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

Optimal. Leaf size=169 \[ -\frac {1}{8} b^3 \cosh (2 a) \text {Chi}(2 b x)+\frac {9}{8} b^3 \cosh (6 a) \text {Chi}(6 b x)-\frac {1}{8} b^3 \sinh (2 a) \text {Shi}(2 b x)+\frac {9}{8} b^3 \sinh (6 a) \text {Shi}(6 b x)+\frac {b^2 \sinh (2 a+2 b x)}{16 x}-\frac {3 b^2 \sinh (6 a+6 b x)}{16 x}+\frac {\sinh (2 a+2 b x)}{32 x^3}-\frac {\sinh (6 a+6 b x)}{96 x^3}+\frac {b \cosh (2 a+2 b x)}{32 x^2}-\frac {b \cosh (6 a+6 b x)}{32 x^2} \]

[Out]

-1/8*b^3*Chi(2*b*x)*cosh(2*a)+9/8*b^3*Chi(6*b*x)*cosh(6*a)+1/32*b*cosh(2*b*x+2*a)/x^2-1/32*b*cosh(6*b*x+6*a)/x
^2-1/8*b^3*Shi(2*b*x)*sinh(2*a)+9/8*b^3*Shi(6*b*x)*sinh(6*a)+1/32*sinh(2*b*x+2*a)/x^3+1/16*b^2*sinh(2*b*x+2*a)
/x-1/96*sinh(6*b*x+6*a)/x^3-3/16*b^2*sinh(6*b*x+6*a)/x

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Rubi [A]  time = 0.32, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 14, 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} \[ -\frac {1}{8} b^3 \cosh (2 a) \text {Chi}(2 b x)+\frac {9}{8} b^3 \cosh (6 a) \text {Chi}(6 b x)-\frac {1}{8} b^3 \sinh (2 a) \text {Shi}(2 b x)+\frac {9}{8} b^3 \sinh (6 a) \text {Shi}(6 b x)+\frac {b^2 \sinh (2 a+2 b x)}{16 x}-\frac {3 b^2 \sinh (6 a+6 b x)}{16 x}+\frac {\sinh (2 a+2 b x)}{32 x^3}-\frac {\sinh (6 a+6 b x)}{96 x^3}+\frac {b \cosh (2 a+2 b x)}{32 x^2}-\frac {b \cosh (6 a+6 b x)}{32 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Cosh[2*a + 2*b*x])/(32*x^2) - (b*Cosh[6*a + 6*b*x])/(32*x^2) - (b^3*Cosh[2*a]*CoshIntegral[2*b*x])/8 + (9*b
^3*Cosh[6*a]*CoshIntegral[6*b*x])/8 + Sinh[2*a + 2*b*x]/(32*x^3) + (b^2*Sinh[2*a + 2*b*x])/(16*x) - Sinh[6*a +
6*b*x]/(96*x^3) - (3*b^2*Sinh[6*a + 6*b*x])/(16*x) - (b^3*Sinh[2*a]*SinhIntegral[2*b*x])/8 + (9*b^3*Sinh[6*a]
*SinhIntegral[6*b*x])/8

