Integrand size = 20, antiderivative size = 73 \[ \int \frac {\cosh ^2(a+b x) \sinh ^3(a+b x)}{x} \, dx=-\frac {1}{8} \text {Chi}(b x) \sinh (a)-\frac {1}{16} \text {Chi}(3 b x) \sinh (3 a)+\frac {1}{16} \text {Chi}(5 b x) \sinh (5 a)-\frac {1}{8} \cosh (a) \text {Shi}(b x)-\frac {1}{16} \cosh (3 a) \text {Shi}(3 b x)+\frac {1}{16} \cosh (5 a) \text {Shi}(5 b x) \] Output:
-1/8*Chi(b*x)*sinh(a)-1/16*Chi(3*b*x)*sinh(3*a)+1/16*Chi(5*b*x)*sinh(5*a)- 1/8*cosh(a)*Shi(b*x)-1/16*cosh(3*a)*Shi(3*b*x)+1/16*cosh(5*a)*Shi(5*b*x)
Time = 0.06 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.86 \[ \int \frac {\cosh ^2(a+b x) \sinh ^3(a+b x)}{x} \, dx=\frac {1}{16} (-2 \text {Chi}(b x) \sinh (a)-\text {Chi}(3 b x) \sinh (3 a)+\text {Chi}(5 b x) \sinh (5 a)-2 \cosh (a) \text {Shi}(b x)-\cosh (3 a) \text {Shi}(3 b x)+\cosh (5 a) \text {Shi}(5 b x)) \] Input:
Integrate[(Cosh[a + b*x]^2*Sinh[a + b*x]^3)/x,x]
Output:
(-2*CoshIntegral[b*x]*Sinh[a] - CoshIntegral[3*b*x]*Sinh[3*a] + CoshIntegr al[5*b*x]*Sinh[5*a] - 2*Cosh[a]*SinhIntegral[b*x] - Cosh[3*a]*SinhIntegral [3*b*x] + Cosh[5*a]*SinhIntegral[5*b*x])/16
Time = 0.39 (sec) , antiderivative size = 73, 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.
\(\displaystyle \int \frac {\sinh ^3(a+b x) \cosh ^2(a+b x)}{x} \, dx\) |
\(\Big \downarrow \) 5971 |
\(\displaystyle \int \left (-\frac {\sinh (a+b x)}{8 x}-\frac {\sinh (3 a+3 b x)}{16 x}+\frac {\sinh (5 a+5 b x)}{16 x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {1}{8} \sinh (a) \text {Chi}(b x)-\frac {1}{16} \sinh (3 a) \text {Chi}(3 b x)+\frac {1}{16} \sinh (5 a) \text {Chi}(5 b x)-\frac {1}{8} \cosh (a) \text {Shi}(b x)-\frac {1}{16} \cosh (3 a) \text {Shi}(3 b x)+\frac {1}{16} \cosh (5 a) \text {Shi}(5 b x)\) |
Input:
Int[(Cosh[a + b*x]^2*Sinh[a + b*x]^3)/x,x]
Output:
-1/8*(CoshIntegral[b*x]*Sinh[a]) - (CoshIntegral[3*b*x]*Sinh[3*a])/16 + (C oshIntegral[5*b*x]*Sinh[5*a])/16 - (Cosh[a]*SinhIntegral[b*x])/8 - (Cosh[3 *a]*SinhIntegral[3*b*x])/16 + (Cosh[5*a]*SinhIntegral[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 = 14.86 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.97
method | result | size |
risch | \(\frac {{\mathrm e}^{-5 a} \operatorname {expIntegral}_{1}\left (5 b x \right )}{32}-\frac {{\mathrm e}^{-3 a} \operatorname {expIntegral}_{1}\left (3 b x \right )}{32}-\frac {{\mathrm e}^{-a} \operatorname {expIntegral}_{1}\left (b x \right )}{16}+\frac {{\mathrm e}^{a} \operatorname {expIntegral}_{1}\left (-b x \right )}{16}+\frac {{\mathrm e}^{3 a} \operatorname {expIntegral}_{1}\left (-3 b x \right )}{32}-\frac {{\mathrm e}^{5 a} \operatorname {expIntegral}_{1}\left (-5 b x \right )}{32}\) | \(71\) |
Input:
int(cosh(b*x+a)^2*sinh(b*x+a)^3/x,x,method=_RETURNVERBOSE)
Output:
1/32*exp(-5*a)*Ei(1,5*b*x)-1/32*exp(-3*a)*Ei(1,3*b*x)-1/16*exp(-a)*Ei(1,b* x)+1/16*exp(a)*Ei(1,-b*x)+1/32*exp(3*a)*Ei(1,-3*b*x)-1/32*exp(5*a)*Ei(1,-5 *b*x)
