### 3.322 $$\int \frac{\cosh ^2(a+b x) \sinh ^3(a+b x)}{x^2} \, dx$$

Optimal. Leaf size=124 $-\frac{1}{8} b \cosh (a) \text{Chi}(b x)-\frac{3}{16} b \cosh (3 a) \text{Chi}(3 b x)+\frac{5}{16} b \cosh (5 a) \text{Chi}(5 b x)-\frac{1}{8} b \sinh (a) \text{Shi}(b x)-\frac{3}{16} b \sinh (3 a) \text{Shi}(3 b x)+\frac{5}{16} b \sinh (5 a) \text{Shi}(5 b x)+\frac{\sinh (a+b x)}{8 x}+\frac{\sinh (3 a+3 b x)}{16 x}-\frac{\sinh (5 a+5 b x)}{16 x}$

[Out]

-(b*Cosh[a]*CoshIntegral[b*x])/8 - (3*b*Cosh[3*a]*CoshIntegral[3*b*x])/16 + (5*b*Cosh[5*a]*CoshIntegral[5*b*x]
)/16 + Sinh[a + b*x]/(8*x) + Sinh[3*a + 3*b*x]/(16*x) - Sinh[5*a + 5*b*x]/(16*x) - (b*Sinh[a]*SinhIntegral[b*x
])/8 - (3*b*Sinh[3*a]*SinhIntegral[3*b*x])/16 + (5*b*Sinh[5*a]*SinhIntegral[5*b*x])/16

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Rubi [A]  time = 0.249725, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 14, 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} $-\frac{1}{8} b \cosh (a) \text{Chi}(b x)-\frac{3}{16} b \cosh (3 a) \text{Chi}(3 b x)+\frac{5}{16} b \cosh (5 a) \text{Chi}(5 b x)-\frac{1}{8} b \sinh (a) \text{Shi}(b x)-\frac{3}{16} b \sinh (3 a) \text{Shi}(3 b x)+\frac{5}{16} b \sinh (5 a) \text{Shi}(5 b x)+\frac{\sinh (a+b x)}{8 x}+\frac{\sinh (3 a+3 b x)}{16 x}-\frac{\sinh (5 a+5 b x)}{16 x}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(b*Cosh[a]*CoshIntegral[b*x])/8 - (3*b*Cosh[3*a]*CoshIntegral[3*b*x])/16 + (5*b*Cosh[5*a]*CoshIntegral[5*b*x]
)/16 + Sinh[a + b*x]/(8*x) + Sinh[3*a + 3*b*x]/(16*x) - Sinh[5*a + 5*b*x]/(16*x) - (b*Sinh[a]*SinhIntegral[b*x
])/8 - (3*b*Sinh[3*a]*SinhIntegral[3*b*x])/16 + (5*b*Sinh[5*a]*SinhIntegral[5*b*x])/16

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

Mathematica [A]  time = 0.264244, size = 106, normalized size = 0.85 $\frac{-2 b x \cosh (a) \text{Chi}(b x)-3 b x \cosh (3 a) \text{Chi}(3 b x)+5 b x \cosh (5 a) \text{Chi}(5 b x)-2 b x \sinh (a) \text{Shi}(b x)-3 b x \sinh (3 a) \text{Shi}(3 b x)+5 b x \sinh (5 a) \text{Shi}(5 b x)+2 \sinh (a+b x)+\sinh (3 (a+b x))-\sinh (5 (a+b x))}{16 x}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.095, size = 158, normalized size = 1.3 \begin{align*}{\frac{{{\rm e}^{-5\,bx-5\,a}}}{32\,x}}-{\frac{5\,b{{\rm e}^{-5\,a}}{\it Ei} \left ( 1,5\,bx \right ) }{32}}-{\frac{{{\rm e}^{-3\,bx-3\,a}}}{32\,x}}+{\frac{3\,b{{\rm e}^{-3\,a}}{\it Ei} \left ( 1,3\,bx \right ) }{32}}-{\frac{{{\rm e}^{-bx-a}}}{16\,x}}+{\frac{b{{\rm e}^{-a}}{\it Ei} \left ( 1,bx \right ) }{16}}+{\frac{{{\rm e}^{bx+a}}}{16\,x}}+{\frac{b{{\rm e}^{a}}{\it Ei} \left ( 1,-bx \right ) }{16}}+{\frac{{{\rm e}^{3\,bx+3\,a}}}{32\,x}}+{\frac{3\,b{{\rm e}^{3\,a}}{\it Ei} \left ( 1,-3\,bx \right ) }{32}}-{\frac{{{\rm e}^{5\,bx+5\,a}}}{32\,x}}-{\frac{5\,b{{\rm e}^{5\,a}}{\it Ei} \left ( 1,-5\,bx \right ) }{32}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*exp(-5*b*x-5*a)/x-5/32*b*exp(-5*a)*Ei(1,5*b*x)-1/32*exp(-3*b*x-3*a)/x+3/32*b*exp(-3*a)*Ei(1,3*b*x)-1/16*e
xp(-b*x-a)/x+1/16*b*exp(-a)*Ei(1,b*x)+1/16/x*exp(b*x+a)+1/16*b*exp(a)*Ei(1,-b*x)+1/32/x*exp(3*b*x+3*a)+3/32*b*
exp(3*a)*Ei(1,-3*b*x)-1/32/x*exp(5*b*x+5*a)-5/32*b*exp(5*a)*Ei(1,-5*b*x)

