3.40 \(\int e^{\text{csch}^{-1}(a x^2)} x^2 \, dx\)

Optimal. Leaf size=86 \[ -\frac{\sqrt{\frac{a^2+\frac{1}{x^4}}{\left (a+\frac{1}{x^2}\right )^2}} \left (a+\frac{1}{x^2}\right ) \text{EllipticF}\left (2 \cot ^{-1}\left (\sqrt{a} x\right ),\frac{1}{2}\right )}{3 a^{5/2} \sqrt{\frac{1}{a^2 x^4}+1}}+\frac{1}{3} x^3 \sqrt{\frac{1}{a^2 x^4}+1}+\frac{x}{a} \]

[Out]

x/a + (Sqrt[1 + 1/(a^2*x^4)]*x^3)/3 - (Sqrt[(a^2 + x^(-4))/(a + x^(-2))^2]*(a + x^(-2))*EllipticF[2*ArcCot[Sqr
t[a]*x], 1/2])/(3*a^(5/2)*Sqrt[1 + 1/(a^2*x^4)])

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Rubi [A]  time = 0.0532904, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {6336, 8, 335, 277, 220} \[ \frac{1}{3} x^3 \sqrt{\frac{1}{a^2 x^4}+1}-\frac{\sqrt{\frac{a^2+\frac{1}{x^4}}{\left (a+\frac{1}{x^2}\right )^2}} \left (a+\frac{1}{x^2}\right ) F\left (2 \cot ^{-1}\left (\sqrt{a} x\right )|\frac{1}{2}\right )}{3 a^{5/2} \sqrt{\frac{1}{a^2 x^4}+1}}+\frac{x}{a} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCsch[a*x^2]*x^2,x]

[Out]

x/a + (Sqrt[1 + 1/(a^2*x^4)]*x^3)/3 - (Sqrt[(a^2 + x^(-4))/(a + x^(-2))^2]*(a + x^(-2))*EllipticF[2*ArcCot[Sqr
t[a]*x], 1/2])/(3*a^(5/2)*Sqrt[1 + 1/(a^2*x^4)])

Rule 6336

Int[E^ArcCsch[(a_.)*(x_)^(p_.)]*(x_)^(m_.), x_Symbol] :> Dist[1/a, Int[x^(m - p), x], x] + Int[x^m*Sqrt[1 + 1/
(a^2*x^(2*p))], x] /; FreeQ[{a, m, p}, x]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rule 277

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

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rubi steps

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

Mathematica [C]  time = 0.214752, size = 113, normalized size = 1.31 \[ -\frac{2 \sqrt{2} x e^{-\text{csch}^{-1}\left (a x^2\right )} \left (\frac{e^{\text{csch}^{-1}\left (a x^2\right )}}{e^{2 \text{csch}^{-1}\left (a x^2\right )}-1}\right )^{3/2} \left (\left (1-e^{2 \text{csch}^{-1}\left (a x^2\right )}\right )^{3/2} \left (-\text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},e^{2 \text{csch}^{-1}\left (a x^2\right )}\right )\right )-2 e^{2 \text{csch}^{-1}\left (a x^2\right )}+1\right )}{3 a \sqrt{a x^2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcCsch[a*x^2]*x^2,x]

[Out]

(-2*Sqrt[2]*(E^ArcCsch[a*x^2]/(-1 + E^(2*ArcCsch[a*x^2])))^(3/2)*x*(1 - 2*E^(2*ArcCsch[a*x^2]) - (1 - E^(2*Arc
Csch[a*x^2]))^(3/2)*Hypergeometric2F1[1/4, 1/2, 5/4, E^(2*ArcCsch[a*x^2])]))/(3*a*E^ArcCsch[a*x^2]*Sqrt[a*x^2]
)

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Maple [C]  time = 0.187, size = 104, normalized size = 1.2 \begin{align*}{\frac{{x}^{2}}{3\,{a}^{2}{x}^{4}+3}\sqrt{{\frac{{a}^{2}{x}^{4}+1}{{a}^{2}{x}^{4}}}} \left ( \sqrt{ia}{x}^{5}{a}^{2}+2\,\sqrt{1-ia{x}^{2}}\sqrt{1+ia{x}^{2}}{\it EllipticF} \left ( x\sqrt{ia},i \right ) +x\sqrt{ia} \right ){\frac{1}{\sqrt{ia}}}}+{\frac{x}{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((a^2*x^4+1)/a^2/x^4)^(1/2)*x^2*((I*a)^(1/2)*x^5*a^2+2*(1-I*a*x^2)^(1/2)*(1+I*a*x^2)^(1/2)*EllipticF(x*(I*
a)^(1/2),I)+x*(I*a)^(1/2))/(a^2*x^4+1)/(I*a)^(1/2)+x/a

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{x}{a} + \frac{\frac{x \Gamma \left (\frac{1}{4}\right ) \,_2F_1\left (\begin{matrix} -\frac{1}{2},\frac{1}{4} \\ \frac{5}{4} \end{matrix} ; -a^{2} x^{4} \right )}{4 \, \Gamma \left (\frac{5}{4}\right )}}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{a x^{2} \sqrt{\frac{a^{2} x^{4} + 1}{a^{2} x^{4}}} + 1}{a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*x^2*sqrt((a^2*x^4 + 1)/(a^2*x^4)) + 1)/a, x)

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Sympy [C]  time = 3.27897, size = 41, normalized size = 0.48 \begin{align*} - \frac{x^{3} \Gamma \left (- \frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{4}, - \frac{1}{2} \\ \frac{1}{4} \end{matrix}\middle |{\frac{e^{i \pi }}{a^{2} x^{4}}} \right )}}{4 \Gamma \left (\frac{1}{4}\right )} + \frac{x}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x**3*gamma(-3/4)*hyper((-3/4, -1/2), (1/4,), exp_polar(I*pi)/(a**2*x**4))/(4*gamma(1/4)) + x/a

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2}{\left (\sqrt{\frac{1}{a^{2} x^{4}} + 1} + \frac{1}{a x^{2}}\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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