Integrand size = 23, antiderivative size = 133 \[ \int \frac {\cosh ^5(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\frac {\left (3 a^2+2 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {b} \sinh (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} b^{5/2} d}-\frac {(a-b) \cosh ^2(c+d x) \sinh (c+d x)}{4 a b d \left (a+b \sinh ^2(c+d x)\right )^2}+\frac {3 \left (\frac {1}{a^2}-\frac {1}{b^2}\right ) \sinh (c+d x)}{8 d \left (a+b \sinh ^2(c+d x)\right )} \]
1/8*(3*a^2+2*a*b+3*b^2)*arctan(sinh(d*x+c)*b^(1/2)/a^(1/2))/a^(5/2)/b^(5/2 )/d-1/4*(a-b)*cosh(d*x+c)^2*sinh(d*x+c)/a/b/d/(a+b*sinh(d*x+c)^2)^2+3/8*(1 /a^2-1/b^2)*sinh(d*x+c)/d/(a+b*sinh(d*x+c)^2)
Time = 0.32 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.12 \[ \int \frac {\cosh ^5(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\frac {-\left (\left (3 a^2+2 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {a} \text {csch}(c+d x)}{\sqrt {b}}\right )\right )+\frac {8 a^{3/2} (a-b)^2 \sqrt {b} \sinh (c+d x)}{(2 a-b+b \cosh (2 (c+d x)))^2}-\frac {2 \sqrt {a} \sqrt {b} \left (5 a^2-2 a b-3 b^2\right ) \sinh (c+d x)}{2 a-b+b \cosh (2 (c+d x))}}{8 a^{5/2} b^{5/2} d} \]
(-((3*a^2 + 2*a*b + 3*b^2)*ArcTan[(Sqrt[a]*Csch[c + d*x])/Sqrt[b]]) + (8*a ^(3/2)*(a - b)^2*Sqrt[b]*Sinh[c + d*x])/(2*a - b + b*Cosh[2*(c + d*x)])^2 - (2*Sqrt[a]*Sqrt[b]*(5*a^2 - 2*a*b - 3*b^2)*Sinh[c + d*x])/(2*a - b + b*C osh[2*(c + d*x)]))/(8*a^(5/2)*b^(5/2)*d)
Time = 0.32 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 3669, 315, 298, 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.
\(\displaystyle \int \frac {\cosh ^5(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (i c+i d x)^5}{\left (a-b \sin (i c+i d x)^2\right )^3}dx\) |
\(\Big \downarrow \) 3669 |
\(\displaystyle \frac {\int \frac {\left (\sinh ^2(c+d x)+1\right )^2}{\left (b \sinh ^2(c+d x)+a\right )^3}d\sinh (c+d x)}{d}\) |
\(\Big \downarrow \) 315 |
\(\displaystyle \frac {\frac {\int \frac {(3 a+b) \sinh ^2(c+d x)+a+3 b}{\left (b \sinh ^2(c+d x)+a\right )^2}d\sinh (c+d x)}{4 a b}-\frac {(a-b) \sinh (c+d x) \left (\sinh ^2(c+d x)+1\right )}{4 a b \left (a+b \sinh ^2(c+d x)\right )^2}}{d}\) |
\(\Big \downarrow \) 298 |
\(\displaystyle \frac {\frac {\frac {1}{2} \left (\frac {3 a}{b}+\frac {3 b}{a}+2\right ) \int \frac {1}{b \sinh ^2(c+d x)+a}d\sinh (c+d x)-\frac {3 \left (\frac {a}{b}-\frac {b}{a}\right ) \sinh (c+d x)}{2 \left (a+b \sinh ^2(c+d x)\right )}}{4 a b}-\frac {(a-b) \sinh (c+d x) \left (\sinh ^2(c+d x)+1\right )}{4 a b \left (a+b \sinh ^2(c+d x)\right )^2}}{d}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {\frac {\left (\frac {3 a}{b}+\frac {3 b}{a}+2\right ) \arctan \left (\frac {\sqrt {b} \sinh (c+d x)}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {b}}-\frac {3 \left (\frac {a}{b}-\frac {b}{a}\right ) \sinh (c+d x)}{2 \left (a+b \sinh ^2(c+d x)\right )}}{4 a b}-\frac {(a-b) \sinh (c+d x) \left (\sinh ^2(c+d x)+1\right )}{4 a b \left (a+b \sinh ^2(c+d x)\right )^2}}{d}\) |
