Integrand size = 18, antiderivative size = 125 \[ \int \frac {\cosh ^3(a+b x) \sinh (a+b x)}{x^3} \, dx=-\frac {b \cosh (2 a+2 b x)}{4 x}-\frac {b \cosh (4 a+4 b x)}{4 x}+\frac {1}{2} b^2 \text {Chi}(2 b x) \sinh (2 a)+b^2 \text {Chi}(4 b x) \sinh (4 a)-\frac {\sinh (2 a+2 b x)}{8 x^2}-\frac {\sinh (4 a+4 b x)}{16 x^2}+\frac {1}{2} b^2 \cosh (2 a) \text {Shi}(2 b x)+b^2 \cosh (4 a) \text {Shi}(4 b x) \]
-1/4*b*cosh(2*b*x+2*a)/x-1/4*b*cosh(4*b*x+4*a)/x+1/2*b^2*cosh(2*a)*Shi(2*b *x)+b^2*cosh(4*a)*Shi(4*b*x)+1/2*b^2*Chi(2*b*x)*sinh(2*a)+b^2*Chi(4*b*x)*s inh(4*a)-1/8*sinh(2*b*x+2*a)/x^2-1/16*sinh(4*b*x+4*a)/x^2
Time = 0.37 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.90 \[ \int \frac {\cosh ^3(a+b x) \sinh (a+b x)}{x^3} \, dx=b^2 \cosh (a) \text {Chi}(2 b x) \sinh (a)+b^2 \text {Chi}(4 b x) \sinh (4 a)-\frac {2 b x \cosh (2 (a+b x))+\sinh (2 (a+b x))}{8 x^2}-\frac {4 b x \cosh (4 (a+b x))+\sinh (4 (a+b x))}{16 x^2}+\frac {1}{2} b^2 \cosh (2 a) \text {Shi}(2 b x)+b^2 \cosh (4 a) \text {Shi}(4 b x) \]
b^2*Cosh[a]*CoshIntegral[2*b*x]*Sinh[a] + b^2*CoshIntegral[4*b*x]*Sinh[4*a ] - (2*b*x*Cosh[2*(a + b*x)] + Sinh[2*(a + b*x)])/(8*x^2) - (4*b*x*Cosh[4* (a + b*x)] + Sinh[4*(a + b*x)])/(16*x^2) + (b^2*Cosh[2*a]*SinhIntegral[2*b *x])/2 + b^2*Cosh[4*a]*SinhIntegral[4*b*x]
Time = 0.46 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, 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.
\(\displaystyle \int \frac {\sinh (a+b x) \cosh ^3(a+b x)}{x^3} \, dx\) |
\(\Big \downarrow \) 5971 |
\(\displaystyle \int \left (\frac {\sinh (2 a+2 b x)}{4 x^3}+\frac {\sinh (4 a+4 b x)}{8 x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} b^2 \sinh (2 a) \text {Chi}(2 b x)+b^2 \sinh (4 a) \text {Chi}(4 b x)+\frac {1}{2} b^2 \cosh (2 a) \text {Shi}(2 b x)+b^2 \cosh (4 a) \text {Shi}(4 b x)-\frac {\sinh (2 a+2 b x)}{8 x^2}-\frac {\sinh (4 a+4 b x)}{16 x^2}-\frac {b \cosh (2 a+2 b x)}{4 x}-\frac {b \cosh (4 a+4 b x)}{4 x}\) |
-1/4*(b*Cosh[2*a + 2*b*x])/x - (b*Cosh[4*a + 4*b*x])/(4*x) + (b^2*CoshInte gral[2*b*x]*Sinh[2*a])/2 + b^2*CoshIntegral[4*b*x]*Sinh[4*a] - Sinh[2*a + 2*b*x]/(8*x^2) - Sinh[4*a + 4*b*x]/(16*x^2) + (b^2*Cosh[2*a]*SinhIntegral[ 2*b*x])/2 + b^2*Cosh[4*a]*SinhIntegral[4*b*x]
3.3.75.3.1 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 = 6.58 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.38
