3.267 \(\int \frac{\cosh ^2(a+b x) \sinh (a+b x)}{x^4} \, dx\)

Optimal. Leaf size=154 \[ \frac{1}{24} b^3 \cosh (a) \text{Chi}(b x)+\frac{9}{8} b^3 \cosh (3 a) \text{Chi}(3 b x)+\frac{1}{24} b^3 \sinh (a) \text{Shi}(b x)+\frac{9}{8} b^3 \sinh (3 a) \text{Shi}(3 b x)-\frac{b^2 \sinh (a+b x)}{24 x}-\frac{3 b^2 \sinh (3 a+3 b x)}{8 x}-\frac{\sinh (a+b x)}{12 x^3}-\frac{\sinh (3 a+3 b x)}{12 x^3}-\frac{b \cosh (a+b x)}{24 x^2}-\frac{b \cosh (3 a+3 b x)}{8 x^2} \]

[Out]

-(b*Cosh[a + b*x])/(24*x^2) - (b*Cosh[3*a + 3*b*x])/(8*x^2) + (b^3*Cosh[a]*CoshIntegral[b*x])/24 + (9*b^3*Cosh
[3*a]*CoshIntegral[3*b*x])/8 - Sinh[a + b*x]/(12*x^3) - (b^2*Sinh[a + b*x])/(24*x) - Sinh[3*a + 3*b*x]/(12*x^3
) - (3*b^2*Sinh[3*a + 3*b*x])/(8*x) + (b^3*Sinh[a]*SinhIntegral[b*x])/24 + (9*b^3*Sinh[3*a]*SinhIntegral[3*b*x
])/8

________________________________________________________________________________________

Rubi [A]  time = 0.289493, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {5448, 3297, 3303, 3298, 3301} \[ \frac{1}{24} b^3 \cosh (a) \text{Chi}(b x)+\frac{9}{8} b^3 \cosh (3 a) \text{Chi}(3 b x)+\frac{1}{24} b^3 \sinh (a) \text{Shi}(b x)+\frac{9}{8} b^3 \sinh (3 a) \text{Shi}(3 b x)-\frac{b^2 \sinh (a+b x)}{24 x}-\frac{3 b^2 \sinh (3 a+3 b x)}{8 x}-\frac{\sinh (a+b x)}{12 x^3}-\frac{\sinh (3 a+3 b x)}{12 x^3}-\frac{b \cosh (a+b x)}{24 x^2}-\frac{b \cosh (3 a+3 b x)}{8 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cosh[a + b*x]^2*Sinh[a + b*x])/x^4,x]

[Out]

-(b*Cosh[a + b*x])/(24*x^2) - (b*Cosh[3*a + 3*b*x])/(8*x^2) + (b^3*Cosh[a]*CoshIntegral[b*x])/24 + (9*b^3*Cosh
[3*a]*CoshIntegral[3*b*x])/8 - Sinh[a + b*x]/(12*x^3) - (b^2*Sinh[a + b*x])/(24*x) - Sinh[3*a + 3*b*x]/(12*x^3
) - (3*b^2*Sinh[3*a + 3*b*x])/(8*x) + (b^3*Sinh[a]*SinhIntegral[b*x])/24 + (9*b^3*Sinh[3*a]*SinhIntegral[3*b*x
])/8

Rule 5448

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]

Rule 3297

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x])/(d*(
m + 1)), x] - Dist[f/(d*(m + 1)), Int[(c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[
m, -1]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3298

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(I*SinhIntegral[(c*f*fz)
/d + f*fz*x])/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rule 3301

