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

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

[Out]

x/a+1/3*x^3*(1+1/a^2/x^4)^(1/2)-1/3*(a+1/x^2)*(cos(2*arccot(x*a^(1/2)))^2)^(1/2)/cos(2*arccot(x*a^(1/2)))*Elli
pticF(sin(2*arccot(x*a^(1/2))),1/2*2^(1/2))*((a^2+1/x^4)/(a+1/x^2)^2)^(1/2)/a^(5/2)/(1+1/a^2/x^4)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 86, normalized size of antiderivative = 1.00, 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 8

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

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 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 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]

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*}

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Mathematica [C]  time = 0.23, size = 113, normalized size = 1.31 \[ -\frac {2 \sqrt {2} x e^{-\text {csch}^{-1}\left (a x^2\right )} \left (\frac {e^{\text {csch}^{-1}\left (a x^2\right )}}{e^{2 \text {csch}^{-1}\left (a x^2\right )}-1}\right )^{3/2} \left (\left (1-e^{2 \text {csch}^{-1}\left (a x^2\right )}\right )^{3/2} \left (-\, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};e^{2 \text {csch}^{-1}\left (a x^2\right )}\right )\right )-2 e^{2 \text {csch}^{-1}\left (a x^2\right )}+1\right )}{3 a \sqrt {a x^2}} \]

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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fricas [F]  time = 2.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a x^{2} \sqrt {\frac {a^{2} x^{4} + 1}{a^{2} x^{4}}} + 1}{a}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} {\left (\sqrt {\frac {1}{a^{2} x^{4}} + 1} + \frac {1}{a x^{2}}\right )}\,{d x} \]

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)

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maple [C]  time = 0.05, size = 104, normalized size = 1.21 \[ \frac {\sqrt {\frac {a^{2} x^{4}+1}{a^{2} x^{4}}}\, x^{2} \left (\sqrt {i a}\, x^{5} a^{2}+2 \sqrt {-i a \,x^{2}+1}\, \sqrt {i a \,x^{2}+1}\, \EllipticF \left (x \sqrt {i a}, i\right )+x \sqrt {i a}\right )}{3 \left (a^{2} x^{4}+1\right ) \sqrt {i a}}+\frac {x}{a} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \frac {x}{a} + \frac {\frac {x \Gamma \left (\frac {1}{4}\right ) \,_2F_1\left (\begin {matrix} -\frac {1}{2},\frac {1}{4} \\ \frac {5}{4} \end {matrix} ; -a^{2} x^{4} \right )}{4 \, \Gamma \left (\frac {5}{4}\right )}}{a} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,\left (\sqrt {\frac {1}{a^2\,x^4}+1}+\frac {1}{a\,x^2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 2.16, size = 41, normalized size = 0.48 \[ - \frac {x^{3} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{a^{2} x^{4}}} \right )}}{4 \Gamma \left (\frac {1}{4}\right )} + \frac {x}{a} \]

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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