### 3.264 $$\int \frac{\cosh ^2(a+b x) \sinh (a+b x)}{x} \, dx$$

Optimal. Leaf size=47 $\frac{1}{4} \sinh (a) \text{Chi}(b x)+\frac{1}{4} \sinh (3 a) \text{Chi}(3 b x)+\frac{1}{4} \cosh (a) \text{Shi}(b x)+\frac{1}{4} \cosh (3 a) \text{Shi}(3 b x)$

[Out]

(CoshIntegral[b*x]*Sinh[a])/4 + (CoshIntegral[3*b*x]*Sinh[3*a])/4 + (Cosh[a]*SinhIntegral[b*x])/4 + (Cosh[3*a]
*SinhIntegral[3*b*x])/4

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Rubi [A]  time = 0.141382, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.222, Rules used = {5448, 3303, 3298, 3301} $\frac{1}{4} \sinh (a) \text{Chi}(b x)+\frac{1}{4} \sinh (3 a) \text{Chi}(3 b x)+\frac{1}{4} \cosh (a) \text{Shi}(b x)+\frac{1}{4} \cosh (3 a) \text{Shi}(3 b x)$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(CoshIntegral[b*x]*Sinh[a])/4 + (CoshIntegral[3*b*x]*Sinh[3*a])/4 + (Cosh[a]*SinhIntegral[b*x])/4 + (Cosh[3*a]
*SinhIntegral[3*b*x])/4

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 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} \, dx &=\int \left (\frac{\sinh (a+b x)}{4 x}+\frac{\sinh (3 a+3 b x)}{4 x}\right ) \, dx\\ &=\frac{1}{4} \int \frac{\sinh (a+b x)}{x} \, dx+\frac{1}{4} \int \frac{\sinh (3 a+3 b x)}{x} \, dx\\ &=\frac{1}{4} \cosh (a) \int \frac{\sinh (b x)}{x} \, dx+\frac{1}{4} \cosh (3 a) \int \frac{\sinh (3 b x)}{x} \, dx+\frac{1}{4} \sinh (a) \int \frac{\cosh (b x)}{x} \, dx+\frac{1}{4} \sinh (3 a) \int \frac{\cosh (3 b x)}{x} \, dx\\ &=\frac{1}{4} \text{Chi}(b x) \sinh (a)+\frac{1}{4} \text{Chi}(3 b x) \sinh (3 a)+\frac{1}{4} \cosh (a) \text{Shi}(b x)+\frac{1}{4} \cosh (3 a) \text{Shi}(3 b x)\\ \end{align*}

Mathematica [A]  time = 0.0702908, size = 39, normalized size = 0.83 $\frac{1}{4} (\sinh (a) \text{Chi}(b x)+\sinh (3 a) \text{Chi}(3 b x)+\cosh (a) \text{Shi}(b x)+\cosh (3 a) \text{Shi}(3 b x))$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(CoshIntegral[b*x]*Sinh[a] + CoshIntegral[3*b*x]*Sinh[3*a] + Cosh[a]*SinhIntegral[b*x] + Cosh[3*a]*SinhIntegra
l[3*b*x])/4

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Maple [A]  time = 0.057, size = 47, normalized size = 1. \begin{align*}{\frac{{{\rm e}^{-3\,a}}{\it Ei} \left ( 1,3\,bx \right ) }{8}}+{\frac{{{\rm e}^{-a}}{\it Ei} \left ( 1,bx \right ) }{8}}-{\frac{{{\rm e}^{a}}{\it Ei} \left ( 1,-bx \right ) }{8}}-{\frac{{{\rm e}^{3\,a}}{\it Ei} \left ( 1,-3\,bx \right ) }{8}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*exp(-3*a)*Ei(1,3*b*x)+1/8*exp(-a)*Ei(1,b*x)-1/8*exp(a)*Ei(1,-b*x)-1/8*exp(3*a)*Ei(1,-3*b*x)

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Maxima [A]  time = 1.36931, size = 57, normalized size = 1.21 \begin{align*} \frac{1}{8} \,{\rm Ei}\left (3 \, b x\right ) e^{\left (3 \, a\right )} - \frac{1}{8} \,{\rm Ei}\left (-b x\right ) e^{\left (-a\right )} - \frac{1}{8} \,{\rm Ei}\left (-3 \, b x\right ) e^{\left (-3 \, a\right )} + \frac{1}{8} \,{\rm Ei}\left (b x\right ) e^{a} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*Ei(3*b*x)*e^(3*a) - 1/8*Ei(-b*x)*e^(-a) - 1/8*Ei(-3*b*x)*e^(-3*a) + 1/8*Ei(b*x)*e^a

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Fricas [A]  time = 2.38121, size = 204, normalized size = 4.34 \begin{align*} \frac{1}{8} \,{\left ({\rm Ei}\left (3 \, b x\right ) -{\rm Ei}\left (-3 \, b x\right )\right )} \cosh \left (3 \, a\right ) + \frac{1}{8} \,{\left ({\rm Ei}\left (b x\right ) -{\rm Ei}\left (-b x\right )\right )} \cosh \left (a\right ) + \frac{1}{8} \,{\left ({\rm Ei}\left (3 \, b x\right ) +{\rm Ei}\left (-3 \, b x\right )\right )} \sinh \left (3 \, a\right ) + \frac{1}{8} \,{\left ({\rm Ei}\left (b x\right ) +{\rm Ei}\left (-b x\right )\right )} \sinh \left (a\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,x, algorithm="fricas")

[Out]

1/8*(Ei(3*b*x) - Ei(-3*b*x))*cosh(3*a) + 1/8*(Ei(b*x) - Ei(-b*x))*cosh(a) + 1/8*(Ei(3*b*x) + Ei(-3*b*x))*sinh(
3*a) + 1/8*(Ei(b*x) + Ei(-b*x))*sinh(a)

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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}\, 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,x)

[Out]

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

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Giac [A]  time = 1.18299, size = 57, normalized size = 1.21 \begin{align*} \frac{1}{8} \,{\rm Ei}\left (3 \, b x\right ) e^{\left (3 \, a\right )} - \frac{1}{8} \,{\rm Ei}\left (-b x\right ) e^{\left (-a\right )} - \frac{1}{8} \,{\rm Ei}\left (-3 \, b x\right ) e^{\left (-3 \, a\right )} + \frac{1}{8} \,{\rm Ei}\left (b x\right ) e^{a} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*Ei(3*b*x)*e^(3*a) - 1/8*Ei(-b*x)*e^(-a) - 1/8*Ei(-3*b*x)*e^(-3*a) + 1/8*Ei(b*x)*e^a