Integrand size = 20, antiderivative size = 124 \[ \int \frac {\cosh ^3(a+b x) \sinh ^2(a+b x)}{x^2} \, dx=\frac {\cosh (a+b x)}{8 x}-\frac {\cosh (3 a+3 b x)}{16 x}-\frac {\cosh (5 a+5 b x)}{16 x}-\frac {1}{8} b \text {Chi}(b x) \sinh (a)+\frac {3}{16} b \text {Chi}(3 b x) \sinh (3 a)+\frac {5}{16} b \text {Chi}(5 b x) \sinh (5 a)-\frac {1}{8} b \cosh (a) \text {Shi}(b x)+\frac {3}{16} b \cosh (3 a) \text {Shi}(3 b x)+\frac {5}{16} b \cosh (5 a) \text {Shi}(5 b x) \]
1/8*cosh(b*x+a)/x-1/16*cosh(3*b*x+3*a)/x-1/16*cosh(5*b*x+5*a)/x-1/8*b*cosh (a)*Shi(b*x)+3/16*b*cosh(3*a)*Shi(3*b*x)+5/16*b*cosh(5*a)*Shi(5*b*x)-1/8*b *Chi(b*x)*sinh(a)+3/16*b*Chi(3*b*x)*sinh(3*a)+5/16*b*Chi(5*b*x)*sinh(5*a)
Time = 0.21 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.84 \[ \int \frac {\cosh ^3(a+b x) \sinh ^2(a+b x)}{x^2} \, dx=-\frac {-2 \cosh (a+b x)+\cosh (3 (a+b x))+\cosh (5 (a+b x))+2 b x \text {Chi}(b x) \sinh (a)-3 b x \text {Chi}(3 b x) \sinh (3 a)-5 b x \text {Chi}(5 b x) \sinh (5 a)+2 b x \cosh (a) \text {Shi}(b x)-3 b x \cosh (3 a) \text {Shi}(3 b x)-5 b x \cosh (5 a) \text {Shi}(5 b x)}{16 x} \]
-1/16*(-2*Cosh[a + b*x] + Cosh[3*(a + b*x)] + Cosh[5*(a + b*x)] + 2*b*x*Co shIntegral[b*x]*Sinh[a] - 3*b*x*CoshIntegral[3*b*x]*Sinh[3*a] - 5*b*x*Cosh Integral[5*b*x]*Sinh[5*a] + 2*b*x*Cosh[a]*SinhIntegral[b*x] - 3*b*x*Cosh[3 *a]*SinhIntegral[3*b*x] - 5*b*x*Cosh[5*a]*SinhIntegral[5*b*x])/x
Time = 0.47 (sec) , antiderivative size = 124, 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.
\(\displaystyle \int \frac {\sinh ^2(a+b x) \cosh ^3(a+b x)}{x^2} \, dx\) |
\(\Big \downarrow \) 5971 |
\(\displaystyle \int \left (-\frac {\cosh (a+b x)}{8 x^2}+\frac {\cosh (3 a+3 b x)}{16 x^2}+\frac {\cosh (5 a+5 b x)}{16 x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {1}{8} b \sinh (a) \text {Chi}(b x)+\frac {3}{16} b \sinh (3 a) \text {Chi}(3 b x)+\frac {5}{16} b \sinh (5 a) \text {Chi}(5 b x)-\frac {1}{8} b \cosh (a) \text {Shi}(b x)+\frac {3}{16} b \cosh (3 a) \text {Shi}(3 b x)+\frac {5}{16} b \cosh (5 a) \text {Shi}(5 b x)+\frac {\cosh (a+b x)}{8 x}-\frac {\cosh (3 a+3 b x)}{16 x}-\frac {\cosh (5 a+5 b x)}{16 x}\) |
Cosh[a + b*x]/(8*x) - Cosh[3*a + 3*b*x]/(16*x) - Cosh[5*a + 5*b*x]/(16*x) - (b*CoshIntegral[b*x]*Sinh[a])/8 + (3*b*CoshIntegral[3*b*x]*Sinh[3*a])/16 + (5*b*CoshIntegral[5*b*x]*Sinh[5*a])/16 - (b*Cosh[a]*SinhIntegral[b*x])/ 8 + (3*b*Cosh[3*a]*SinhIntegral[3*b*x])/16 + (5*b*Cosh[5*a]*SinhIntegral[5 *b*x])/16
