \log \left (c x^{n}\right ) + a\right )^{3} \,d x } \]

[In]

integrate(csch(a+b*log(c*x^n))^3,x, algorithm="fricas")

[Out]

integral(csch(b*log(c*x^n) + a)^3, x)

Sympy [F]

\[ \int \text {csch}^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\int \operatorname {csch}^{3}{\left (a + b \log {\left (c x^{n} \right )} \right )}\, dx \]

[In]

integrate(csch(a+b*ln(c*x**n))**3,x)

[Out]

Integral(csch(a + b*log(c*x**n))**3, x)

Maxima [F]

\[ \int \text {csch}^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\int { \operatorname {csch}\left (b \log \left (c x^{n}\right ) + a\right )^{3} \,d x } \]

[In]

integrate(csch(a+b*log(c*x^n))^3,x, algorithm="maxima")

[Out]

-8*(b^2*n^2 - 1)*integrate(1/16/(b^2*c^b*n^2*e^(b*log(x^n) + a) + b^2*n^2), x) - 8*(b^2*n^2 - 1)*integrate(1/1
6/(b^2*c^b*n^2*e^(b*log(x^n) + a) - b^2*n^2), x) - ((b*c^(3*b)*n + c^(3*b))*x*e^(3*b*log(x^n) + 3*a) + (b*c^b*
n - c^b)*x*e^(b*log(x^n) + a))/(b^2*c^(4*b)*n^2*e^(4*b*log(x^n) + 4*a) - 2*b^2*c^(2*b)*n^2*e^(2*b*log(x^n) + 2
*a) + b^2*n^2)

Giac [F]

\[ \int \text {csch}^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\int { \operatorname {csch}\left (b \log \left (c x^{n}\right ) + a\right )^{3} \,d x } \]

[In]

integrate(csch(a+b*log(c*x^n))^3,x, algorithm="giac")

[Out]

integrate(csch(b*log(c*x^n) + a)^3, x)

Mupad [F(-1)]

Timed out. \[ \int \text {csch}^3\left (a+b \log \left (c x^n\right )\right ) \, dx=\int \frac {1}{{\mathrm {sinh}\left (a+b\,\ln \left (c\,x^n\right )\right )}^3} \,d x \]

[In]

int(1/sinh(a + b*log(c*x^n))^3,x)

[Out]

int(1/sinh(a + b*log(c*x^n))^3, x)