Integrand size = 18, antiderivative size = 89 \[ \int \frac {\cosh (a+b x) \sinh ^3(a+b x)}{x^2} \, dx=-\frac {1}{2} b \cosh (2 a) \text {Chi}(2 b x)+\frac {1}{2} b \cosh (4 a) \text {Chi}(4 b x)+\frac {\sinh (2 a+2 b x)}{4 x}-\frac {\sinh (4 a+4 b x)}{8 x}-\frac {1}{2} b \sinh (2 a) \text {Shi}(2 b x)+\frac {1}{2} b \sinh (4 a) \text {Shi}(4 b x) \] Output:
-1/2*b*cosh(2*a)*Chi(2*b*x)+1/2*b*cosh(4*a)*Chi(4*b*x)+1/4*sinh(2*b*x+2*a) /x-1/8*sinh(4*b*x+4*a)/x-1/2*b*sinh(2*a)*Shi(2*b*x)+1/2*b*sinh(4*a)*Shi(4* b*x)
Time = 0.15 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.88 \[ \int \frac {\cosh (a+b x) \sinh ^3(a+b x)}{x^2} \, dx=-\frac {4 b x \cosh (2 a) \text {Chi}(2 b x)-4 b x \cosh (4 a) \text {Chi}(4 b x)-2 \sinh (2 (a+b x))+\sinh (4 (a+b x))+4 b x \sinh (2 a) \text {Shi}(2 b x)-4 b x \sinh (4 a) \text {Shi}(4 b x)}{8 x} \] Input:
Integrate[(Cosh[a + b*x]*Sinh[a + b*x]^3)/x^2,x]
Output:
-1/8*(4*b*x*Cosh[2*a]*CoshIntegral[2*b*x] - 4*b*x*Cosh[4*a]*CoshIntegral[4 *b*x] - 2*Sinh[2*(a + b*x)] + Sinh[4*(a + b*x)] + 4*b*x*Sinh[2*a]*SinhInte gral[2*b*x] - 4*b*x*Sinh[4*a]*SinhIntegral[4*b*x])/x
Time = 0.53 (sec) , antiderivative size = 89, 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.
\(\displaystyle \int \frac {\sinh ^3(a+b x) \cosh (a+b x)}{x^2} \, dx\) |
\(\Big \downarrow \) 5971 |
\(\displaystyle \int \left (\frac {\sinh (4 a+4 b x)}{8 x^2}-\frac {\sinh (2 a+2 b x)}{4 x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {1}{2} b \cosh (2 a) \text {Chi}(2 b x)+\frac {1}{2} b \cosh (4 a) \text {Chi}(4 b x)-\frac {1}{2} b \sinh (2 a) \text {Shi}(2 b x)+\frac {1}{2} b \sinh (4 a) \text {Shi}(4 b x)+\frac {\sinh (2 a+2 b x)}{4 x}-\frac {\sinh (4 a+4 b x)}{8 x}\) |
Input:
Int[(Cosh[a + b*x]*Sinh[a + b*x]^3)/x^2,x]
Output:
-1/2*(b*Cosh[2*a]*CoshIntegral[2*b*x]) + (b*Cosh[4*a]*CoshIntegral[4*b*x]) /2 + Sinh[2*a + 2*b*x]/(4*x) - Sinh[4*a + 4*b*x]/(8*x) - (b*Sinh[2*a]*Sinh Integral[2*b*x])/2 + (b*Sinh[4*a]*SinhIntegral[4*b*x])/2
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 = 8.20 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.18
method | result | size |
risch | \(\frac {4 \,{\mathrm e}^{2 a} \operatorname {expIntegral}_{1}\left (-2 b x \right ) b x -4 \,{\mathrm e}^{4 a} \operatorname {expIntegral}_{1}\left (-4 b x \right ) b x -4 \,{\mathrm e}^{-4 a} \operatorname {expIntegral}_{1}\left (4 b x \right ) b x +4 \,{\mathrm e}^{-2 a} \operatorname {expIntegral}_{1}\left (2 b x \right ) b x +2 \,{\mathrm e}^{2 b x +2 a}-{\mathrm e}^{4 b x +4 a}+{\mathrm e}^{-4 b x -4 a}-2 \,{\mathrm e}^{-2 b x -2 a}}{16 x}\) | \(105\) |
Input:
int(cosh(b*x+a)*sinh(b*x+a)^3/x^2,x,method=_RETURNVERBOSE)
Output:
1/16*(4*exp(2*a)*Ei(1,-2*b*x)*b*x-4*exp(4*a)*Ei(1,-4*b*x)*b*x-4*exp(-4*a)* Ei(1,4*b*x)*b*x+4*exp(-2*a)*Ei(1,2*b*x)*b*x+2*exp(2*b*x+2*a)-exp(4*b*x+4*a )+exp(-4*b*x-4*a)-2*exp(-2*b*x-2*a))/x
Time = 0.08 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.56 \[ \int \frac {\cosh (a+b x) \sinh ^3(a+b x)}{x^2} \, dx=-\frac {2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} - {\left (b x {\rm Ei}\left (4 \, b x\right ) + b x {\rm Ei}\left (-4 \, b x\right )\right )} \cosh \left (4 \, a\right ) + {\left (b x {\rm Ei}\left (2 \, b x\right ) + b x {\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (2 \, a\right ) + 2 \, {\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) - {\left (b x {\rm Ei}\left (4 \, b x\right ) - b x {\rm Ei}\left (-4 \, b x\right )\right )} \sinh \left (4 \, a\right ) + {\left (b x {\rm Ei}\left (2 \, b x\right ) - b x {\rm Ei}\left (-2 \, b x\right )\right )} \sinh \left (2 \, a\right )}{4 \, x} \] Input:
