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]

-1/16*b^2*Chi(b*x)*cosh(a)+9/32*b^2*Chi(3*b*x)*cosh(3*a)+25/32*b^2*Chi(5*b*x)*cosh(5*a)+1/16*cosh(b*x+a)/x^2-1
/32*cosh(3*b*x+3*a)/x^2-1/32*cosh(5*b*x+5*a)/x^2-1/16*b^2*Shi(b*x)*sinh(a)+9/32*b^2*Shi(3*b*x)*sinh(3*a)+25/32
*b^2*Shi(5*b*x)*sinh(5*a)+1/16*b*sinh(b*x+a)/x-3/32*b*sinh(3*b*x+3*a)/x-5/32*b*sinh(5*b*x+5*a)/x

________________________________________________________________________________________

Rubi [A]  time = 0.34, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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 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]

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 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]

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 = 0.56, 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]

-1/32*(-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*Cosh[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*Sin
h[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]*SinhInteg
ral[3*b*x] - 25*b^2*x^2*Sinh[5*a]*SinhIntegral[5*b*x])/x^2

________________________________________________________________________________________

fricas [B]  time = 0.59, size = 338, normalized size = 1.84 \[ -\frac {10 \, b x \sinh \left (b x + a\right )^{5} + 2 \, \cosh \left (b x + a\right )^{5} + 10 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} + 2 \, {\left (50 \, b x \cosh \left (b x + a\right )^{2} + 3 \, b x\right )} \sinh \left (b x + a\right )^{3} + 2 \, \cosh \left (b x + a\right )^{3} + 2 \, {\left (10 \, \cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 25 \, {\left (b^{2} x^{2} {\rm Ei}\left (5 \, b x\right ) + b^{2} x^{2} {\rm Ei}\left (-5 \, b x\right )\right )} \cosh \left (5 \, a\right ) - 9 \, {\left (b^{2} x^{2} {\rm Ei}\left (3 \, b x\right ) + b^{2} x^{2} {\rm Ei}\left (-3 \, b x\right )\right )} \cosh \left (3 \, a\right ) + 2 \, {\left (b^{2} x^{2} {\rm Ei}\left (b x\right ) + b^{2} x^{2} {\rm Ei}\left (-b x\right )\right )} \cosh \relax (a) + 2 \, {\left (25 \, b x \cosh \left (b x + a\right )^{4} + 9 \, b x \cosh \left (b x + a\right )^{2} - 2 \, b x\right )} \sinh \left (b x + a\right ) - 25 \, {\left (b^{2} x^{2} {\rm Ei}\left (5 \, b x\right ) - b^{2} x^{2} {\rm Ei}\left (-5 \, b x\right )\right )} \sinh \left (5 \, a\right ) - 9 \, {\left (b^{2} x^{2} {\rm Ei}\left (3 \, b x\right ) - b^{2} x^{2} {\rm Ei}\left (-3 \, b x\right )\right )} \sinh \left (3 \, a\right ) + 2 \, {\left (b^{2} x^{2} {\rm Ei}\left (b x\right ) - b^{2} x^{2} {\rm Ei}\left (-b x\right )\right )} \sinh \relax (a) - 4 \, \cosh \left (b x + a\right )}{64 \, x^{2}} \]

