Integrand size = 20, antiderivative size = 184 \[ \int \frac {\cosh ^3(a+b x) \sinh ^2(a+b x)}{x^3} \, dx=\frac {\cosh (a+b x)}{16 x^2}-\frac {\cosh (3 a+3 b x)}{32 x^2}-\frac {\cosh (5 a+5 b x)}{32 x^2}-\frac {1}{16} b^2 \cosh (a) \text {Chi}(b x)+\frac {9}{32} b^2 \cosh (3 a) \text {Chi}(3 b x)+\frac {25}{32} b^2 \cosh (5 a) \text {Chi}(5 b x)+\frac {b \sinh (a+b x)}{16 x}-\frac {3 b \sinh (3 a+3 b x)}{32 x}-\frac {5 b \sinh (5 a+5 b x)}{32 x}-\frac {1}{16} b^2 \sinh (a) \text {Shi}(b x)+\frac {9}{32} b^2 \sinh (3 a) \text {Shi}(3 b x)+\frac {25}{32} b^2 \sinh (5 a) \text {Shi}(5 b x) \]
-1/16*b^2*Chi(b*x)*cosh(a)+9/32*b^2*Chi(3*b*x)*cosh(3*a)+25/32*b^2*Chi(5*b *x)*cosh(5*a)+1/16*cosh(b*x+a)/x^2-1/32*cosh(3*b*x+3*a)/x^2-1/32*cosh(5*b* x+5*a)/x^2-1/16*b^2*Shi(b*x)*sinh(a)+9/32*b^2*Shi(3*b*x)*sinh(3*a)+25/32*b ^2*Shi(5*b*x)*sinh(5*a)+1/16*b*sinh(b*x+a)/x-3/32*b*sinh(3*b*x+3*a)/x-5/32 *b*sinh(5*b*x+5*a)/x
Time = 0.30 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.88 \[ \int \frac {\cosh ^3(a+b x) \sinh ^2(a+b x)}{x^3} \, dx=-\frac {-2 \cosh (a+b x)+\cosh (3 (a+b x))+\cosh (5 (a+b x))+2 b^2 x^2 \cosh (a) \text {Chi}(b x)-9 b^2 x^2 \cosh (3 a) \text {Chi}(3 b x)-25 b^2 x^2 \cosh (5 a) \text {Chi}(5 b x)-2 b x \sinh (a+b x)+3 b x \sinh (3 (a+b x))+5 b x \sinh (5 (a+b x))+2 b^2 x^2 \sinh (a) \text {Shi}(b x)-9 b^2 x^2 \sinh (3 a) \text {Shi}(3 b x)-25 b^2 x^2 \sinh (5 a) \text {Shi}(5 b x)}{32 x^2} \]
-1/32*(-2*Cosh[a + b*x] + Cosh[3*(a + b*x)] + Cosh[5*(a + b*x)] + 2*b^2*x^ 2*Cosh[a]*CoshIntegral[b*x] - 9*b^2*x^2*Cosh[3*a]*CoshIntegral[3*b*x] - 25 *b^2*x^2*Cosh[5*a]*CoshIntegral[5*b*x] - 2*b*x*Sinh[a + b*x] + 3*b*x*Sinh[ 3*(a + b*x)] + 5*b*x*Sinh[5*(a + b*x)] + 2*b^2*x^2*Sinh[a]*SinhIntegral[b* x] - 9*b^2*x^2*Sinh[3*a]*SinhIntegral[3*b*x] - 25*b^2*x^2*Sinh[5*a]*SinhIn tegral[5*b*x])/x^2
Time = 0.55 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {5971, 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.