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 ^3(a+b x)}{x^4} \, dx &=\int \left (-\frac {3 \sinh (2 a+2 b x)}{32 x^4}+\frac {\sinh (6 a+6 b x)}{32 x^4}\right ) \, dx\\ &=\frac {1}{32} \int \frac {\sinh (6 a+6 b x)}{x^4} \, dx-\frac {3}{32} \int \frac {\sinh (2 a+2 b x)}{x^4} \, dx\\ &=\frac {\sinh (2 a+2 b x)}{32 x^3}-\frac {\sinh (6 a+6 b x)}{96 x^3}-\frac {1}{16} b \int \frac {\cosh (2 a+2 b x)}{x^3} \, dx+\frac {1}{16} b \int \frac {\cosh (6 a+6 b x)}{x^3} \, dx\\ &=\frac {b \cosh (2 a+2 b x)}{32 x^2}-\frac {b \cosh (6 a+6 b x)}{32 x^2}+\frac {\sinh (2 a+2 b x)}{32 x^3}-\frac {\sinh (6 a+6 b x)}{96 x^3}-\frac {1}{16} b^2 \int \frac {\sinh (2 a+2 b x)}{x^2} \, dx+\frac {1}{16} \left (3 b^2\right ) \int \frac {\sinh (6 a+6 b x)}{x^2} \, dx\\ &=\frac {b \cosh (2 a+2 b x)}{32 x^2}-\frac {b \cosh (6 a+6 b x)}{32 x^2}+\frac {\sinh (2 a+2 b x)}{32 x^3}+\frac {b^2 \sinh (2 a+2 b x)}{16 x}-\frac {\sinh (6 a+6 b x)}{96 x^3}-\frac {3 b^2 \sinh (6 a+6 b x)}{16 x}-\frac {1}{8} b^3 \int \frac {\cosh (2 a+2 b x)}{x} \, dx+\frac {1}{8} \left (9 b^3\right ) \int \frac {\cosh (6 a+6 b x)}{x} \, dx\\ &=\frac {b \cosh (2 a+2 b x)}{32 x^2}-\frac {b \cosh (6 a+6 b x)}{32 x^2}+\frac {\sinh (2 a+2 b x)}{32 x^3}+\frac {b^2 \sinh (2 a+2 b x)}{16 x}-\frac {\sinh (6 a+6 b x)}{96 x^3}-\frac {3 b^2 \sinh (6 a+6 b x)}{16 x}-\frac {1}{8} \left (b^3 \cosh (2 a)\right ) \int \frac {\cosh (2 b x)}{x} \, dx+\frac {1}{8} \left (9 b^3 \cosh (6 a)\right ) \int \frac {\cosh (6 b x)}{x} \, dx-\frac {1}{8} \left (b^3 \sinh (2 a)\right ) \int \frac {\sinh (2 b x)}{x} \, dx+\frac {1}{8} \left (9 b^3 \sinh (6 a)\right ) \int \frac {\sinh (6 b x)}{x} \, dx\\ &=\frac {b \cosh (2 a+2 b x)}{32 x^2}-\frac {b \cosh (6 a+6 b x)}{32 x^2}-\frac {1}{8} b^3 \cosh (2 a) \text {Chi}(2 b x)+\frac {9}{8} b^3 \cosh (6 a) \text {Chi}(6 b x)+\frac {\sinh (2 a+2 b x)}{32 x^3}+\frac {b^2 \sinh (2 a+2 b x)}{16 x}-\frac {\sinh (6 a+6 b x)}{96 x^3}-\frac {3 b^2 \sinh (6 a+6 b x)}{16 x}-\frac {1}{8} b^3 \sinh (2 a) \text {Shi}(2 b x)+\frac {9}{8} b^3 \sinh (6 a) \text {Shi}(6 b x)\\ \end {align*}

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Mathematica [A]  time = 0.34, size = 150, normalized size = 0.89 \[ -\frac {12 b^3 x^3 \cosh (2 a) \text {Chi}(2 b x)-108 b^3 x^3 \cosh (6 a) \text {Chi}(6 b x)+12 b^3 x^3 \sinh (2 a) \text {Shi}(2 b x)-108 b^3 x^3 \sinh (6 a) \text {Shi}(6 b x)-6 b^2 x^2 \sinh (2 (a+b x))+18 b^2 x^2 \sinh (6 (a+b x))-3 \sinh (2 (a+b x))+\sinh (6 (a+b x))-3 b x \cosh (2 (a+b x))+3 b x \cosh (6 (a+b x))}{96 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/96*(-3*b*x*Cosh[2*(a + b*x)] + 3*b*x*Cosh[6*(a + b*x)] + 12*b^3*x^3*Cosh[2*a]*CoshIntegral[2*b*x] - 108*b^3
*x^3*Cosh[6*a]*CoshIntegral[6*b*x] - 3*Sinh[2*(a + b*x)] - 6*b^2*x^2*Sinh[2*(a + b*x)] + Sinh[6*(a + b*x)] + 1
8*b^2*x^2*Sinh[6*(a + b*x)] + 12*b^3*x^3*Sinh[2*a]*SinhIntegral[2*b*x] - 108*b^3*x^3*Sinh[6*a]*SinhIntegral[6*
b*x])/x^3

