### 3.297 $$\int \frac{\cosh ^2(a+b x) \sinh ^2(a+b x)}{x^4} \, dx$$

Optimal. Leaf size=92 $\frac{4}{3} b^3 \sinh (4 a) \text{Chi}(4 b x)+\frac{4}{3} b^3 \cosh (4 a) \text{Shi}(4 b x)-\frac{b^2 \cosh (4 a+4 b x)}{3 x}-\frac{b \sinh (4 a+4 b x)}{12 x^2}-\frac{\cosh (4 a+4 b x)}{24 x^3}+\frac{1}{24 x^3}$

[Out]

1/(24*x^3) - Cosh[4*a + 4*b*x]/(24*x^3) - (b^2*Cosh[4*a + 4*b*x])/(3*x) + (4*b^3*CoshIntegral[4*b*x]*Sinh[4*a]
)/3 - (b*Sinh[4*a + 4*b*x])/(12*x^2) + (4*b^3*Cosh[4*a]*SinhIntegral[4*b*x])/3

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Rubi [A]  time = 0.169529, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 8, 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} $\frac{4}{3} b^3 \sinh (4 a) \text{Chi}(4 b x)+\frac{4}{3} b^3 \cosh (4 a) \text{Shi}(4 b x)-\frac{b^2 \cosh (4 a+4 b x)}{3 x}-\frac{b \sinh (4 a+4 b x)}{12 x^2}-\frac{\cosh (4 a+4 b x)}{24 x^3}+\frac{1}{24 x^3}$

Antiderivative was successfully veriﬁed.

[In]

Int[(Cosh[a + b*x]^2*Sinh[a + b*x]^2)/x^4,x]

[Out]

1/(24*x^3) - Cosh[4*a + 4*b*x]/(24*x^3) - (b^2*Cosh[4*a + 4*b*x])/(3*x) + (4*b^3*CoshIntegral[4*b*x]*Sinh[4*a]
)/3 - (b*Sinh[4*a + 4*b*x])/(12*x^2) + (4*b^3*Cosh[4*a]*SinhIntegral[4*b*x])/3

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^4} \, dx &=\int \left (-\frac{1}{8 x^4}+\frac{\cosh (4 a+4 b x)}{8 x^4}\right ) \, dx\\ &=\frac{1}{24 x^3}+\frac{1}{8} \int \frac{\cosh (4 a+4 b x)}{x^4} \, dx\\ &=\frac{1}{24 x^3}-\frac{\cosh (4 a+4 b x)}{24 x^3}+\frac{1}{6} b \int \frac{\sinh (4 a+4 b x)}{x^3} \, dx\\ &=\frac{1}{24 x^3}-\frac{\cosh (4 a+4 b x)}{24 x^3}-\frac{b \sinh (4 a+4 b x)}{12 x^2}+\frac{1}{3} b^2 \int \frac{\cosh (4 a+4 b x)}{x^2} \, dx\\ &=\frac{1}{24 x^3}-\frac{\cosh (4 a+4 b x)}{24 x^3}-\frac{b^2 \cosh (4 a+4 b x)}{3 x}-\frac{b \sinh (4 a+4 b x)}{12 x^2}+\frac{1}{3} \left (4 b^3\right ) \int \frac{\sinh (4 a+4 b x)}{x} \, dx\\ &=\frac{1}{24 x^3}-\frac{\cosh (4 a+4 b x)}{24 x^3}-\frac{b^2 \cosh (4 a+4 b x)}{3 x}-\frac{b \sinh (4 a+4 b x)}{12 x^2}+\frac{1}{3} \left (4 b^3 \cosh (4 a)\right ) \int \frac{\sinh (4 b x)}{x} \, dx+\frac{1}{3} \left (4 b^3 \sinh (4 a)\right ) \int \frac{\cosh (4 b x)}{x} \, dx\\ &=\frac{1}{24 x^3}-\frac{\cosh (4 a+4 b x)}{24 x^3}-\frac{b^2 \cosh (4 a+4 b x)}{3 x}+\frac{4}{3} b^3 \text{Chi}(4 b x) \sinh (4 a)-\frac{b \sinh (4 a+4 b x)}{12 x^2}+\frac{4}{3} b^3 \cosh (4 a) \text{Shi}(4 b x)\\ \end{align*}

Mathematica [A]  time = 0.191029, size = 79, normalized size = 0.86 $-\frac{-32 b^3 x^3 \sinh (4 a) \text{Chi}(4 b x)-32 b^3 x^3 \cosh (4 a) \text{Shi}(4 b x)+8 b^2 x^2 \cosh (4 (a+b x))+2 b x \sinh (4 (a+b x))+\cosh (4 (a+b x))-1}{24 x^3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(Cosh[a + b*x]^2*Sinh[a + b*x]^2)/x^4,x]

[Out]

-(-1 + Cosh[4*(a + b*x)] + 8*b^2*x^2*Cosh[4*(a + b*x)] - 32*b^3*x^3*CoshIntegral[4*b*x]*Sinh[4*a] + 2*b*x*Sinh
[4*(a + b*x)] - 32*b^3*x^3*Cosh[4*a]*SinhIntegral[4*b*x])/(24*x^3)

