### 3.305 $$\int \frac{\cosh ^3(a+b x) \sinh ^2(a+b x)}{x^3} \, dx$$

Optimal. Leaf size=184 $-\frac{1}{16} b^2 \cosh (a) \text{Chi}(b x)+\frac{9}{32} b^2 \cosh (3 a) \text{Chi}(3 b x)+\frac{25}{32} b^2 \cosh (5 a) \text{Chi}(5 b x)-\frac{1}{16} b^2 \sinh (a) \text{Shi}(b x)+\frac{9}{32} b^2 \sinh (3 a) \text{Shi}(3 b x)+\frac{25}{32} b^2 \sinh (5 a) \text{Shi}(5 b x)+\frac{\cosh (a+b x)}{16 x^2}-\frac{\cosh (3 a+3 b x)}{32 x^2}-\frac{\cosh (5 a+5 b x)}{32 x^2}+\frac{b \sinh (a+b x)}{16 x}-\frac{3 b \sinh (3 a+3 b x)}{32 x}-\frac{5 b \sinh (5 a+5 b x)}{32 x}$

[Out]

Cosh[a + b*x]/(16*x^2) - Cosh[3*a + 3*b*x]/(32*x^2) - Cosh[5*a + 5*b*x]/(32*x^2) - (b^2*Cosh[a]*CoshIntegral[b
*x])/16 + (9*b^2*Cosh[3*a]*CoshIntegral[3*b*x])/32 + (25*b^2*Cosh[5*a]*CoshIntegral[5*b*x])/32 + (b*Sinh[a + b
*x])/(16*x) - (3*b*Sinh[3*a + 3*b*x])/(32*x) - (5*b*Sinh[5*a + 5*b*x])/(32*x) - (b^2*Sinh[a]*SinhIntegral[b*x]
)/16 + (9*b^2*Sinh[3*a]*SinhIntegral[3*b*x])/32 + (25*b^2*Sinh[5*a]*SinhIntegral[5*b*x])/32

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Rubi [A]  time = 0.343936, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 5, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.25, Rules used = {5448, 3297, 3303, 3298, 3301} $-\frac{1}{16} b^2 \cosh (a) \text{Chi}(b x)+\frac{9}{32} b^2 \cosh (3 a) \text{Chi}(3 b x)+\frac{25}{32} b^2 \cosh (5 a) \text{Chi}(5 b x)-\frac{1}{16} b^2 \sinh (a) \text{Shi}(b x)+\frac{9}{32} b^2 \sinh (3 a) \text{Shi}(3 b x)+\frac{25}{32} b^2 \sinh (5 a) \text{Shi}(5 b x)+\frac{\cosh (a+b x)}{16 x^2}-\frac{\cosh (3 a+3 b x)}{32 x^2}-\frac{\cosh (5 a+5 b x)}{32 x^2}+\frac{b \sinh (a+b x)}{16 x}-\frac{3 b \sinh (3 a+3 b x)}{32 x}-\frac{5 b \sinh (5 a+5 b x)}{32 x}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

Cosh[a + b*x]/(16*x^2) - Cosh[3*a + 3*b*x]/(32*x^2) - Cosh[5*a + 5*b*x]/(32*x^2) - (b^2*Cosh[a]*CoshIntegral[b
*x])/16 + (9*b^2*Cosh[3*a]*CoshIntegral[3*b*x])/32 + (25*b^2*Cosh[5*a]*CoshIntegral[5*b*x])/32 + (b*Sinh[a + b
*x])/(16*x) - (3*b*Sinh[3*a + 3*b*x])/(32*x) - (5*b*Sinh[5*a + 5*b*x])/(32*x) - (b^2*Sinh[a]*SinhIntegral[b*x]
)/16 + (9*b^2*Sinh[3*a]*SinhIntegral[3*b*x])/32 + (25*b^2*Sinh[5*a]*SinhIntegral[5*b*x])/32

