Integrand size = 23, antiderivative size = 117 \[ \int \frac {\cosh ^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\frac {(a+3 b) \arctan \left (\frac {\sqrt {b} \sinh (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} b^{3/2} d}-\frac {(a-b) \sinh (c+d x)}{4 a b d \left (a+b \sinh ^2(c+d x)\right )^2}+\frac {(a+3 b) \sinh (c+d x)}{8 a^2 b d \left (a+b \sinh ^2(c+d x)\right )} \]
1/8*(a+3*b)*arctan(sinh(d*x+c)*b^(1/2)/a^(1/2))/a^(5/2)/b^(3/2)/d-1/4*(a-b )*sinh(d*x+c)/a/b/d/(a+b*sinh(d*x+c)^2)^2+1/8*(a+3*b)*sinh(d*x+c)/a^2/b/d/ (a+b*sinh(d*x+c)^2)
Time = 0.41 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.97 \[ \int \frac {\cosh ^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\frac {-\frac {\sinh (c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2}+(a+3 b) \left (\frac {3 \arctan \left (\frac {\sqrt {b} \sinh (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {b}}+\frac {\sinh (c+d x) \left (5 a+3 b \sinh ^2(c+d x)\right )}{8 a^2 \left (a+b \sinh ^2(c+d x)\right )^2}\right )}{3 b d} \]
(-(Sinh[c + d*x]/(a + b*Sinh[c + d*x]^2)^2) + (a + 3*b)*((3*ArcTan[(Sqrt[b ]*Sinh[c + d*x])/Sqrt[a]])/(8*a^(5/2)*Sqrt[b]) + (Sinh[c + d*x]*(5*a + 3*b *Sinh[c + d*x]^2))/(8*a^2*(a + b*Sinh[c + d*x]^2)^2)))/(3*b*d)
Time = 0.28 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 3669, 298, 215, 218}
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.
\(\displaystyle \int \frac {\cosh ^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (i c+i d x)^3}{\left (a-b \sin (i c+i d x)^2\right )^3}dx\) |
\(\Big \downarrow \) 3669 |
\(\displaystyle \frac {\int \frac {\sinh ^2(c+d x)+1}{\left (b \sinh ^2(c+d x)+a\right )^3}d\sinh (c+d x)}{d}\) |
\(\Big \downarrow \) 298 |
\(\displaystyle \frac {\frac {(a+3 b) \int \frac {1}{\left (b \sinh ^2(c+d x)+a\right )^2}d\sinh (c+d x)}{4 a b}-\frac {(a-b) \sinh (c+d x)}{4 a b \left (a+b \sinh ^2(c+d x)\right )^2}}{d}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {\frac {(a+3 b) \left (\frac {\int \frac {1}{b \sinh ^2(c+d x)+a}d\sinh (c+d x)}{2 a}+\frac {\sinh (c+d x)}{2 a \left (a+b \sinh ^2(c+d x)\right )}\right )}{4 a b}-\frac {(a-b) \sinh (c+d x)}{4 a b \left (a+b \sinh ^2(c+d x)\right )^2}}{d}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {(a+3 b) \left (\frac {\arctan \left (\frac {\sqrt {b} \sinh (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b}}+\frac {\sinh (c+d x)}{2 a \left (a+b \sinh ^2(c+d x)\right )}\right )}{4 a b}-\frac {(a-b) \sinh (c+d x)}{4 a b \left (a+b \sinh ^2(c+d x)\right )^2}}{d}\) |
(-1/4*((a - b)*Sinh[c + d*x])/(a*b*(a + b*Sinh[c + d*x]^2)^2) + ((a + 3*b) *(ArcTan[(Sqrt[b]*Sinh[c + d*x])/Sqrt[a]]/(2*a^(3/2)*Sqrt[b]) + Sinh[c + d *x]/(2*a*(a + b*Sinh[c + d*x]^2))))/(4*a*b))/d
3.4.42.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f S ubst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e + f*x] /ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Time = 0.07 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.78
\[\frac {\frac {\frac {\left (a +3 b \right ) \sinh \left (d x +c \right )^{3}}{8 a^{2}}-\frac {\left (a -5 b \right ) \sinh \left (d x +c \right )}{8 a b}}{\left (a +b \sinh \left (d x +c \right )^{2}\right )^{2}}+\frac {\left (a +3 b \right ) \arctan \left (\frac {b \sinh \left (d x +c \right )}{\sqrt {a b}}\right )}{8 a^{2} b \sqrt {a b}}}{d}\]
1/d*((1/8*(a+3*b)/a^2*sinh(d*x+c)^3-1/8*(a-5*b)/a/b*sinh(d*x+c))/(a+b*sinh (d*x+c)^2)^2+1/8*(a+3*b)/a^2/b/(a*b)^(1/2)*arctan(b*sinh(d*x+c)/(a*b)^(1/2 )))
Leaf count of result is larger than twice the leaf count of optimal. 2696 vs. \(2 (103) = 206\).
