### 3.50 $$\int e^{2 \text{csch}^{-1}(a x)} x^3 \, dx$$

Optimal. Leaf size=38 $\frac{2 x^3 \left (\frac{1}{a^2 x^2}+1\right )^{3/2}}{3 a}+\frac{x^2}{a^2}+\frac{x^4}{4}$

[Out]

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

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Rubi [A]  time = 0.224786, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 12, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.25, Rules used = {6338, 6742, 264} $\frac{2 x^3 \left (\frac{1}{a^2 x^2}+1\right )^{3/2}}{3 a}+\frac{x^2}{a^2}+\frac{x^4}{4}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^(2*ArcCsch[a*x])*x^3,x]

[Out]

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

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 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0467991, size = 44, normalized size = 1.16 $\frac{x^2}{a^2}+\frac{2 \sqrt{\frac{1}{a^2 x^2}+1} \left (a^2 x^3+x\right )}{3 a^3}+\frac{x^4}{4}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[E^(2*ArcCsch[a*x])*x^3,x]

[Out]

x^2/a^2 + x^4/4 + (2*Sqrt[1 + 1/(a^2*x^2)]*(x + a^2*x^3))/(3*a^3)

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Maple [A]  time = 0.183, size = 61, normalized size = 1.6 \begin{align*}{\frac{1}{{a}^{2}} \left ({\frac{{a}^{2}{x}^{4}}{4}}+{\frac{{x}^{2}}{2}} \right ) }+{\frac{2\,x \left ({a}^{2}{x}^{2}+1 \right ) }{3\,{a}^{3}}\sqrt{{\frac{{a}^{2}{x}^{2}+1}{{a}^{2}{x}^{2}}}}}+{\frac{{x}^{2}}{2\,{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^3,x)

[Out]

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

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Maxima [A]  time = 0.991612, size = 43, normalized size = 1.13 \begin{align*} \frac{1}{4} \, x^{4} + \frac{2 \, x^{3}{\left (\frac{1}{a^{2} x^{2}} + 1\right )}^{\frac{3}{2}}}{3 \, a} + \frac{x^{2}}{a^{2}} \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^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.5358, size = 109, normalized size = 2.87 \begin{align*} \frac{3 \, a^{3} x^{4} + 12 \, a x^{2} + 8 \,{\left (a^{2} x^{3} + x\right )} \sqrt{\frac{a^{2} x^{2} + 1}{a^{2} x^{2}}}}{12 \, a^{3}} \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^3,x, algorithm="fricas")

[Out]

1/12*(3*a^3*x^4 + 12*a*x^2 + 8*(a^2*x^3 + x)*sqrt((a^2*x^2 + 1)/(a^2*x^2)))/a^3

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Sympy [A]  time = 3.48901, size = 51, normalized size = 1.34 \begin{align*} \frac{x^{4}}{4} + \frac{2 x^{2} \sqrt{a^{2} x^{2} + 1}}{3 a^{2}} + \frac{x^{2}}{a^{2}} + \frac{2 \sqrt{a^{2} x^{2} + 1}}{3 a^{4}} \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**3,x)

[Out]

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

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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^3,x, algorithm="giac")

[Out]

Exception raised: TypeError