Integrand size = 23, antiderivative size = 130 \[ \int \frac {\cosh ^5(c+d x)}{\left (a+b \sinh ^n(c+d x)\right )^2} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (2,\frac {1}{n},1+\frac {1}{n},-\frac {b \sinh ^n(c+d x)}{a}\right ) \sinh (c+d x)}{a^2 d}+\frac {2 \operatorname {Hypergeometric2F1}\left (2,\frac {3}{n},\frac {3+n}{n},-\frac {b \sinh ^n(c+d x)}{a}\right ) \sinh ^3(c+d x)}{3 a^2 d}+\frac {\operatorname {Hypergeometric2F1}\left (2,\frac {5}{n},\frac {5+n}{n},-\frac {b \sinh ^n(c+d x)}{a}\right ) \sinh ^5(c+d x)}{5 a^2 d} \]
hypergeom([2, 1/n],[1+1/n],-b*sinh(d*x+c)^n/a)*sinh(d*x+c)/a^2/d+2/3*hyper geom([2, 3/n],[(3+n)/n],-b*sinh(d*x+c)^n/a)*sinh(d*x+c)^3/a^2/d+1/5*hyperg eom([2, 5/n],[(5+n)/n],-b*sinh(d*x+c)^n/a)*sinh(d*x+c)^5/a^2/d
Time = 0.09 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.92 \[ \int \frac {\cosh ^5(c+d x)}{\left (a+b \sinh ^n(c+d x)\right )^2} \, dx=\frac {15 \operatorname {Hypergeometric2F1}\left (2,\frac {1}{n},1+\frac {1}{n},-\frac {b \sinh ^n(c+d x)}{a}\right ) \sinh (c+d x)+10 \operatorname {Hypergeometric2F1}\left (2,\frac {3}{n},\frac {3+n}{n},-\frac {b \sinh ^n(c+d x)}{a}\right ) \sinh ^3(c+d x)+3 \operatorname {Hypergeometric2F1}\left (2,\frac {5}{n},\frac {5+n}{n},-\frac {b \sinh ^n(c+d x)}{a}\right ) \sinh ^5(c+d x)}{15 a^2 d} \]
(15*Hypergeometric2F1[2, n^(-1), 1 + n^(-1), -((b*Sinh[c + d*x]^n)/a)]*Sin h[c + d*x] + 10*Hypergeometric2F1[2, 3/n, (3 + n)/n, -((b*Sinh[c + d*x]^n) /a)]*Sinh[c + d*x]^3 + 3*Hypergeometric2F1[2, 5/n, (5 + n)/n, -((b*Sinh[c + d*x]^n)/a)]*Sinh[c + d*x]^5)/(15*a^2*d)
Time = 0.37 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3702, 2432, 2009}
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.
\(\displaystyle \int \frac {\cosh ^5(c+d x)}{\left (a+b \sinh ^n(c+d x)\right )^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (i c+i d x)^5}{\left (a+b (-i \sin (i c+i d x))^n\right )^2}dx\) |
\(\Big \downarrow \) 3702 |
\(\displaystyle \frac {\int \frac {\left (\sinh ^2(c+d x)+1\right )^2}{\left (b \sinh ^n(c+d x)+a\right )^2}d\sinh (c+d x)}{d}\) |
\(\Big \downarrow \) 2432 |
\(\displaystyle \frac {\int \left (\frac {\sinh ^4(c+d x)}{\left (b \sinh ^n(c+d x)+a\right )^2}+\frac {2 \sinh ^2(c+d x)}{\left (b \sinh ^n(c+d x)+a\right )^2}+\frac {1}{\left (b \sinh ^n(c+d x)+a\right )^2}\right )d\sinh (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {\sinh (c+d x) \operatorname {Hypergeometric2F1}\left (2,\frac {1}{n},1+\frac {1}{n},-\frac {b \sinh ^n(c+d x)}{a}\right )}{a^2}+\frac {\sinh ^5(c+d x) \operatorname {Hypergeometric2F1}\left (2,\frac {5}{n},\frac {n+5}{n},-\frac {b \sinh ^n(c+d x)}{a}\right )}{5 a^2}+\frac {2 \sinh ^3(c+d x) \operatorname {Hypergeometric2F1}\left (2,\frac {3}{n},\frac {n+3}{n},-\frac {b \sinh ^n(c+d x)}{a}\right )}{3 a^2}}{d}\) |