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[(c*f*fz)/d
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rubi steps

\begin{align*} \int \frac{\cosh ^2(a+b x) \sinh (a+b x)}{x^4} \, dx &=\int \left (\frac{\sinh (a+b x)}{4 x^4}+\frac{\sinh (3 a+3 b x)}{4 x^4}\right ) \, dx\\ &=\frac{1}{4} \int \frac{\sinh (a+b x)}{x^4} \, dx+\frac{1}{4} \int \frac{\sinh (3 a+3 b x)}{x^4} \, dx\\ &=-\frac{\sinh (a+b x)}{12 x^3}-\frac{\sinh (3 a+3 b x)}{12 x^3}+\frac{1}{12} b \int \frac{\cosh (a+b x)}{x^3} \, dx+\frac{1}{4} b \int \frac{\cosh (3 a+3 b x)}{x^3} \, dx\\ &=-\frac{b \cosh (a+b x)}{24 x^2}-\frac{b \cosh (3 a+3 b x)}{8 x^2}-\frac{\sinh (a+b x)}{12 x^3}-\frac{\sinh (3 a+3 b x)}{12 x^3}+\frac{1}{24} b^2 \int \frac{\sinh (a+b x)}{x^2} \, dx+\frac{1}{8} \left (3 b^2\right ) \int \frac{\sinh (3 a+3 b x)}{x^2} \, dx\\ &=-\frac{b \cosh (a+b x)}{24 x^2}-\frac{b \cosh (3 a+3 b x)}{8 x^2}-\frac{\sinh (a+b x)}{12 x^3}-\frac{b^2 \sinh (a+b x)}{24 x}-\frac{\sinh (3 a+3 b x)}{12 x^3}-\frac{3 b^2 \sinh (3 a+3 b x)}{8 x}+\frac{1}{24} b^3 \int \frac{\cosh (a+b x)}{x} \, dx+\frac{1}{8} \left (9 b^3\right ) \int \frac{\cosh (3 a+3 b x)}{x} \, dx\\ &=-\frac{b \cosh (a+b x)}{24 x^2}-\frac{b \cosh (3 a+3 b x)}{8 x^2}-\frac{\sinh (a+b x)}{12 x^3}-\frac{b^2 \sinh (a+b x)}{24 x}-\frac{\sinh (3 a+3 b x)}{12 x^3}-\frac{3 b^2 \sinh (3 a+3 b x)}{8 x}+\frac{1}{24} \left (b^3 \cosh (a)\right ) \int \frac{\cosh (b x)}{x} \, dx+\frac{1}{8} \left (9 b^3 \cosh (3 a)\right ) \int \frac{\cosh (3 b x)}{x} \, dx+\frac{1}{24} \left (b^3 \sinh (a)\right ) \int \frac{\sinh (b x)}{x} \, dx+\frac{1}{8} \left (9 b^3 \sinh (3 a)\right ) \int \frac{\sinh (3 b x)}{x} \, dx\\ &=-\frac{b \cosh (a+b x)}{24 x^2}-\frac{b \cosh (3 a+3 b x)}{8 x^2}+\frac{1}{24} b^3 \cosh (a) \text{Chi}(b x)+\frac{9}{8} b^3 \cosh (3 a) \text{Chi}(3 b x)-\frac{\sinh (a+b x)}{12 x^3}-\frac{b^2 \sinh (a+b x)}{24 x}-\frac{\sinh (3 a+3 b x)}{12 x^3}-\frac{3 b^2 \sinh (3 a+3 b x)}{8 x}+\frac{1}{24} b^3 \sinh (a) \text{Shi}(b x)+\frac{9}{8} b^3 \sinh (3 a) \text{Shi}(3 b x)\\ \end{align*}

Mathematica [A]  time = 0.349048, size = 138, normalized size = 0.9 \[ -\frac{-b^3 x^3 \cosh (a) \text{Chi}(b x)-27 b^3 x^3 \cosh (3 a) \text{Chi}(3 b x)-b^3 x^3 \sinh (a) \text{Shi}(b x)-27 b^3 x^3 \sinh (3 a) \text{Shi}(3 b x)+b^2 x^2 \sinh (a+b x)+9 b^2 x^2 \sinh (3 (a+b x))+2 \sinh (a+b x)+2 \sinh (3 (a+b x))+b x \cosh (a+b x)+3 b x \cosh (3 (a+b x))}{24 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cosh[a + b*x]^2*Sinh[a + b*x])/x^4,x]

[Out]