method | result | size |
risch | \(-\frac {-16 \,{\mathrm e}^{-4 a} \operatorname {Ei}_{1}\left (4 b x \right ) x^{2} b^{2}+8 \,{\mathrm e}^{2 a} \operatorname {Ei}_{1}\left (-2 b x \right ) x^{2} b^{2}-8 \,{\mathrm e}^{-2 a} \operatorname {Ei}_{1}\left (2 b x \right ) x^{2} b^{2}+16 \,{\mathrm e}^{4 a} \operatorname {Ei}_{1}\left (-4 b x \right ) x^{2} b^{2}+4 \,{\mathrm e}^{-4 b x -4 a} b x +4 \,{\mathrm e}^{2 b x +2 a} b x +4 \,{\mathrm e}^{-2 b x -2 a} b x +4 \,{\mathrm e}^{4 b x +4 a} b x -{\mathrm e}^{-4 b x -4 a}+2 \,{\mathrm e}^{2 b x +2 a}-2 \,{\mathrm e}^{-2 b x -2 a}+{\mathrm e}^{4 b x +4 a}}{32 x^{2}}\) | \(173\) |
-1/32*(-16*exp(-4*a)*Ei(1,4*b*x)*x^2*b^2+8*exp(2*a)*Ei(1,-2*b*x)*x^2*b^2-8 *exp(-2*a)*Ei(1,2*b*x)*x^2*b^2+16*exp(4*a)*Ei(1,-4*b*x)*x^2*b^2+4*exp(-4*b *x-4*a)*b*x+4*exp(2*b*x+2*a)*b*x+4*exp(-2*b*x-2*a)*b*x+4*exp(4*b*x+4*a)*b* x-exp(-4*b*x-4*a)+2*exp(2*b*x+2*a)-2*exp(-2*b*x-2*a)+exp(4*b*x+4*a))/x^2
Leaf count of result is larger than twice the leaf count of optimal. 227 vs. \(2 (113) = 226\).
Time = 0.25 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.82 \[ \int \frac {\cosh ^3(a+b x) \sinh (a+b x)}{x^3} \, dx=-\frac {b x \cosh \left (b x + a\right )^{4} + b x \sinh \left (b x + a\right )^{4} + b x \cosh \left (b x + a\right )^{2} + \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + {\left (6 \, b x \cosh \left (b x + a\right )^{2} + b x\right )} \sinh \left (b x + a\right )^{2} - 2 \, {\left (b^{2} x^{2} {\rm Ei}\left (4 \, b x\right ) - b^{2} x^{2} {\rm Ei}\left (-4 \, b x\right )\right )} \cosh \left (4 \, a\right ) - {\left (b^{2} x^{2} {\rm Ei}\left (2 \, b x\right ) - b^{2} x^{2} {\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (2 \, a\right ) + {\left (\cosh \left (b x + a\right )^{3} + \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) - 2 \, {\left (b^{2} x^{2} {\rm Ei}\left (4 \, b x\right ) + b^{2} x^{2} {\rm Ei}\left (-4 \, b x\right )\right )} \sinh \left (4 \, a\right ) - {\left (b^{2} x^{2} {\rm Ei}\left (2 \, b x\right ) + b^{2} x^{2} {\rm Ei}\left (-2 \, b x\right )\right )} \sinh \left (2 \, a\right )}{4 \, x^{2}} \]