integrate(cosh(b*x+a)*sinh(b*x+a)^3/x^2,x, algorithm="fricas")
Output:
-1/4*(2*cosh(b*x + a)*sinh(b*x + a)^3 - (b*x*Ei(4*b*x) + b*x*Ei(-4*b*x))*c osh(4*a) + (b*x*Ei(2*b*x) + b*x*Ei(-2*b*x))*cosh(2*a) + 2*(cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) - (b*x*Ei(4*b*x) - b*x*Ei(-4*b*x))*sinh(4* a) + (b*x*Ei(2*b*x) - b*x*Ei(-2*b*x))*sinh(2*a))/x
\[ \int \frac {\cosh (a+b x) \sinh ^3(a+b x)}{x^2} \, dx=\int \frac {\sinh ^{3}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{x^{2}}\, dx \] Input:
integrate(cosh(b*x+a)*sinh(b*x+a)**3/x**2,x)
Output:
Integral(sinh(a + b*x)**3*cosh(a + b*x)/x**2, x)
Time = 0.15 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.60 \[ \int \frac {\cosh (a+b x) \sinh ^3(a+b x)}{x^2} \, dx=\frac {1}{4} \, b e^{\left (-4 \, a\right )} \Gamma \left (-1, 4 \, b x\right ) - \frac {1}{4} \, b e^{\left (-2 \, a\right )} \Gamma \left (-1, 2 \, b x\right ) - \frac {1}{4} \, b e^{\left (2 \, a\right )} \Gamma \left (-1, -2 \, b x\right ) + \frac {1}{4} \, b e^{\left (4 \, a\right )} \Gamma \left (-1, -4 \, b x\right ) \] Input:
integrate(cosh(b*x+a)*sinh(b*x+a)^3/x^2,x, algorithm="maxima")
Output:
1/4*b*e^(-4*a)*gamma(-1, 4*b*x) - 1/4*b*e^(-2*a)*gamma(-1, 2*b*x) - 1/4*b* e^(2*a)*gamma(-1, -2*b*x) + 1/4*b*e^(4*a)*gamma(-1, -4*b*x)
Time = 0.12 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.12 \[ \int \frac {\cosh (a+b x) \sinh ^3(a+b x)}{x^2} \, dx=\frac {4 \, b x {\rm Ei}\left (4 \, b x\right ) e^{\left (4 \, a\right )} - 4 \, b x {\rm Ei}\left (2 \, b x\right ) e^{\left (2 \, a\right )} - 4 \, b x {\rm Ei}\left (-2 \, b x\right ) e^{\left (-2 \, a\right )} + 4 \, b x {\rm Ei}\left (-4 \, b x\right ) e^{\left (-4 \, a\right )} - e^{\left (4 \, b x + 4 \, a\right )} + 2 \, e^{\left (2 \, b x + 2 \, a\right )} - 2 \, e^{\left (-2 \, b x - 2 \, a\right )} + e^{\left (-4 \, b x - 4 \, a\right )}}{16 \, x} \] Input:
integrate(cosh(b*x+a)*sinh(b*x+a)^3/x^2,x, algorithm="giac")
Output:
1/16*(4*b*x*Ei(4*b*x)*e^(4*a) - 4*b*x*Ei(2*b*x)*e^(2*a) - 4*b*x*Ei(-2*b*x) *e^(-2*a) + 4*b*x*Ei(-4*b*x)*e^(-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
Timed out. \[ \int \frac {\cosh (a+b x) \sinh ^3(a+b x)}{x^2} \, dx=\int \frac {\mathrm {cosh}\left (a+b\,x\right )\,{\mathrm {sinh}\left (a+b\,x\right )}^3}{x^2} \,d x \] Input:
int((cosh(a + b*x)*sinh(a + b*x)^3)/x^2,x)
Output:
int((cosh(a + b*x)*sinh(a + b*x)^3)/x^2, x)
Time = 0.21 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.43 \[ \int \frac {\cosh (a+b x) \sinh ^3(a+b x)}{x^2} \, dx=\frac {4 e^{4 b x} \mathit {ei} \left (-4 b x \right ) b x -4 e^{4 b x +2 a} \mathit {ei} \left (-2 b x \right ) b x +4 e^{4 b x +8 a} \mathit {ei} \left (4 b x \right ) b x -4 e^{4 b x +6 a} \mathit {ei} \left (2 b x \right ) b x -e^{8 b x +8 a}+2 e^{6 b x +6 a}-2 e^{2 b x +2 a}+1}{16 e^{4 b x +4 a} x} \] Input:
int(cosh(b*x+a)*sinh(b*x+a)^3/x^2,x)
Output:
(4*e**(4*b*x)*ei( - 4*b*x)*b*x - 4*e**(2*a + 4*b*x)*ei( - 2*b*x)*b*x + 4*e **(8*a + 4*b*x)*ei(4*b*x)*b*x - 4*e**(6*a + 4*b*x)*ei(2*b*x)*b*x - e**(8*a + 8*b*x) + 2*e**(6*a + 6*b*x) - 2*e**(2*a + 2*b*x) + 1)/(16*e**(4*a + 4*b *x)*x)