### 3.296 $$\int \frac{\cosh ^2(a+b x) \sinh ^2(a+b x)}{x^3} \, dx$$

Optimal. Leaf size=67 $b^2 \cosh (4 a) \text{Chi}(4 b x)+b^2 \sinh (4 a) \text{Shi}(4 b x)-\frac{\cosh (4 a+4 b x)}{16 x^2}-\frac{b \sinh (4 a+4 b x)}{4 x}+\frac{1}{16 x^2}$

[Out]

1/(16*x^2) - Cosh[4*a + 4*b*x]/(16*x^2) + b^2*Cosh[4*a]*CoshIntegral[4*b*x] - (b*Sinh[4*a + 4*b*x])/(4*x) + b^
2*Sinh[4*a]*SinhIntegral[4*b*x]

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Rubi [A]  time = 0.135611, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 7, 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} $b^2 \cosh (4 a) \text{Chi}(4 b x)+b^2 \sinh (4 a) \text{Shi}(4 b x)-\frac{\cosh (4 a+4 b x)}{16 x^2}-\frac{b \sinh (4 a+4 b x)}{4 x}+\frac{1}{16 x^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[(Cosh[a + b*x]^2*Sinh[a + b*x]^2)/x^3,x]

[Out]

1/(16*x^2) - Cosh[4*a + 4*b*x]/(16*x^2) + b^2*Cosh[4*a]*CoshIntegral[4*b*x] - (b*Sinh[4*a + 4*b*x])/(4*x) + b^
2*Sinh[4*a]*SinhIntegral[4*b*x]

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

Mathematica [A]  time = 0.10449, size = 65, normalized size = 0.97 $\frac{16 b^2 x^2 \cosh (4 a) \text{Chi}(4 b x)+16 b^2 x^2 \sinh (4 a) \text{Shi}(4 b x)-4 b x \sinh (4 (a+b x))-\cosh (4 (a+b x))+1}{16 x^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(Cosh[a + b*x]^2*Sinh[a + b*x]^2)/x^3,x]

[Out]

(1 - Cosh[4*(a + b*x)] + 16*b^2*x^2*Cosh[4*a]*CoshIntegral[4*b*x] - 4*b*x*Sinh[4*(a + b*x)] + 16*b^2*x^2*Sinh[
4*a]*SinhIntegral[4*b*x])/(16*x^2)

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Maple [A]  time = 0.055, size = 95, normalized size = 1.4 \begin{align*}{\frac{1}{16\,{x}^{2}}}+{\frac{b{{\rm e}^{-4\,bx-4\,a}}}{8\,x}}-{\frac{{{\rm e}^{-4\,bx-4\,a}}}{32\,{x}^{2}}}-{\frac{{b}^{2}{{\rm e}^{-4\,a}}{\it Ei} \left ( 1,4\,bx \right ) }{2}}-{\frac{{{\rm e}^{4\,bx+4\,a}}}{32\,{x}^{2}}}-{\frac{b{{\rm e}^{4\,bx+4\,a}}}{8\,x}}-{\frac{{b}^{2}{{\rm e}^{4\,a}}{\it Ei} \left ( 1,-4\,bx \right ) }{2}} \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^3,x)

[Out]

1/16/x^2+1/8*b*exp(-4*b*x-4*a)/x-1/32*exp(-4*b*x-4*a)/x^2-1/2*b^2*exp(-4*a)*Ei(1,4*b*x)-1/32/x^2*exp(4*b*x+4*a
)-1/8*b/x*exp(4*b*x+4*a)-1/2*b^2*exp(4*a)*Ei(1,-4*b*x)

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Maxima [A]  time = 1.20587, size = 49, normalized size = 0.73 \begin{align*} -b^{2} e^{\left (-4 \, a\right )} \Gamma \left (-2, 4 \, b x\right ) - b^{2} e^{\left (4 \, a\right )} \Gamma \left (-2, -4 \, b x\right ) + \frac{1}{16 \, x^{2}} \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^3,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.76163, size = 371, normalized size = 5.54 \begin{align*} -\frac{16 \, b x \cosh \left (b x + a\right )^{3} \sinh \left (b x + a\right ) + 16 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \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} - 8 \,{\left (b^{2} x^{2}{\rm Ei}\left (4 \, b x\right ) + b^{2} x^{2}{\rm Ei}\left (-4 \, b x\right )\right )} \cosh \left (4 \, a\right ) - 8 \,{\left (b^{2} x^{2}{\rm Ei}\left (4 \, b x\right ) - b^{2} x^{2}{\rm Ei}\left (-4 \, b x\right )\right )} \sinh \left (4 \, a\right ) - 1}{16 \, x^{2}} \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^3,x, algorithm="fricas")

[Out]

-1/16*(16*b*x*cosh(b*x + a)^3*sinh(b*x + a) + 16*b*x*cosh(b*x + a)*sinh(b*x + a)^3 + cosh(b*x + a)^4 + 6*cosh(
b*x + a)^2*sinh(b*x + a)^2 + sinh(b*x + a)^4 - 8*(b^2*x^2*Ei(4*b*x) + b^2*x^2*Ei(-4*b*x))*cosh(4*a) - 8*(b^2*x
^2*Ei(4*b*x) - b^2*x^2*Ei(-4*b*x))*sinh(4*a) - 1)/x^2

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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^{3}}\, 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**3,x)

[Out]

Integral(sinh(a + b*x)**2*cosh(a + b*x)**2/x**3, x)

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Giac [A]  time = 1.18005, size = 120, normalized size = 1.79 \begin{align*} \frac{16 \, b^{2} x^{2}{\rm Ei}\left (4 \, b x\right ) e^{\left (4 \, a\right )} + 16 \, b^{2} x^{2}{\rm Ei}\left (-4 \, b x\right ) e^{\left (-4 \, a\right )} - 4 \, b x e^{\left (4 \, b x + 4 \, a\right )} + 4 \, b x e^{\left (-4 \, b x - 4 \, a\right )} - e^{\left (4 \, b x + 4 \, a\right )} - e^{\left (-4 \, b x - 4 \, a\right )} + 2}{32 \, x^{2}} \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^3,x, algorithm="giac")

[Out]

1/32*(16*b^2*x^2*Ei(4*b*x)*e^(4*a) + 16*b^2*x^2*Ei(-4*b*x)*e^(-4*a) - 4*b*x*e^(4*b*x + 4*a) + 4*b*x*e^(-4*b*x
- 4*a) - e^(4*b*x + 4*a) - e^(-4*b*x - 4*a) + 2)/x^2