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

Optimal. Leaf size=131 \[ -\frac {3}{16} b^2 \sinh (2 a) \text {Chi}(2 b x)+\frac {9}{16} b^2 \sinh (6 a) \text {Chi}(6 b x)-\frac {3}{16} b^2 \cosh (2 a) \text {Shi}(2 b x)+\frac {9}{16} b^2 \cosh (6 a) \text {Shi}(6 b x)+\frac {3 \sinh (2 a+2 b x)}{64 x^2}-\frac {\sinh (6 a+6 b x)}{64 x^2}+\frac {3 b \cosh (2 a+2 b x)}{32 x}-\frac {3 b \cosh (6 a+6 b x)}{32 x} \]

[Out]

3/32*b*cosh(2*b*x+2*a)/x-3/32*b*cosh(6*b*x+6*a)/x-3/16*b^2*cosh(2*a)*Shi(2*b*x)+9/16*b^2*cosh(6*a)*Shi(6*b*x)-
3/16*b^2*Chi(2*b*x)*sinh(2*a)+9/16*b^2*Chi(6*b*x)*sinh(6*a)+3/64*sinh(2*b*x+2*a)/x^2-1/64*sinh(6*b*x+6*a)/x^2

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Rubi [A]  time = 0.26, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 12, 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} \[ -\frac {3}{16} b^2 \sinh (2 a) \text {Chi}(2 b x)+\frac {9}{16} b^2 \sinh (6 a) \text {Chi}(6 b x)-\frac {3}{16} b^2 \cosh (2 a) \text {Shi}(2 b x)+\frac {9}{16} b^2 \cosh (6 a) \text {Shi}(6 b x)+\frac {3 \sinh (2 a+2 b x)}{64 x^2}-\frac {\sinh (6 a+6 b x)}{64 x^2}+\frac {3 b \cosh (2 a+2 b x)}{32 x}-\frac {3 b \cosh (6 a+6 b x)}{32 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*Cosh[2*a + 2*b*x])/(32*x) - (3*b*Cosh[6*a + 6*b*x])/(32*x) - (3*b^2*CoshIntegral[2*b*x]*Sinh[2*a])/16 + (
9*b^2*CoshIntegral[6*b*x]*Sinh[6*a])/16 + (3*Sinh[2*a + 2*b*x])/(64*x^2) - Sinh[6*a + 6*b*x]/(64*x^2) - (3*b^2
*Cosh[2*a]*SinhIntegral[2*b*x])/16 + (9*b^2*Cosh[6*a]*SinhIntegral[6*b*x])/16

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^3} \, dx &=\int \left (-\frac {3 \sinh (2 a+2 b x)}{32 x^3}+\frac {\sinh (6 a+6 b x)}{32 x^3}\right ) \, dx\\ &=\frac {1}{32} \int \frac {\sinh (6 a+6 b x)}{x^3} \, dx-\frac {3}{32} \int \frac {\sinh (2 a+2 b x)}{x^3} \, dx\\ &=\frac {3 \sinh (2 a+2 b x)}{64 x^2}-\frac {\sinh (6 a+6 b x)}{64 x^2}-\frac {1}{32} (3 b) \int \frac {\cosh (2 a+2 b x)}{x^2} \, dx+\frac {1}{32} (3 b) \int \frac {\cosh (6 a+6 b x)}{x^2} \, dx\\ &=\frac {3 b \cosh (2 a+2 b x)}{32 x}-\frac {3 b \cosh (6 a+6 b x)}{32 x}+\frac {3 \sinh (2 a+2 b x)}{64 x^2}-\frac {\sinh (6 a+6 b x)}{64 x^2}-\frac {1}{16} \left (3 b^2\right ) \int \frac {\sinh (2 a+2 b x)}{x} \, dx+\frac {1}{16} \left (9 b^2\right ) \int \frac {\sinh (6 a+6 b x)}{x} \, dx\\ &=\frac {3 b \cosh (2 a+2 b x)}{32 x}-\frac {3 b \cosh (6 a+6 b x)}{32 x}+\frac {3 \sinh (2 a+2 b x)}{64 x^2}-\frac {\sinh (6 a+6 b x)}{64 x^2}-\frac {1}{16} \left (3 b^2 \cosh (2 a)\right ) \int \frac {\sinh (2 b x)}{x} \, dx+\frac {1}{16} \left (9 b^2 \cosh (6 a)\right ) \int \frac {\sinh (6 b x)}{x} \, dx-\frac {1}{16} \left (3 b^2 \sinh (2 a)\right ) \int \frac {\cosh (2 b x)}{x} \, dx+\frac {1}{16} \left (9 b^2 \sinh (6 a)\right ) \int \frac {\cosh (6 b x)}{x} \, dx\\ &=\frac {3 b \cosh (2 a+2 b x)}{32 x}-\frac {3 b \cosh (6 a+6 b x)}{32 x}-\frac {3}{16} b^2 \text {Chi}(2 b x) \sinh (2 a)+\frac {9}{16} b^2 \text {Chi}(6 b x) \sinh (6 a)+\frac {3 \sinh (2 a+2 b x)}{64 x^2}-\frac {\sinh (6 a+6 b x)}{64 x^2}-\frac {3}{16} b^2 \cosh (2 a) \text {Shi}(2 b x)+\frac {9}{16} b^2 \cosh (6 a) \text {Shi}(6 b x)\\ \end {align*}