Time = 0.32 (sec) , antiderivative size = 4907, normalized size of antiderivative = 41.94 \[ \int \frac {\cosh ^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]
[1/16*(4*(a^2*b^2 + 3*a*b^3)*cosh(d*x + c)^7 + 28*(a^2*b^2 + 3*a*b^3)*cosh (d*x + c)*sinh(d*x + c)^6 + 4*(a^2*b^2 + 3*a*b^3)*sinh(d*x + c)^7 - 4*(4*a ^3*b - 17*a^2*b^2 + 9*a*b^3)*cosh(d*x + c)^5 - 4*(4*a^3*b - 17*a^2*b^2 + 9 *a*b^3 - 21*(a^2*b^2 + 3*a*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^5 + 20*(7*( a^2*b^2 + 3*a*b^3)*cosh(d*x + c)^3 - (4*a^3*b - 17*a^2*b^2 + 9*a*b^3)*cosh (d*x + c))*sinh(d*x + c)^4 + 4*(4*a^3*b - 17*a^2*b^2 + 9*a*b^3)*cosh(d*x + c)^3 + 4*(35*(a^2*b^2 + 3*a*b^3)*cosh(d*x + c)^4 + 4*a^3*b - 17*a^2*b^2 + 9*a*b^3 - 10*(4*a^3*b - 17*a^2*b^2 + 9*a*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^3 + 4*(21*(a^2*b^2 + 3*a*b^3)*cosh(d*x + c)^5 - 10*(4*a^3*b - 17*a^2*b ^2 + 9*a*b^3)*cosh(d*x + c)^3 + 3*(4*a^3*b - 17*a^2*b^2 + 9*a*b^3)*cosh(d* x + c))*sinh(d*x + c)^2 - ((a*b^2 + 3*b^3)*cosh(d*x + c)^8 + 8*(a*b^2 + 3* b^3)*cosh(d*x + c)*sinh(d*x + c)^7 + (a*b^2 + 3*b^3)*sinh(d*x + c)^8 + 4*( 2*a^2*b + 5*a*b^2 - 3*b^3)*cosh(d*x + c)^6 + 4*(2*a^2*b + 5*a*b^2 - 3*b^3 + 7*(a*b^2 + 3*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^6 + 8*(7*(a*b^2 + 3*b^3 )*cosh(d*x + c)^3 + 3*(2*a^2*b + 5*a*b^2 - 3*b^3)*cosh(d*x + c))*sinh(d*x + c)^5 + 2*(8*a^3 + 16*a^2*b - 21*a*b^2 + 9*b^3)*cosh(d*x + c)^4 + 2*(35*( a*b^2 + 3*b^3)*cosh(d*x + c)^4 + 8*a^3 + 16*a^2*b - 21*a*b^2 + 9*b^3 + 30* (2*a^2*b + 5*a*b^2 - 3*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 8*(7*(a*b^2 + 3*b^3)*cosh(d*x + c)^5 + 10*(2*a^2*b + 5*a*b^2 - 3*b^3)*cosh(d*x + c)^3 + (8*a^3 + 16*a^2*b - 21*a*b^2 + 9*b^3)*cosh(d*x + c))*sinh(d*x + c)^3...
Timed out. \[ \int \frac {\cosh ^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\text {Timed out} \]
\[ \int \frac {\cosh ^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\int { \frac {\cosh \left (d x + c\right )^{3}}{{\left (b \sinh \left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]
1/4*((a*b*e^(7*c) + 3*b^2*e^(7*c))*e^(7*d*x) - (4*a^2*e^(5*c) - 17*a*b*e^( 5*c) + 9*b^2*e^(5*c))*e^(5*d*x) + (4*a^2*e^(3*c) - 17*a*b*e^(3*c) + 9*b^2* e^(3*c))*e^(3*d*x) - (a*b*e^c + 3*b^2*e^c)*e^(d*x))/(a^2*b^3*d*e^(8*d*x + 8*c) + a^2*b^3*d + 4*(2*a^3*b^2*d*e^(6*c) - a^2*b^3*d*e^(6*c))*e^(6*d*x) + 2*(8*a^4*b*d*e^(4*c) - 8*a^3*b^2*d*e^(4*c) + 3*a^2*b^3*d*e^(4*c))*e^(4*d* x) + 4*(2*a^3*b^2*d*e^(2*c) - a^2*b^3*d*e^(2*c))*e^(2*d*x)) + 1/8*integrat e(2*((a*e^(3*c) + 3*b*e^(3*c))*e^(3*d*x) + (a*e^c + 3*b*e^c)*e^(d*x))/(a^2 *b^2*e^(4*d*x + 4*c) + a^2*b^2 + 2*(2*a^3*b*e^(2*c) - a^2*b^2*e^(2*c))*e^( 2*d*x)), x)
\[ \int \frac {\cosh ^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\int { \frac {\cosh \left (d x + c\right )^{3}}{{\left (b \sinh \left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {\cosh ^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\int \frac {{\mathrm {cosh}\left (c+d\,x\right )}^3}{{\left (b\,{\mathrm {sinh}\left (c+d\,x\right )}^2+a\right )}^3} \,d x \]