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

Optimal. Leaf size=53 \[ -\frac {3}{32} \sinh (2 a) \text {Chi}(2 b x)+\frac {1}{32} \sinh (6 a) \text {Chi}(6 b x)-\frac {3}{32} \cosh (2 a) \text {Shi}(2 b x)+\frac {1}{32} \cosh (6 a) \text {Shi}(6 b x) \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5448, 3303, 3298, 3301} \[ -\frac {3}{32} \sinh (2 a) \text {Chi}(2 b x)+\frac {1}{32} \sinh (6 a) \text {Chi}(6 b x)-\frac {3}{32} \cosh (2 a) \text {Shi}(2 b x)+\frac {1}{32} \cosh (6 a) \text {Shi}(6 b x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.18, size = 47, normalized size = 0.89 \[ \frac {1}{32} (\sinh (6 a) \text {Chi}(6 b x)-6 \sinh (a) \cosh (a) \text {Chi}(2 b x)-3 \cosh (2 a) \text {Shi}(2 b x)+\cosh (6 a) \text {Shi}(6 b x)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*Cosh[a]*CoshIntegral[2*b*x]*Sinh[a] + CoshIntegral[6*b*x]*Sinh[6*a] - 3*Cosh[2*a]*SinhIntegral[2*b*x] + Co
sh[6*a]*SinhIntegral[6*b*x])/32

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fricas [A]  time = 0.68, size = 73, normalized size = 1.38 \[ \frac {1}{64} \, {\left ({\rm Ei}\left (6 \, b x\right ) - {\rm Ei}\left (-6 \, b x\right )\right )} \cosh \left (6 \, a\right ) - \frac {3}{64} \, {\left ({\rm Ei}\left (2 \, b x\right ) - {\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (2 \, a\right ) + \frac {1}{64} \, {\left ({\rm Ei}\left (6 \, b x\right ) + {\rm Ei}\left (-6 \, b x\right )\right )} \sinh \left (6 \, a\right ) - \frac {3}{64} \, {\left ({\rm Ei}\left (2 \, b x\right ) + {\rm Ei}\left (-2 \, b x\right )\right )} \sinh \left (2 \, a\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(Ei(6*b*x) - Ei(-6*b*x))*cosh(6*a) - 3/64*(Ei(2*b*x) - Ei(-2*b*x))*cosh(2*a) + 1/64*(Ei(6*b*x) + Ei(-6*b*
x))*sinh(6*a) - 3/64*(Ei(2*b*x) + Ei(-2*b*x))*sinh(2*a)

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giac [A]  time = 0.13, size = 45, normalized size = 0.85 \[ \frac {1}{64} \, {\rm Ei}\left (6 \, b x\right ) e^{\left (6 \, a\right )} - \frac {3}{64} \, {\rm Ei}\left (2 \, b x\right ) e^{\left (2 \, a\right )} + \frac {3}{64} \, {\rm Ei}\left (-2 \, b x\right ) e^{\left (-2 \, a\right )} - \frac {1}{64} \, {\rm Ei}\left (-6 \, b x\right ) e^{\left (-6 \, a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*Ei(6*b*x)*e^(6*a) - 3/64*Ei(2*b*x)*e^(2*a) + 3/64*Ei(-2*b*x)*e^(-2*a) - 1/64*Ei(-6*b*x)*e^(-6*a)

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maple [A]  time = 0.68, size = 50, normalized size = 0.94 \[ \frac {{\mathrm e}^{-6 a} \Ei \left (1, 6 b x \right )}{64}-\frac {3 \,{\mathrm e}^{-2 a} \Ei \left (1, 2 b x \right )}{64}+\frac {3 \,{\mathrm e}^{2 a} \Ei \left (1, -2 b x \right )}{64}-\frac {{\mathrm e}^{6 a} \Ei \left (1, -6 b x \right )}{64} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.53, size = 45, normalized size = 0.85 \[ \frac {1}{64} \, {\rm Ei}\left (6 \, b x\right ) e^{\left (6 \, a\right )} - \frac {3}{64} \, {\rm Ei}\left (2 \, b x\right ) e^{\left (2 \, a\right )} + \frac {3}{64} \, {\rm Ei}\left (-2 \, b x\right ) e^{\left (-2 \, a\right )} - \frac {1}{64} \, {\rm Ei}\left (-6 \, b x\right ) e^{\left (-6 \, a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*Ei(6*b*x)*e^(6*a) - 3/64*Ei(2*b*x)*e^(2*a) + 3/64*Ei(-2*b*x)*e^(-2*a) - 1/64*Ei(-6*b*x)*e^(-6*a)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\mathrm {cosh}\left (a+b\,x\right )}^3\,{\mathrm {sinh}\left (a+b\,x\right )}^3}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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