3.324 \(\int \frac{\cosh ^2(a+b x) \sinh ^3(a+b x)}{x^4} \, dx\)

Optimal. Leaf size=238 \[ -\frac{1}{48} b^3 \cosh (a) \text{Chi}(b x)-\frac{9}{32} b^3 \cosh (3 a) \text{Chi}(3 b x)+\frac{125}{96} b^3 \cosh (5 a) \text{Chi}(5 b x)-\frac{1}{48} b^3 \sinh (a) \text{Shi}(b x)-\frac{9}{32} b^3 \sinh (3 a) \text{Shi}(3 b x)+\frac{125}{96} b^3 \sinh (5 a) \text{Shi}(5 b x)+\frac{b^2 \sinh (a+b x)}{48 x}+\frac{3 b^2 \sinh (3 a+3 b x)}{32 x}-\frac{25 b^2 \sinh (5 a+5 b x)}{96 x}+\frac{\sinh (a+b x)}{24 x^3}+\frac{\sinh (3 a+3 b x)}{48 x^3}-\frac{\sinh (5 a+5 b x)}{48 x^3}+\frac{b \cosh (a+b x)}{48 x^2}+\frac{b \cosh (3 a+3 b x)}{32 x^2}-\frac{5 b \cosh (5 a+5 b x)}{96 x^2} \]

[Out]

(b*Cosh[a + b*x])/(48*x^2) + (b*Cosh[3*a + 3*b*x])/(32*x^2) - (5*b*Cosh[5*a + 5*b*x])/(96*x^2) - (b^3*Cosh[a]*
CoshIntegral[b*x])/48 - (9*b^3*Cosh[3*a]*CoshIntegral[3*b*x])/32 + (125*b^3*Cosh[5*a]*CoshIntegral[5*b*x])/96
+ Sinh[a + b*x]/(24*x^3) + (b^2*Sinh[a + b*x])/(48*x) + Sinh[3*a + 3*b*x]/(48*x^3) + (3*b^2*Sinh[3*a + 3*b*x])
/(32*x) - Sinh[5*a + 5*b*x]/(48*x^3) - (25*b^2*Sinh[5*a + 5*b*x])/(96*x) - (b^3*Sinh[a]*SinhIntegral[b*x])/48
- (9*b^3*Sinh[3*a]*SinhIntegral[3*b*x])/32 + (125*b^3*Sinh[5*a]*SinhIntegral[5*b*x])/96

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Rubi [A]  time = 0.413544, antiderivative size = 238, normalized size of antiderivative = 1., number of steps used = 20, 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} \[ -\frac{1}{48} b^3 \cosh (a) \text{Chi}(b x)-\frac{9}{32} b^3 \cosh (3 a) \text{Chi}(3 b x)+\frac{125}{96} b^3 \cosh (5 a) \text{Chi}(5 b x)-\frac{1}{48} b^3 \sinh (a) \text{Shi}(b x)-\frac{9}{32} b^3 \sinh (3 a) \text{Shi}(3 b x)+\frac{125}{96} b^3 \sinh (5 a) \text{Shi}(5 b x)+\frac{b^2 \sinh (a+b x)}{48 x}+\frac{3 b^2 \sinh (3 a+3 b x)}{32 x}-\frac{25 b^2 \sinh (5 a+5 b x)}{96 x}+\frac{\sinh (a+b x)}{24 x^3}+\frac{\sinh (3 a+3 b x)}{48 x^3}-\frac{\sinh (5 a+5 b x)}{48 x^3}+\frac{b \cosh (a+b x)}{48 x^2}+\frac{b \cosh (3 a+3 b x)}{32 x^2}-\frac{5 b \cosh (5 a+5 b x)}{96 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cosh[a + b*x]^2*Sinh[a + b*x]^3)/x^4,x]

[Out]