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fricas [B]  time = 0.78, size = 315, normalized size = 1.86 \[ -\frac {3 \, b x \cosh \left (b x + a\right )^{6} + 45 \, b x \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{4} + 3 \, b x \sinh \left (b x + a\right )^{6} + 20 \, {\left (18 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )^{3} \sinh \left (b x + a\right )^{3} + 6 \, {\left (18 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{5} - 3 \, b x \cosh \left (b x + a\right )^{2} + 3 \, {\left (15 \, b x \cosh \left (b x + a\right )^{4} - b x\right )} \sinh \left (b x + a\right )^{2} - 54 \, {\left (b^{3} x^{3} {\rm Ei}\left (6 \, b x\right ) + b^{3} x^{3} {\rm Ei}\left (-6 \, b x\right )\right )} \cosh \left (6 \, a\right ) + 6 \, {\left (b^{3} x^{3} {\rm Ei}\left (2 \, b x\right ) + b^{3} x^{3} {\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (2 \, a\right ) + 6 \, {\left ({\left (18 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )^{5} - {\left (2 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) - 54 \, {\left (b^{3} x^{3} {\rm Ei}\left (6 \, b x\right ) - b^{3} x^{3} {\rm Ei}\left (-6 \, b x\right )\right )} \sinh \left (6 \, a\right ) + 6 \, {\left (b^{3} x^{3} {\rm Ei}\left (2 \, b x\right ) - b^{3} x^{3} {\rm Ei}\left (-2 \, b x\right )\right )} \sinh \left (2 \, a\right )}{96 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(3*b*x*cosh(b*x + a)^6 + 45*b*x*cosh(b*x + a)^2*sinh(b*x + a)^4 + 3*b*x*sinh(b*x + a)^6 + 20*(18*b^2*x^2
+ 1)*cosh(b*x + a)^3*sinh(b*x + a)^3 + 6*(18*b^2*x^2 + 1)*cosh(b*x + a)*sinh(b*x + a)^5 - 3*b*x*cosh(b*x + a)
^2 + 3*(15*b*x*cosh(b*x + a)^4 - b*x)*sinh(b*x + a)^2 - 54*(b^3*x^3*Ei(6*b*x) + b^3*x^3*Ei(-6*b*x))*cosh(6*a)
+ 6*(b^3*x^3*Ei(2*b*x) + b^3*x^3*Ei(-2*b*x))*cosh(2*a) + 6*((18*b^2*x^2 + 1)*cosh(b*x + a)^5 - (2*b^2*x^2 + 1)
*cosh(b*x + a))*sinh(b*x + a) - 54*(b^3*x^3*Ei(6*b*x) - b^3*x^3*Ei(-6*b*x))*sinh(6*a) + 6*(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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giac [A]  time = 0.13, size = 236, normalized size = 1.40 \[ \frac {108 \, b^{3} x^{3} {\rm Ei}\left (6 \, b x\right ) e^{\left (6 \, a\right )} - 12 \, b^{3} x^{3} {\rm Ei}\left (2 \, b x\right ) e^{\left (2 \, a\right )} - 12 \, b^{3} x^{3} {\rm Ei}\left (-2 \, b x\right ) e^{\left (-2 \, a\right )} + 108 \, b^{3} x^{3} {\rm Ei}\left (-6 \, b x\right ) e^{\left (-6 \, a\right )} - 18 \, b^{2} x^{2} e^{\left (6 \, b x + 6 \, a\right )} + 6 \, b^{2} x^{2} e^{\left (2 \, b x + 2 \, a\right )} - 6 \, b^{2} x^{2} e^{\left (-2 \, b x - 2 \, a\right )} + 18 \, b^{2} x^{2} e^{\left (-6 \, b x - 6 \, a\right )} - 3 \, b x e^{\left (6 \, b x + 6 \, a\right )} + 3 \, b x e^{\left (2 \, b x + 2 \, a\right )} + 3 \, b x e^{\left (-2 \, b x - 2 \, a\right )} - 3 \, b x e^{\left (-6 \, b x - 6 \, a\right )} - e^{\left (6 \, b x + 6 \, a\right )} + 3 \, e^{\left (2 \, b x + 2 \, a\right )} - 3 \, e^{\left (-2 \, b x - 2 \, a\right )} + e^{\left (-6 \, b x - 6 \, a\right )}}{192 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(108*b^3*x^3*Ei(6*b*x)*e^(6*a) - 12*b^3*x^3*Ei(2*b*x)*e^(2*a) - 12*b^3*x^3*Ei(-2*b*x)*e^(-2*a) + 108*b^3
*x^3*Ei(-6*b*x)*e^(-6*a) - 18*b^2*x^2*e^(6*b*x + 6*a) + 6*b^2*x^2*e^(2*b*x + 2*a) - 6*b^2*x^2*e^(-2*b*x - 2*a)
+ 18*b^2*x^2*e^(-6*b*x - 6*a) - 3*b*x*e^(6*b*x + 6*a) + 3*b*x*e^(2*b*x + 2*a) + 3*b*x*e^(-2*b*x - 2*a) - 3*b*
x*e^(-6*b*x - 6*a) - e^(6*b*x + 6*a) + 3*e^(2*b*x + 2*a) - 3*e^(-2*b*x - 2*a) + e^(-6*b*x - 6*a))/x^3

