Integrand size = 20, antiderivative size = 238 \[ \int \frac {\cosh ^2(a+b x) \sinh ^3(a+b x)}{x^4} \, dx=\frac {b \cosh (a+b x)}{48 x^2}+\frac {b \cosh (3 a+3 b x)}{32 x^2}-\frac {5 b \cosh (5 a+5 b x)}{96 x^2}-\frac {1}{48} b^3 \cosh (a) \text {Chi}(b x)-\frac {9}{32} b^3 \cosh (3 a) \text {Chi}(3 b x)+\frac {125}{96} b^3 \cosh (5 a) \text {Chi}(5 b x)+\frac {\sinh (a+b x)}{24 x^3}+\frac {b^2 \sinh (a+b x)}{48 x}+\frac {\sinh (3 a+3 b x)}{48 x^3}+\frac {3 b^2 \sinh (3 a+3 b x)}{32 x}-\frac {\sinh (5 a+5 b x)}{48 x^3}-\frac {25 b^2 \sinh (5 a+5 b x)}{96 x}-\frac {1}{48} b^3 \sinh (a) \text {Shi}(b x)-\frac {9}{32} b^3 \sinh (3 a) \text {Shi}(3 b x)+\frac {125}{96} b^3 \sinh (5 a) \text {Shi}(5 b x) \]
-1/48*b^3*Chi(b*x)*cosh(a)-9/32*b^3*Chi(3*b*x)*cosh(3*a)+125/96*b^3*Chi(5* b*x)*cosh(5*a)+1/48*b*cosh(b*x+a)/x^2+1/32*b*cosh(3*b*x+3*a)/x^2-5/96*b*co sh(5*b*x+5*a)/x^2-1/48*b^3*Shi(b*x)*sinh(a)-9/32*b^3*Shi(3*b*x)*sinh(3*a)+ 125/96*b^3*Shi(5*b*x)*sinh(5*a)+1/24*sinh(b*x+a)/x^3+1/48*b^2*sinh(b*x+a)/ x+1/48*sinh(3*b*x+3*a)/x^3+3/32*b^2*sinh(3*b*x+3*a)/x-1/48*sinh(5*b*x+5*a) /x^3-25/96*b^2*sinh(5*b*x+5*a)/x
Time = 0.29 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.89 \[ \int \frac {\cosh ^2(a+b x) \sinh ^3(a+b x)}{x^4} \, dx=\frac {2 b x \cosh (a+b x)+3 b x \cosh (3 (a+b x))-5 b x \cosh (5 (a+b x))-2 b^3 x^3 \cosh (a) \text {Chi}(b x)-27 b^3 x^3 \cosh (3 a) \text {Chi}(3 b x)+125 b^3 x^3 \cosh (5 a) \text {Chi}(5 b x)+4 \sinh (a+b x)+2 b^2 x^2 \sinh (a+b x)+2 \sinh (3 (a+b x))+9 b^2 x^2 \sinh (3 (a+b x))-2 \sinh (5 (a+b x))-25 b^2 x^2 \sinh (5 (a+b x))-2 b^3 x^3 \sinh (a) \text {Shi}(b x)-27 b^3 x^3 \sinh (3 a) \text {Shi}(3 b x)+125 b^3 x^3 \sinh (5 a) \text {Shi}(5 b x)}{96 x^3} \]
(2*b*x*Cosh[a + b*x] + 3*b*x*Cosh[3*(a + b*x)] - 5*b*x*Cosh[5*(a + b*x)] - 2*b^3*x^3*Cosh[a]*CoshIntegral[b*x] - 27*b^3*x^3*Cosh[3*a]*CoshIntegral[3 *b*x] + 125*b^3*x^3*Cosh[5*a]*CoshIntegral[5*b*x] + 4*Sinh[a + b*x] + 2*b^ 2*x^2*Sinh[a + b*x] + 2*Sinh[3*(a + b*x)] + 9*b^2*x^2*Sinh[3*(a + b*x)] - 2*Sinh[5*(a + b*x)] - 25*b^2*x^2*Sinh[5*(a + b*x)] - 2*b^3*x^3*Sinh[a]*Sin hIntegral[b*x] - 27*b^3*x^3*Sinh[3*a]*SinhIntegral[3*b*x] + 125*b^3*x^3*Si nh[5*a]*SinhIntegral[5*b*x])/(96*x^3)
Time = 0.63 (sec) , antiderivative size = 238, 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.