| \(\displaystyle \int \frac {\sinh ^2(a+b x) \cosh ^3(a+b x)}{x^3} \, dx\) |
| \(\Big \downarrow \) 5971 |
| \(\displaystyle \int \left (-\frac {\cosh (a+b x)}{8 x^3}+\frac {\cosh (3 a+3 b x)}{16 x^3}+\frac {\cosh (5 a+5 b x)}{16 x^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {1}{16} b^2 \cosh (a) \text {Chi}(b x)+\frac {9}{32} b^2 \cosh (3 a) \text {Chi}(3 b x)+\frac {25}{32} b^2 \cosh (5 a) \text {Chi}(5 b x)-\frac {1}{16} b^2 \sinh (a) \text {Shi}(b x)+\frac {9}{32} b^2 \sinh (3 a) \text {Shi}(3 b x)+\frac {25}{32} b^2 \sinh (5 a) \text {Shi}(5 b x)+\frac {\cosh (a+b x)}{16 x^2}-\frac {\cosh (3 a+3 b x)}{32 x^2}-\frac {\cosh (5 a+5 b x)}{32 x^2}+\frac {b \sinh (a+b x)}{16 x}-\frac {3 b \sinh (3 a+3 b x)}{32 x}-\frac {5 b \sinh (5 a+5 b x)}{32 x}\) |
Cosh[a + b*x]/(16*x^2) - Cosh[3*a + 3*b*x]/(32*x^2) - Cosh[5*a + 5*b*x]/(3 2*x^2) - (b^2*Cosh[a]*CoshIntegral[b*x])/16 + (9*b^2*Cosh[3*a]*CoshIntegra l[3*b*x])/32 + (25*b^2*Cosh[5*a]*CoshIntegral[5*b*x])/32 + (b*Sinh[a + b*x ])/(16*x) - (3*b*Sinh[3*a + 3*b*x])/(32*x) - (5*b*Sinh[5*a + 5*b*x])/(32*x ) - (b^2*Sinh[a]*SinhIntegral[b*x])/16 + (9*b^2*Sinh[3*a]*SinhIntegral[3*b *x])/32 + (25*b^2*Sinh[5*a]*SinhIntegral[5*b*x])/32
3.4.5.3.1 Defintions of rubi rules used
Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] & & IGtQ[p, 0]
Time = 14.24 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.36
| method | result | size |
| risch | \(\frac {-9 \,{\mathrm e}^{-3 a} \operatorname {Ei}_{1}\left (3 b x \right ) x^{2} b^{2}+2 \,{\mathrm e}^{-a} \operatorname {Ei}_{1}\left (b x \right ) x^{2} b^{2}+2 \,{\mathrm e}^{a} \operatorname {Ei}_{1}\left (-b x \right ) x^{2} b^{2}-25 \,{\mathrm e}^{-5 a} \operatorname {Ei}_{1}\left (5 b x \right ) x^{2} b^{2}-25 \,{\mathrm e}^{5 a} \operatorname {Ei}_{1}\left (-5 b x \right ) x^{2} b^{2}-9 \,{\mathrm e}^{3 a} \operatorname {Ei}_{1}\left (-3 b x \right ) x^{2} b^{2}+2 \,{\mathrm e}^{b x +a} b x -2 \,{\mathrm e}^{-b x -a} b x +3 \,{\mathrm e}^{-3 b x -3 a} b x +5 \,{\mathrm e}^{-5 b x -5 a} b x -5 \,{\mathrm e}^{5 b x +5 a} b x -3 \,{\mathrm e}^{3 b x +3 a} b x +2 \,{\mathrm e}^{b x +a}+2 \,{\mathrm e}^{-b x -a}-{\mathrm e}^{-3 b x -3 a}-{\mathrm e}^{-5 b x -5 a}-{\mathrm e}^{5 b x +5 a}-{\mathrm e}^{3 b x +3 a}}{64 x^{2}}\) | \(250\) |
1/64*(-9*exp(-3*a)*Ei(1,3*b*x)*x^2*b^2+2*exp(-a)*Ei(1,b*x)*x^2*b^2+2*exp(a )*Ei(1,-b*x)*x^2*b^2-25*exp(-5*a)*Ei(1,5*b*x)*x^2*b^2-25*exp(5*a)*Ei(1,-5* b*x)*x^2*b^2-9*exp(3*a)*Ei(1,-3*b*x)*x^2*b^2+2*exp(b*x+a)*b*x-2*exp(-b*x-a )*b*x+3*exp(-3*b*x-3*a)*b*x+5*exp(-5*b*x-5*a)*b*x-5*exp(5*b*x+5*a)*b*x-3*e xp(3*b*x+3*a)*b*x+2*exp(b*x+a)+2*exp(-b*x-a)-exp(-3*b*x-3*a)-exp(-5*b*x-5* a)-exp(5*b*x+5*a)-exp(3*b*x+3*a))/x^2
Leaf count of result is larger than twice the leaf count of optimal. 338 vs. \(2 (160) = 320\).