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 ^2(a+b x)}{x^3} \, dx &=\int \left (-\frac{\cosh (a+b x)}{8 x^3}+\frac{\cosh (3 a+3 b x)}{16 x^3}+\frac{\cosh (5 a+5 b x)}{16 x^3}\right ) \, dx\\ &=\frac{1}{16} \int \frac{\cosh (3 a+3 b x)}{x^3} \, dx+\frac{1}{16} \int \frac{\cosh (5 a+5 b x)}{x^3} \, dx-\frac{1}{8} \int \frac{\cosh (a+b x)}{x^3} \, dx\\ &=\frac{\cosh (a+b x)}{16 x^2}-\frac{\cosh (3 a+3 b x)}{32 x^2}-\frac{\cosh (5 a+5 b x)}{32 x^2}-\frac{1}{16} b \int \frac{\sinh (a+b x)}{x^2} \, dx+\frac{1}{32} (3 b) \int \frac{\sinh (3 a+3 b x)}{x^2} \, dx+\frac{1}{32} (5 b) \int \frac{\sinh (5 a+5 b x)}{x^2} \, dx\\ &=\frac{\cosh (a+b x)}{16 x^2}-\frac{\cosh (3 a+3 b x)}{32 x^2}-\frac{\cosh (5 a+5 b x)}{32 x^2}+\frac{b \sinh (a+b x)}{16 x}-\frac{3 b \sinh (3 a+3 b x)}{32 x}-\frac{5 b \sinh (5 a+5 b x)}{32 x}-\frac{1}{16} b^2 \int \frac{\cosh (a+b x)}{x} \, dx+\frac{1}{32} \left (9 b^2\right ) \int \frac{\cosh (3 a+3 b x)}{x} \, dx+\frac{1}{32} \left (25 b^2\right ) \int \frac{\cosh (5 a+5 b x)}{x} \, dx\\ &=\frac{\cosh (a+b x)}{16 x^2}-\frac{\cosh (3 a+3 b x)}{32 x^2}-\frac{\cosh (5 a+5 b x)}{32 x^2}+\frac{b \sinh (a+b x)}{16 x}-\frac{3 b \sinh (3 a+3 b x)}{32 x}-\frac{5 b \sinh (5 a+5 b x)}{32 x}-\frac{1}{16} \left (b^2 \cosh (a)\right ) \int \frac{\cosh (b x)}{x} \, dx+\frac{1}{32} \left (9 b^2 \cosh (3 a)\right ) \int \frac{\cosh (3 b x)}{x} \, dx+\frac{1}{32} \left (25 b^2 \cosh (5 a)\right ) \int \frac{\cosh (5 b x)}{x} \, dx-\frac{1}{16} \left (b^2 \sinh (a)\right ) \int \frac{\sinh (b x)}{x} \, dx+\frac{1}{32} \left (9 b^2 \sinh (3 a)\right ) \int \frac{\sinh (3 b x)}{x} \, dx+\frac{1}{32} \left (25 b^2 \sinh (5 a)\right ) \int \frac{\sinh (5 b x)}{x} \, dx\\ &=\frac{\cosh (a+b x)}{16 x^2}-\frac{\cosh (3 a+3 b x)}{32 x^2}-\frac{\cosh (5 a+5 b x)}{32 x^2}-\frac{1}{16} b^2 \cosh (a) \text{Chi}(b x)+\frac{9}{32} b^2 \cosh (3 a) \text{Chi}(3 b x)+\frac{25}{32} b^2 \cosh (5 a) \text{Chi}(5 b x)+\frac{b \sinh (a+b x)}{16 x}-\frac{3 b \sinh (3 a+3 b x)}{32 x}-\frac{5 b \sinh (5 a+5 b x)}{32 x}-\frac{1}{16} b^2 \sinh (a) \text{Shi}(b x)+\frac{9}{32} b^2 \sinh (3 a) \text{Shi}(3 b x)+\frac{25}{32} b^2 \sinh (5 a) \text{Shi}(5 b x)\\ \end{align*}