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Mathematica [A]  time = 0.25, size = 118, normalized size = 0.90 \[ -\frac {12 b^2 x^2 \sinh (2 a) \text {Chi}(2 b x)-36 b^2 x^2 \sinh (6 a) \text {Chi}(6 b x)+12 b^2 x^2 \cosh (2 a) \text {Shi}(2 b x)-36 b^2 x^2 \cosh (6 a) \text {Shi}(6 b x)-3 \sinh (2 (a+b x))+\sinh (6 (a+b x))-6 b x \cosh (2 (a+b x))+6 b x \cosh (6 (a+b x))}{64 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/64*(-6*b*x*Cosh[2*(a + b*x)] + 6*b*x*Cosh[6*(a + b*x)] + 12*b^2*x^2*CoshIntegral[2*b*x]*Sinh[2*a] - 36*b^2*
x^2*CoshIntegral[6*b*x]*Sinh[6*a] - 3*Sinh[2*(a + b*x)] + Sinh[6*(a + b*x)] + 12*b^2*x^2*Cosh[2*a]*SinhIntegra
l[2*b*x] - 36*b^2*x^2*Cosh[6*a]*SinhIntegral[6*b*x])/x^2

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fricas [B]  time = 1.09, size = 274, normalized size = 2.09 \[ -\frac {3 \, b x \cosh \left (b x + a\right )^{6} + 45 \, b x \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{4} + 3 \, b x \sinh \left (b x + a\right )^{6} + 10 \, \cosh \left (b x + a\right )^{3} \sinh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{5} - 3 \, b x \cosh \left (b x + a\right )^{2} + 3 \, {\left (15 \, b x \cosh \left (b x + a\right )^{4} - b x\right )} \sinh \left (b x + a\right )^{2} - 9 \, {\left (b^{2} x^{2} {\rm Ei}\left (6 \, b x\right ) - b^{2} x^{2} {\rm Ei}\left (-6 \, b x\right )\right )} \cosh \left (6 \, a\right ) + 3 \, {\left (b^{2} x^{2} {\rm Ei}\left (2 \, b x\right ) - b^{2} x^{2} {\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (2 \, a\right ) + 3 \, {\left (\cosh \left (b x + a\right )^{5} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) - 9 \, {\left (b^{2} x^{2} {\rm Ei}\left (6 \, b x\right ) + b^{2} x^{2} {\rm Ei}\left (-6 \, b x\right )\right )} \sinh \left (6 \, a\right ) + 3 \, {\left (b^{2} x^{2} {\rm Ei}\left (2 \, b x\right ) + b^{2} x^{2} {\rm Ei}\left (-2 \, b x\right )\right )} \sinh \left (2 \, a\right )}{32 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(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 + 10*cosh(b*x +
a)^3*sinh(b*x + a)^3 + 3*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 - 9*(b^2*x^2*Ei(6*b*x) - b^2*x^2*Ei(-6*b*x))*cosh(6*a) + 3*(b^2*x^2*Ei(2*b*x) - b^2*x^2*Ei
(-2*b*x))*cosh(2*a) + 3*(cosh(b*x + a)^5 - cosh(b*x + a))*sinh(b*x + a) - 9*(b^2*x^2*Ei(6*b*x) + b^2*x^2*Ei(-6
*b*x))*sinh(6*a) + 3*(b^2*x^2*Ei(2*b*x) + b^2*x^2*Ei(-2*b*x))*sinh(2*a))/x^2