| \(\displaystyle \int \frac {\sinh ^3(a+b x) \cosh ^2(a+b x)}{x^4} \, dx\) |
| \(\Big \downarrow \) 5971 |
| \(\displaystyle \int \left (-\frac {\sinh (a+b x)}{8 x^4}-\frac {\sinh (3 a+3 b x)}{16 x^4}+\frac {\sinh (5 a+5 b x)}{16 x^4}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {1}{48} b^3 \cosh (a) \text {Chi}(b x)-\frac {9}{32} b^3 \cosh (3 a) \text {Chi}(3 b x)+\frac {125}{96} b^3 \cosh (5 a) \text {Chi}(5 b x)-\frac {1}{48} b^3 \sinh (a) \text {Shi}(b x)-\frac {9}{32} b^3 \sinh (3 a) \text {Shi}(3 b x)+\frac {125}{96} b^3 \sinh (5 a) \text {Shi}(5 b x)+\frac {b^2 \sinh (a+b x)}{48 x}+\frac {3 b^2 \sinh (3 a+3 b x)}{32 x}-\frac {25 b^2 \sinh (5 a+5 b x)}{96 x}+\frac {\sinh (a+b x)}{24 x^3}+\frac {\sinh (3 a+3 b x)}{48 x^3}-\frac {\sinh (5 a+5 b x)}{48 x^3}+\frac {b \cosh (a+b x)}{48 x^2}+\frac {b \cosh (3 a+3 b x)}{32 x^2}-\frac {5 b \cosh (5 a+5 b x)}{96 x^2}\) |
(b*Cosh[a + b*x])/(48*x^2) + (b*Cosh[3*a + 3*b*x])/(32*x^2) - (5*b*Cosh[5* a + 5*b*x])/(96*x^2) - (b^3*Cosh[a]*CoshIntegral[b*x])/48 - (9*b^3*Cosh[3* a]*CoshIntegral[3*b*x])/32 + (125*b^3*Cosh[5*a]*CoshIntegral[5*b*x])/96 + Sinh[a + b*x]/(24*x^3) + (b^2*Sinh[a + b*x])/(48*x) + Sinh[3*a + 3*b*x]/(4 8*x^3) + (3*b^2*Sinh[3*a + 3*b*x])/(32*x) - Sinh[5*a + 5*b*x]/(48*x^3) - ( 25*b^2*Sinh[5*a + 5*b*x])/(96*x) - (b^3*Sinh[a]*SinhIntegral[b*x])/48 - (9 *b^3*Sinh[3*a]*SinhIntegral[3*b*x])/32 + (125*b^3*Sinh[5*a]*SinhIntegral[5 *b*x])/96
3.4.24.3.1 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 = 21.62 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.47
| method | result | size |
| risch | \(-\frac {125 \,{\mathrm e}^{5 a} \operatorname {Ei}_{1}\left (-5 b x \right ) x^{3} b^{3}+125 \,{\mathrm e}^{-5 a} \operatorname {Ei}_{1}\left (5 b x \right ) x^{3} b^{3}-27 \,{\mathrm e}^{-3 a} \operatorname {Ei}_{1}\left (3 b x \right ) x^{3} b^{3}-2 \,{\mathrm e}^{-a} \operatorname {Ei}_{1}\left (b x \right ) x^{3} b^{3}-2 \,{\mathrm e}^{a} \operatorname {Ei}_{1}\left (-b x \right ) x^{3} b^{3}-27 \,{\mathrm e}^{3 a} \operatorname {Ei}_{1}\left (-3 b x \right ) x^{3} b^{3}+25 \,{\mathrm e}^{5 b x +5 a} b^{2} x^{2}-2 \,{\mathrm e}^{b x +a} b^{2} x^{2}-25 \,{\mathrm e}^{-5 b x -5 a} b^{2} x^{2}+9 \,{\mathrm e}^{-3 b x -3 a} b^{2} x^{2}+2 \,{\mathrm e}^{-b x -a} b^{2} x^{2}-9 \,{\mathrm e}^{3 b x +3 a} b^{2} x^{2}+5 \,{\mathrm e}^{5 b x +5 a} b x -2 \,{\mathrm e}^{b x +a} b x +5 \,{\mathrm e}^{-5 b x -5 a} b x -3 \,{\mathrm e}^{-3 b x -3 a} b x -2 \,{\mathrm e}^{-b x -a} b x -3 \,{\mathrm e}^{3 b x +3 a} b x +2 \,{\mathrm e}^{5 b x +5 a}-4 \,{\mathrm e}^{b x +a}-2 \,{\mathrm e}^{-5 b x -5 a}+2 \,{\mathrm e}^{-3 b x -3 a}+4 \,{\mathrm e}^{-b x -a}-2 \,{\mathrm e}^{3 b x +3 a}}{192 x^{3}}\) | \(349\) |
-1/192*(125*exp(5*a)*Ei(1,-5*b*x)*x^3*b^3+125*exp(-5*a)*Ei(1,5*b*x)*x^3*b^ 3-27*exp(-3*a)*Ei(1,3*b*x)*x^3*b^3-2*exp(-a)*Ei(1,b*x)*x^3*b^3-2*exp(a)*Ei (1,-b*x)*x^3*b^3-27*exp(3*a)*Ei(1,-3*b*x)*x^3*b^3+25*exp(5*b*x+5*a)*b^2*x^ 2-2*exp(b*x+a)*b^2*x^2-25*exp(-5*b*x-5*a)*b^2*x^2+9*exp(-3*b*x-3*a)*b^2*x^ 2+2*exp(-b*x-a)*b^2*x^2-9*exp(3*b*x+3*a)*b^2*x^2+5*exp(5*b*x+5*a)*b*x-2*ex p(b*x+a)*b*x+5*exp(-5*b*x-5*a)*b*x-3*exp(-3*b*x-3*a)*b*x-2*exp(-b*x-a)*b*x -3*exp(3*b*x+3*a)*b*x+2*exp(5*b*x+5*a)-4*exp(b*x+a)-2*exp(-5*b*x-5*a)+2*ex p(-3*b*x-3*a)+4*exp(-b*x-a)-2*exp(3*b*x+3*a))/x^3
Time = 0.24 (sec) , antiderivative size = 392, normalized size of antiderivative = 1.65 \[ \int \frac {\cosh ^2(a+b x) \sinh ^3(a+b x)}{x^4} \, dx=-\frac {10 \, b x \cosh \left (b x + a\right )^{5} + 50 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} + 2 \, {\left (25 \, b^{2} x^{2} + 2\right )} \sinh \left (b x + a\right )^{5} - 6 \, b x \cosh \left (b x + a\right )^{3} - 2 \, {\left (9 \, b^{2} x^{2} - 10 \, {\left (25 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )^{2} + 2\right )} \sinh \left (b x + a\right )^{3} - 4 \, b x \cosh \left (b x + a\right ) + 2 \, {\left (50 \, b x \cosh \left (b x + a\right )^{3} - 9 \, b x \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 125 \, {\left (b^{3} x^{3} {\rm Ei}\left (5 \, b x\right ) + b^{3} x^{3} {\rm Ei}\left (-5 \, b x\right )\right )} \cosh \left (5 \, a\right ) + 27 \, {\left (b^{3} x^{3} {\rm Ei}\left (3 \, b x\right ) + b^{3} x^{3} {\rm Ei}\left (-3 \, b x\right )\right )} \cosh \left (3 \, a\right ) + 2 \, {\left (b^{3} x^{3} {\rm Ei}\left (b x\right ) + b^{3} x^{3} {\rm Ei}\left (-b x\right )\right )} \cosh \left (a\right ) + 2 \, {\left (5 \, {\left (25 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )^{4} - 