### 3.57 $$\int \frac{e^{2 \text{csch}^{-1}(a x)}}{x^4} \, dx$$

Optimal. Leaf size=58 $-\frac{2}{5} a^3 \left (\frac{1}{a^2 x^2}+1\right )^{5/2}+\frac{2}{3} a^3 \left (\frac{1}{a^2 x^2}+1\right )^{3/2}-\frac{2}{5 a^2 x^5}-\frac{1}{3 x^3}$

[Out]

(2*a^3*(1 + 1/(a^2*x^2))^(3/2))/3 - (2*a^3*(1 + 1/(a^2*x^2))^(5/2))/5 - 2/(5*a^2*x^5) - 1/(3*x^3)

________________________________________________________________________________________

Rubi [A]  time = 0.233002, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 12, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.333, Rules used = {6338, 6742, 266, 43} $-\frac{2}{5} a^3 \left (\frac{1}{a^2 x^2}+1\right )^{5/2}+\frac{2}{3} a^3 \left (\frac{1}{a^2 x^2}+1\right )^{3/2}-\frac{2}{5 a^2 x^5}-\frac{1}{3 x^3}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^(2*ArcCsch[a*x])/x^4,x]

[Out]

(2*a^3*(1 + 1/(a^2*x^2))^(3/2))/3 - (2*a^3*(1 + 1/(a^2*x^2))^(5/2))/5 - 2/(5*a^2*x^5) - 1/(3*x^3)

Rule 6338

Int[E^(ArcCsch[u_]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*(1/u + Sqrt[1 + 1/u^2])^n, x] /; FreeQ[m, x] && Int
egerQ[n]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{e^{2 \text{csch}^{-1}(a x)}}{x^4} \, dx &=\int \frac{\left (\sqrt{1+\frac{1}{a^2 x^2}}+\frac{1}{a x}\right )^2}{x^4} \, dx\\ &=\int \left (\frac{2}{a^2 x^6}+\frac{2 \sqrt{1+\frac{1}{a^2 x^2}}}{a x^5}+\frac{1}{x^4}\right ) \, dx\\ &=-\frac{2}{5 a^2 x^5}-\frac{1}{3 x^3}+\frac{2 \int \frac{\sqrt{1+\frac{1}{a^2 x^2}}}{x^5} \, dx}{a}\\ &=-\frac{2}{5 a^2 x^5}-\frac{1}{3 x^3}-\frac{\operatorname{Subst}\left (\int x \sqrt{1+\frac{x}{a^2}} \, dx,x,\frac{1}{x^2}\right )}{a}\\ &=-\frac{2}{5 a^2 x^5}-\frac{1}{3 x^3}-\frac{\operatorname{Subst}\left (\int \left (-a^2 \sqrt{1+\frac{x}{a^2}}+a^2 \left (1+\frac{x}{a^2}\right )^{3/2}\right ) \, dx,x,\frac{1}{x^2}\right )}{a}\\ &=\frac{2}{3} a^3 \left (1+\frac{1}{a^2 x^2}\right )^{3/2}-\frac{2}{5} a^3 \left (1+\frac{1}{a^2 x^2}\right )^{5/2}-\frac{2}{5 a^2 x^5}-\frac{1}{3 x^3}\\ \end{align*}

Mathematica [A]  time = 0.0536828, size = 54, normalized size = 0.93 $-\frac{5 a^2 x^2+2 a x \sqrt{\frac{1}{a^2 x^2}+1} \left (-2 a^4 x^4+a^2 x^2+3\right )+6}{15 a^2 x^5}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[E^(2*ArcCsch[a*x])/x^4,x]

[Out]

-(6 + 5*a^2*x^2 + 2*a*Sqrt[1 + 1/(a^2*x^2)]*x*(3 + a^2*x^2 - 2*a^4*x^4))/(15*a^2*x^5)

________________________________________________________________________________________

Maple [A]  time = 0.184, size = 73, normalized size = 1.3 \begin{align*}{\frac{1}{{a}^{2}} \left ( -{\frac{1}{5\,{x}^{5}}}-{\frac{{a}^{2}}{3\,{x}^{3}}} \right ) }+{\frac{ \left ( 2\,{a}^{2}{x}^{2}+2 \right ) \left ( 2\,{a}^{2}{x}^{2}-3 \right ) }{15\,{x}^{4}a}\sqrt{{\frac{{a}^{2}{x}^{2}+1}{{a}^{2}{x}^{2}}}}}-{\frac{1}{5\,{x}^{5}{a}^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((1/a/x+(1+1/a^2/x^2)^(1/2))^2/x^4,x)

[Out]

1/a^2*(-1/5/x^5-1/3*a^2/x^3)+2/15/a*((a^2*x^2+1)/a^2/x^2)^(1/2)/x^4*(a^2*x^2+1)*(2*a^2*x^2-3)-1/5/x^5/a^2

________________________________________________________________________________________

Maxima [A]  time = 1.00086, size = 70, normalized size = 1.21 \begin{align*} -\frac{2 \,{\left (3 \, a^{4}{\left (\frac{1}{a^{2} x^{2}} + 1\right )}^{\frac{5}{2}} - 5 \, a^{4}{\left (\frac{1}{a^{2} x^{2}} + 1\right )}^{\frac{3}{2}}\right )}}{15 \, a} - \frac{1}{3 \, x^{3}} - \frac{2}{5 \, a^{2} x^{5}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x+(1+1/a^2/x^2)^(1/2))^2/x^4,x, algorithm="maxima")

[Out]

-2/15*(3*a^4*(1/(a^2*x^2) + 1)^(5/2) - 5*a^4*(1/(a^2*x^2) + 1)^(3/2))/a - 1/3/x^3 - 2/5/(a^2*x^5)

________________________________________________________________________________________

Fricas [A]  time = 2.68057, size = 146, normalized size = 2.52 \begin{align*} \frac{4 \, a^{5} x^{5} - 5 \, a^{2} x^{2} + 2 \,{\left (2 \, a^{5} x^{5} - a^{3} x^{3} - 3 \, a x\right )} \sqrt{\frac{a^{2} x^{2} + 1}{a^{2} x^{2}}} - 6}{15 \, a^{2} x^{5}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x+(1+1/a^2/x^2)^(1/2))^2/x^4,x, algorithm="fricas")

[Out]

1/15*(4*a^5*x^5 - 5*a^2*x^2 + 2*(2*a^5*x^5 - a^3*x^3 - 3*a*x)*sqrt((a^2*x^2 + 1)/(a^2*x^2)) - 6)/(a^2*x^5)

________________________________________________________________________________________

Sympy [A]  time = 3.76731, size = 76, normalized size = 1.31 \begin{align*} \frac{4 a^{2} \sqrt{a^{2} x^{2} + 1}}{15 x} - \frac{2 \sqrt{a^{2} x^{2} + 1}}{15 x^{3}} - \frac{1}{3 x^{3}} - \frac{2 \sqrt{a^{2} x^{2} + 1}}{5 a^{2} x^{5}} - \frac{2}{5 a^{2} x^{5}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x+(1+1/a**2/x**2)**(1/2))**2/x**4,x)

[Out]

4*a**2*sqrt(a**2*x**2 + 1)/(15*x) - 2*sqrt(a**2*x**2 + 1)/(15*x**3) - 1/(3*x**3) - 2*sqrt(a**2*x**2 + 1)/(5*a*
*2*x**5) - 2/(5*a**2*x**5)

________________________________________________________________________________________

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x+(1+1/a^2/x^2)^(1/2))^2/x^4,x, algorithm="giac")

[Out]

Exception raised: TypeError