### 3.276 $$\int \frac{\cosh ^3(a+b x) \sinh (a+b x)}{x^4} \, dx$$

Optimal. Leaf size=169 $\frac{1}{3} b^3 \cosh (2 a) \text{Chi}(2 b x)+\frac{4}{3} b^3 \cosh (4 a) \text{Chi}(4 b x)+\frac{1}{3} b^3 \sinh (2 a) \text{Shi}(2 b x)+\frac{4}{3} b^3 \sinh (4 a) \text{Shi}(4 b x)-\frac{b^2 \sinh (2 a+2 b x)}{6 x}-\frac{b^2 \sinh (4 a+4 b x)}{3 x}-\frac{\sinh (2 a+2 b x)}{12 x^3}-\frac{\sinh (4 a+4 b x)}{24 x^3}-\frac{b \cosh (2 a+2 b x)}{12 x^2}-\frac{b \cosh (4 a+4 b x)}{12 x^2}$

[Out]

-(b*Cosh[2*a + 2*b*x])/(12*x^2) - (b*Cosh[4*a + 4*b*x])/(12*x^2) + (b^3*Cosh[2*a]*CoshIntegral[2*b*x])/3 + (4*
b^3*Cosh[4*a]*CoshIntegral[4*b*x])/3 - Sinh[2*a + 2*b*x]/(12*x^3) - (b^2*Sinh[2*a + 2*b*x])/(6*x) - Sinh[4*a +
4*b*x]/(24*x^3) - (b^2*Sinh[4*a + 4*b*x])/(3*x) + (b^3*Sinh[2*a]*SinhIntegral[2*b*x])/3 + (4*b^3*Sinh[4*a]*Si
nhIntegral[4*b*x])/3

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Rubi [A]  time = 0.29719, antiderivative size = 169, 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} $\frac{1}{3} b^3 \cosh (2 a) \text{Chi}(2 b x)+\frac{4}{3} b^3 \cosh (4 a) \text{Chi}(4 b x)+\frac{1}{3} b^3 \sinh (2 a) \text{Shi}(2 b x)+\frac{4}{3} b^3 \sinh (4 a) \text{Shi}(4 b x)-\frac{b^2 \sinh (2 a+2 b x)}{6 x}-\frac{b^2 \sinh (4 a+4 b x)}{3 x}-\frac{\sinh (2 a+2 b x)}{12 x^3}-\frac{\sinh (4 a+4 b x)}{24 x^3}-\frac{b \cosh (2 a+2 b x)}{12 x^2}-\frac{b \cosh (4 a+4 b x)}{12 x^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[(Cosh[a + b*x]^3*Sinh[a + b*x])/x^4,x]

[Out]

-(b*Cosh[2*a + 2*b*x])/(12*x^2) - (b*Cosh[4*a + 4*b*x])/(12*x^2) + (b^3*Cosh[2*a]*CoshIntegral[2*b*x])/3 + (4*
b^3*Cosh[4*a]*CoshIntegral[4*b*x])/3 - Sinh[2*a + 2*b*x]/(12*x^3) - (b^2*Sinh[2*a + 2*b*x])/(6*x) - Sinh[4*a +
4*b*x]/(24*x^3) - (b^2*Sinh[4*a + 4*b*x])/(3*x) + (b^3*Sinh[2*a]*SinhIntegral[2*b*x])/3 + (4*b^3*Sinh[4*a]*Si
nhIntegral[4*b*x])/3

