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

Optimal. Leaf size=169 \[ \frac{(a x+1)^3 \left (\sqrt{\frac{1-a x}{a x+1}}+1\right )^4}{12 a^3}-\frac{\sqrt{\frac{1-a x}{a x+1}} (a x+1)^2 \left (\sqrt{\frac{1-a x}{a x+1}}+1\right )^3}{6 a^3}+\frac{(a x+1) \left (1-\sqrt{\frac{1-a x}{a x+1}}\right ) \left (\sqrt{\frac{1-a x}{a x+1}}+1\right )}{2 a^3}-\frac{2 \tan ^{-1}\left (\sqrt{\frac{1-a x}{a x+1}}\right )}{a^3} \]

[Out]

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

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Rubi [A]  time = 0.46505, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6337, 821, 12, 729, 723, 203} \[ \frac{(a x+1)^3 \left (\sqrt{\frac{1-a x}{a x+1}}+1\right )^4}{12 a^3}-\frac{\sqrt{\frac{1-a x}{a x+1}} (a x+1)^2 \left (\sqrt{\frac{1-a x}{a x+1}}+1\right )^3}{6 a^3}+\frac{(a x+1) \left (1-\sqrt{\frac{1-a x}{a x+1}}\right ) \left (\sqrt{\frac{1-a x}{a x+1}}+1\right )}{2 a^3}-\frac{2 \tan ^{-1}\left (\sqrt{\frac{1-a x}{a x+1}}\right )}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6337

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

Rule 821

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^m*
(a + c*x^2)^(p + 1)*(a*g - c*f*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*
x^2)^(p + 1)*Simp[a*e*g*m - c*d*f*(2*p + 3) - c*e*f*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x
] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 729

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^m*(2*c*x)*(a + c*x^2)^(
p + 1))/(4*a*c*(p + 1)), x] - Dist[(m*(2*c*d))/(4*a*c*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1), x],
x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 3, 0] && LtQ[p, -1]

Rule 723

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(a*e - c*d*x)*(a
+ c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] + Dist[((2*p + 3)*(c*d^2 + a*e^2))/(2*a*c*(p + 1)), Int[(d + e*x)^(m -
2)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 2, 0] && Lt
Q[p, -1]

Rule 203

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

Rubi steps

\begin{align*} \int e^{2 \text{sech}^{-1}(a x)} x^2 \, dx &=\int x^2 \left (\frac{1}{a x}+\sqrt{\frac{1-a x}{1+a x}}+\frac{\sqrt{\frac{1-a x}{1+a x}}}{a x}\right )^2 \, dx\\ &=-\frac{4 \operatorname{Subst}\left (\int \frac{x (1+x)^4}{\left (1+x^2\right )^4} \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )}{a^3}\\ &=\frac{(1+a x)^3 \left (1+\sqrt{\frac{1-a x}{1+a x}}\right )^4}{12 a^3}-\frac{2 \operatorname{Subst}\left (\int \frac{4 (1+x)^3}{\left (1+x^2\right )^3} \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )}{3 a^3}\\ &=\frac{(1+a x)^3 \left (1+\sqrt{\frac{1-a x}{1+a x}}\right )^4}{12 a^3}-\frac{8 \operatorname{Subst}\left (\int \frac{(1+x)^3}{\left (1+x^2\right )^3} \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )}{3 a^3}\\ &=-\frac{\sqrt{\frac{1-a x}{1+a x}} (1+a x)^2 \left (1+\sqrt{\frac{1-a x}{1+a x}}\right )^3}{6 a^3}+\frac{(1+a x)^3 \left (1+\sqrt{\frac{1-a x}{1+a x}}\right )^4}{12 a^3}-\frac{2 \operatorname{Subst}\left (\int \frac{(1+x)^2}{\left (1+x^2\right )^2} \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )}{a^3}\\ &=\frac{(1+a x) \left (1-\sqrt{\frac{1-a x}{1+a x}}\right ) \left (1+\sqrt{\frac{1-a x}{1+a x}}\right )}{2 a^3}-\frac{\sqrt{\frac{1-a x}{1+a x}} (1+a x)^2 \left (1+\sqrt{\frac{1-a x}{1+a x}}\right )^3}{6 a^3}+\frac{(1+a x)^3 \left (1+\sqrt{\frac{1-a x}{1+a x}}\right )^4}{12 a^3}-\frac{2 \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )}{a^3}\\ &=\frac{(1+a x) \left (1-\sqrt{\frac{1-a x}{1+a x}}\right ) \left (1+\sqrt{\frac{1-a x}{1+a x}}\right )}{2 a^3}-\frac{\sqrt{\frac{1-a x}{1+a x}} (1+a x)^2 \left (1+\sqrt{\frac{1-a x}{1+a x}}\right )^3}{6 a^3}+\frac{(1+a x)^3 \left (1+\sqrt{\frac{1-a x}{1+a x}}\right )^4}{12 a^3}-\frac{2 \tan ^{-1}\left (\sqrt{\frac{1-a x}{1+a x}}\right )}{a^3}\\ \end{align*}

Mathematica [C]  time = 0.0703845, size = 86, normalized size = 0.51 \[ \sqrt{\frac{1-a x}{a x+1}} \left (\frac{x}{a^2}+\frac{x^2}{a}\right )+\frac{2 x}{a^2}+\frac{i \log \left (2 \sqrt{\frac{1-a x}{a x+1}} (a x+1)-2 i a x\right )}{a^3}-\frac{x^3}{3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*x)/a^2 - x^3/3 + Sqrt[(1 - a*x)/(1 + a*x)]*(x/a^2 + x^2/a) + (I*Log[(-2*I)*a*x + 2*Sqrt[(1 - a*x)/(1 + a*x)
]*(1 + a*x)])/a^3

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Maple [C]  time = 0.182, size = 97, normalized size = 0.6 \begin{align*} -{\frac{{x}^{3}}{3}}+2\,{\frac{x}{{a}^{2}}}+{\frac{x{\it csgn} \left ( a \right ) }{{a}^{2}}\sqrt{-{\frac{ax-1}{ax}}}\sqrt{{\frac{ax+1}{ax}}} \left ( x\sqrt{-{a}^{2}{x}^{2}+1}{\it csgn} \left ( a \right ) a+\arctan \left ({x{\it csgn} \left ( a \right ) a{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) \right ){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.11647, size = 192, normalized size = 1.14 \begin{align*} -\frac{a^{3} x^{3} - 3 \, a^{2} x^{2} \sqrt{\frac{a x + 1}{a x}} \sqrt{-\frac{a x - 1}{a x}} - 6 \, a x + 3 \, \arctan \left (\sqrt{\frac{a x + 1}{a x}} \sqrt{-\frac{a x - 1}{a x}}\right )}{3 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(a^3*x^3 - 3*a^2*x^2*sqrt((a*x + 1)/(a*x))*sqrt(-(a*x - 1)/(a*x)) - 6*a*x + 3*arctan(sqrt((a*x + 1)/(a*x)
)*sqrt(-(a*x - 1)/(a*x))))/a^3

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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