Time = 0.25 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.84 \[ \int \frac {\cosh ^3(a+b x) \sinh ^2(a+b x)}{x^3} \, dx=-\frac {10 \, b x \sinh \left (b x + a\right )^{5} + 2 \, \cosh \left (b x + a\right )^{5} + 10 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} + 2 \, {\left (50 \, b x \cosh \left (b x + a\right )^{2} + 3 \, b x\right )} \sinh \left (b x + a\right )^{3} + 2 \, \cosh \left (b x + a\right )^{3} + 2 \, {\left (10 \, \cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 25 \, {\left (b^{2} x^{2} {\rm Ei}\left (5 \, b x\right ) + b^{2} x^{2} {\rm Ei}\left (-5 \, b x\right )\right )} \cosh \left (5 \, a\right ) - 9 \, {\left (b^{2} x^{2} {\rm Ei}\left (3 \, b x\right ) + b^{2} x^{2} {\rm Ei}\left (-3 \, b x\right )\right )} \cosh \left (3 \, a\right ) + 2 \, {\left (b^{2} x^{2} {\rm Ei}\left (b x\right ) + b^{2} x^{2} {\rm Ei}\left (-b x\right )\right )} \cosh \left (a\right ) + 2 \, {\left (25 \, b x \cosh \left (b x + a\right )^{4} + 9 \, b x \cosh \left (b x + a\right )^{2} - 2 \, b x\right )} \sinh \left (b x + a\right ) - 25 \, {\left (b^{2} x^{2} {\rm Ei}\left (5 \, b x\right ) - b^{2} x^{2} {\rm Ei}\left (-5 \, b x\right )\right )} \sinh \left (5 \, a\right ) - 9 \, {\left (b^{2} x^{2} {\rm Ei}\left (3 \, b x\right ) - b^{2} x^{2} {\rm Ei}\left (-3 \, b x\right )\right )} \sinh \left (3 \, a\right ) + 2 \, {\left (b^{2} x^{2} {\rm Ei}\left (b x\right ) - b^{2} x^{2} {\rm Ei}\left (-b x\right )\right )} \sinh \left (a\right ) - 4 \, \cosh \left (b x + a\right )}{64 \, x^{2}} \]
-1/64*(10*b*x*sinh(b*x + a)^5 + 2*cosh(b*x + a)^5 + 10*cosh(b*x + a)*sinh( b*x + a)^4 + 2*(50*b*x*cosh(b*x + a)^2 + 3*b*x)*sinh(b*x + a)^3 + 2*cosh(b *x + a)^3 + 2*(10*cosh(b*x + a)^3 + 3*cosh(b*x + a))*sinh(b*x + a)^2 - 25* (b^2*x^2*Ei(5*b*x) + b^2*x^2*Ei(-5*b*x))*cosh(5*a) - 9*(b^2*x^2*Ei(3*b*x) + b^2*x^2*Ei(-3*b*x))*cosh(3*a) + 2*(b^2*x^2*Ei(b*x) + b^2*x^2*Ei(-b*x))*c osh(a) + 2*(25*b*x*cosh(b*x + a)^4 + 9*b*x*cosh(b*x + a)^2 - 2*b*x)*sinh(b *x + a) - 25*(b^2*x^2*Ei(5*b*x) - b^2*x^2*Ei(-5*b*x))*sinh(5*a) - 9*(b^2*x ^2*Ei(3*b*x) - b^2*x^2*Ei(-3*b*x))*sinh(3*a) + 2*(b^2*x^2*Ei(b*x) - b^2*x^ 2*Ei(-b*x))*sinh(a) - 4*cosh(b*x + a))/x^2
