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

Optimal. Leaf size=33 \[ \frac {1}{8} \cosh (4 a) \text {Chi}(4 b x)+\frac {1}{8} \sinh (4 a) \text {Shi}(4 b x)-\frac {\log (x)}{8} \]

[Out]

1/8*Chi(4*b*x)*cosh(4*a)-1/8*ln(x)+1/8*Shi(4*b*x)*sinh(4*a)

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Rubi [A]  time = 0.08, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5448, 3303, 3298, 3301} \[ \frac {1}{8} \cosh (4 a) \text {Chi}(4 b x)+\frac {1}{8} \sinh (4 a) \text {Shi}(4 b x)-\frac {\log (x)}{8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cosh[4*a]*CoshIntegral[4*b*x])/8 - Log[x]/8 + (Sinh[4*a]*SinhIntegral[4*b*x])/8

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

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Mathematica [A]  time = 0.10, size = 32, normalized size = 0.97 \[ \frac {1}{8} (\cosh (4 a) \text {Chi}(4 b x)+\sinh (4 a) \text {Shi}(4 b x)-\log (2 b x)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cosh[4*a]*CoshIntegral[4*b*x] - Log[2*b*x] + Sinh[4*a]*SinhIntegral[4*b*x])/8

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fricas [A]  time = 0.73, size = 41, normalized size = 1.24 \[ \frac {1}{16} \, {\left ({\rm Ei}\left (4 \, b x\right ) + {\rm Ei}\left (-4 \, b x\right )\right )} \cosh \left (4 \, a\right ) + \frac {1}{16} \, {\left ({\rm Ei}\left (4 \, b x\right ) - {\rm Ei}\left (-4 \, b x\right )\right )} \sinh \left (4 \, a\right ) - \frac {1}{8} \, \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^2*sinh(b*x+a)^2/x,x, algorithm="fricas")

[Out]

1/16*(Ei(4*b*x) + Ei(-4*b*x))*cosh(4*a) + 1/16*(Ei(4*b*x) - Ei(-4*b*x))*sinh(4*a) - 1/8*log(x)

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giac [A]  time = 0.12, size = 27, normalized size = 0.82 \[ \frac {1}{16} \, {\rm Ei}\left (4 \, b x\right ) e^{\left (4 \, a\right )} + \frac {1}{16} \, {\rm Ei}\left (-4 \, b x\right ) e^{\left (-4 \, a\right )} - \frac {1}{8} \, \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^2*sinh(b*x+a)^2/x,x, algorithm="giac")

[Out]

1/16*Ei(4*b*x)*e^(4*a) + 1/16*Ei(-4*b*x)*e^(-4*a) - 1/8*log(x)

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maple [A]  time = 0.34, size = 30, normalized size = 0.91 \[ -\frac {\ln \relax (x )}{8}-\frac {{\mathrm e}^{-4 a} \Ei \left (1, 4 b x \right )}{16}-\frac {{\mathrm e}^{4 a} \Ei \left (1, -4 b x \right )}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*ln(x)-1/16*exp(-4*a)*Ei(1,4*b*x)-1/16*exp(4*a)*Ei(1,-4*b*x)

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maxima [A]  time = 0.80, size = 27, normalized size = 0.82 \[ \frac {1}{16} \, {\rm Ei}\left (4 \, b x\right ) e^{\left (4 \, a\right )} + \frac {1}{16} \, {\rm Ei}\left (-4 \, b x\right ) e^{\left (-4 \, a\right )} - \frac {1}{8} \, \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^2*sinh(b*x+a)^2/x,x, algorithm="maxima")

[Out]

1/16*Ei(4*b*x)*e^(4*a) + 1/16*Ei(-4*b*x)*e^(-4*a) - 1/8*log(x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {{\mathrm {cosh}\left (a+b\,x\right )}^2\,{\mathrm {sinh}\left (a+b\,x\right )}^2}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)**2*sinh(b*x+a)**2/x,x)

[Out]

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

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