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maple [A]  time = 0.69, size = 246, normalized size = 1.46 \[ \frac {3 b^{2} {\mathrm e}^{-6 b x -6 a}}{32 x}-\frac {b \,{\mathrm e}^{-6 b x -6 a}}{64 x^{2}}+\frac {{\mathrm e}^{-6 b x -6 a}}{192 x^{3}}-\frac {9 b^{3} {\mathrm e}^{-6 a} \Ei \left (1, 6 b x \right )}{16}-\frac {b^{2} {\mathrm e}^{-2 b x -2 a}}{32 x}+\frac {b \,{\mathrm e}^{-2 b x -2 a}}{64 x^{2}}-\frac {{\mathrm e}^{-2 b x -2 a}}{64 x^{3}}+\frac {b^{3} {\mathrm e}^{-2 a} \Ei \left (1, 2 b x \right )}{16}+\frac {{\mathrm e}^{2 b x +2 a}}{64 x^{3}}+\frac {b \,{\mathrm e}^{2 b x +2 a}}{64 x^{2}}+\frac {b^{2} {\mathrm e}^{2 b x +2 a}}{32 x}+\frac {b^{3} {\mathrm e}^{2 a} \Ei \left (1, -2 b x \right )}{16}-\frac {{\mathrm e}^{6 b x +6 a}}{192 x^{3}}-\frac {b \,{\mathrm e}^{6 b x +6 a}}{64 x^{2}}-\frac {3 b^{2} {\mathrm e}^{6 b x +6 a}}{32 x}-\frac {9 b^{3} {\mathrm e}^{6 a} \Ei \left (1, -6 b x \right )}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/32*b^2*exp(-6*b*x-6*a)/x-1/64*b*exp(-6*b*x-6*a)/x^2+1/192*exp(-6*b*x-6*a)/x^3-9/16*b^3*exp(-6*a)*Ei(1,6*b*x)
-1/32*b^2*exp(-2*b*x-2*a)/x+1/64*b*exp(-2*b*x-2*a)/x^2-1/64*exp(-2*b*x-2*a)/x^3+1/16*b^3*exp(-2*a)*Ei(1,2*b*x)
+1/64*exp(2*b*x+2*a)/x^3+1/64*b*exp(2*b*x+2*a)/x^2+1/32*b^2*exp(2*b*x+2*a)/x+1/16*b^3*exp(2*a)*Ei(1,-2*b*x)-1/
192/x^3*exp(6*b*x+6*a)-1/64*b/x^2*exp(6*b*x+6*a)-3/32*b^2/x*exp(6*b*x+6*a)-9/16*b^3*exp(6*a)*Ei(1,-6*b*x)

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maxima [A]  time = 0.43, size = 61, normalized size = 0.36 \[ \frac {27}{8} \, b^{3} e^{\left (-6 \, a\right )} \Gamma \left (-3, 6 \, b x\right ) - \frac {3}{8} \, b^{3} e^{\left (-2 \, a\right )} \Gamma \left (-3, 2 \, b x\right ) - \frac {3}{8} \, b^{3} e^{\left (2 \, a\right )} \Gamma \left (-3, -2 \, b x\right ) + \frac {27}{8} \, b^{3} e^{\left (6 \, a\right )} \Gamma \left (-3, -6 \, b x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

27/8*b^3*e^(-6*a)*gamma(-3, 6*b*x) - 3/8*b^3*e^(-2*a)*gamma(-3, 2*b*x) - 3/8*b^3*e^(2*a)*gamma(-3, -2*b*x) + 2
7/8*b^3*e^(6*a)*gamma(-3, -6*b*x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {cosh}\left (a+b\,x\right )}^3\,{\mathrm {sinh}\left (a+b\,x\right )}^3}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh ^{3}{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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