(b*Cosh[a + b*x])/(48*x^2) + (b*Cosh[3*a + 3*b*x])/(32*x^2) - (5*b*Cosh[5*a + 5*b*x])/(96*x^2) - (b^3*Cosh[a]*
CoshIntegral[b*x])/48 - (9*b^3*Cosh[3*a]*CoshIntegral[3*b*x])/32 + (125*b^3*Cosh[5*a]*CoshIntegral[5*b*x])/96
+ Sinh[a + b*x]/(24*x^3) + (b^2*Sinh[a + b*x])/(48*x) + Sinh[3*a + 3*b*x]/(48*x^3) + (3*b^2*Sinh[3*a + 3*b*x])
/(32*x) - Sinh[5*a + 5*b*x]/(48*x^3) - (25*b^2*Sinh[5*a + 5*b*x])/(96*x) - (b^3*Sinh[a]*SinhIntegral[b*x])/48
- (9*b^3*Sinh[3*a]*SinhIntegral[3*b*x])/32 + (125*b^3*Sinh[5*a]*SinhIntegral[5*b*x])/96

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^4} \, dx &=\int \left (-\frac{\sinh (a+b x)}{8 x^4}-\frac{\sinh (3 a+3 b x)}{16 x^4}+\frac{\sinh (5 a+5 b x)}{16 x^4}\right ) \, dx\\ &=-\left (\frac{1}{16} \int \frac{\sinh (3 a+3 b x)}{x^4} \, dx\right )+\frac{1}{16} \int \frac{\sinh (5 a+5 b x)}{x^4} \, dx-\frac{1}{8} \int \frac{\sinh (a+b x)}{x^4} \, dx\\ &=\frac{\sinh (a+b x)}{24 x^3}+\frac{\sinh (3 a+3 b x)}{48 x^3}-\frac{\sinh (5 a+5 b x)}{48 x^3}-\frac{1}{24} b \int \frac{\cosh (a+b x)}{x^3} \, dx-\frac{1}{16} b \int \frac{\cosh (3 a+3 b x)}{x^3} \, dx+\frac{1}{48} (5 b) \int \frac{\cosh (5 a+5 b x)}{x^3} \, dx\\ &=\frac{b \cosh (a+b x)}{48 x^2}+\frac{b \cosh (3 a+3 b x)}{32 x^2}-\frac{5 b \cosh (5 a+5 b x)}{96 x^2}+\frac{\sinh (a+b x)}{24 x^3}+\frac{\sinh (3 a+3 b x)}{48 x^3}-\frac{\sinh (5 a+5 b x)}{48 x^3}-\frac{1}{48} b^2 \int \frac{\sinh (a+b x)}{x^2} \, dx-\frac{1}{32} \left (3 b^2\right ) \int \frac{\sinh (3 a+3 b x)}{x^2} \, dx+\frac{1}{96} \left (25 b^2\right ) \int \frac{\sinh (5 a+5 b x)}{x^2} \, dx\\ &=\frac{b \cosh (a+b x)}{48 x^2}+\frac{b \cosh (3 a+3 b x)}{32 x^2}-\frac{5 b \cosh (5 a+5 b x)}{96 x^2}+\frac{\sinh (a+b x)}{24 x^3}+\frac{b^2 \sinh (a+b x)}{48 x}+\frac{\sinh (3 a+3 b x)}{48 x^3}+\frac{3 b^2 \sinh (3 a+3 b x)}{32 x}-\frac{\sinh (5 a+5 b x)}{48 x^3}-\frac{25 b^2 \sinh (5 a+5 b x)}{96 x}-\frac{1}{48} b^3 \int \frac{\cosh (a+b x)}{x} \, dx-\frac{1}{32} \left (9 b^3\right ) \int \frac{\cosh (3 a+3 b x)}{x} \, dx+\frac{1}{96} \left (125 b^3\right ) \int \frac{\cosh (5 a+5 b x)}{x} \, dx\\ &=\frac{b \cosh (a+b x)}{48 x^2}+\frac{b \cosh (3 a+3 b x)}{32 x^2}-\frac{5 b \cosh (5 a+5 b x)}{96 x^2}+\frac{\sinh (a+b x)}{24 x^3}+\frac{b^2 \sinh (a+b x)}{48 x}+\frac{\sinh (3 a+3 b x)}{48 x^3}+\frac{3 b^2 \sinh (3 a+3 b x)}{32 x}-\frac{\sinh (5 a+5 b x)}{48 x^3}-\frac{25 b^2 \sinh (5 a+5 b x)}{96 x}-\frac{1}{48} \left (b^3 \cosh (a)\right ) \int \frac{\cosh (b x)}{x} \, dx-\frac{1}{32} \left (9 b^3 \cosh (3 a)\right ) \int \frac{\cosh (3 b x)}{x} \, dx+\frac{1}{96} \left (125 b^3 \cosh (5 a)\right ) \int \frac{\cosh (5 b x)}{x} \, dx-\frac{1}{48} \left (b^3 \sinh (a)\right ) \int \frac{\sinh (b x)}{x} \, dx-\frac{1}{32} \left (9 b^3 \sinh (3 a)\right ) \int \frac{\sinh (3 b x)}{x} \, dx+\frac{1}{96} \left (125 b^3 \sinh (5 a)\right ) \int \frac{\sinh (5 b x)}{x} \, dx\\ &=\frac{b \cosh (a+b x)}{48 x^2}+\frac{b \cosh (3 a+3 b x)}{32 x^2}-\frac{5 b \cosh (5 a+5 b x)}{96 x^2}-\frac{1}{48} b^3 \cosh (a) \text{Chi}(b x)-\frac{9}{32} b^3 \cosh (3 a) \text{Chi}(3 b x)+\frac{125}{96} b^3 \cosh (5 a) \text{Chi}(5 b x)+\frac{\sinh (a+b x)}{24 x^3}+\frac{b^2 \sinh (a+b x)}{48 x}+\frac{\sinh (3 a+3 b x)}{48 x^3}+\frac{3 b^2 \sinh (3 a+3 b x)}{32 x}-\frac{\sinh (5 a+5 b x)}{48 x^3}-\frac{25 b^2 \sinh (5 a+5 b x)}{96 x}-\frac{1}{48} b^3 \sinh (a) \text{Shi}(b x)-\frac{9}{32} b^3 \sinh (3 a) \text{Shi}(3 b x)+\frac{125}{96} b^3 \sinh (5 a) \text{Shi}(5 b x)\\ \end{align*}