-(b*x*Cosh[a + b*x] + 3*b*x*Cosh[3*(a + b*x)] - b^3*x^3*Cosh[a]*CoshIntegral[b*x] - 27*b^3*x^3*Cosh[3*a]*CoshI
ntegral[3*b*x] + 2*Sinh[a + b*x] + b^2*x^2*Sinh[a + b*x] + 2*Sinh[3*(a + b*x)] + 9*b^2*x^2*Sinh[3*(a + b*x)] -
b^3*x^3*Sinh[a]*SinhIntegral[b*x] - 27*b^3*x^3*Sinh[3*a]*SinhIntegral[3*b*x])/(24*x^3)

________________________________________________________________________________________

Maple [A]  time = 0.076, size = 234, normalized size = 1.5 \begin{align*}{\frac{3\,{b}^{2}{{\rm e}^{-3\,bx-3\,a}}}{16\,x}}-{\frac{b{{\rm e}^{-3\,bx-3\,a}}}{16\,{x}^{2}}}+{\frac{{{\rm e}^{-3\,bx-3\,a}}}{24\,{x}^{3}}}-{\frac{9\,{b}^{3}{{\rm e}^{-3\,a}}{\it Ei} \left ( 1,3\,bx \right ) }{16}}+{\frac{{b}^{2}{{\rm e}^{-bx-a}}}{48\,x}}-{\frac{b{{\rm e}^{-bx-a}}}{48\,{x}^{2}}}+{\frac{{{\rm e}^{-bx-a}}}{24\,{x}^{3}}}-{\frac{{b}^{3}{{\rm e}^{-a}}{\it Ei} \left ( 1,bx \right ) }{48}}-{\frac{{{\rm e}^{bx+a}}}{24\,{x}^{3}}}-{\frac{b{{\rm e}^{bx+a}}}{48\,{x}^{2}}}-{\frac{{b}^{2}{{\rm e}^{bx+a}}}{48\,x}}-{\frac{{b}^{3}{{\rm e}^{a}}{\it Ei} \left ( 1,-bx \right ) }{48}}-{\frac{{{\rm e}^{3\,bx+3\,a}}}{24\,{x}^{3}}}-{\frac{b{{\rm e}^{3\,bx+3\,a}}}{16\,{x}^{2}}}-{\frac{3\,{b}^{2}{{\rm e}^{3\,bx+3\,a}}}{16\,x}}-{\frac{9\,{b}^{3}{{\rm e}^{3\,a}}{\it Ei} \left ( 1,-3\,bx \right ) }{16}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(b*x+a)^2*sinh(b*x+a)/x^4,x)

[Out]

3/16*b^2*exp(-3*b*x-3*a)/x-1/16*b*exp(-3*b*x-3*a)/x^2+1/24*exp(-3*b*x-3*a)/x^3-9/16*b^3*exp(-3*a)*Ei(1,3*b*x)+
1/48*b^2*exp(-b*x-a)/x-1/48*b*exp(-b*x-a)/x^2+1/24*exp(-b*x-a)/x^3-1/48*b^3*exp(-a)*Ei(1,b*x)-1/24/x^3*exp(b*x
+a)-1/48*b/x^2*exp(b*x+a)-1/48*b^2/x*exp(b*x+a)-1/48*b^3*exp(a)*Ei(1,-b*x)-1/24/x^3*exp(3*b*x+3*a)-1/16*b/x^2*
exp(3*b*x+3*a)-3/16*b^2/x*exp(3*b*x+3*a)-9/16*b^3*exp(3*a)*Ei(1,-3*b*x)

________________________________________________________________________________________

Maxima [A]  time = 1.26281, size = 78, normalized size = 0.51 \begin{align*} \frac{27}{8} \, b^{3} e^{\left (-3 \, a\right )} \Gamma \left (-3, 3 \, b x\right ) + \frac{1}{8} \, b^{3} e^{\left (-a\right )} \Gamma \left (-3, b x\right ) + \frac{1}{8} \, b^{3} e^{a} \Gamma \left (-3, -b x\right ) + \frac{27}{8} \, b^{3} e^{\left (3 \, a\right )} \Gamma \left (-3, -3 \, b x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^2*sinh(b*x+a)/x^4,x, algorithm="maxima")

[Out]