Mathematica [A]  time = 0.558273, size = 162, normalized size = 0.88 $-\frac{2 b^2 x^2 \cosh (a) \text{Chi}(b x)-9 b^2 x^2 \cosh (3 a) \text{Chi}(3 b x)-25 b^2 x^2 \cosh (5 a) \text{Chi}(5 b x)+2 b^2 x^2 \sinh (a) \text{Shi}(b x)-9 b^2 x^2 \sinh (3 a) \text{Shi}(3 b x)-25 b^2 x^2 \sinh (5 a) \text{Shi}(5 b x)-2 b x \sinh (a+b x)+3 b x \sinh (3 (a+b x))+5 b x \sinh (5 (a+b x))-2 \cosh (a+b x)+\cosh (3 (a+b x))+\cosh (5 (a+b x))}{32 x^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(-2*Cosh[a + b*x] + Cosh[3*(a + b*x)] + Cosh[5*(a + b*x)] + 2*b^2*x^2*Cosh[a]*CoshIntegral[b*x] - 9*b^2*x^2*C
osh[3*a]*CoshIntegral[3*b*x] - 25*b^2*x^2*Cosh[5*a]*CoshIntegral[5*b*x] - 2*b*x*Sinh[a + b*x] + 3*b*x*Sinh[3*(
a + b*x)] + 5*b*x*Sinh[5*(a + b*x)] + 2*b^2*x^2*Sinh[a]*SinhIntegral[b*x] - 9*b^2*x^2*Sinh[3*a]*SinhIntegral[3
*b*x] - 25*b^2*x^2*Sinh[5*a]*SinhIntegral[5*b*x])/(32*x^2)

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Maple [A]  time = 0.115, size = 257, normalized size = 1.4 \begin{align*}{\frac{5\,b{{\rm e}^{-5\,bx-5\,a}}}{64\,x}}-{\frac{{{\rm e}^{-5\,bx-5\,a}}}{64\,{x}^{2}}}-{\frac{25\,{b}^{2}{{\rm e}^{-5\,a}}{\it Ei} \left ( 1,5\,bx \right ) }{64}}+{\frac{3\,b{{\rm e}^{-3\,bx-3\,a}}}{64\,x}}-{\frac{{{\rm e}^{-3\,bx-3\,a}}}{64\,{x}^{2}}}-{\frac{9\,{b}^{2}{{\rm e}^{-3\,a}}{\it Ei} \left ( 1,3\,bx \right ) }{64}}-{\frac{b{{\rm e}^{-bx-a}}}{32\,x}}+{\frac{{{\rm e}^{-bx-a}}}{32\,{x}^{2}}}+{\frac{{b}^{2}{{\rm e}^{-a}}{\it Ei} \left ( 1,bx \right ) }{32}}+{\frac{{{\rm e}^{bx+a}}}{32\,{x}^{2}}}+{\frac{b{{\rm e}^{bx+a}}}{32\,x}}+{\frac{{b}^{2}{{\rm e}^{a}}{\it Ei} \left ( 1,-bx \right ) }{32}}-{\frac{{{\rm e}^{3\,bx+3\,a}}}{64\,{x}^{2}}}-{\frac{3\,b{{\rm e}^{3\,bx+3\,a}}}{64\,x}}-{\frac{9\,{b}^{2}{{\rm e}^{3\,a}}{\it Ei} \left ( 1,-3\,bx \right ) }{64}}-{\frac{{{\rm e}^{5\,bx+5\,a}}}{64\,{x}^{2}}}-{\frac{5\,b{{\rm e}^{5\,bx+5\,a}}}{64\,x}}-{\frac{25\,{b}^{2}{{\rm e}^{5\,a}}{\it Ei} \left ( 1,-5\,bx \right ) }{64}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/64*b*exp(-5*b*x-5*a)/x-1/64*exp(-5*b*x-5*a)/x^2-25/64*b^2*exp(-5*a)*Ei(1,5*b*x)+3/64*b*exp(-3*b*x-3*a)/x-1/6
4*exp(-3*b*x-3*a)/x^2-9/64*b^2*exp(-3*a)*Ei(1,3*b*x)-1/32*b*exp(-b*x-a)/x+1/32*exp(-b*x-a)/x^2+1/32*b^2*exp(-a
)*Ei(1,b*x)+1/32/x^2*exp(b*x+a)+1/32*b/x*exp(b*x+a)+1/32*b^2*exp(a)*Ei(1,-b*x)-1/64/x^2*exp(3*b*x+3*a)-3/64*b/
x*exp(3*b*x+3*a)-9/64*b^2*exp(3*a)*Ei(1,-3*b*x)-1/64/x^2*exp(5*b*x+5*a)-5/64*b/x*exp(5*b*x+5*a)-25/64*b^2*exp(
5*a)*Ei(1,-5*b*x)

