Integrand size = 20, antiderivative size = 124 \[ \int \frac {\cosh ^2(a+b x) \sinh ^3(a+b x)}{x^2} \, dx=-\frac {1}{8} b \cosh (a) \text {Chi}(b x)-\frac {3}{16} b \cosh (3 a) \text {Chi}(3 b x)+\frac {5}{16} b \cosh (5 a) \text {Chi}(5 b x)+\frac {\sinh (a+b x)}{8 x}+\frac {\sinh (3 a+3 b x)}{16 x}-\frac {\sinh (5 a+5 b x)}{16 x}-\frac {1}{8} b \sinh (a) \text {Shi}(b x)-\frac {3}{16} b \sinh (3 a) \text {Shi}(3 b x)+\frac {5}{16} b \sinh (5 a) \text {Shi}(5 b x) \] Output:
-1/8*b*cosh(a)*Chi(b*x)-3/16*b*cosh(3*a)*Chi(3*b*x)+5/16*b*cosh(5*a)*Chi(5 *b*x)+1/8*sinh(b*x+a)/x+1/16*sinh(3*b*x+3*a)/x-1/16*sinh(5*b*x+5*a)/x-1/8* b*sinh(a)*Shi(b*x)-3/16*b*sinh(3*a)*Shi(3*b*x)+5/16*b*sinh(5*a)*Shi(5*b*x)
Time = 0.16 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.85 \[ \int \frac {\cosh ^2(a+b x) \sinh ^3(a+b x)}{x^2} \, dx=\frac {-2 b x \cosh (a) \text {Chi}(b x)-3 b x \cosh (3 a) \text {Chi}(3 b x)+5 b x \cosh (5 a) \text {Chi}(5 b x)+2 \sinh (a+b x)+\sinh (3 (a+b x))-\sinh (5 (a+b x))-2 b x \sinh (a) \text {Shi}(b x)-3 b x \sinh (3 a) \text {Shi}(3 b x)+5 b x \sinh (5 a) \text {Shi}(5 b x)}{16 x} \] Input:
Integrate[(Cosh[a + b*x]^2*Sinh[a + b*x]^3)/x^2,x]
Output:
(-2*b*x*Cosh[a]*CoshIntegral[b*x] - 3*b*x*Cosh[3*a]*CoshIntegral[3*b*x] + 5*b*x*Cosh[5*a]*CoshIntegral[5*b*x] + 2*Sinh[a + b*x] + Sinh[3*(a + b*x)] - Sinh[5*(a + b*x)] - 2*b*x*Sinh[a]*SinhIntegral[b*x] - 3*b*x*Sinh[3*a]*Si nhIntegral[3*b*x] + 5*b*x*Sinh[5*a]*SinhIntegral[5*b*x])/(16*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.
\(\displaystyle \int \frac {\sinh ^3(a+b x) \cosh ^2(a+b x)}{x^2} \, dx\) |
\(\Big \downarrow \) 5971 |
\(\displaystyle \int \left (-\frac {\sinh (a+b x)}{8 x^2}-\frac {\sinh (3 a+3 b x)}{16 x^2}+\frac {\sinh (5 a+5 b x)}{16 x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {1}{8} b \cosh (a) \text {Chi}(b x)-\frac {3}{16} b \cosh (3 a) \text {Chi}(3 b x)+\frac {5}{16} b \cosh (5 a) \text {Chi}(5 b x)-\frac {1}{8} b \sinh (a) \text {Shi}(b x)-\frac {3}{16} b \sinh (3 a) \text {Shi}(3 b x)+\frac {5}{16} b \sinh (5 a) \text {Shi}(5 b x)+\frac {\sinh (a+b x)}{8 x}+\frac {\sinh (3 a+3 b x)}{16 x}-\frac {\sinh (5 a+5 b x)}{16 x}\) |
Input:
Int[(Cosh[a + b*x]^2*Sinh[a + b*x]^3)/x^2,x]
Output:
-1/8*(b*Cosh[a]*CoshIntegral[b*x]) - (3*b*Cosh[3*a]*CoshIntegral[3*b*x])/1 6 + (5*b*Cosh[5*a]*CoshIntegral[5*b*x])/16 + Sinh[a + b*x]/(8*x) + Sinh[3* a + 3*b*x]/(16*x) - Sinh[5*a + 5*b*x]/(16*x) - (b*Sinh[a]*SinhIntegral[b*x ])/8 - (3*b*Sinh[3*a]*SinhIntegral[3*b*x])/16 + (5*b*Sinh[5*a]*SinhIntegra l[5*b*x])/16
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 = 16.40 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.19
method | result | size |