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 ^3(a+b x) \sinh (a+b x)}{x^4} \, dx &=\int \left (\frac{\sinh (2 a+2 b x)}{4 x^4}+\frac{\sinh (4 a+4 b x)}{8 x^4}\right ) \, dx\\ &=\frac{1}{8} \int \frac{\sinh (4 a+4 b x)}{x^4} \, dx+\frac{1}{4} \int \frac{\sinh (2 a+2 b x)}{x^4} \, dx\\ &=-\frac{\sinh (2 a+2 b x)}{12 x^3}-\frac{\sinh (4 a+4 b x)}{24 x^3}+\frac{1}{6} b \int \frac{\cosh (2 a+2 b x)}{x^3} \, dx+\frac{1}{6} b \int \frac{\cosh (4 a+4 b x)}{x^3} \, dx\\ &=-\frac{b \cosh (2 a+2 b x)}{12 x^2}-\frac{b \cosh (4 a+4 b x)}{12 x^2}-\frac{\sinh (2 a+2 b x)}{12 x^3}-\frac{\sinh (4 a+4 b x)}{24 x^3}+\frac{1}{6} b^2 \int \frac{\sinh (2 a+2 b x)}{x^2} \, dx+\frac{1}{3} b^2 \int \frac{\sinh (4 a+4 b x)}{x^2} \, dx\\ &=-\frac{b \cosh (2 a+2 b x)}{12 x^2}-\frac{b \cosh (4 a+4 b x)}{12 x^2}-\frac{\sinh (2 a+2 b x)}{12 x^3}-\frac{b^2 \sinh (2 a+2 b x)}{6 x}-\frac{\sinh (4 a+4 b x)}{24 x^3}-\frac{b^2 \sinh (4 a+4 b x)}{3 x}+\frac{1}{3} b^3 \int \frac{\cosh (2 a+2 b x)}{x} \, dx+\frac{1}{3} \left (4 b^3\right ) \int \frac{\cosh (4 a+4 b x)}{x} \, dx\\ &=-\frac{b \cosh (2 a+2 b x)}{12 x^2}-\frac{b \cosh (4 a+4 b x)}{12 x^2}-\frac{\sinh (2 a+2 b x)}{12 x^3}-\frac{b^2 \sinh (2 a+2 b x)}{6 x}-\frac{\sinh (4 a+4 b x)}{24 x^3}-\frac{b^2 \sinh (4 a+4 b x)}{3 x}+\frac{1}{3} \left (b^3 \cosh (2 a)\right ) \int \frac{\cosh (2 b x)}{x} \, dx+\frac{1}{3} \left (4 b^3 \cosh (4 a)\right ) \int \frac{\cosh (4 b x)}{x} \, dx+\frac{1}{3} \left (b^3 \sinh (2 a)\right ) \int \frac{\sinh (2 b x)}{x} \, dx+\frac{1}{3} \left (4 b^3 \sinh (4 a)\right ) \int \frac{\sinh (4 b x)}{x} \, dx\\ &=-\frac{b \cosh (2 a+2 b x)}{12 x^2}-\frac{b \cosh (4 a+4 b x)}{12 x^2}+\frac{1}{3} b^3 \cosh (2 a) \text{Chi}(2 b x)+\frac{4}{3} b^3 \cosh (4 a) \text{Chi}(4 b x)-\frac{\sinh (2 a+2 b x)}{12 x^3}-\frac{b^2 \sinh (2 a+2 b x)}{6 x}-\frac{\sinh (4 a+4 b x)}{24 x^3}-\frac{b^2 \sinh (4 a+4 b x)}{3 x}+\frac{1}{3} b^3 \sinh (2 a) \text{Shi}(2 b x)+\frac{4}{3} b^3 \sinh (4 a) \text{Shi}(4 b x)\\ \end{align*}

Mathematica [A]  time = 0.534986, size = 150, normalized size = 0.89 $-\frac{-8 b^3 x^3 \cosh (2 a) \text{Chi}(2 b x)-32 b^3 x^3 \cosh (4 a) \text{Chi}(4 b x)-8 b^3 x^3 \sinh (2 a) \text{Shi}(2 b x)-32 b^3 x^3 \sinh (4 a) \text{Shi}(4 b x)+4 b^2 x^2 \sinh (2 (a+b x))+8 b^2 x^2 \sinh (4 (a+b x))+2 \sinh (2 (a+b x))+\sinh (4 (a+b x))+2 b x \cosh (2 (a+b x))+2 b x \cosh (4 (a+b x))}{24 x^3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(Cosh[a + b*x]^3*Sinh[a + b*x])/x^4,x]

[Out]

-(2*b*x*Cosh[2*(a + b*x)] + 2*b*x*Cosh[4*(a + b*x)] - 8*b^3*x^3*Cosh[2*a]*CoshIntegral[2*b*x] - 32*b^3*x^3*Cos
h[4*a]*CoshIntegral[4*b*x] + 2*Sinh[2*(a + b*x)] + 4*b^2*x^2*Sinh[2*(a + b*x)] + Sinh[4*(a + b*x)] + 8*b^2*x^2
*Sinh[4*(a + b*x)] - 8*b^3*x^3*Sinh[2*a]*SinhIntegral[2*b*x] - 32*b^3*x^3*Sinh[4*a]*SinhIntegral[4*b*x])/(24*x
^3)

