### 3.312 $$\int \frac{\cosh (a+b x) \sinh ^3(a+b x)}{x} \, dx$$

Optimal. Leaf size=53 $-\frac{1}{4} \sinh (2 a) \text{Chi}(2 b x)+\frac{1}{8} \sinh (4 a) \text{Chi}(4 b x)-\frac{1}{4} \cosh (2 a) \text{Shi}(2 b x)+\frac{1}{8} \cosh (4 a) \text{Shi}(4 b x)$

[Out]

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

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Rubi [A]  time = 0.132692, antiderivative size = 53, 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} $-\frac{1}{4} \sinh (2 a) \text{Chi}(2 b x)+\frac{1}{8} \sinh (4 a) \text{Chi}(4 b x)-\frac{1}{4} \cosh (2 a) \text{Shi}(2 b x)+\frac{1}{8} \cosh (4 a) \text{Shi}(4 b x)$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(CoshIntegral[2*b*x]*Sinh[2*a])/4 + (CoshIntegral[4*b*x]*Sinh[4*a])/8 - (Cosh[2*a]*SinhIntegral[2*b*x])/4 + (
Cosh[4*a]*SinhIntegral[4*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 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} \, dx &=\int \left (-\frac{\sinh (2 a+2 b x)}{4 x}+\frac{\sinh (4 a+4 b x)}{8 x}\right ) \, dx\\ &=\frac{1}{8} \int \frac{\sinh (4 a+4 b x)}{x} \, dx-\frac{1}{4} \int \frac{\sinh (2 a+2 b x)}{x} \, dx\\ &=-\left (\frac{1}{4} \cosh (2 a) \int \frac{\sinh (2 b x)}{x} \, dx\right )+\frac{1}{8} \cosh (4 a) \int \frac{\sinh (4 b x)}{x} \, dx-\frac{1}{4} \sinh (2 a) \int \frac{\cosh (2 b x)}{x} \, dx+\frac{1}{8} \sinh (4 a) \int \frac{\cosh (4 b x)}{x} \, dx\\ &=-\frac{1}{4} \text{Chi}(2 b x) \sinh (2 a)+\frac{1}{8} \text{Chi}(4 b x) \sinh (4 a)-\frac{1}{4} \cosh (2 a) \text{Shi}(2 b x)+\frac{1}{8} \cosh (4 a) \text{Shi}(4 b x)\\ \end{align*}

Mathematica [A]  time = 0.0798878, size = 47, normalized size = 0.89 $\frac{1}{8} (-2 \sinh (2 a) \text{Chi}(2 b x)+\sinh (4 a) \text{Chi}(4 b x)-2 \cosh (2 a) \text{Shi}(2 b x)+\cosh (4 a) \text{Shi}(4 b x))$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.067, size = 50, normalized size = 0.9 \begin{align*}{\frac{{{\rm e}^{-4\,a}}{\it Ei} \left ( 1,4\,bx \right ) }{16}}-{\frac{{{\rm e}^{-2\,a}}{\it Ei} \left ( 1,2\,bx \right ) }{8}}-{\frac{{{\rm e}^{4\,a}}{\it Ei} \left ( 1,-4\,bx \right ) }{16}}+{\frac{{{\rm e}^{2\,a}}{\it Ei} \left ( 1,-2\,bx \right ) }{8}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.21239, size = 61, normalized size = 1.15 \begin{align*} \frac{1}{16} \,{\rm Ei}\left (4 \, b x\right ) e^{\left (4 \, a\right )} - \frac{1}{8} \,{\rm Ei}\left (2 \, b x\right ) e^{\left (2 \, a\right )} + \frac{1}{8} \,{\rm Ei}\left (-2 \, b x\right ) e^{\left (-2 \, a\right )} - \frac{1}{16} \,{\rm Ei}\left (-4 \, b x\right ) e^{\left (-4 \, a\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,x, algorithm="maxima")

[Out]

1/16*Ei(4*b*x)*e^(4*a) - 1/8*Ei(2*b*x)*e^(2*a) + 1/8*Ei(-2*b*x)*e^(-2*a) - 1/16*Ei(-4*b*x)*e^(-4*a)

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Fricas [A]  time = 1.79897, size = 223, normalized size = 4.21 \begin{align*} \frac{1}{16} \,{\left ({\rm Ei}\left (4 \, b x\right ) -{\rm Ei}\left (-4 \, b x\right )\right )} \cosh \left (4 \, a\right ) - \frac{1}{8} \,{\left ({\rm Ei}\left (2 \, b x\right ) -{\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (2 \, a\right ) + \frac{1}{16} \,{\left ({\rm Ei}\left (4 \, b x\right ) +{\rm Ei}\left (-4 \, b x\right )\right )} \sinh \left (4 \, a\right ) - \frac{1}{8} \,{\left ({\rm Ei}\left (2 \, b x\right ) +{\rm Ei}\left (-2 \, b x\right )\right )} \sinh \left (2 \, a\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,x, algorithm="fricas")

[Out]

1/16*(Ei(4*b*x) - Ei(-4*b*x))*cosh(4*a) - 1/8*(Ei(2*b*x) - Ei(-2*b*x))*cosh(2*a) + 1/16*(Ei(4*b*x) + Ei(-4*b*x
))*sinh(4*a) - 1/8*(Ei(2*b*x) + Ei(-2*b*x))*sinh(2*a)

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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}\, 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,x)

[Out]

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

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Giac [A]  time = 1.21327, size = 61, normalized size = 1.15 \begin{align*} \frac{1}{16} \,{\rm Ei}\left (4 \, b x\right ) e^{\left (4 \, a\right )} - \frac{1}{8} \,{\rm Ei}\left (2 \, b x\right ) e^{\left (2 \, a\right )} + \frac{1}{8} \,{\rm Ei}\left (-2 \, b x\right ) e^{\left (-2 \, a\right )} - \frac{1}{16} \,{\rm Ei}\left (-4 \, b x\right ) e^{\left (-4 \, a\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,x, algorithm="giac")

[Out]

1/16*Ei(4*b*x)*e^(4*a) - 1/8*Ei(2*b*x)*e^(2*a) + 1/8*Ei(-2*b*x)*e^(-2*a) - 1/16*Ei(-4*b*x)*e^(-4*a)