2 \, b^{2} x^{2} - 3 \, {\left (9 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )^{2} - 4\right )} \sinh \left (b x + a\right ) - 125 \, {\left (b^{3} x^{3} {\rm Ei}\left (5 \, b x\right ) - b^{3} x^{3} {\rm Ei}\left (-5 \, b x\right )\right )} \sinh \left (5 \, a\right ) + 27 \, {\left (b^{3} x^{3} {\rm Ei}\left (3 \, b x\right ) - b^{3} x^{3} {\rm Ei}\left (-3 \, b x\right )\right )} \sinh \left (3 \, a\right ) + 2 \, {\left (b^{3} x^{3} {\rm Ei}\left (b x\right ) - b^{3} x^{3} {\rm Ei}\left (-b x\right )\right )} \sinh \left (a\right )}{192 \, x^{3}} \]
-1/192*(10*b*x*cosh(b*x + a)^5 + 50*b*x*cosh(b*x + a)*sinh(b*x + a)^4 + 2* (25*b^2*x^2 + 2)*sinh(b*x + a)^5 - 6*b*x*cosh(b*x + a)^3 - 2*(9*b^2*x^2 - 10*(25*b^2*x^2 + 2)*cosh(b*x + a)^2 + 2)*sinh(b*x + a)^3 - 4*b*x*cosh(b*x + a) + 2*(50*b*x*cosh(b*x + a)^3 - 9*b*x*cosh(b*x + a))*sinh(b*x + a)^2 - 125*(b^3*x^3*Ei(5*b*x) + b^3*x^3*Ei(-5*b*x))*cosh(5*a) + 27*(b^3*x^3*Ei(3* b*x) + b^3*x^3*Ei(-3*b*x))*cosh(3*a) + 2*(b^3*x^3*Ei(b*x) + b^3*x^3*Ei(-b* x))*cosh(a) + 2*(5*(25*b^2*x^2 + 2)*cosh(b*x + a)^4 - 2*b^2*x^2 - 3*(9*b^2 *x^2 + 2)*cosh(b*x + a)^2 - 4)*sinh(b*x + a) - 125*(b^3*x^3*Ei(5*b*x) - b^ 3*x^3*Ei(-5*b*x))*sinh(5*a) + 27*(b^3*x^3*Ei(3*b*x) - b^3*x^3*Ei(-3*b*x))* sinh(3*a) + 2*(b^3*x^3*Ei(b*x) - b^3*x^3*Ei(-b*x))*sinh(a))/x^3
\[ \int \frac {\cosh ^2(a+b x) \sinh ^3(a+b x)}{x^4} \, dx=\int \frac {\sinh ^{3}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{x^{4}}\, dx \]
Time = 0.33 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.37 \[ \int \frac {\cosh ^2(a+b x) \sinh ^3(a+b x)}{x^4} \, dx=\frac {125}{32} \, b^{3} e^{\left (-5 \, a\right )} \Gamma \left (-3, 5 \, b x\right ) - \frac {27}{32} \, b^{3} e^{\left (-3 \, a\right )} \Gamma \left (-3, 3 \, b x\right ) - \frac {1}{16} \, b^{3} e^{\left (-a\right )} \Gamma \left (-3, b x\right ) - \frac {1}{16} \, b^{3} e^{a} \Gamma \left (-3, -b x\right ) - \frac {27}{32} \, b^{3} e^{\left (3 \, a\right )} \Gamma \left (-3, -3 \, b x\right ) + \frac {125}{32} \, b^{3} e^{\left (5 \, a\right )} \Gamma \left (-3, -5 \, b x\right ) \]
125/32*b^3*e^(-5*a)*gamma(-3, 5*b*x) - 27/32*b^3*e^(-3*a)*gamma(-3, 3*b*x) - 1/16*b^3*e^(-a)*gamma(-3, b*x) - 1/16*b^3*e^a*gamma(-3, -b*x) - 27/32*b ^3*e^(3*a)*gamma(-3, -3*b*x) + 125/32*b^3*e^(5*a)*gamma(-3, -5*b*x)