3.4.4.3.1 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 = 9.30 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.22
method | result | size |
risch | \(\frac {3 \,{\mathrm e}^{-3 a} \operatorname {Ei}_{1}\left (3 b x \right ) b x -2 \,{\mathrm e}^{-a} \operatorname {Ei}_{1}\left (b x \right ) b x +2 \,{\mathrm e}^{a} \operatorname {Ei}_{1}\left (-b x \right ) b x -3 \,{\mathrm e}^{3 a} \operatorname {Ei}_{1}\left (-3 b x \right ) b x +5 \,{\mathrm e}^{-5 a} \operatorname {Ei}_{1}\left (5 b x \right ) b x -5 \,{\mathrm e}^{5 a} \operatorname {Ei}_{1}\left (-5 b x \right ) b x +2 \,{\mathrm e}^{b x +a}-{\mathrm e}^{-3 b x -3 a}+2 \,{\mathrm e}^{-b x -a}-{\mathrm e}^{3 b x +3 a}-{\mathrm e}^{-5 b x -5 a}-{\mathrm e}^{5 b x +5 a}}{32 x}\) | \(151\) |
1/32*(3*exp(-3*a)*Ei(1,3*b*x)*b*x-2*exp(-a)*Ei(1,b*x)*b*x+2*exp(a)*Ei(1,-b *x)*b*x-3*exp(3*a)*Ei(1,-3*b*x)*b*x+5*exp(-5*a)*Ei(1,5*b*x)*b*x-5*exp(5*a) *Ei(1,-5*b*x)*b*x+2*exp(b*x+a)-exp(-3*b*x-3*a)+2*exp(-b*x-a)-exp(3*b*x+3*a )-exp(-5*b*x-5*a)-exp(5*b*x+5*a))/x
Leaf count of result is larger than twice the leaf count of optimal. 214 vs. \(2 (106) = 212\).
Time = 0.29 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.73 \[ \int \frac {\cosh ^3(a+b x) \sinh ^2(a+b x)}{x^2} \, dx=-\frac {2 \, \cosh \left (b x + a\right )^{5} + 10 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} + 2 \, \cosh \left (b x + a\right )^{3} + 2 \, {\left (10 \, \cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 5 \, {\left (b x {\rm Ei}\left (5 \, b x\right ) - b x {\rm Ei}\left (-5 \, b x\right )\right )} \cosh \left (5 \, a\right ) - 3 \, {\left (b x {\rm Ei}\left (3 \, b x\right ) - b x {\rm Ei}\left (-3 \, b x\right )\right )} \cosh \left (3 \, a\right ) + 2 \, {\left (b x {\rm Ei}\left (b x\right ) - b x {\rm Ei}\left (-b x\right )\right )} \cosh \left (a\right ) - 5 \, {\left (b x {\rm Ei}\left (5 \, b x\right ) + b x {\rm Ei}\left (-5 \, b x\right )\right )} \sinh \left (5 \, a\right ) - 3 \, {\left (b x {\rm Ei}\left (3 \, b x\right ) + b x {\rm Ei}\left (-3 \, b x\right )\right )} \sinh \left (3 \, a\right ) + 2 \, {\left (b x {\rm Ei}\left (b x\right ) + b x {\rm Ei}\left (-b x\right )\right )} \sinh \left (a\right ) - 4 \, \cosh \left (b x + a\right )}{32 \, x} \]