risch | \(\frac {-5 \,{\mathrm e}^{-5 a} \operatorname {expIntegral}_{1}\left (5 b x \right ) b x +3 \,{\mathrm e}^{-3 a} \operatorname {expIntegral}_{1}\left (3 b x \right ) b x +2 \,{\mathrm e}^{-a} \operatorname {expIntegral}_{1}\left (b x \right ) b x +2 \,{\mathrm e}^{a} \operatorname {expIntegral}_{1}\left (-b x \right ) b x +3 \,{\mathrm e}^{3 a} \operatorname {expIntegral}_{1}\left (-3 b x \right ) b x -5 \,{\mathrm e}^{5 a} \operatorname {expIntegral}_{1}\left (-5 b x \right ) b x +2 \,{\mathrm e}^{b x +a}+{\mathrm e}^{-5 b x -5 a}-{\mathrm e}^{-3 b x -3 a}-2 \,{\mathrm e}^{-b x -a}+{\mathrm e}^{3 b x +3 a}-{\mathrm e}^{5 b x +5 a}}{32 x}\) | \(147\) |
Input:
int(cosh(b*x+a)^2*sinh(b*x+a)^3/x^2,x,method=_RETURNVERBOSE)
Output:
1/32*(-5*exp(-5*a)*Ei(1,5*b*x)*b*x+3*exp(-3*a)*Ei(1,3*b*x)*b*x+2*exp(-a)*E i(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+2*exp(b*x+a)+exp(-5*b*x-5*a)-exp(-3*b*x-3*a)-2*exp(-b*x -a)+exp(3*b*x+3*a)-exp(5*b*x+5*a))/x
Time = 0.09 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.64 \[ \int \frac {\cosh ^2(a+b x) \sinh ^3(a+b x)}{x^2} \, dx=-\frac {2 \, \sinh \left (b x + a\right )^{5} + 2 \, {\left (10 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{3} - 5 \, {\left (b x {\rm Ei}\left (5 \, b x\right ) + b x {\rm Ei}\left (-5 \, b x\right )\right )} \cosh \left (5 \, a\right ) + 3 \, {\left (b x {\rm Ei}\left (3 \, b x\right ) + b x {\rm Ei}\left (-3 \, b x\right )\right )} \cosh \left (3 \, a\right ) + 2 \, {\left (b x {\rm Ei}\left (b x\right ) + b x {\rm Ei}\left (-b x\right )\right )} \cosh \left (a\right ) + 2 \, {\left (5 \, \cosh \left (b x + a\right )^{4} - 3 \, \cosh \left (b x + a\right )^{2} - 2\right )} \sinh \left (b x + a\right ) - 5 \, {\left (b x {\rm Ei}\left (5 \, b x\right ) - b x {\rm Ei}\left (-5 \, b x\right )\right )} \sinh \left (5 \, a\right ) + 3 \, {\left (b x {\rm Ei}\left (3 \, b x\right ) - b x {\rm Ei}\left (-3 \, b x\right )\right )} \sinh \left (3 \, a\right ) + 2 \, {\left (b x {\rm Ei}\left (b x\right ) - b x {\rm Ei}\left (-b x\right )\right )} \sinh \left (a\right )}{32 \, x} \] Input:
integrate(cosh(b*x+a)^2*sinh(b*x+a)^3/x^2,x, algorithm="fricas")
Output:
-1/32*(2*sinh(b*x + a)^5 + 2*(10*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^3 - 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) + 2*(5*cosh(b*x + a)^4 - 3*cosh(b*x + a)^2 - 2)*sinh(b*x + 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))/x
\[ \int \frac {\cosh ^2(a+b x) \sinh ^3(a+b x)}{x^2} \, dx=\int \frac {\sinh ^{3}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{x^{2}}\, dx \] Input:
integrate(cosh(b*x+a)**2*sinh(b*x+a)**3/x**2,x)
Output:
Integral(sinh(a + b*x)**3*cosh(a + b*x)**2/x**2, x)
