3.19 \(\int e^{3 \tanh ^{-1}(a x)} x^2 \, dx\)

Optimal. Leaf size=92 \[ \frac{(a x+1)^3}{a^3 \sqrt{1-a^2 x^2}}+\frac{(a x+3)^2 \sqrt{1-a^2 x^2}}{3 a^3}+\frac{(3 a x+28) \sqrt{1-a^2 x^2}}{6 a^3}-\frac{11 \sin ^{-1}(a x)}{2 a^3} \]

[Out]

(1 + a*x)^3/(a^3*Sqrt[1 - a^2*x^2]) + ((3 + a*x)^2*Sqrt[1 - a^2*x^2])/(3*a^3) + ((28 + 3*a*x)*Sqrt[1 - a^2*x^2
])/(6*a^3) - (11*ArcSin[a*x])/(2*a^3)

________________________________________________________________________________________

Rubi [A]  time = 0.649267, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {6124, 1633, 1593, 12, 852, 1635, 1654, 780, 216} \[ \frac{(a x+1)^3}{a^3 \sqrt{1-a^2 x^2}}+\frac{(a x+3)^2 \sqrt{1-a^2 x^2}}{3 a^3}+\frac{(3 a x+28) \sqrt{1-a^2 x^2}}{6 a^3}-\frac{11 \sin ^{-1}(a x)}{2 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 + a*x)^3/(a^3*Sqrt[1 - a^2*x^2]) + ((3 + a*x)^2*Sqrt[1 - a^2*x^2])/(3*a^3) + ((28 + 3*a*x)*Sqrt[1 - a^2*x^2
])/(6*a^3) - (11*ArcSin[a*x])/(2*a^3)

Rule 6124

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 + a*x)^((n + 1)/2)/((1 - a*x)^((n - 1)/
2)*Sqrt[1 - a^2*x^2])), x] /; FreeQ[{a, m}, x] && IntegerQ[(n - 1)/2]

Rule 1633

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d*e, Int[(d + e*x)^(m - 1)*
PolynomialQuotient[Pq, a*e + c*d*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq
, x] && EqQ[c*d^2 + a*e^2, 0] && EqQ[PolynomialRemainder[Pq, a*e + c*d*x, x], 0]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - 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 852

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

Rule 1635

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, -Simp[(d*f*(d + e*x)^m*(a + c*x^2)^(p + 1))/(2*
a*e*(p + 1)), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)*Q
+ f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p +
1/2, 0] && GtQ[m, 0]

Rule 1654

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff
[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + c*x^2)^(p + 1))/(c*e^(q - 1)*(m + q + 2*p + 1)), x]
+ Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*
f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(a*e^2*(m + q - 1) - c*d^2*(m + q + 2*p + 1) - 2*c*d*e*(
m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq
, x] && NeQ[c*d^2 + a*e^2, 0] &&  !(EqQ[d, 0] && True) &&  !(IGtQ[m, 0] && RationalQ[a, c, d, e] && (IntegerQ[
p] || ILtQ[p + 1/2, 0]))

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

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^{3 \tanh ^{-1}(a x)} x^2 \, dx &=\int \frac{x^2 (1+a x)^2}{(1-a x) \sqrt{1-a^2 x^2}} \, dx\\ &=-\left (a \int \frac{\sqrt{1-a^2 x^2} \left (-\frac{x^2}{a}-x^3\right )}{(1-a x)^2} \, dx\right )\\ &=-\left (a \int \frac{\left (-\frac{1}{a}-x\right ) x^2 \sqrt{1-a^2 x^2}}{(1-a x)^2} \, dx\right )\\ &=a^2 \int \frac{x^2 \left (1-a^2 x^2\right )^{3/2}}{a^2 (1-a x)^3} \, dx\\ &=\int \frac{x^2 \left (1-a^2 x^2\right )^{3/2}}{(1-a x)^3} \, dx\\ &=\int \frac{x^2 (1+a x)^3}{\left (1-a^2 x^2\right )^{3/2}} \, dx\\ &=\frac{(1+a x)^3}{a^3 \sqrt{1-a^2 x^2}}-\int \frac{\left (\frac{3}{a^2}+\frac{x}{a}\right ) (1+a x)^2}{\sqrt{1-a^2 x^2}} \, dx\\ &=\frac{(1+a x)^3}{a^3 \sqrt{1-a^2 x^2}}+\frac{(3+a x)^2 \sqrt{1-a^2 x^2}}{3 a^3}+\frac{1}{3} \int \frac{\left (\frac{3}{a^2}+\frac{x}{a}\right ) (-5-3 a x)}{\sqrt{1-a^2 x^2}} \, dx\\ &=\frac{(1+a x)^3}{a^3 \sqrt{1-a^2 x^2}}+\frac{(3+a x)^2 \sqrt{1-a^2 x^2}}{3 a^3}+\frac{(28+3 a x) \sqrt{1-a^2 x^2}}{6 a^3}-\frac{11 \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{2 a^2}\\ &=\frac{(1+a x)^3}{a^3 \sqrt{1-a^2 x^2}}+\frac{(3+a x)^2 \sqrt{1-a^2 x^2}}{3 a^3}+\frac{(28+3 a x) \sqrt{1-a^2 x^2}}{6 a^3}-\frac{11 \sin ^{-1}(a x)}{2 a^3}\\ \end{align*}