-1/4*(b*x*cosh(b*x + a)^4 + b*x*sinh(b*x + a)^4 + b*x*cosh(b*x + a)^2 + co sh(b*x + a)*sinh(b*x + a)^3 + (6*b*x*cosh(b*x + a)^2 + b*x)*sinh(b*x + a)^ 2 - 2*(b^2*x^2*Ei(4*b*x) - b^2*x^2*Ei(-4*b*x))*cosh(4*a) - (b^2*x^2*Ei(2*b *x) - b^2*x^2*Ei(-2*b*x))*cosh(2*a) + (cosh(b*x + a)^3 + cosh(b*x + a))*si nh(b*x + a) - 2*(b^2*x^2*Ei(4*b*x) + b^2*x^2*Ei(-4*b*x))*sinh(4*a) - (b^2* x^2*Ei(2*b*x) + b^2*x^2*Ei(-2*b*x))*sinh(2*a))/x^2
\[ \int \frac {\cosh ^3(a+b x) \sinh (a+b x)}{x^3} \, dx=\int \frac {\sinh {\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{x^{3}}\, dx \]
Time = 0.33 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.48 \[ \int \frac {\cosh ^3(a+b x) \sinh (a+b x)}{x^3} \, dx=b^{2} e^{\left (-4 \, a\right )} \Gamma \left (-2, 4 \, b x\right ) + \frac {1}{2} \, b^{2} e^{\left (-2 \, a\right )} \Gamma \left (-2, 2 \, b x\right ) - \frac {1}{2} \, b^{2} e^{\left (2 \, a\right )} \Gamma \left (-2, -2 \, b x\right ) - b^{2} e^{\left (4 \, a\right )} \Gamma \left (-2, -4 \, b x\right ) \]
b^2*e^(-4*a)*gamma(-2, 4*b*x) + 1/2*b^2*e^(-2*a)*gamma(-2, 2*b*x) - 1/2*b^ 2*e^(2*a)*gamma(-2, -2*b*x) - b^2*e^(4*a)*gamma(-2, -4*b*x)
Time = 0.26 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.34 \[ \int \frac {\cosh ^3(a+b x) \sinh (a+b x)}{x^3} \, dx=\frac {16 \, b^{2} x^{2} {\rm Ei}\left (4 \, b x\right ) e^{\left (4 \, a\right )} + 8 \, b^{2} x^{2} {\rm Ei}\left (2 \, b x\right ) e^{\left (2 \, a\right )} - 8 \, b^{2} x^{2} {\rm Ei}\left (-2 \, b x\right ) e^{\left (-2 \, a\right )} - 16 \, b^{2} x^{2} {\rm Ei}\left (-4 \, b x\right ) e^{\left (-4 \, a\right )} - 4 \, b x e^{\left (4 \, b x + 4 \, a\right )} - 4 \, b x e^{\left (2 \, b x + 2 \, a\right )} - 4 \, b x e^{\left (-2 \, b x - 2 \, a\right )} - 4 \, b x e^{\left (-4 \, b x - 4 \, a\right )} - e^{\left (4 \, b x + 4 \, a\right )} - 2 \, e^{\left (2 \, b x + 2 \, a\right )} + 2 \, e^{\left (-2 \, b x - 2 \, a\right )} + e^{\left (-4 \, b x - 4 \, a\right )}}{32 \, x^{2}} \]
1/32*(16*b^2*x^2*Ei(4*b*x)*e^(4*a) + 8*b^2*x^2*Ei(2*b*x)*e^(2*a) - 8*b^2*x ^2*Ei(-2*b*x)*e^(-2*a) - 16*b^2*x^2*Ei(-4*b*x)*e^(-4*a) - 4*b*x*e^(4*b*x + 4*a) - 4*b*x*e^(2*b*x + 2*a) - 4*b*x*e^(-2*b*x - 2*a) - 4*b*x*e^(-4*b*x - 4*a) - e^(4*b*x + 4*a) - 2*e^(2*b*x + 2*a) + 2*e^(-2*b*x - 2*a) + e^(-4*b *x - 4*a))/x^2
Timed out. \[ \int \frac {\cosh ^3(a+b x) \sinh (a+b x)}{x^3} \, dx=\int \frac {{\mathrm {cosh}\left (a+b\,x\right )}^3\,\mathrm {sinh}\left (a+b\,x\right )}{x^3} \,d x \]