Time = 0.16 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.61 \[ \int \frac {\cosh ^2(a+b x) \sinh ^3(a+b x)}{x^2} \, dx=\frac {5}{32} \, b e^{\left (-5 \, a\right )} \Gamma \left (-1, 5 \, b x\right ) - \frac {3}{32} \, b e^{\left (-3 \, a\right )} \Gamma \left (-1, 3 \, b x\right ) - \frac {1}{16} \, b e^{\left (-a\right )} \Gamma \left (-1, b x\right ) - \frac {1}{16} \, b e^{a} \Gamma \left (-1, -b x\right ) - \frac {3}{32} \, b e^{\left (3 \, a\right )} \Gamma \left (-1, -3 \, b x\right ) + \frac {5}{32} \, b e^{\left (5 \, a\right )} \Gamma \left (-1, -5 \, b x\right ) \] Input:
integrate(cosh(b*x+a)^2*sinh(b*x+a)^3/x^2,x, algorithm="maxima")
Output:
5/32*b*e^(-5*a)*gamma(-1, 5*b*x) - 3/32*b*e^(-3*a)*gamma(-1, 3*b*x) - 1/16 *b*e^(-a)*gamma(-1, b*x) - 1/16*b*e^a*gamma(-1, -b*x) - 3/32*b*e^(3*a)*gam ma(-1, -3*b*x) + 5/32*b*e^(5*a)*gamma(-1, -5*b*x)
Time = 0.12 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.13 \[ \int \frac {\cosh ^2(a+b x) \sinh ^3(a+b x)}{x^2} \, dx=\frac {5 \, b x {\rm Ei}\left (5 \, b x\right ) e^{\left (5 \, a\right )} - 3 \, b x {\rm Ei}\left (3 \, b x\right ) e^{\left (3 \, a\right )} - 2 \, b x {\rm Ei}\left (-b x\right ) e^{\left (-a\right )} - 3 \, b x {\rm Ei}\left (-3 \, b x\right ) e^{\left (-3 \, a\right )} + 5 \, b x {\rm Ei}\left (-5 \, b x\right ) e^{\left (-5 \, a\right )} - 2 \, b x {\rm Ei}\left (b x\right ) e^{a} - e^{\left (5 \, b x + 5 \, a\right )} + e^{\left (3 \, b x + 3 \, a\right )} + 2 \, e^{\left (b x + a\right )} - 2 \, e^{\left (-b x - a\right )} - e^{\left (-3 \, b x - 3 \, a\right )} + e^{\left (-5 \, b x - 5 \, a\right )}}{32 \, x} \] Input:
integrate(cosh(b*x+a)^2*sinh(b*x+a)^3/x^2,x, algorithm="giac")
Output:
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. \[ \int \frac {\cosh ^2(a+b x) \sinh ^3(a+b x)}{x^2} \, dx=\int \frac {{\mathrm {cosh}\left (a+b\,x\right )}^2\,{\mathrm {sinh}\left (a+b\,x\right )}^3}{x^2} \,d x \] Input:
int((cosh(a + b*x)^2*sinh(a + b*x)^3)/x^2,x)
Output:
int((cosh(a + b*x)^2*sinh(a + b*x)^3)/x^2, x)
Time = 0.23 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.50 \[ \int \frac {\cosh ^2(a+b x) \sinh ^3(a+b x)}{x^2} \, dx=\frac {-2 e^{5 b x +4 a} \mathit {ei} \left (-b x \right ) b x +5 e^{5 b x} \mathit {ei} \left (-5 b x \right ) b x -3 e^{5 b x +2 a} \mathit {ei} \left (-3 b x \right ) b x -2 e^{5 b x +6 a} \mathit {ei} \left (b x \right ) b x +5 e^{5 b x +10 a} \mathit {ei} \left (5 b x \right ) b x -3 e^{5 b x +8 a} \mathit {ei} \left (3 b x \right ) b x -e^{10 b x +10 a}+e^{8 b x +8 a}+2 e^{6 b x +6 a}-2 e^{4 b x +4 a}-e^{2 b x +2 a}+1}{32 e^{5 b x +5 a} x} \] Input:
int(cosh(b*x+a)^2*sinh(b*x+a)^3/x^2,x)
Output:
( - 2*e**(4*a + 5*b*x)*ei( - b*x)*b*x + 5*e**(5*b*x)*ei( - 5*b*x)*b*x - 3* e**(2*a + 5*b*x)*ei( - 3*b*x)*b*x - 2*e**(6*a + 5*b*x)*ei(b*x)*b*x + 5*e** (10*a + 5*b*x)*ei(5*b*x)*b*x - 3*e**(8*a + 5*b*x)*ei(3*b*x)*b*x - e**(10*a + 10*b*x) + e**(8*a + 8*b*x) + 2*e**(6*a + 6*b*x) - 2*e**(4*a + 4*b*x) - e**(2*a + 2*b*x) + 1)/(32*e**(5*a + 5*b*x)*x)