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Maxima [A]  time = 1.2915, size = 103, normalized size = 0.83 \begin{align*} \frac{5}{32} \, b e^{\left (-5 \, a\right )} \Gamma \left (-1, 5 \, b x\right ) - \frac{3}{32} \, b e^{\left (-3 \, a\right )} \Gamma \left (-1, 3 \, b x\right ) - \frac{1}{16} \, b e^{\left (-a\right )} \Gamma \left (-1, b x\right ) - \frac{1}{16} \, b e^{a} \Gamma \left (-1, -b x\right ) - \frac{3}{32} \, b e^{\left (3 \, a\right )} \Gamma \left (-1, -3 \, b x\right ) + \frac{5}{32} \, b e^{\left (5 \, a\right )} \Gamma \left (-1, -5 \, b x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.83772, size = 548, normalized size = 4.42 \begin{align*} -\frac{2 \, \sinh \left (b x + a\right )^{5} + 2 \,{\left (10 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{3} - 5 \,{\left (b x{\rm Ei}\left (5 \, b x\right ) + b x{\rm Ei}\left (-5 \, b x\right )\right )} \cosh \left (5 \, a\right ) + 3 \,{\left (b x{\rm Ei}\left (3 \, b x\right ) + b x{\rm Ei}\left (-3 \, b x\right )\right )} \cosh \left (3 \, a\right ) + 2 \,{\left (b x{\rm Ei}\left (b x\right ) + b x{\rm Ei}\left (-b x\right )\right )} \cosh \left (a\right ) + 2 \,{\left (5 \, \cosh \left (b x + a\right )^{4} - 3 \, \cosh \left (b x + a\right )^{2} - 2\right )} \sinh \left (b x + a\right ) - 5 \,{\left (b x{\rm Ei}\left (5 \, b x\right ) - b x{\rm Ei}\left (-5 \, b x\right )\right )} \sinh \left (5 \, a\right ) + 3 \,{\left (b x{\rm Ei}\left (3 \, b x\right ) - b x{\rm Ei}\left (-3 \, b x\right )\right )} \sinh \left (3 \, a\right ) + 2 \,{\left (b x{\rm Ei}\left (b x\right ) - b x{\rm Ei}\left (-b x\right )\right )} \sinh \left (a\right )}{32 \, x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(2*sinh(b*x + a)^5 + 2*(10*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^3 - 5*(b*x*Ei(5*b*x) + b*x*Ei(-5*b*x))*cos
h(5*a) + 3*(b*x*Ei(3*b*x) + b*x*Ei(-3*b*x))*cosh(3*a) + 2*(b*x*Ei(b*x) + b*x*Ei(-b*x))*cosh(a) + 2*(5*cosh(b*x
+ a)^4 - 3*cosh(b*x + a)^2 - 2)*sinh(b*x + a) - 5*(b*x*Ei(5*b*x) - b*x*Ei(-5*b*x))*sinh(5*a) + 3*(b*x*Ei(3*b*
x) - b*x*Ei(-3*b*x))*sinh(3*a) + 2*(b*x*Ei(b*x) - b*x*Ei(-b*x))*sinh(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 ^{2}{\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)**2*sinh(b*x+a)**3/x**2,x)

[Out]

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

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Giac [A]  time = 1.19002, size = 189, normalized size = 1.52 \begin{align*} \frac{5 \, b x{\rm Ei}\left (5 \, b x\right ) e^{\left (5 \, a\right )} - 3 \, b x{\rm Ei}\left (3 \, b x\right ) e^{\left (3 \, a\right )} - 2 \, b x{\rm Ei}\left (-b x\right ) e^{\left (-a\right )} - 3 \, b x{\rm Ei}\left (-3 \, b x\right ) e^{\left (-3 \, a\right )} + 5 \, b x{\rm Ei}\left (-5 \, b x\right ) e^{\left (-5 \, a\right )} - 2 \, b x{\rm Ei}\left (b x\right ) e^{a} - 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 )}}{32 \, x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(5*b*x*Ei(5*b*x)*e^(5*a) - 3*b*x*Ei(3*b*x)*e^(3*a) - 2*b*x*Ei(-b*x)*e^(-a) - 3*b*x*Ei(-3*b*x)*e^(-3*a) +
5*b*x*Ei(-5*b*x)*e^(-5*a) - 2*b*x*Ei(b*x)*e^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