Integrand size = 13, antiderivative size = 68 \[ \int \text {csch}^4\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {16 e^{4 a} x \left (c x^n\right )^{4 b} \operatorname {Hypergeometric2F1}\left (4,\frac {1}{2} \left (4+\frac {1}{b n}\right ),\frac {1}{2} \left (6+\frac {1}{b n}\right ),e^{2 a} \left (c x^n\right )^{2 b}\right )}{1+4 b n} \]
16*exp(4*a)*x*(c*x^n)^(4*b)*hypergeom([4, 2+1/2/b/n],[3+1/2/b/n],exp(2*a)* (c*x^n)^(2*b))/(4*b*n+1)
Leaf count is larger than twice the leaf count of optimal. \(200\) vs. \(2(68)=136\).
Time = 5.65 (sec) , antiderivative size = 200, normalized size of antiderivative = 2.94 \[ \int \text {csch}^4\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x \left (4 e^{2 a} (-1+2 b n) \left (c x^n\right )^{2 b} \operatorname {Hypergeometric2F1}\left (1,1+\frac {1}{2 b n},2+\frac {1}{2 b n},e^{2 \left (a+b \log \left (c x^n\right )\right )}\right )+4 \left (-1+4 b^2 n^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 b n},1+\frac {1}{2 b n},e^{2 \left (a+b \log \left (c x^n\right )\right )}\right )+\text {csch}^3\left (a+b \log \left (c x^n\right )\right ) \left (\left (1-12 b^2 n^2\right ) \cosh \left (a+b \log \left (c x^n\right )\right )+\left (-1+4 b^2 n^2\right ) \cosh \left (3 \left (a+b \log \left (c x^n\right )\right )\right )-4 b n \sinh \left (a+b \log \left (c x^n\right )\right )\right )\right )}{24 b^3 n^3} \]
(x*(4*E^(2*a)*(-1 + 2*b*n)*(c*x^n)^(2*b)*Hypergeometric2F1[1, 1 + 1/(2*b*n ), 2 + 1/(2*b*n), E^(2*(a + b*Log[c*x^n]))] + 4*(-1 + 4*b^2*n^2)*Hypergeom etric2F1[1, 1/(2*b*n), 1 + 1/(2*b*n), E^(2*(a + b*Log[c*x^n]))] + Csch[a + b*Log[c*x^n]]^3*((1 - 12*b^2*n^2)*Cosh[a + b*Log[c*x^n]] + (-1 + 4*b^2*n^ 2)*Cosh[3*(a + b*Log[c*x^n])] - 4*b*n*Sinh[a + b*Log[c*x^n]])))/(24*b^3*n^ 3)
Time = 0.32 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {6080, 6082, 795, 888}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \text {csch}^4\left (a+b \log \left (c x^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 6080 |
\(\displaystyle \frac {x \left (c x^n\right )^{-1/n} \int \left (c x^n\right )^{\frac {1}{n}-1} \text {csch}^4\left (a+b \log \left (c x^n\right )\right )d\left (c x^n\right )}{n}\) |
\(\Big \downarrow \) 6082 |
\(\displaystyle \frac {16 e^{-4 a} x \left (c x^n\right )^{-1/n} \int \frac {\left (c x^n\right )^{-4 b+\frac {1}{n}-1}}{\left (1-e^{-2 a} \left (c x^n\right )^{-2 b}\right )^4}d\left (c x^n\right )}{n}\) |
\(\Big \downarrow \) 795 |
\(\displaystyle \frac {16 e^{-4 a} x \left (c x^n\right )^{-1/n} \int \frac {\left (c x^n\right )^{4 b+\frac {1}{n}-1}}{\left (\left (c x^n\right )^{2 b}-e^{-2 a}\right )^4}d\left (c x^n\right )}{n}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {16 e^{4 a} x \left (c x^n\right )^{4 b} \operatorname {Hypergeometric2F1}\left (4,\frac {1}{2} \left (4+\frac {1}{b n}\right ),\frac {1}{2} \left (6+\frac {1}{b n}\right ),e^{2 a} \left (c x^n\right )^{2 b}\right )}{4 b n+1}\) |
(16*E^(4*a)*x*(c*x^n)^(4*b)*Hypergeometric2F1[4, (4 + 1/(b*n))/2, (6 + 1/( b*n))/2, E^(2*a)*(c*x^n)^(2*b)])/(1 + 4*b*n)
3.2.59.3.1 Defintions of rubi rules used
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]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p *((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 , (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0] && (ILt Q[p, 0] || GtQ[a, 0])
Int[Csch[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> S imp[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] )
Int[Csch[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Simp[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]
\[\int {\operatorname {csch}\left (a +b \ln \left (c \,x^{n}\right )\right )}^{4}d x\]
\[ \int \text {csch}^4\left (a+b \log \left (c x^n\right )\right ) \, dx=\int { \operatorname {csch}\left (b \log \left (c x^{n}\right ) + a\right )^{4} \,d x } \]
\[ \int \text {csch}^4\left (a+b \log \left (c x^n\right )\right ) \, dx=\int \operatorname {csch}^{4}{\left (a + b \log {\left (c x^{n} \right )} \right )}\, dx \]
\[ \int \text {csch}^4\left (a+b \log \left (c x^n\right )\right ) \, dx=\int { \operatorname {csch}\left (b \log \left (c x^{n}\right ) + a\right )^{4} \,d x } \]
16*(4*b^2*n^2 - 1)*integrate(1/96/(b^3*c^b*n^3*e^(b*log(x^n) + a) + b^3*n^ 3), x) - 16*(4*b^2*n^2 - 1)*integrate(1/96/(b^3*c^b*n^3*e^(b*log(x^n) + a) - b^3*n^3), x) - 1/3*((2*b*c^(4*b)*n + c^(4*b))*x*e^(4*b*log(x^n) + 4*a) + 2*(6*b^2*c^(2*b)*n^2 - b*c^(2*b)*n - c^(2*b))*x*e^(2*b*log(x^n) + 2*a) - (4*b^2*n^2 - 1)*x)/(b^3*c^(6*b)*n^3*e^(6*b*log(x^n) + 6*a) - 3*b^3*c^(4*b )*n^3*e^(4*b*log(x^n) + 4*a) + 3*b^3*c^(2*b)*n^3*e^(2*b*log(x^n) + 2*a) - b^3*n^3)
\[ \int \text {csch}^4\left (a+b \log \left (c x^n\right )\right ) \, dx=\int { \operatorname {csch}\left (b \log \left (c x^{n}\right ) + a\right )^{4} \,d x } \]
Timed out. \[ \int \text {csch}^4\left (a+b \log \left (c x^n\right )\right ) \, dx=\int \frac {1}{{\mathrm {sinh}\left (a+b\,\ln \left (c\,x^n\right )\right )}^4} \,d x \]