### 3.313 $$\int \frac{\cosh (a+b x) \sinh ^3(a+b x)}{x^2} \, dx$$

Optimal. Leaf size=89 $-\frac{1}{2} b \cosh (2 a) \text{Chi}(2 b x)+\frac{1}{2} b \cosh (4 a) \text{Chi}(4 b x)-\frac{1}{2} b \sinh (2 a) \text{Shi}(2 b x)+\frac{1}{2} b \sinh (4 a) \text{Shi}(4 b x)+\frac{\sinh (2 a+2 b x)}{4 x}-\frac{\sinh (4 a+4 b x)}{8 x}$

[Out]

-(b*Cosh[2*a]*CoshIntegral[2*b*x])/2 + (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]*SinhIntegral[2*b*x])/2 + (b*Sinh[4*a]*SinhIntegral[4*b*x])/2

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Rubi [A]  time = 0.175784, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 10, 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} $-\frac{1}{2} b \cosh (2 a) \text{Chi}(2 b x)+\frac{1}{2} b \cosh (4 a) \text{Chi}(4 b x)-\frac{1}{2} b \sinh (2 a) \text{Shi}(2 b x)+\frac{1}{2} b \sinh (4 a) \text{Shi}(4 b x)+\frac{\sinh (2 a+2 b x)}{4 x}-\frac{\sinh (4 a+4 b x)}{8 x}$

Antiderivative was successfully veriﬁed.

[In]

Int[(Cosh[a + b*x]*Sinh[a + b*x]^3)/x^2,x]

[Out]

-(b*Cosh[2*a]*CoshIntegral[2*b*x])/2 + (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]*SinhIntegral[2*b*x])/2 + (b*Sinh[4*a]*SinhIntegral[4*b*x])/2

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

Mathematica [A]  time = 0.249716, size = 78, normalized size = 0.88 $-\frac{4 b x \cosh (2 a) \text{Chi}(2 b x)-4 b x \cosh (4 a) \text{Chi}(4 b x)+4 b x \sinh (2 a) \text{Shi}(2 b x)-4 b x \sinh (4 a) \text{Shi}(4 b x)-2 \sinh (2 (a+b x))+\sinh (4 (a+b x))}{8 x}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(Cosh[a + b*x]*Sinh[a + b*x]^3)/x^2,x]

[Out]

-(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]*SinhIntegral[2*b*x] - 4*b*x*Sinh[4*a]*SinhIntegral[4*b*x])/(8*x)

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Maple [A]  time = 0.066, size = 110, normalized size = 1.2 \begin{align*}{\frac{{{\rm e}^{-4\,bx-4\,a}}}{16\,x}}-{\frac{b{{\rm e}^{-4\,a}}{\it Ei} \left ( 1,4\,bx \right ) }{4}}-{\frac{{{\rm e}^{-2\,bx-2\,a}}}{8\,x}}+{\frac{b{{\rm e}^{-2\,a}}{\it Ei} \left ( 1,2\,bx \right ) }{4}}-{\frac{{{\rm e}^{4\,bx+4\,a}}}{16\,x}}-{\frac{b{{\rm e}^{4\,a}}{\it Ei} \left ( 1,-4\,bx \right ) }{4}}+{\frac{{{\rm e}^{2\,bx+2\,a}}}{8\,x}}+{\frac{b{{\rm e}^{2\,a}}{\it Ei} \left ( 1,-2\,bx \right ) }{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(b*x+a)*sinh(b*x+a)^3/x^2,x)

[Out]

1/16*exp(-4*b*x-4*a)/x-1/4*b*exp(-4*a)*Ei(1,4*b*x)-1/8*exp(-2*b*x-2*a)/x+1/4*b*exp(-2*a)*Ei(1,2*b*x)-1/16/x*ex
p(4*b*x+4*a)-1/4*b*exp(4*a)*Ei(1,-4*b*x)+1/8*exp(2*b*x+2*a)/x+1/4*b*exp(2*a)*Ei(1,-2*b*x)

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Maxima [A]  time = 1.24912, size = 72, normalized size = 0.81 \begin{align*} \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 ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*sinh(b*x+a)^3/x^2,x, algorithm="maxima")

[Out]

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)

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Fricas [A]  time = 1.80503, size = 370, normalized size = 4.16 \begin{align*} -\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} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*sinh(b*x+a)^3/x^2,x, algorithm="fricas")

[Out]

-1/4*(2*cosh(b*x + a)*sinh(b*x + a)^3 - (b*x*Ei(4*b*x) + b*x*Ei(-4*b*x))*cosh(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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh ^{3}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{x^{2}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*sinh(b*x+a)**3/x**2,x)

[Out]

Integral(sinh(a + b*x)**3*cosh(a + b*x)/x**2, x)

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Giac [A]  time = 1.1774, size = 135, normalized size = 1.52 \begin{align*} \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} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*sinh(b*x+a)^3/x^2,x, algorithm="giac")

[Out]

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