((Hypergeometric2F1[2, n^(-1), 1 + n^(-1), -((b*Sinh[c + d*x]^n)/a)]*Sinh[ c + d*x])/a^2 + (2*Hypergeometric2F1[2, 3/n, (3 + n)/n, -((b*Sinh[c + d*x] ^n)/a)]*Sinh[c + d*x]^3)/(3*a^2) + (Hypergeometric2F1[2, 5/n, (5 + n)/n, - ((b*Sinh[c + d*x]^n)/a)]*Sinh[c + d*x]^5)/(5*a^2))/d
3.5.22.3.1 Defintions of rubi rules used
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[ Pq*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, n, p}, x] && (PolyQ[Pq, x] || Poly Q[Pq, x^n])
Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x _)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Si mp[ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^n)^p, x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (EqQ[n, 4] || GtQ[m, 0] || IGtQ[p, 0] || IntegersQ[m, p])
\[\int \frac {\cosh \left (d x +c \right )^{5}}{\left (a +b \sinh \left (d x +c \right )^{n}\right )^{2}}d x\]
\[ \int \frac {\cosh ^5(c+d x)}{\left (a+b \sinh ^n(c+d x)\right )^2} \, dx=\int { \frac {\cosh \left (d x + c\right )^{5}}{{\left (b \sinh \left (d x + c\right )^{n} + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\cosh ^5(c+d x)}{\left (a+b \sinh ^n(c+d x)\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {\cosh ^5(c+d x)}{\left (a+b \sinh ^n(c+d x)\right )^2} \, dx=\int { \frac {\cosh \left (d x + c\right )^{5}}{{\left (b \sinh \left (d x + c\right )^{n} + a\right )}^{2}} \,d x } \]
1/32*(2^n*e^(c*n + 10*d*x + 10*c) + 3*2^n*e^(c*n + 8*d*x + 8*c) + 2^(n + 1 )*e^(c*n + 6*d*x + 6*c) - 2^(n + 1)*e^(c*n + 4*d*x + 4*c) - 3*2^n*e^(c*n + 2*d*x + 2*c) - 2^n*e^(c*n))*e^(d*n*x)/(2^n*a^2*d*n*e^(d*n*x + c*n + 5*d*x + 5*c) + a*b*d*n*e^(5*d*x + n*log(e^(d*x + c) + 1) + n*log(e^(d*x + c) - 1) + 5*c)) + 1/32*integrate((2^n*n*e^(c*n) - 5*2^n*e^(c*n) + (2^n*n*e^(c*n ) - 5*2^n*e^(c*n))*e^(10*d*x + 10*c) + (5*2^n*n*e^(c*n) - 9*2^n*e^(c*n))*e ^(8*d*x + 8*c) + (5*2^(n + 1)*n*e^(c*n) - 2^(n + 1)*e^(c*n))*e^(6*d*x + 6* c) + (5*2^(n + 1)*n*e^(c*n) - 2^(n + 1)*e^(c*n))*e^(4*d*x + 4*c) + (5*2^n* n*e^(c*n) - 9*2^n*e^(c*n))*e^(2*d*x + 2*c))*e^(d*n*x)/(2^n*a^2*n*e^(d*n*x + c*n + 5*d*x + 5*c) + a*b*n*e^(5*d*x + n*log(e^(d*x + c) + 1) + n*log(e^( d*x + c) - 1) + 5*c)), x)
\[ \int \frac {\cosh ^5(c+d x)}{\left (a+b \sinh ^n(c+d x)\right )^2} \, dx=\int { \frac {\cosh \left (d x + c\right )^{5}}{{\left (b \sinh \left (d x + c\right )^{n} + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\cosh ^5(c+d x)}{\left (a+b \sinh ^n(c+d x)\right )^2} \, dx=\int \frac {{\mathrm {cosh}\left (c+d\,x\right )}^5}{{\left (a+b\,{\mathrm {sinh}\left (c+d\,x\right )}^n\right )}^2} \,d x \]