Mathematica [A]  time = 0.520538, size = 212, normalized size = 0.89 \[ \frac{-2 b^3 x^3 \cosh (a) \text{Chi}(b x)-27 b^3 x^3 \cosh (3 a) \text{Chi}(3 b x)+125 b^3 x^3 \cosh (5 a) \text{Chi}(5 b x)-2 b^3 x^3 \sinh (a) \text{Shi}(b x)-27 b^3 x^3 \sinh (3 a) \text{Shi}(3 b x)+125 b^3 x^3 \sinh (5 a) \text{Shi}(5 b x)+2 b^2 x^2 \sinh (a+b x)+9 b^2 x^2 \sinh (3 (a+b x))-25 b^2 x^2 \sinh (5 (a+b x))+4 \sinh (a+b x)+2 \sinh (3 (a+b x))-2 \sinh (5 (a+b x))+2 b x \cosh (a+b x)+3 b x \cosh (3 (a+b x))-5 b x \cosh (5 (a+b x))}{96 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cosh[a + b*x]^2*Sinh[a + b*x]^3)/x^4,x]

[Out]

(2*b*x*Cosh[a + b*x] + 3*b*x*Cosh[3*(a + b*x)] - 5*b*x*Cosh[5*(a + b*x)] - 2*b^3*x^3*Cosh[a]*CoshIntegral[b*x]
- 27*b^3*x^3*Cosh[3*a]*CoshIntegral[3*b*x] + 125*b^3*x^3*Cosh[5*a]*CoshIntegral[5*b*x] + 4*Sinh[a + b*x] + 2*
b^2*x^2*Sinh[a + b*x] + 2*Sinh[3*(a + b*x)] + 9*b^2*x^2*Sinh[3*(a + b*x)] - 2*Sinh[5*(a + b*x)] - 25*b^2*x^2*S
inh[5*(a + b*x)] - 2*b^3*x^3*Sinh[a]*SinhIntegral[b*x] - 27*b^3*x^3*Sinh[3*a]*SinhIntegral[3*b*x] + 125*b^3*x^
3*Sinh[5*a]*SinhIntegral[5*b*x])/(96*x^3)