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Maxima [A]  time = 1.31934, size = 119, normalized size = 0.65 \begin{align*} -\frac{25}{32} \, b^{2} e^{\left (-5 \, a\right )} \Gamma \left (-2, 5 \, b x\right ) - \frac{9}{32} \, b^{2} e^{\left (-3 \, a\right )} \Gamma \left (-2, 3 \, b x\right ) + \frac{1}{16} \, b^{2} e^{\left (-a\right )} \Gamma \left (-2, b x\right ) + \frac{1}{16} \, b^{2} e^{a} \Gamma \left (-2, -b x\right ) - \frac{9}{32} \, b^{2} e^{\left (3 \, a\right )} \Gamma \left (-2, -3 \, b x\right ) - \frac{25}{32} \, b^{2} e^{\left (5 \, a\right )} \Gamma \left (-2, -5 \, 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)^2/x^3,x, algorithm="maxima")

[Out]

-25/32*b^2*e^(-5*a)*gamma(-2, 5*b*x) - 9/32*b^2*e^(-3*a)*gamma(-2, 3*b*x) + 1/16*b^2*e^(-a)*gamma(-2, b*x) + 1
/16*b^2*e^a*gamma(-2, -b*x) - 9/32*b^2*e^(3*a)*gamma(-2, -3*b*x) - 25/32*b^2*e^(5*a)*gamma(-2, -5*b*x)

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Fricas [B]  time = 1.80532, size = 859, normalized size = 4.67 \begin{align*} -\frac{10 \, b x \sinh \left (b x + a\right )^{5} + 2 \, \cosh \left (b x + a\right )^{5} + 10 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} + 2 \,{\left (50 \, b x \cosh \left (b x + a\right )^{2} + 3 \, b x\right )} \sinh \left (b x + a\right )^{3} + 2 \, \cosh \left (b x + a\right )^{3} + 2 \,{\left (10 \, \cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 25 \,{\left (b^{2} x^{2}{\rm Ei}\left (5 \, b x\right ) + b^{2} x^{2}{\rm Ei}\left (-5 \, b x\right )\right )} \cosh \left (5 \, a\right ) - 9 \,{\left (b^{2} x^{2}{\rm Ei}\left (3 \, b x\right ) + b^{2} x^{2}{\rm Ei}\left (-3 \, b x\right )\right )} \cosh \left (3 \, a\right ) + 2 \,{\left (b^{2} x^{2}{\rm Ei}\left (b x\right ) + b^{2} x^{2}{\rm Ei}\left (-b x\right )\right )} \cosh \left (a\right ) + 2 \,{\left (25 \, b x \cosh \left (b x + a\right )^{4} + 9 \, b x \cosh \left (b x + a\right )^{2} - 2 \, b x\right )} \sinh \left (b x + a\right ) - 25 \,{\left (b^{2} x^{2}{\rm Ei}\left (5 \, b x\right ) - b^{2} x^{2}{\rm Ei}\left (-5 \, b x\right )\right )} \sinh \left (5 \, a\right ) - 9 \,{\left (b^{2} x^{2}{\rm Ei}\left (3 \, b x\right ) - b^{2} x^{2}{\rm Ei}\left (-3 \, b x\right )\right )} \sinh \left (3 \, a\right ) + 2 \,{\left (b^{2} x^{2}{\rm Ei}\left (b x\right ) - b^{2} x^{2}{\rm Ei}\left (-b x\right )\right )} \sinh \left (a\right ) - 4 \, \cosh \left (b x + a\right )}{64 \, x^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(10*b*x*sinh(b*x + a)^5 + 2*cosh(b*x + a)^5 + 10*cosh(b*x + a)*sinh(b*x + a)^4 + 2*(50*b*x*cosh(b*x + a)
^2 + 3*b*x)*sinh(b*x + a)^3 + 2*cosh(b*x + a)^3 + 2*(10*cosh(b*x + a)^3 + 3*cosh(b*x + a))*sinh(b*x + a)^2 - 2
5*(b^2*x^2*Ei(5*b*x) + b^2*x^2*Ei(-5*b*x))*cosh(5*a) - 9*(b^2*x^2*Ei(3*b*x) + b^2*x^2*Ei(-3*b*x))*cosh(3*a) +
2*(b^2*x^2*Ei(b*x) + b^2*x^2*Ei(-b*x))*cosh(a) + 2*(25*b*x*cosh(b*x + a)^4 + 9*b*x*cosh(b*x + a)^2 - 2*b*x)*si
nh(b*x + a) - 25*(b^2*x^2*Ei(5*b*x) - b^2*x^2*Ei(-5*b*x))*sinh(5*a) - 9*(b^2*x^2*Ei(3*b*x) - b^2*x^2*Ei(-3*b*x
))*sinh(3*a) + 2*(b^2*x^2*Ei(b*x) - b^2*x^2*Ei(-b*x))*sinh(a) - 4*cosh(b*x + a))/x^2

