### 3.258 $$\int \frac{\cosh (a+b x) \sinh (a+b x)}{x^4} \, dx$$

Optimal. Leaf size=85 $\frac{2}{3} b^3 \cosh (2 a) \text{Chi}(2 b x)+\frac{2}{3} b^3 \sinh (2 a) \text{Shi}(2 b x)-\frac{b^2 \sinh (2 a+2 b x)}{3 x}-\frac{\sinh (2 a+2 b x)}{6 x^3}-\frac{b \cosh (2 a+2 b x)}{6 x^2}$

[Out]

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

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Rubi [A]  time = 0.150576, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 16, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.375, Rules used = {5448, 12, 3297, 3303, 3298, 3301} $\frac{2}{3} b^3 \cosh (2 a) \text{Chi}(2 b x)+\frac{2}{3} b^3 \sinh (2 a) \text{Shi}(2 b x)-\frac{b^2 \sinh (2 a+2 b x)}{3 x}-\frac{\sinh (2 a+2 b x)}{6 x^3}-\frac{b \cosh (2 a+2 b x)}{6 x^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(b*Cosh[2*a + 2*b*x])/(6*x^2) + (2*b^3*Cosh[2*a]*CoshIntegral[2*b*x])/3 - Sinh[2*a + 2*b*x]/(6*x^3) - (b^2*Si
nh[2*a + 2*b*x])/(3*x) + (2*b^3*Sinh[2*a]*SinhIntegral[2*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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A]  time = 0.155667, size = 77, normalized size = 0.91 $-\frac{-4 b^3 x^3 \cosh (2 a) \text{Chi}(2 b x)-4 b^3 x^3 \sinh (2 a) \text{Shi}(2 b x)+2 b^2 x^2 \sinh (2 (a+b x))+\sinh (2 (a+b x))+b x \cosh (2 (a+b x))}{6 x^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(b*x*Cosh[2*(a + b*x)] - 4*b^3*x^3*Cosh[2*a]*CoshIntegral[2*b*x] + Sinh[2*(a + b*x)] + 2*b^2*x^2*Sinh[2*(a +
b*x)] - 4*b^3*x^3*Sinh[2*a]*SinhIntegral[2*b*x])/(6*x^3)

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Maple [A]  time = 0.034, size = 124, normalized size = 1.5 \begin{align*}{\frac{{b}^{2}{{\rm e}^{-2\,bx-2\,a}}}{6\,x}}-{\frac{b{{\rm e}^{-2\,bx-2\,a}}}{12\,{x}^{2}}}+{\frac{{{\rm e}^{-2\,bx-2\,a}}}{12\,{x}^{3}}}-{\frac{{b}^{3}{{\rm e}^{-2\,a}}{\it Ei} \left ( 1,2\,bx \right ) }{3}}-{\frac{{{\rm e}^{2\,bx+2\,a}}}{12\,{x}^{3}}}-{\frac{b{{\rm e}^{2\,bx+2\,a}}}{12\,{x}^{2}}}-{\frac{{b}^{2}{{\rm e}^{2\,bx+2\,a}}}{6\,x}}-{\frac{{b}^{3}{{\rm e}^{2\,a}}{\it Ei} \left ( 1,-2\,bx \right ) }{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(b*x+a)*sinh(b*x+a)/x^4,x)

[Out]

1/6*b^2*exp(-2*b*x-2*a)/x-1/12*b*exp(-2*b*x-2*a)/x^2+1/12*exp(-2*b*x-2*a)/x^3-1/3*b^3*exp(-2*a)*Ei(1,2*b*x)-1/
12*exp(2*b*x+2*a)/x^3-1/12*b*exp(2*b*x+2*a)/x^2-1/6*b^2*exp(2*b*x+2*a)/x-1/3*b^3*exp(2*a)*Ei(1,-2*b*x)

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Maxima [A]  time = 1.39122, size = 42, normalized size = 0.49 \begin{align*} 2 \, b^{3} e^{\left (-2 \, a\right )} \Gamma \left (-3, 2 \, b x\right ) + 2 \, b^{3} e^{\left (2 \, a\right )} \Gamma \left (-3, -2 \, b x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*sinh(b*x+a)/x^4,x, algorithm="maxima")

[Out]

2*b^3*e^(-2*a)*gamma(-3, 2*b*x) + 2*b^3*e^(2*a)*gamma(-3, -2*b*x)

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Fricas [A]  time = 1.96662, size = 286, normalized size = 3.36 \begin{align*} -\frac{b x \cosh \left (b x + a\right )^{2} + b x \sinh \left (b x + a\right )^{2} + 2 \,{\left (2 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) - 2 \,{\left (b^{3} x^{3}{\rm Ei}\left (2 \, b x\right ) + b^{3} x^{3}{\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (2 \, a\right ) - 2 \,{\left (b^{3} x^{3}{\rm Ei}\left (2 \, b x\right ) - b^{3} x^{3}{\rm Ei}\left (-2 \, b x\right )\right )} \sinh \left (2 \, a\right )}{6 \, x^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*sinh(b*x+a)/x^4,x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Giac [A]  time = 1.11924, size = 162, normalized size = 1.91 \begin{align*} \frac{4 \, b^{3} x^{3}{\rm Ei}\left (2 \, b x\right ) e^{\left (2 \, a\right )} + 4 \, b^{3} x^{3}{\rm Ei}\left (-2 \, b x\right ) e^{\left (-2 \, a\right )} - 2 \, b^{2} x^{2} e^{\left (2 \, b x + 2 \, a\right )} + 2 \, b^{2} x^{2} e^{\left (-2 \, b x - 2 \, a\right )} - b x e^{\left (2 \, b x + 2 \, a\right )} - b x e^{\left (-2 \, b x - 2 \, a\right )} - e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )}}{12 \, x^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*sinh(b*x+a)/x^4,x, algorithm="giac")

[Out]

1/12*(4*b^3*x^3*Ei(2*b*x)*e^(2*a) + 4*b^3*x^3*Ei(-2*b*x)*e^(-2*a) - 2*b^2*x^2*e^(2*b*x + 2*a) + 2*b^2*x^2*e^(-
2*b*x - 2*a) - b*x*e^(2*b*x + 2*a) - b*x*e^(-2*b*x - 2*a) - e^(2*b*x + 2*a) + e^(-2*b*x - 2*a))/x^3