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Maple [A]  time = 0.102, size = 356, normalized size = 1.5 \begin{align*}{\frac{25\,{b}^{2}{{\rm e}^{-5\,bx-5\,a}}}{192\,x}}-{\frac{5\,b{{\rm e}^{-5\,bx-5\,a}}}{192\,{x}^{2}}}+{\frac{{{\rm e}^{-5\,bx-5\,a}}}{96\,{x}^{3}}}-{\frac{125\,{b}^{3}{{\rm e}^{-5\,a}}{\it Ei} \left ( 1,5\,bx \right ) }{192}}-{\frac{3\,{b}^{2}{{\rm e}^{-3\,bx-3\,a}}}{64\,x}}+{\frac{b{{\rm e}^{-3\,bx-3\,a}}}{64\,{x}^{2}}}-{\frac{{{\rm e}^{-3\,bx-3\,a}}}{96\,{x}^{3}}}+{\frac{9\,{b}^{3}{{\rm e}^{-3\,a}}{\it Ei} \left ( 1,3\,bx \right ) }{64}}-{\frac{{b}^{2}{{\rm e}^{-bx-a}}}{96\,x}}+{\frac{b{{\rm e}^{-bx-a}}}{96\,{x}^{2}}}-{\frac{{{\rm e}^{-bx-a}}}{48\,{x}^{3}}}+{\frac{{b}^{3}{{\rm e}^{-a}}{\it Ei} \left ( 1,bx \right ) }{96}}+{\frac{{{\rm e}^{bx+a}}}{48\,{x}^{3}}}+{\frac{b{{\rm e}^{bx+a}}}{96\,{x}^{2}}}+{\frac{{b}^{2}{{\rm e}^{bx+a}}}{96\,x}}+{\frac{{b}^{3}{{\rm e}^{a}}{\it Ei} \left ( 1,-bx \right ) }{96}}+{\frac{{{\rm e}^{3\,bx+3\,a}}}{96\,{x}^{3}}}+{\frac{b{{\rm e}^{3\,bx+3\,a}}}{64\,{x}^{2}}}+{\frac{3\,{b}^{2}{{\rm e}^{3\,bx+3\,a}}}{64\,x}}+{\frac{9\,{b}^{3}{{\rm e}^{3\,a}}{\it Ei} \left ( 1,-3\,bx \right ) }{64}}-{\frac{{{\rm e}^{5\,bx+5\,a}}}{96\,{x}^{3}}}-{\frac{5\,b{{\rm e}^{5\,bx+5\,a}}}{192\,{x}^{2}}}-{\frac{25\,{b}^{2}{{\rm e}^{5\,bx+5\,a}}}{192\,x}}-{\frac{125\,{b}^{3}{{\rm e}^{5\,a}}{\it Ei} \left ( 1,-5\,bx \right ) }{192}} \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^4,x)

[Out]

25/192*b^2*exp(-5*b*x-5*a)/x-5/192*b*exp(-5*b*x-5*a)/x^2+1/96*exp(-5*b*x-5*a)/x^3-125/192*b^3*exp(-5*a)*Ei(1,5
*b*x)-3/64*b^2*exp(-3*b*x-3*a)/x+1/64*b*exp(-3*b*x-3*a)/x^2-1/96*exp(-3*b*x-3*a)/x^3+9/64*b^3*exp(-3*a)*Ei(1,3
*b*x)-1/96*b^2*exp(-b*x-a)/x+1/96*b*exp(-b*x-a)/x^2-1/48*exp(-b*x-a)/x^3+1/96*b^3*exp(-a)*Ei(1,b*x)+1/48/x^3*e
xp(b*x+a)+1/96*b/x^2*exp(b*x+a)+1/96*b^2/x*exp(b*x+a)+1/96*b^3*exp(a)*Ei(1,-b*x)+1/96/x^3*exp(3*b*x+3*a)+1/64*
b/x^2*exp(3*b*x+3*a)+3/64*b^2/x*exp(3*b*x+3*a)+9/64*b^3*exp(3*a)*Ei(1,-3*b*x)-1/96/x^3*exp(5*b*x+5*a)-5/192*b/
x^2*exp(5*b*x+5*a)-25/192*b^2/x*exp(5*b*x+5*a)-125/192*b^3*exp(5*a)*Ei(1,-5*b*x)