27/8*b^3*e^(-3*a)*gamma(-3, 3*b*x) + 1/8*b^3*e^(-a)*gamma(-3, b*x) + 1/8*b^3*e^a*gamma(-3, -b*x) + 27/8*b^3*e^
(3*a)*gamma(-3, -3*b*x)

________________________________________________________________________________________

Fricas [A]  time = 1.75587, size = 548, normalized size = 3.56 \begin{align*} -\frac{6 \, b x \cosh \left (b x + a\right )^{3} + 18 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + 2 \,{\left (9 \, b^{2} x^{2} + 2\right )} \sinh \left (b x + a\right )^{3} + 2 \, b x \cosh \left (b x + 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 ) -{\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 (b^{2} x^{2} + 3 \,{\left (9 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )^{2} + 2\right )} \sinh \left (b x + 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 ) -{\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 )}{48 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^2*sinh(b*x+a)/x^4,x, algorithm="fricas")

[Out]

-1/48*(6*b*x*cosh(b*x + a)^3 + 18*b*x*cosh(b*x + a)*sinh(b*x + a)^2 + 2*(9*b^2*x^2 + 2)*sinh(b*x + a)^3 + 2*b*
x*cosh(b*x + a) - 27*(b^3*x^3*Ei(3*b*x) + b^3*x^3*Ei(-3*b*x))*cosh(3*a) - (b^3*x^3*Ei(b*x) + b^3*x^3*Ei(-b*x))
*cosh(a) + 2*(b^2*x^2 + 3*(9*b^2*x^2 + 2)*cosh(b*x + a)^2 + 2)*sinh(b*x + a) - 27*(b^3*x^3*Ei(3*b*x) - b^3*x^3
*Ei(-3*b*x))*sinh(3*a) - (b^3*x^3*Ei(b*x) - b^3*x^3*Ei(-b*x))*sinh(a))/x^3

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)**2*sinh(b*x+a)/x**4,x)

[Out]

Integral(sinh(a + b*x)*cosh(a + b*x)**2/x**4, x)

________________________________________________________________________________________

Giac [A]  time = 1.16996, size = 301, normalized size = 1.95 \begin{align*} \frac{27 \, b^{3} x^{3}{\rm Ei}\left (3 \, b x\right ) e^{\left (3 \, a\right )} + 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 )} + b^{3} x^{3}{\rm Ei}\left (b x\right ) e^{a} - 9 \, b^{2} x^{2} e^{\left (3 \, b x + 3 \, a\right )} - b^{2} x^{2} e^{\left (b x + a\right )} + b^{2} x^{2} e^{\left (-b x - a\right )} + 9 \, b^{2} x^{2} e^{\left (-3 \, b x - 3 \, a\right )} - 3 \, b x e^{\left (3 \, b x + 3 \, a\right )} - b x e^{\left (b x + a\right )} - b x e^{\left (-b x - a\right )} - 3 \, b x e^{\left (-3 \, b x - 3 \, a\right )} - 2 \, e^{\left (3 \, b x + 3 \, a\right )} - 2 \, e^{\left (b x + a\right )} + 2 \, e^{\left (-b x - a\right )} + 2 \, e^{\left (-3 \, b x - 3 \, a\right )}}{48 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^2*sinh(b*x+a)/x^4,x, algorithm="giac")

[Out]

1/48*(27*b^3*x^3*Ei(3*b*x)*e^(3*a) + b^3*x^3*Ei(-b*x)*e^(-a) + 27*b^3*x^3*Ei(-3*b*x)*e^(-3*a) + b^3*x^3*Ei(b*x
)*e^a - 9*b^2*x^2*e^(3*b*x + 3*a) - b^2*x^2*e^(b*x + a) + b^2*x^2*e^(-b*x - a) + 9*b^2*x^2*e^(-3*b*x - 3*a) -
3*b*x*e^(3*b*x + 3*a) - b*x*e^(b*x + a) - b*x*e^(-b*x - a) - 3*b*x*e^(-3*b*x - 3*a) - 2*e^(3*b*x + 3*a) - 2*e^
(b*x + a) + 2*e^(-b*x - a) + 2*e^(-3*b*x - 3*a))/x^3