(-1/4*((a - b)*Sinh[c + d*x]*(1 + Sinh[c + d*x]^2))/(a*b*(a + b*Sinh[c + d *x]^2)^2) + (((2 + (3*a)/b + (3*b)/a)*ArcTan[(Sqrt[b]*Sinh[c + d*x])/Sqrt[ a]])/(2*Sqrt[a]*Sqrt[b]) - (3*(a/b - b/a)*Sinh[c + d*x])/(2*(a + b*Sinh[c + d*x]^2)))/(4*a*b))/d
3.4.40.3.1 Defintions of rubi rules used
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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(a*d - c*b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(2*a*b*(p + 1))), x] - Simp[1/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*S imp[c*(a*d - c*b*(2*p + 3)) + d*(a*d*(2*(q - 1) + 1) - b*c*(2*(p + q) + 1)) *x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, - 1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
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 = 0.06 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.93
\[\frac {\frac {-\frac {\left (5 a^{2}-2 a b -3 b^{2}\right ) \sinh \left (d x +c \right )^{3}}{8 a^{2} b}-\frac {\left (3 a^{2}+2 a b -5 b^{2}\right ) \sinh \left (d x +c \right )}{8 a \,b^{2}}}{\left (a +b \sinh \left (d x +c \right )^{2}\right )^{2}}+\frac {\left (3 a^{2}+2 a b +3 b^{2}\right ) \arctan \left (\frac {b \sinh \left (d x +c \right )}{\sqrt {a b}}\right )}{8 a^{2} b^{2} \sqrt {a b}}}{d}\]
1/d*((-1/8*(5*a^2-2*a*b-3*b^2)/a^2/b*sinh(d*x+c)^3-1/8*(3*a^2+2*a*b-5*b^2) /a/b^2*sinh(d*x+c))/(a+b*sinh(d*x+c)^2)^2+1/8*(3*a^2+2*a*b+3*b^2)/a^2/b^2/ (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. 3266 vs. \(2 (119) = 238\).
Time = 0.34 (sec) , antiderivative size = 5844, normalized size of antiderivative = 43.94 \[ \int \frac {\cosh ^5(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {\cosh ^5(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\text {Timed out} \]
\[ \int \frac {\cosh ^5(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\int { \frac {\cosh \left (d x + c\right )^{5}}{{\left (b \sinh \left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]
-1/4*((5*a^2*b*e^(7*c) - 2*a*b^2*e^(7*c) - 3*b^3*e^(7*c))*e^(7*d*x) + (12* a^3*e^(5*c) - 7*a^2*b*e^(5*c) - 14*a*b^2*e^(5*c) + 9*b^3*e^(5*c))*e^(5*d*x ) - (12*a^3*e^(3*c) - 7*a^2*b*e^(3*c) - 14*a*b^2*e^(3*c) + 9*b^3*e^(3*c))* e^(3*d*x) - (5*a^2*b*e^c - 2*a*b^2*e^c - 3*b^3*e^c)*e^(d*x))/(a^2*b^4*d*e^ (8*d*x + 8*c) + a^2*b^4*d + 4*(2*a^3*b^3*d*e^(6*c) - a^2*b^4*d*e^(6*c))*e^ (6*d*x) + 2*(8*a^4*b^2*d*e^(4*c) - 8*a^3*b^3*d*e^(4*c) + 3*a^2*b^4*d*e^(4* c))*e^(4*d*x) + 4*(2*a^3*b^3*d*e^(2*c) - a^2*b^4*d*e^(2*c))*e^(2*d*x)) + 1 /32*integrate(8*((3*a^2*e^(3*c) + 2*a*b*e^(3*c) + 3*b^2*e^(3*c))*e^(3*d*x) + (3*a^2*e^c + 2*a*b*e^c + 3*b^2*e^c)*e^(d*x))/(a^2*b^3*e^(4*d*x + 4*c) + a^2*b^3 + 2*(2*a^3*b^2*e^(2*c) - a^2*b^3*e^(2*c))*e^(2*d*x)), x)
\[ \int \frac {\cosh ^5(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\int { \frac {\cosh \left (d x + c\right )^{5}}{{\left (b \sinh \left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {\cosh ^5(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\int \frac {{\mathrm {cosh}\left (c+d\,x\right )}^5}{{\left (b\,{\mathrm {sinh}\left (c+d\,x\right )}^2+a\right )}^3} \,d x \]