\[ \int \frac {\cosh ^3(a+b x) \sinh ^2(a+b x)}{x^3} \, dx=\int \frac {\sinh ^{2}{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{x^{3}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.48 \[ \int \frac {\cosh ^3(a+b x) \sinh ^2(a+b x)}{x^3} \, dx=-\frac {25}{32} \, b^{2} e^{\left (-5 \, a\right )} \Gamma \left (-2, 5 \, b x\right ) - \frac {9}{32} \, b^{2} e^{\left (-3 \, a\right )} \Gamma \left (-2, 3 \, b x\right ) + \frac {1}{16} \, b^{2} e^{\left (-a\right )} \Gamma \left (-2, b x\right ) + \frac {1}{16} \, b^{2} e^{a} \Gamma \left (-2, -b x\right ) - \frac {9}{32} \, b^{2} e^{\left (3 \, a\right )} \Gamma \left (-2, -3 \, b x\right ) - \frac {25}{32} \, b^{2} e^{\left (5 \, a\right )} \Gamma \left (-2, -5 \, b x\right ) \]
-25/32*b^2*e^(-5*a)*gamma(-2, 5*b*x) - 9/32*b^2*e^(-3*a)*gamma(-2, 3*b*x) + 1/16*b^2*e^(-a)*gamma(-2, b*x) + 1/16*b^2*e^a*gamma(-2, -b*x) - 9/32*b^2 *e^(3*a)*gamma(-2, -3*b*x) - 25/32*b^2*e^(5*a)*gamma(-2, -5*b*x)
Time = 0.28 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.32 \[ \int \frac {\cosh ^3(a+b x) \sinh ^2(a+b x)}{x^3} \, dx=\frac {25 \, b^{2} x^{2} {\rm Ei}\left (5 \, b x\right ) e^{\left (5 \, a\right )} + 9 \, b^{2} x^{2} {\rm Ei}\left (3 \, b x\right ) e^{\left (3 \, a\right )} - 2 \, b^{2} x^{2} {\rm Ei}\left (-b x\right ) e^{\left (-a\right )} + 9 \, b^{2} x^{2} {\rm Ei}\left (-3 \, b x\right ) e^{\left (-3 \, a\right )} + 25 \, b^{2} x^{2} {\rm Ei}\left (-5 \, b x\right ) e^{\left (-5 \, a\right )} - 2 \, b^{2} x^{2} {\rm Ei}\left (b x\right ) e^{a} - 5 \, b x e^{\left (5 \, b x + 5 \, a\right )} - 3 \, b x e^{\left (3 \, b x + 3 \, a\right )} + 2 \, b x e^{\left (b x + a\right )} - 2 \, b x e^{\left (-b x - a\right )} + 3 \, b x e^{\left (-3 \, b x - 3 \, a\right )} + 5 \, b x e^{\left (-5 \, b x - 5 \, a\right )} - e^{\left (5 \, b x + 5 \, a\right )} - e^{\left (3 \, b x + 3 \, a\right )} + 2 \, e^{\left (b x + a\right )} + 2 \, e^{\left (-b x - a\right )} - e^{\left (-3 \, b x - 3 \, a\right )} - e^{\left (-5 \, b x - 5 \, a\right )}}{64 \, x^{2}} \]
1/64*(25*b^2*x^2*Ei(5*b*x)*e^(5*a) + 9*b^2*x^2*Ei(3*b*x)*e^(3*a) - 2*b^2*x ^2*Ei(-b*x)*e^(-a) + 9*b^2*x^2*Ei(-3*b*x)*e^(-3*a) + 25*b^2*x^2*Ei(-5*b*x) *e^(-5*a) - 2*b^2*x^2*Ei(b*x)*e^a - 5*b*x*e^(5*b*x + 5*a) - 3*b*x*e^(3*b*x + 3*a) + 2*b*x*e^(b*x + a) - 2*b*x*e^(-b*x - a) + 3*b*x*e^(-3*b*x - 3*a) + 5*b*x*e^(-5*b*x - 5*a) - e^(5*b*x + 5*a) - e^(3*b*x + 3*a) + 2*e^(b*x + a) + 2*e^(-b*x - a) - e^(-3*b*x - 3*a) - e^(-5*b*x - 5*a))/x^2
Timed out. \[ \int \frac {\cosh ^3(a+b x) \sinh ^2(a+b x)}{x^3} \, dx=\int \frac {{\mathrm {cosh}\left (a+b\,x\right )}^3\,{\mathrm {sinh}\left (a+b\,x\right )}^2}{x^3} \,d x \]