3.49 \(\int e^{\text{sech}^{-1}(a x^2)} x^3 \, dx\)

Optimal. Leaf size=63 \[ \frac{\sqrt{\frac{1}{a x^2+1}} \sqrt{a x^2+1} \sin ^{-1}\left (a x^2\right )}{4 a^2}+\frac{x^2}{4 a}+\frac{1}{4} x^4 e^{\text{sech}^{-1}\left (a x^2\right )} \]

[Out]

x^2/(4*a) + (E^ArcSech[a*x^2]*x^4)/4 + (Sqrt[(1 + a*x^2)^(-1)]*Sqrt[1 + a*x^2]*ArcSin[a*x^2])/(4*a^2)

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Rubi [A]  time = 0.0368681, antiderivative size = 63, 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 = {6335, 30, 259, 275, 216} \[ \frac{\sqrt{\frac{1}{a x^2+1}} \sqrt{a x^2+1} \sin ^{-1}\left (a x^2\right )}{4 a^2}+\frac{x^2}{4 a}+\frac{1}{4} x^4 e^{\text{sech}^{-1}\left (a x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcSech[a*x^2]*x^3,x]

[Out]

x^2/(4*a) + (E^ArcSech[a*x^2]*x^4)/4 + (Sqrt[(1 + a*x^2)^(-1)]*Sqrt[1 + a*x^2]*ArcSin[a*x^2])/(4*a^2)

Rule 6335

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 259

Int[((c_.)*(x_))^(m_.)*((a1_) + (b1_.)*(x_)^(n_))^(p_)*((a2_) + (b2_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[(c*x)
^m*(a1*a2 + b1*b2*x^(2*n))^p, x] /; FreeQ[{a1, b1, a2, b2, c, m, n, p}, x] && EqQ[a2*b1 + a1*b2, 0] && (Intege
rQ[p] || (GtQ[a1, 0] && GtQ[a2, 0]))

Rule 275

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

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

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

Mathematica [C]  time = 0.103778, size = 92, normalized size = 1.46 \[ \frac{2 a x^2+a \sqrt{\frac{1-a x^2}{a x^2+1}} \left (a x^4+x^2\right )+i \log \left (2 \sqrt{\frac{1-a x^2}{a x^2+1}} \left (a x^2+1\right )-2 i a x^2\right )}{4 a^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcSech[a*x^2]*x^3,x]

[Out]

(2*a*x^2 + a*Sqrt[(1 - a*x^2)/(1 + a*x^2)]*(x^2 + a*x^4) + I*Log[(-2*I)*a*x^2 + 2*Sqrt[(1 - a*x^2)/(1 + a*x^2)
]*(1 + a*x^2)])/(4*a^2)

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Maple [A]  time = 0.33, size = 112, normalized size = 1.8 \begin{align*}{\frac{{x}^{2}}{4\,{a}^{2}}\sqrt{-{\frac{a{x}^{2}-1}{a{x}^{2}}}}\sqrt{{\frac{a{x}^{2}+1}{a{x}^{2}}}} \left ({x}^{2}\sqrt{-{\frac{{a}^{2}{x}^{4}-1}{{a}^{2}}}}{a}^{2}+\arctan \left ({{x}^{2}{\frac{1}{\sqrt{-{\frac{{a}^{2}{x}^{4}-1}{{a}^{2}}}}}}} \right ) \right ){\frac{1}{\sqrt{-{\frac{{a}^{2}{x}^{4}-1}{{a}^{2}}}}}}}+{\frac{{x}^{2}}{2\,a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(-(a*x^2-1)/a/x^2)^(1/2)*x^2*((a*x^2+1)/a/x^2)^(1/2)*(x^2*(-(a^2*x^4-1)/a^2)^(1/2)*a^2+arctan(x^2/(-(a^2*x
^4-1)/a^2)^(1/2)))/(-(a^2*x^4-1)/a^2)^(1/2)/a^2+1/2*x^2/a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x^2+(1/a/x^2-1)^(1/2)*(1/a/x^2+1)^(1/2))*x^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.05342, size = 225, normalized size = 3.57 \begin{align*} \frac{a^{2} x^{4} \sqrt{\frac{a x^{2} + 1}{a x^{2}}} \sqrt{-\frac{a x^{2} - 1}{a x^{2}}} + 2 \, a x^{2} - 2 \, \arctan \left (\frac{a x^{2} \sqrt{\frac{a x^{2} + 1}{a x^{2}}} \sqrt{-\frac{a x^{2} - 1}{a x^{2}}} - 1}{a x^{2}}\right )}{4 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x^2+(1/a/x^2-1)^(1/2)*(1/a/x^2+1)^(1/2))*x^3,x, algorithm="fricas")

[Out]

1/4*(a^2*x^4*sqrt((a*x^2 + 1)/(a*x^2))*sqrt(-(a*x^2 - 1)/(a*x^2)) + 2*a*x^2 - 2*arctan((a*x^2*sqrt((a*x^2 + 1)
/(a*x^2))*sqrt(-(a*x^2 - 1)/(a*x^2)) - 1)/(a*x^2)))/a^2

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int x\, dx + \int a x^{3} \sqrt{-1 + \frac{1}{a x^{2}}} \sqrt{1 + \frac{1}{a x^{2}}}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x**2+(1/a/x**2-1)**(1/2)*(1/a/x**2+1)**(1/2))*x**3,x)

[Out]

(Integral(x, x) + Integral(a*x**3*sqrt(-1 + 1/(a*x**2))*sqrt(1 + 1/(a*x**2)), x))/a

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Giac [A]  time = 1.16296, size = 107, normalized size = 1.7 \begin{align*} \frac{2 \,{\left (a^{2} x^{2} + a\right )} a^{2} +{\left (\sqrt{a^{2} x^{2} + a} \sqrt{-a^{2} x^{2} + a} a^{2} x^{2} - 2 \, a^{2} \arcsin \left (\frac{\sqrt{2} \sqrt{-a^{2} x^{2} + a}}{2 \, \sqrt{a}}\right )\right )} a}{4 \, a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x^2+(1/a/x^2-1)^(1/2)*(1/a/x^2+1)^(1/2))*x^3,x, algorithm="giac")

[Out]

1/4*(2*(a^2*x^2 + a)*a^2 + (sqrt(a^2*x^2 + a)*sqrt(-a^2*x^2 + a)*a^2*x^2 - 2*a^2*arcsin(1/2*sqrt(2)*sqrt(-a^2*
x^2 + a)/sqrt(a)))*a)/a^5