Time = 0.09 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.41 \[ \int \frac {\cosh ^2(a+b x) \sinh ^3(a+b x)}{x} \, dx=\frac {1}{32} \, {\left ({\rm Ei}\left (5 \, b x\right ) - {\rm Ei}\left (-5 \, b x\right )\right )} \cosh \left (5 \, a\right ) - \frac {1}{32} \, {\left ({\rm Ei}\left (3 \, b x\right ) - {\rm Ei}\left (-3 \, b x\right )\right )} \cosh \left (3 \, a\right ) - \frac {1}{16} \, {\left ({\rm Ei}\left (b x\right ) - {\rm Ei}\left (-b x\right )\right )} \cosh \left (a\right ) + \frac {1}{32} \, {\left ({\rm Ei}\left (5 \, b x\right ) + {\rm Ei}\left (-5 \, b x\right )\right )} \sinh \left (5 \, a\right ) - \frac {1}{32} \, {\left ({\rm Ei}\left (3 \, b x\right ) + {\rm Ei}\left (-3 \, b x\right )\right )} \sinh \left (3 \, a\right ) - \frac {1}{16} \, {\left ({\rm Ei}\left (b x\right ) + {\rm Ei}\left (-b x\right )\right )} \sinh \left (a\right ) \] Input:
integrate(cosh(b*x+a)^2*sinh(b*x+a)^3/x,x, algorithm="fricas")
Output:
1/32*(Ei(5*b*x) - Ei(-5*b*x))*cosh(5*a) - 1/32*(Ei(3*b*x) - Ei(-3*b*x))*co sh(3*a) - 1/16*(Ei(b*x) - Ei(-b*x))*cosh(a) + 1/32*(Ei(5*b*x) + Ei(-5*b*x) )*sinh(5*a) - 1/32*(Ei(3*b*x) + Ei(-3*b*x))*sinh(3*a) - 1/16*(Ei(b*x) + Ei (-b*x))*sinh(a)
\[ \int \frac {\cosh ^2(a+b x) \sinh ^3(a+b x)}{x} \, dx=\int \frac {\sinh ^{3}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{x}\, dx \] Input:
integrate(cosh(b*x+a)**2*sinh(b*x+a)**3/x,x)
Output:
Integral(sinh(a + b*x)**3*cosh(a + b*x)**2/x, x)
Time = 0.18 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.88 \[ \int \frac {\cosh ^2(a+b x) \sinh ^3(a+b x)}{x} \, dx=\frac {1}{32} \, {\rm Ei}\left (5 \, b x\right ) e^{\left (5 \, a\right )} - \frac {1}{32} \, {\rm Ei}\left (3 \, b x\right ) e^{\left (3 \, a\right )} + \frac {1}{16} \, {\rm Ei}\left (-b x\right ) e^{\left (-a\right )} + \frac {1}{32} \, {\rm Ei}\left (-3 \, b x\right ) e^{\left (-3 \, a\right )} - \frac {1}{32} \, {\rm Ei}\left (-5 \, b x\right ) e^{\left (-5 \, a\right )} - \frac {1}{16} \, {\rm Ei}\left (b x\right ) e^{a} \] Input:
integrate(cosh(b*x+a)^2*sinh(b*x+a)^3/x,x, algorithm="maxima")
Output:
1/32*Ei(5*b*x)*e^(5*a) - 1/32*Ei(3*b*x)*e^(3*a) + 1/16*Ei(-b*x)*e^(-a) + 1 /32*Ei(-3*b*x)*e^(-3*a) - 1/32*Ei(-5*b*x)*e^(-5*a) - 1/16*Ei(b*x)*e^a
Time = 0.12 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.90 \[ \int \frac {\cosh ^2(a+b x) \sinh ^3(a+b x)}{x} \, dx=\frac {1}{32} \, {\left ({\rm Ei}\left (5 \, b x\right ) e^{\left (10 \, a\right )} - {\rm Ei}\left (3 \, b x\right ) e^{\left (8 \, a\right )} - 2 \, {\rm Ei}\left (b x\right ) e^{\left (6 \, a\right )} + 2 \, {\rm Ei}\left (-b x\right ) e^{\left (4 \, a\right )} + {\rm Ei}\left (-3 \, b x\right ) e^{\left (2 \, a\right )} - {\rm Ei}\left (-5 \, b x\right )\right )} e^{\left (-5 \, a\right )} \] Input:
integrate(cosh(b*x+a)^2*sinh(b*x+a)^3/x,x, algorithm="giac")
Output:
1/32*(Ei(5*b*x)*e^(10*a) - Ei(3*b*x)*e^(8*a) - 2*Ei(b*x)*e^(6*a) + 2*Ei(-b *x)*e^(4*a) + Ei(-3*b*x)*e^(2*a) - Ei(-5*b*x))*e^(-5*a)
Timed out. \[ \int \frac {\cosh ^2(a+b x) \sinh ^3(a+b x)}{x} \, dx=\int \frac {{\mathrm {cosh}\left (a+b\,x\right )}^2\,{\mathrm {sinh}\left (a+b\,x\right )}^3}{x} \,d x \] Input:
int((cosh(a + b*x)^2*sinh(a + b*x)^3)/x,x)
Output:
int((cosh(a + b*x)^2*sinh(a + b*x)^3)/x, x)
\[ \int \frac {\cosh ^2(a+b x) \sinh ^3(a+b x)}{x} \, dx=\int \frac {\cosh \left (b x +a \right )^{2} \sinh \left (b x +a \right )^{3}}{x}d x \] Input:
int(cosh(b*x+a)^2*sinh(b*x+a)^3/x,x)
Output:
int((cosh(a + b*x)**2*sinh(a + b*x)**3)/x,x)