-1/32*(2*cosh(b*x + a)^5 + 10*cosh(b*x + a)*sinh(b*x + a)^4 + 2*cosh(b*x + a)^3 + 2*(10*cosh(b*x + a)^3 + 3*cosh(b*x + a))*sinh(b*x + a)^2 - 5*(b*x* Ei(5*b*x) - b*x*Ei(-5*b*x))*cosh(5*a) - 3*(b*x*Ei(3*b*x) - b*x*Ei(-3*b*x)) *cosh(3*a) + 2*(b*x*Ei(b*x) - b*x*Ei(-b*x))*cosh(a) - 5*(b*x*Ei(5*b*x) + b *x*Ei(-5*b*x))*sinh(5*a) - 3*(b*x*Ei(3*b*x) + b*x*Ei(-3*b*x))*sinh(3*a) + 2*(b*x*Ei(b*x) + b*x*Ei(-b*x))*sinh(a) - 4*cosh(b*x + a))/x
\[ \int \frac {\cosh ^3(a+b x) \sinh ^2(a+b x)}{x^2} \, dx=\int \frac {\sinh ^{2}{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{x^{2}}\, dx \]
Time = 0.32 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.61 \[ \int \frac {\cosh ^3(a+b x) \sinh ^2(a+b x)}{x^2} \, dx=-\frac {5}{32} \, b e^{\left (-5 \, a\right )} \Gamma \left (-1, 5 \, b x\right ) - \frac {3}{32} \, b e^{\left (-3 \, a\right )} \Gamma \left (-1, 3 \, b x\right ) + \frac {1}{16} \, b e^{\left (-a\right )} \Gamma \left (-1, b x\right ) - \frac {1}{16} \, b e^{a} \Gamma \left (-1, -b x\right ) + \frac {3}{32} \, b e^{\left (3 \, a\right )} \Gamma \left (-1, -3 \, b x\right ) + \frac {5}{32} \, b e^{\left (5 \, a\right )} \Gamma \left (-1, -5 \, b x\right ) \]
-5/32*b*e^(-5*a)*gamma(-1, 5*b*x) - 3/32*b*e^(-3*a)*gamma(-1, 3*b*x) + 1/1 6*b*e^(-a)*gamma(-1, b*x) - 1/16*b*e^a*gamma(-1, -b*x) + 3/32*b*e^(3*a)*ga mma(-1, -3*b*x) + 5/32*b*e^(5*a)*gamma(-1, -5*b*x)
Time = 0.27 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.16 \[ \int \frac {\cosh ^3(a+b x) \sinh ^2(a+b x)}{x^2} \, dx=\frac {5 \, b x {\rm Ei}\left (5 \, b x\right ) e^{\left (5 \, a\right )} + 3 \, b x {\rm Ei}\left (3 \, b x\right ) e^{\left (3 \, a\right )} + 2 \, b x {\rm Ei}\left (-b x\right ) e^{\left (-a\right )} - 3 \, b x {\rm Ei}\left (-3 \, b x\right ) e^{\left (-3 \, a\right )} - 5 \, b x {\rm Ei}\left (-5 \, b x\right ) e^{\left (-5 \, a\right )} - 2 \, b x {\rm Ei}\left (b x\right ) e^{a} - e^{\left (5 \, b x + 5 \, a\right )} - e^{\left (3 \, b x + 3 \, a\right )} + 2 \, e^{\left (b x + a\right )} + 2 \, e^{\left (-b x - a\right )} - e^{\left (-3 \, b x - 3 \, a\right )} - e^{\left (-5 \, b x - 5 \, a\right )}}{32 \, x} \]
1/32*(5*b*x*Ei(5*b*x)*e^(5*a) + 3*b*x*Ei(3*b*x)*e^(3*a) + 2*b*x*Ei(-b*x)*e ^(-a) - 3*b*x*Ei(-3*b*x)*e^(-3*a) - 5*b*x*Ei(-5*b*x)*e^(-5*a) - 2*b*x*Ei(b *x)*e^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
Timed out. \[ \int \frac {\cosh ^3(a+b x) \sinh ^2(a+b x)}{x^2} \, dx=\int \frac {{\mathrm {cosh}\left (a+b\,x\right )}^3\,{\mathrm {sinh}\left (a+b\,x\right )}^2}{x^2} \,d x \]