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

________________________________________________________________________________________

giac [A]  time = 0.13, size = 243, normalized size = 1.32 \[ \frac {25 \, b^{2} x^{2} {\rm Ei}\left (5 \, b x\right ) e^{\left (5 \, a\right )} + 9 \, b^{2} x^{2} {\rm Ei}\left (3 \, b x\right ) e^{\left (3 \, a\right )} - 2 \, b^{2} x^{2} {\rm Ei}\left (-b x\right ) e^{\left (-a\right )} + 9 \, b^{2} x^{2} {\rm Ei}\left (-3 \, b x\right ) e^{\left (-3 \, a\right )} + 25 \, b^{2} x^{2} {\rm Ei}\left (-5 \, b x\right ) e^{\left (-5 \, a\right )} - 2 \, b^{2} x^{2} {\rm Ei}\left (b x\right ) e^{a} - 5 \, b x e^{\left (5 \, b x + 5 \, a\right )} - 3 \, b x e^{\left (3 \, b x + 3 \, a\right )} + 2 \, b x e^{\left (b x + a\right )} - 2 \, b x e^{\left (-b x - a\right )} + 3 \, b x e^{\left (-3 \, b x - 3 \, a\right )} + 5 \, b x e^{\left (-5 \, b x - 5 \, a\right )} - e^{\left (5 \, b x + 5 \, a\right )} - e^{\left (3 \, b x + 3 \, a\right )} + 2 \, e^{\left (b x + a\right )} + 2 \, e^{\left (-b x - a\right )} - e^{\left (-3 \, b x - 3 \, a\right )} - e^{\left (-5 \, b x - 5 \, a\right )}}{64 \, x^{2}} \]

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

________________________________________________________________________________________

maple [A]  time = 0.53, size = 257, normalized size = 1.40 \[ \frac {5 b \,{\mathrm e}^{-5 b x -5 a}}{64 x}-\frac {{\mathrm e}^{-5 b x -5 a}}{64 x^{2}}-\frac {25 b^{2} {\mathrm e}^{-5 a} \Ei \left (1, 5 b x \right )}{64}+\frac {3 b \,{\mathrm e}^{-3 b x -3 a}}{64 x}-\frac {{\mathrm e}^{-3 b x -3 a}}{64 x^{2}}-\frac {9 b^{2} {\mathrm e}^{-3 a} \Ei \left (1, 3 b x \right )}{64}-\frac {b \,{\mathrm e}^{-b x -a}}{32 x}+\frac {{\mathrm e}^{-b x -a}}{32 x^{2}}+\frac {b^{2} {\mathrm e}^{-a} \Ei \left (1, b x \right )}{32}+\frac {{\mathrm e}^{b x +a}}{32 x^{2}}+\frac {b \,{\mathrm e}^{b x +a}}{32 x}+\frac {b^{2} {\mathrm e}^{a} \Ei \left (1, -b x \right )}{32}-\frac {{\mathrm e}^{3 b x +3 a}}{64 x^{2}}-\frac {3 b \,{\mathrm e}^{3 b x +3 a}}{64 x}-\frac {9 b^{2} {\mathrm e}^{3 a} \Ei \left (1, -3 b x \right )}{64}-\frac {{\mathrm e}^{5 b x +5 a}}{64 x^{2}}-\frac {5 b \,{\mathrm e}^{5 b x +5 a}}{64 x}-\frac {25 b^{2} {\mathrm e}^{5 a} \Ei \left (1, -5 b x \right )}{64} \]

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)

________________________________________________________________________________________

maxima [A]  time = 0.84, size = 88, normalized size = 0.48 \[ -\frac {25}{32} \, b^{2} e^{\left (-5 \, a\right )} \Gamma \left (-2, 5 \, b x\right ) - \frac {9}{32} \, b^{2} e^{\left (-3 \, a\right )} \Gamma \left (-2, 3 \, b x\right ) + \frac {1}{16} \, b^{2} e^{\left (-a\right )} \Gamma \left (-2, b x\right ) + \frac {1}{16} \, b^{2} e^{a} \Gamma \left (-2, -b x\right ) - \frac {9}{32} \, b^{2} e^{\left (3 \, a\right )} \Gamma \left (-2, -3 \, b x\right ) - \frac {25}{32} \, b^{2} e^{\left (5 \, a\right )} \Gamma \left (-2, -5 \, b x\right ) \]

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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {cosh}\left (a+b\,x\right )}^3\,{\mathrm {sinh}\left (a+b\,x\right )}^2}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh ^{2}{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{x^{3}}\, dx \]

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)

________________________________________________________________________________________