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Maxima [A]  time = 1.30185, size = 119, normalized size = 0.5 \begin{align*} \frac{125}{32} \, b^{3} e^{\left (-5 \, a\right )} \Gamma \left (-3, 5 \, b x\right ) - \frac{27}{32} \, b^{3} e^{\left (-3 \, a\right )} \Gamma \left (-3, 3 \, b x\right ) - \frac{1}{16} \, b^{3} e^{\left (-a\right )} \Gamma \left (-3, b x\right ) - \frac{1}{16} \, b^{3} e^{a} \Gamma \left (-3, -b x\right ) - \frac{27}{32} \, b^{3} e^{\left (3 \, a\right )} \Gamma \left (-3, -3 \, b x\right ) + \frac{125}{32} \, b^{3} e^{\left (5 \, a\right )} \Gamma \left (-3, -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^4,x, algorithm="maxima")

[Out]

125/32*b^3*e^(-5*a)*gamma(-3, 5*b*x) - 27/32*b^3*e^(-3*a)*gamma(-3, 3*b*x) - 1/16*b^3*e^(-a)*gamma(-3, b*x) -
1/16*b^3*e^a*gamma(-3, -b*x) - 27/32*b^3*e^(3*a)*gamma(-3, -3*b*x) + 125/32*b^3*e^(5*a)*gamma(-3, -5*b*x)

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Fricas [A]  time = 1.87401, size = 987, normalized size = 4.15 \begin{align*} -\frac{10 \, b x \cosh \left (b x + a\right )^{5} + 50 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} + 2 \,{\left (25 \, b^{2} x^{2} + 2\right )} \sinh \left (b x + a\right )^{5} - 6 \, b x \cosh \left (b x + a\right )^{3} - 2 \,{\left (9 \, b^{2} x^{2} - 10 \,{\left (25 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )^{2} + 2\right )} \sinh \left (b x + a\right )^{3} - 4 \, b x \cosh \left (b x + a\right ) + 2 \,{\left (50 \, b x \cosh \left (b x + a\right )^{3} - 9 \, b x \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 125 \,{\left (b^{3} x^{3}{\rm Ei}\left (5 \, b x\right ) + b^{3} x^{3}{\rm Ei}\left (-5 \, b x\right )\right )} \cosh \left (5 \, a\right ) + 27 \,{\left (b^{3} x^{3}{\rm Ei}\left (3 \, b x\right ) + b^{3} x^{3}{\rm Ei}\left (-3 \, b x\right )\right )} \cosh \left (3 \, a\right ) + 2 \,{\left (b^{3} x^{3}{\rm Ei}\left (b x\right ) + b^{3} x^{3}{\rm Ei}\left (-b x\right )\right )} \cosh \left (a\right ) + 2 \,{\left (5 \,{\left (25 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )^{4} - 2 \, b^{2} x^{2} - 3 \,{\left (9 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )^{2} - 4\right )} \sinh \left (b x + a\right ) - 125 \,{\left (b^{3} x^{3}{\rm Ei}\left (5 \, b x\right ) - b^{3} x^{3}{\rm Ei}\left (-5 \, b x\right )\right )} \sinh \left (5 \, a\right ) + 27 \,{\left (b^{3} x^{3}{\rm Ei}\left (3 \, b x\right ) - b^{3} x^{3}{\rm Ei}\left (-3 \, b x\right )\right )} \sinh \left (3 \, a\right ) + 2 \,{\left (b^{3} x^{3}{\rm Ei}\left (b x\right ) - b^{3} x^{3}{\rm Ei}\left (-b x\right )\right )} \sinh \left (a\right )}{192 \, x^{3}} \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^4,x, algorithm="fricas")

[Out]