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.24275, size = 328, normalized size = 1.78 \begin{align*} \frac{25 \, b^{2} x^{2}{\rm Ei}\left (5 \, b x\right ) e^{\left (5 \, a\right )} + 9 \, b^{2} x^{2}{\rm Ei}\left (3 \, b x\right ) e^{\left (3 \, a\right )} - 2 \, b^{2} x^{2}{\rm Ei}\left (-b x\right ) e^{\left (-a\right )} + 9 \, b^{2} x^{2}{\rm Ei}\left (-3 \, b x\right ) e^{\left (-3 \, a\right )} + 25 \, b^{2} x^{2}{\rm Ei}\left (-5 \, b x\right ) e^{\left (-5 \, a\right )} - 2 \, b^{2} x^{2}{\rm Ei}\left (b x\right ) e^{a} - 5 \, b x e^{\left (5 \, b x + 5 \, a\right )} - 3 \, b x e^{\left (3 \, b x + 3 \, a\right )} + 2 \, b x e^{\left (b x + a\right )} - 2 \, b x e^{\left (-b x - a\right )} + 3 \, b x e^{\left (-3 \, b x - 3 \, a\right )} + 5 \, b x e^{\left (-5 \, b x - 5 \, a\right )} - e^{\left (5 \, b x + 5 \, a\right )} - e^{\left (3 \, b x + 3 \, a\right )} + 2 \, e^{\left (b x + a\right )} + 2 \, e^{\left (-b x - a\right )} - e^{\left (-3 \, b x - 3 \, a\right )} - e^{\left (-5 \, b x - 5 \, a\right )}}{64 \, x^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(25*b^2*x^2*Ei(5*b*x)*e^(5*a) + 9*b^2*x^2*Ei(3*b*x)*e^(3*a) - 2*b^2*x^2*Ei(-b*x)*e^(-a) + 9*b^2*x^2*Ei(-3
*b*x)*e^(-3*a) + 25*b^2*x^2*Ei(-5*b*x)*e^(-5*a) - 2*b^2*x^2*Ei(b*x)*e^a - 5*b*x*e^(5*b*x + 5*a) - 3*b*x*e^(3*b
*x + 3*a) + 2*b*x*e^(b*x + a) - 2*b*x*e^(-b*x - a) + 3*b*x*e^(-3*b*x - 3*a) + 5*b*x*e^(-5*b*x - 5*a) - e^(5*b*
x + 5*a) - e^(3*b*x + 3*a) + 2*e^(b*x + a) + 2*e^(-b*x - a) - e^(-3*b*x - 3*a) - e^(-5*b*x - 5*a))/x^2