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Maple [A]  time = 0.054, size = 129, normalized size = 1.4 \begin{align*}{\frac{1}{24\,{x}^{3}}}-{\frac{{b}^{2}{{\rm e}^{-4\,bx-4\,a}}}{6\,x}}+{\frac{b{{\rm e}^{-4\,bx-4\,a}}}{24\,{x}^{2}}}-{\frac{{{\rm e}^{-4\,bx-4\,a}}}{48\,{x}^{3}}}+{\frac{2\,{b}^{3}{{\rm e}^{-4\,a}}{\it Ei} \left ( 1,4\,bx \right ) }{3}}-{\frac{{{\rm e}^{4\,bx+4\,a}}}{48\,{x}^{3}}}-{\frac{b{{\rm e}^{4\,bx+4\,a}}}{24\,{x}^{2}}}-{\frac{{b}^{2}{{\rm e}^{4\,bx+4\,a}}}{6\,x}}-{\frac{2\,{b}^{3}{{\rm e}^{4\,a}}{\it Ei} \left ( 1,-4\,bx \right ) }{3}} \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^4,x)

[Out]

1/24/x^3-1/6*b^2*exp(-4*b*x-4*a)/x+1/24*b*exp(-4*b*x-4*a)/x^2-1/48*exp(-4*b*x-4*a)/x^3+2/3*b^3*exp(-4*a)*Ei(1,
4*b*x)-1/48/x^3*exp(4*b*x+4*a)-1/24*b/x^2*exp(4*b*x+4*a)-1/6*b^2/x*exp(4*b*x+4*a)-2/3*b^3*exp(4*a)*Ei(1,-4*b*x
)

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Maxima [A]  time = 1.33638, size = 49, normalized size = 0.53 \begin{align*} -4 \, b^{3} e^{\left (-4 \, a\right )} \Gamma \left (-3, 4 \, b x\right ) + 4 \, b^{3} e^{\left (4 \, a\right )} \Gamma \left (-3, -4 \, b x\right ) + \frac{1}{24 \, x^{3}} \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^4,x, algorithm="maxima")

[Out]

-4*b^3*e^(-4*a)*gamma(-3, 4*b*x) + 4*b^3*e^(4*a)*gamma(-3, -4*b*x) + 1/24/x^3

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Fricas [B]  time = 1.8123, size = 436, normalized size = 4.74 \begin{align*} -\frac{8 \, b x \cosh \left (b x + a\right )^{3} \sinh \left (b x + a\right ) + 8 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} +{\left (8 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )^{4} + 6 \,{\left (8 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{2} +{\left (8 \, b^{2} x^{2} + 1\right )} \sinh \left (b x + a\right )^{4} - 16 \,{\left (b^{3} x^{3}{\rm Ei}\left (4 \, b x\right ) - b^{3} x^{3}{\rm Ei}\left (-4 \, b x\right )\right )} \cosh \left (4 \, a\right ) - 16 \,{\left (b^{3} x^{3}{\rm Ei}\left (4 \, b x\right ) + b^{3} x^{3}{\rm Ei}\left (-4 \, b x\right )\right )} \sinh \left (4 \, a\right ) - 1}{24 \, x^{3}} \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^4,x, algorithm="fricas")

[Out]

-1/24*(8*b*x*cosh(b*x + a)^3*sinh(b*x + a) + 8*b*x*cosh(b*x + a)*sinh(b*x + a)^3 + (8*b^2*x^2 + 1)*cosh(b*x +
a)^4 + 6*(8*b^2*x^2 + 1)*cosh(b*x + a)^2*sinh(b*x + a)^2 + (8*b^2*x^2 + 1)*sinh(b*x + a)^4 - 16*(b^3*x^3*Ei(4*
b*x) - b^3*x^3*Ei(-4*b*x))*cosh(4*a) - 16*(b^3*x^3*Ei(4*b*x) + b^3*x^3*Ei(-4*b*x))*sinh(4*a) - 1)/x^3

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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^{4}}\, 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**4,x)

[Out]

Integral(sinh(a + b*x)**2*cosh(a + b*x)**2/x**4, x)

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Giac [A]  time = 1.1609, size = 166, normalized size = 1.8 \begin{align*} \frac{32 \, b^{3} x^{3}{\rm Ei}\left (4 \, b x\right ) e^{\left (4 \, a\right )} - 32 \, b^{3} x^{3}{\rm Ei}\left (-4 \, b x\right ) e^{\left (-4 \, a\right )} - 8 \, b^{2} x^{2} e^{\left (4 \, b x + 4 \, a\right )} - 8 \, b^{2} x^{2} e^{\left (-4 \, b x - 4 \, a\right )} - 2 \, b x e^{\left (4 \, b x + 4 \, a\right )} + 2 \, 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}{48 \, x^{3}} \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^4,x, algorithm="giac")

[Out]

1/48*(32*b^3*x^3*Ei(4*b*x)*e^(4*a) - 32*b^3*x^3*Ei(-4*b*x)*e^(-4*a) - 8*b^2*x^2*e^(4*b*x + 4*a) - 8*b^2*x^2*e^
(-4*b*x - 4*a) - 2*b*x*e^(4*b*x + 4*a) + 2*b*x*e^(-4*b*x - 4*a) - e^(4*b*x + 4*a) - e^(-4*b*x - 4*a) + 2)/x^3