-1/192*(10*b*x*cosh(b*x + a)^5 + 50*b*x*cosh(b*x + a)*sinh(b*x + a)^4 + 2*(25*b^2*x^2 + 2)*sinh(b*x + a)^5 - 6
*b*x*cosh(b*x + a)^3 - 2*(9*b^2*x^2 - 10*(25*b^2*x^2 + 2)*cosh(b*x + a)^2 + 2)*sinh(b*x + a)^3 - 4*b*x*cosh(b*
x + a) + 2*(50*b*x*cosh(b*x + a)^3 - 9*b*x*cosh(b*x + a))*sinh(b*x + a)^2 - 125*(b^3*x^3*Ei(5*b*x) + b^3*x^3*E
i(-5*b*x))*cosh(5*a) + 27*(b^3*x^3*Ei(3*b*x) + b^3*x^3*Ei(-3*b*x))*cosh(3*a) + 2*(b^3*x^3*Ei(b*x) + b^3*x^3*Ei
(-b*x))*cosh(a) + 2*(5*(25*b^2*x^2 + 2)*cosh(b*x + a)^4 - 2*b^2*x^2 - 3*(9*b^2*x^2 + 2)*cosh(b*x + a)^2 - 4)*s
inh(b*x + a) - 125*(b^3*x^3*Ei(5*b*x) - b^3*x^3*Ei(-5*b*x))*sinh(5*a) + 27*(b^3*x^3*Ei(3*b*x) - b^3*x^3*Ei(-3*
b*x))*sinh(3*a) + 2*(b^3*x^3*Ei(b*x) - b^3*x^3*Ei(-b*x))*sinh(a))/x^3

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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^{4}}\, 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**4,x)

[Out]

Integral(sinh(a + b*x)**3*cosh(a + b*x)**2/x**4, x)

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Giac [A]  time = 1.17844, size = 462, normalized size = 1.94 \begin{align*} \frac{125 \, b^{3} x^{3}{\rm Ei}\left (5 \, b x\right ) e^{\left (5 \, a\right )} - 27 \, b^{3} x^{3}{\rm Ei}\left (3 \, b x\right ) e^{\left (3 \, a\right )} - 2 \, b^{3} x^{3}{\rm Ei}\left (-b x\right ) e^{\left (-a\right )} - 27 \, b^{3} x^{3}{\rm Ei}\left (-3 \, b x\right ) e^{\left (-3 \, a\right )} + 125 \, b^{3} x^{3}{\rm Ei}\left (-5 \, b x\right ) e^{\left (-5 \, a\right )} - 2 \, b^{3} x^{3}{\rm Ei}\left (b x\right ) e^{a} - 25 \, b^{2} x^{2} e^{\left (5 \, b x + 5 \, a\right )} + 9 \, b^{2} x^{2} e^{\left (3 \, b x + 3 \, a\right )} + 2 \, b^{2} x^{2} e^{\left (b x + a\right )} - 2 \, b^{2} x^{2} e^{\left (-b x - a\right )} - 9 \, b^{2} x^{2} e^{\left (-3 \, b x - 3 \, a\right )} + 25 \, b^{2} x^{2} e^{\left (-5 \, b x - 5 \, a\right )} - 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 )} - 2 \, e^{\left (5 \, b x + 5 \, a\right )} + 2 \, e^{\left (3 \, b x + 3 \, a\right )} + 4 \, e^{\left (b x + a\right )} - 4 \, e^{\left (-b x - a\right )} - 2 \, e^{\left (-3 \, b x - 3 \, a\right )} + 2 \, e^{\left (-5 \, b x - 5 \, a\right )}}{192 \, x^{3}} \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^4,x, algorithm="giac")

[Out]

1/192*(125*b^3*x^3*Ei(5*b*x)*e^(5*a) - 27*b^3*x^3*Ei(3*b*x)*e^(3*a) - 2*b^3*x^3*Ei(-b*x)*e^(-a) - 27*b^3*x^3*E
i(-3*b*x)*e^(-3*a) + 125*b^3*x^3*Ei(-5*b*x)*e^(-5*a) - 2*b^3*x^3*Ei(b*x)*e^a - 25*b^2*x^2*e^(5*b*x + 5*a) + 9*
b^2*x^2*e^(3*b*x + 3*a) + 2*b^2*x^2*e^(b*x + a) - 2*b^2*x^2*e^(-b*x - a) - 9*b^2*x^2*e^(-3*b*x - 3*a) + 25*b^2
*x^2*e^(-5*b*x - 5*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) - 2*e^(5*b*x + 5*a) + 2*e^(3*b*x + 3*a) + 4*e^(b*x + a) - 4
*e^(-b*x - a) - 2*e^(-3*b*x - 3*a) + 2*e^(-5*b*x - 5*a))/x^3