Mathematica [A]  time = 0.0552105, size = 58, normalized size = 0.63 \[ \frac{\frac{\sqrt{1-a^2 x^2} \left (2 a^3 x^3+7 a^2 x^2+19 a x-52\right )}{a x-1}-33 \sin ^{-1}(a x)}{6 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[1 - a^2*x^2]*(-52 + 19*a*x + 7*a^2*x^2 + 2*a^3*x^3))/(-1 + a*x) - 33*ArcSin[a*x])/(6*a^3)

________________________________________________________________________________________

Maple [A]  time = 0.047, size = 122, normalized size = 1.3 \begin{align*} -{\frac{{x}^{4}a}{3}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}-{\frac{13\,{x}^{2}}{3\,a}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}+{\frac{26}{3\,{a}^{3}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}-{\frac{3\,{x}^{3}}{2}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}+{\frac{11\,x}{2\,{a}^{2}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}-{\frac{11}{2\,{a}^{2}}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*x^4*a/(-a^2*x^2+1)^(1/2)-13/3/a*x^2/(-a^2*x^2+1)^(1/2)+26/3/a^3/(-a^2*x^2+1)^(1/2)-3/2*x^3/(-a^2*x^2+1)^(
1/2)+11/2*x/a^2/(-a^2*x^2+1)^(1/2)-11/2/a^2/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))

________________________________________________________________________________________

Maxima [A]  time = 1.44558, size = 151, normalized size = 1.64 \begin{align*} -\frac{a x^{4}}{3 \, \sqrt{-a^{2} x^{2} + 1}} - \frac{3 \, x^{3}}{2 \, \sqrt{-a^{2} x^{2} + 1}} - \frac{13 \, x^{2}}{3 \, \sqrt{-a^{2} x^{2} + 1} a} + \frac{11 \, x}{2 \, \sqrt{-a^{2} x^{2} + 1} a^{2}} - \frac{11 \, \arcsin \left (\frac{a^{2} x}{\sqrt{a^{2}}}\right )}{2 \, \sqrt{a^{2}} a^{2}} + \frac{26}{3 \, \sqrt{-a^{2} x^{2} + 1} a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a*x^4/sqrt(-a^2*x^2 + 1) - 3/2*x^3/sqrt(-a^2*x^2 + 1) - 13/3*x^2/(sqrt(-a^2*x^2 + 1)*a) + 11/2*x/(sqrt(-a
^2*x^2 + 1)*a^2) - 11/2*arcsin(a^2*x/sqrt(a^2))/(sqrt(a^2)*a^2) + 26/3/(sqrt(-a^2*x^2 + 1)*a^3)

________________________________________________________________________________________

Fricas [A]  time = 1.79943, size = 197, normalized size = 2.14 \begin{align*} \frac{52 \, a x + 66 \,{\left (a x - 1\right )} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) +{\left (2 \, a^{3} x^{3} + 7 \, a^{2} x^{2} + 19 \, a x - 52\right )} \sqrt{-a^{2} x^{2} + 1} - 52}{6 \,{\left (a^{4} x - a^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(52*a*x + 66*(a*x - 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (2*a^3*x^3 + 7*a^2*x^2 + 19*a*x - 52)*sqrt
(-a^2*x^2 + 1) - 52)/(a^4*x - a^3)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.1822, size = 117, normalized size = 1.27 \begin{align*} \frac{1}{6} \, \sqrt{-a^{2} x^{2} + 1}{\left (x{\left (\frac{2 \, x}{a} + \frac{9}{a^{2}}\right )} + \frac{28}{a^{3}}\right )} - \frac{11 \, \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{2 \, a^{2}{\left | a \right |}} + \frac{8}{a^{2}{\left (\frac{\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a}{a^{2} x} - 1\right )}{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*sqrt(-a^2*x^2 + 1)*(x*(2*x/a + 9/a^2) + 28/a^3) - 11/2*arcsin(a*x)*sgn(a)/(a^2*abs(a)) + 8/(a^2*((sqrt(-a^
2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)*abs(a))