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Maple [A]  time = 0.072, size = 246, normalized size = 1.5 \begin{align*}{\frac{{b}^{2}{{\rm e}^{-4\,bx-4\,a}}}{6\,x}}-{\frac{b{{\rm e}^{-4\,bx-4\,a}}}{24\,{x}^{2}}}+{\frac{{{\rm e}^{-4\,bx-4\,a}}}{48\,{x}^{3}}}-{\frac{2\,{b}^{3}{{\rm e}^{-4\,a}}{\it Ei} \left ( 1,4\,bx \right ) }{3}}+{\frac{{b}^{2}{{\rm e}^{-2\,bx-2\,a}}}{12\,x}}-{\frac{b{{\rm e}^{-2\,bx-2\,a}}}{24\,{x}^{2}}}+{\frac{{{\rm e}^{-2\,bx-2\,a}}}{24\,{x}^{3}}}-{\frac{{b}^{3}{{\rm e}^{-2\,a}}{\it Ei} \left ( 1,2\,bx \right ) }{6}}-{\frac{{{\rm e}^{4\,bx+4\,a}}}{48\,{x}^{3}}}-{\frac{b{{\rm e}^{4\,bx+4\,a}}}{24\,{x}^{2}}}-{\frac{{b}^{2}{{\rm e}^{4\,bx+4\,a}}}{6\,x}}-{\frac{2\,{b}^{3}{{\rm e}^{4\,a}}{\it Ei} \left ( 1,-4\,bx \right ) }{3}}-{\frac{{{\rm e}^{2\,bx+2\,a}}}{24\,{x}^{3}}}-{\frac{b{{\rm e}^{2\,bx+2\,a}}}{24\,{x}^{2}}}-{\frac{{b}^{2}{{\rm e}^{2\,bx+2\,a}}}{12\,x}}-{\frac{{b}^{3}{{\rm e}^{2\,a}}{\it Ei} \left ( 1,-2\,bx \right ) }{6}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(b*x+a)^3*sinh(b*x+a)/x^4,x)

[Out]

1/6*b^2*exp(-4*b*x-4*a)/x-1/24*b*exp(-4*b*x-4*a)/x^2+1/48*exp(-4*b*x-4*a)/x^3-2/3*b^3*exp(-4*a)*Ei(1,4*b*x)+1/
12*b^2*exp(-2*b*x-2*a)/x-1/24*b*exp(-2*b*x-2*a)/x^2+1/24*exp(-2*b*x-2*a)/x^3-1/6*b^3*exp(-2*a)*Ei(1,2*b*x)-1/4
8/x^3*exp(4*b*x+4*a)-1/24*b/x^2*exp(4*b*x+4*a)-1/6*b^2/x*exp(4*b*x+4*a)-2/3*b^3*exp(4*a)*Ei(1,-4*b*x)-1/24*exp
(2*b*x+2*a)/x^3-1/24*b*exp(2*b*x+2*a)/x^2-1/12*b^2*exp(2*b*x+2*a)/x-1/6*b^3*exp(2*a)*Ei(1,-2*b*x)

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Maxima [A]  time = 1.38147, size = 80, normalized size = 0.47 \begin{align*} 4 \, b^{3} e^{\left (-4 \, a\right )} \Gamma \left (-3, 4 \, b x\right ) + b^{3} e^{\left (-2 \, a\right )} \Gamma \left (-3, 2 \, b x\right ) + b^{3} e^{\left (2 \, a\right )} \Gamma \left (-3, -2 \, b x\right ) + 4 \, b^{3} e^{\left (4 \, a\right )} \Gamma \left (-3, -4 \, b x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^3*sinh(b*x+a)/x^4,x, algorithm="maxima")

[Out]

4*b^3*e^(-4*a)*gamma(-3, 4*b*x) + b^3*e^(-2*a)*gamma(-3, 2*b*x) + b^3*e^(2*a)*gamma(-3, -2*b*x) + 4*b^3*e^(4*a
)*gamma(-3, -4*b*x)