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giac [A]  time = 0.13, size = 168, normalized size = 1.28 \[ \frac {36 \, b^{2} x^{2} {\rm Ei}\left (6 \, b x\right ) e^{\left (6 \, a\right )} - 12 \, b^{2} x^{2} {\rm Ei}\left (2 \, b x\right ) e^{\left (2 \, a\right )} + 12 \, b^{2} x^{2} {\rm Ei}\left (-2 \, b x\right ) e^{\left (-2 \, a\right )} - 36 \, b^{2} x^{2} {\rm Ei}\left (-6 \, b x\right ) e^{\left (-6 \, a\right )} - 6 \, b x e^{\left (6 \, b x + 6 \, a\right )} + 6 \, b x e^{\left (2 \, b x + 2 \, a\right )} + 6 \, b x e^{\left (-2 \, b x - 2 \, a\right )} - 6 \, b x e^{\left (-6 \, b x - 6 \, a\right )} - e^{\left (6 \, b x + 6 \, a\right )} + 3 \, e^{\left (2 \, b x + 2 \, a\right )} - 3 \, e^{\left (-2 \, b x - 2 \, a\right )} + e^{\left (-6 \, b x - 6 \, a\right )}}{128 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128*(36*b^2*x^2*Ei(6*b*x)*e^(6*a) - 12*b^2*x^2*Ei(2*b*x)*e^(2*a) + 12*b^2*x^2*Ei(-2*b*x)*e^(-2*a) - 36*b^2*x
^2*Ei(-6*b*x)*e^(-6*a) - 6*b*x*e^(6*b*x + 6*a) + 6*b*x*e^(2*b*x + 2*a) + 6*b*x*e^(-2*b*x - 2*a) - 6*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^2

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maple [A]  time = 0.69, size = 178, normalized size = 1.36 \[ -\frac {3 b \,{\mathrm e}^{-6 b x -6 a}}{64 x}+\frac {{\mathrm e}^{-6 b x -6 a}}{128 x^{2}}+\frac {9 b^{2} {\mathrm e}^{-6 a} \Ei \left (1, 6 b x \right )}{32}+\frac {3 b \,{\mathrm e}^{-2 b x -2 a}}{64 x}-\frac {3 \,{\mathrm e}^{-2 b x -2 a}}{128 x^{2}}-\frac {3 b^{2} {\mathrm e}^{-2 a} \Ei \left (1, 2 b x \right )}{32}+\frac {3 \,{\mathrm e}^{2 b x +2 a}}{128 x^{2}}+\frac {3 b \,{\mathrm e}^{2 b x +2 a}}{64 x}+\frac {3 b^{2} {\mathrm e}^{2 a} \Ei \left (1, -2 b x \right )}{32}-\frac {{\mathrm e}^{6 b x +6 a}}{128 x^{2}}-\frac {3 b \,{\mathrm e}^{6 b x +6 a}}{64 x}-\frac {9 b^{2} {\mathrm e}^{6 a} \Ei \left (1, -6 b x \right )}{32} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/64*b*exp(-6*b*x-6*a)/x+1/128*exp(-6*b*x-6*a)/x^2+9/32*b^2*exp(-6*a)*Ei(1,6*b*x)+3/64*b*exp(-2*b*x-2*a)/x-3/
128*exp(-2*b*x-2*a)/x^2-3/32*b^2*exp(-2*a)*Ei(1,2*b*x)+3/128*exp(2*b*x+2*a)/x^2+3/64*b*exp(2*b*x+2*a)/x+3/32*b
^2*exp(2*a)*Ei(1,-2*b*x)-1/128/x^2*exp(6*b*x+6*a)-3/64*b/x*exp(6*b*x+6*a)-9/32*b^2*exp(6*a)*Ei(1,-6*b*x)

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maxima [A]  time = 0.44, size = 61, normalized size = 0.47 \[ \frac {9}{16} \, b^{2} e^{\left (-6 \, a\right )} \Gamma \left (-2, 6 \, b x\right ) - \frac {3}{16} \, b^{2} e^{\left (-2 \, a\right )} \Gamma \left (-2, 2 \, b x\right ) + \frac {3}{16} \, b^{2} e^{\left (2 \, a\right )} \Gamma \left (-2, -2 \, b x\right ) - \frac {9}{16} \, b^{2} e^{\left (6 \, a\right )} \Gamma \left (-2, -6 \, b x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

9/16*b^2*e^(-6*a)*gamma(-2, 6*b*x) - 3/16*b^2*e^(-2*a)*gamma(-2, 2*b*x) + 3/16*b^2*e^(2*a)*gamma(-2, -2*b*x) -
9/16*b^2*e^(6*a)*gamma(-2, -6*b*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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