### 3.295 $$\int \frac{\cosh ^2(a+b x) \sinh ^2(a+b x)}{x^2} \, dx$$

Optimal. Leaf size=52 $\frac{1}{2} b \sinh (4 a) \text{Chi}(4 b x)+\frac{1}{2} b \cosh (4 a) \text{Shi}(4 b x)-\frac{\cosh (4 a+4 b x)}{8 x}+\frac{1}{8 x}$

[Out]

1/(8*x) - Cosh[4*a + 4*b*x]/(8*x) + (b*CoshIntegral[4*b*x]*Sinh[4*a])/2 + (b*Cosh[4*a]*SinhIntegral[4*b*x])/2

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Rubi [A]  time = 0.111748, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.25, Rules used = {5448, 3297, 3303, 3298, 3301} $\frac{1}{2} b \sinh (4 a) \text{Chi}(4 b x)+\frac{1}{2} b \cosh (4 a) \text{Shi}(4 b x)-\frac{\cosh (4 a+4 b x)}{8 x}+\frac{1}{8 x}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

1/(8*x) - Cosh[4*a + 4*b*x]/(8*x) + (b*CoshIntegral[4*b*x]*Sinh[4*a])/2 + (b*Cosh[4*a]*SinhIntegral[4*b*x])/2

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]

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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.0932268, size = 45, normalized size = 0.87 $\frac{4 b x \sinh (4 a) \text{Chi}(4 b x)+4 b x \cosh (4 a) \text{Shi}(4 b x)-\cosh (4 (a+b x))+1}{8 x}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.049, size = 61, normalized size = 1.2 \begin{align*}{\frac{1}{8\,x}}-{\frac{{{\rm e}^{-4\,bx-4\,a}}}{16\,x}}+{\frac{b{{\rm e}^{-4\,a}}{\it Ei} \left ( 1,4\,bx \right ) }{4}}-{\frac{{{\rm e}^{4\,bx+4\,a}}}{16\,x}}-{\frac{b{{\rm e}^{4\,a}}{\it Ei} \left ( 1,-4\,bx \right ) }{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.3941, size = 43, normalized size = 0.83 \begin{align*} -\frac{1}{4} \, b e^{\left (-4 \, a\right )} \Gamma \left (-1, 4 \, b x\right ) + \frac{1}{4} \, b e^{\left (4 \, a\right )} \Gamma \left (-1, -4 \, b x\right ) + \frac{1}{8 \, x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*b*e^(-4*a)*gamma(-1, 4*b*x) + 1/4*b*e^(4*a)*gamma(-1, -4*b*x) + 1/8/x

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Fricas [B]  time = 1.87895, size = 240, normalized size = 4.62 \begin{align*} -\frac{\cosh \left (b x + a\right )^{4} + 6 \, \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{2} + \sinh \left (b x + a\right )^{4} - 2 \,{\left (b x{\rm Ei}\left (4 \, b x\right ) - b x{\rm Ei}\left (-4 \, b x\right )\right )} \cosh \left (4 \, a\right ) - 2 \,{\left (b x{\rm Ei}\left (4 \, b x\right ) + b x{\rm Ei}\left (-4 \, b x\right )\right )} \sinh \left (4 \, a\right ) - 1}{8 \, x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(cosh(b*x + a)^4 + 6*cosh(b*x + a)^2*sinh(b*x + a)^2 + sinh(b*x + a)^4 - 2*(b*x*Ei(4*b*x) - b*x*Ei(-4*b*x
))*cosh(4*a) - 2*(b*x*Ei(4*b*x) + b*x*Ei(-4*b*x))*sinh(4*a) - 1)/x

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{x^{2}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.1669, size = 74, normalized size = 1.42 \begin{align*} \frac{4 \, b x{\rm Ei}\left (4 \, b x\right ) e^{\left (4 \, a\right )} - 4 \, b x{\rm Ei}\left (-4 \, b x\right ) e^{\left (-4 \, a\right )} - e^{\left (4 \, b x + 4 \, a\right )} - e^{\left (-4 \, b x - 4 \, a\right )} + 2}{16 \, x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(4*b*x*Ei(4*b*x)*e^(4*a) - 4*b*x*Ei(-4*b*x)*e^(-4*a) - e^(4*b*x + 4*a) - e^(-4*b*x - 4*a) + 2)/x