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Fricas [A]  time = 1.85127, size = 647, normalized size = 3.83 \begin{align*} -\frac{b x \cosh \left (b x + a\right )^{4} + b x \sinh \left (b x + a\right )^{4} + 2 \,{\left (8 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b x \cosh \left (b x + a\right )^{2} +{\left (6 \, b x \cosh \left (b x + a\right )^{2} + b x\right )} \sinh \left (b x + a\right )^{2} - 8 \,{\left (b^{3} x^{3}{\rm Ei}\left (4 \, b x\right ) + b^{3} x^{3}{\rm Ei}\left (-4 \, b x\right )\right )} \cosh \left (4 \, a\right ) - 2 \,{\left (b^{3} x^{3}{\rm Ei}\left (2 \, b x\right ) + b^{3} x^{3}{\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (2 \, a\right ) + 2 \,{\left ({\left (8 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )^{3} +{\left (2 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) - 8 \,{\left (b^{3} x^{3}{\rm Ei}\left (4 \, b x\right ) - b^{3} x^{3}{\rm Ei}\left (-4 \, b x\right )\right )} \sinh \left (4 \, a\right ) - 2 \,{\left (b^{3} x^{3}{\rm Ei}\left (2 \, b x\right ) - b^{3} x^{3}{\rm Ei}\left (-2 \, b x\right )\right )} \sinh \left (2 \, a\right )}{12 \, x^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^3*sinh(b*x+a)/x^4,x, algorithm="fricas")

[Out]

-1/12*(b*x*cosh(b*x + a)^4 + b*x*sinh(b*x + a)^4 + 2*(8*b^2*x^2 + 1)*cosh(b*x + a)*sinh(b*x + a)^3 + b*x*cosh(
b*x + a)^2 + (6*b*x*cosh(b*x + a)^2 + b*x)*sinh(b*x + a)^2 - 8*(b^3*x^3*Ei(4*b*x) + b^3*x^3*Ei(-4*b*x))*cosh(4
*a) - 2*(b^3*x^3*Ei(2*b*x) + b^3*x^3*Ei(-2*b*x))*cosh(2*a) + 2*((8*b^2*x^2 + 1)*cosh(b*x + a)^3 + (2*b^2*x^2 +
1)*cosh(b*x + a))*sinh(b*x + a) - 8*(b^3*x^3*Ei(4*b*x) - b^3*x^3*Ei(-4*b*x))*sinh(4*a) - 2*(b^3*x^3*Ei(2*b*x)
- b^3*x^3*Ei(-2*b*x))*sinh(2*a))/x^3

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh{\left (a + b x \right )} \cosh ^{3}{\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)**3*sinh(b*x+a)/x**4,x)

[Out]

Integral(sinh(a + b*x)*cosh(a + b*x)**3/x**4, x)

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Giac [A]  time = 1.15062, size = 319, normalized size = 1.89 \begin{align*} \frac{32 \, b^{3} x^{3}{\rm Ei}\left (4 \, b x\right ) e^{\left (4 \, a\right )} + 8 \, b^{3} x^{3}{\rm Ei}\left (2 \, b x\right ) e^{\left (2 \, a\right )} + 8 \, b^{3} x^{3}{\rm Ei}\left (-2 \, b x\right ) e^{\left (-2 \, a\right )} + 32 \, b^{3} x^{3}{\rm Ei}\left (-4 \, b x\right ) e^{\left (-4 \, a\right )} - 8 \, b^{2} x^{2} e^{\left (4 \, b x + 4 \, a\right )} - 4 \, b^{2} x^{2} e^{\left (2 \, b x + 2 \, a\right )} + 4 \, b^{2} x^{2} e^{\left (-2 \, b x - 2 \, a\right )} + 8 \, b^{2} x^{2} e^{\left (-4 \, b x - 4 \, a\right )} - 2 \, b x e^{\left (4 \, b x + 4 \, a\right )} - 2 \, b x e^{\left (2 \, b x + 2 \, a\right )} - 2 \, b x e^{\left (-2 \, b x - 2 \, a\right )} - 2 \, b x e^{\left (-4 \, b x - 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 )}}{48 \, x^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^3*sinh(b*x+a)/x^4,x, algorithm="giac")

[Out]

1/48*(32*b^3*x^3*Ei(4*b*x)*e^(4*a) + 8*b^3*x^3*Ei(2*b*x)*e^(2*a) + 8*b^3*x^3*Ei(-2*b*x)*e^(-2*a) + 32*b^3*x^3*
Ei(-4*b*x)*e^(-4*a) - 8*b^2*x^2*e^(4*b*x + 4*a) - 4*b^2*x^2*e^(2*b*x + 2*a) + 4*b^2*x^2*e^(-2*b*x - 2*a) + 8*b
^2*x^2*e^(-4*b*x - 4*a) - 2*b*x*e^(4*b*x + 4*a) - 2*b*x*e^(2*b*x + 2*a) - 2*b*x*e^(-2*b*x - 2*a) - 2*b*x*e^(-4
*b*x - 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^3