Time = 0.27 (sec) , antiderivative size = 342, normalized size of antiderivative = 1.44 \[ \int \frac {\cosh ^2(a+b x) \sinh ^3(a+b x)}{x^4} \, dx=\frac {125 \, b^{3} x^{3} {\rm Ei}\left (5 \, b x\right ) e^{\left (5 \, a\right )} - 27 \, b^{3} x^{3} {\rm Ei}\left (3 \, b x\right ) e^{\left (3 \, a\right )} - 2 \, b^{3} x^{3} {\rm Ei}\left (-b x\right ) e^{\left (-a\right )} - 27 \, b^{3} x^{3} {\rm Ei}\left (-3 \, b x\right ) e^{\left (-3 \, a\right )} + 125 \, b^{3} x^{3} {\rm Ei}\left (-5 \, b x\right ) e^{\left (-5 \, a\right )} - 2 \, b^{3} x^{3} {\rm Ei}\left (b x\right ) e^{a} - 25 \, b^{2} x^{2} e^{\left (5 \, b x + 5 \, a\right )} + 9 \, b^{2} x^{2} e^{\left (3 \, b x + 3 \, a\right )} + 2 \, b^{2} x^{2} e^{\left (b x + a\right )} - 2 \, b^{2} x^{2} e^{\left (-b x - a\right )} - 9 \, b^{2} x^{2} e^{\left (-3 \, b x - 3 \, a\right )} + 25 \, b^{2} x^{2} e^{\left (-5 \, b x - 5 \, a\right )} - 5 \, b x e^{\left (5 \, b x + 5 \, a\right )} + 3 \, b x e^{\left (3 \, b x + 3 \, a\right )} + 2 \, b x e^{\left (b x + a\right )} + 2 \, b x e^{\left (-b x - a\right )} + 3 \, b x e^{\left (-3 \, b x - 3 \, a\right )} - 5 \, b x e^{\left (-5 \, b x - 5 \, a\right )} - 2 \, e^{\left (5 \, b x + 5 \, a\right )} + 2 \, e^{\left (3 \, b x + 3 \, a\right )} + 4 \, e^{\left (b x + a\right )} - 4 \, e^{\left (-b x - a\right )} - 2 \, e^{\left (-3 \, b x - 3 \, a\right )} + 2 \, e^{\left (-5 \, b x - 5 \, a\right )}}{192 \, x^{3}} \]
1/192*(125*b^3*x^3*Ei(5*b*x)*e^(5*a) - 27*b^3*x^3*Ei(3*b*x)*e^(3*a) - 2*b^ 3*x^3*Ei(-b*x)*e^(-a) - 27*b^3*x^3*Ei(-3*b*x)*e^(-3*a) + 125*b^3*x^3*Ei(-5 *b*x)*e^(-5*a) - 2*b^3*x^3*Ei(b*x)*e^a - 25*b^2*x^2*e^(5*b*x + 5*a) + 9*b^ 2*x^2*e^(3*b*x + 3*a) + 2*b^2*x^2*e^(b*x + a) - 2*b^2*x^2*e^(-b*x - a) - 9 *b^2*x^2*e^(-3*b*x - 3*a) + 25*b^2*x^2*e^(-5*b*x - 5*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) - 2*e^(5*b*x + 5*a) + 2*e^( 3*b*x + 3*a) + 4*e^(b*x + a) - 4*e^(-b*x - a) - 2*e^(-3*b*x - 3*a) + 2*e^( -5*b*x - 5*a))/x^3
Timed out. \[ \int \frac {\cosh ^2(a+b x) \sinh ^3(a+b x)}{x^4} \, dx=\int \frac {{\mathrm {cosh}\left (a+b\,x\right )}^2\,{\mathrm {sinh}\left (a+b\,x\right )}^3}{x^4} \,d x \]