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

Optimal. Leaf size=95 \[ -\frac{(1-a x)^3}{a^3 \sqrt{1-a^2 x^2}}-\frac{(3-a x)^2 \sqrt{1-a^2 x^2}}{3 a^3}-\frac{(28-3 a x) \sqrt{1-a^2 x^2}}{6 a^3}-\frac{11 \sin ^{-1}(a x)}{2 a^3} \]

[Out]

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

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Rubi [A]  time = 0.637627, antiderivative size = 95, 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} \[ -\frac{(1-a x)^3}{a^3 \sqrt{1-a^2 x^2}}-\frac{(3-a x)^2 \sqrt{1-a^2 x^2}}{3 a^3}-\frac{(28-3 a x) \sqrt{1-a^2 x^2}}{6 a^3}-\frac{11 \sin ^{-1}(a x)}{2 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0688988, size = 58, normalized size = 0.61 \[ -\frac{\frac{\sqrt{1-a^2 x^2} \left (2 a^3 x^3-7 a^2 x^2+19 a x+52\right )}{a x+1}+33 \sin ^{-1}(a x)}{6 a^3} \]

Antiderivative was successfully verified.

[In]

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

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Maple [B]  time = 0.048, size = 170, normalized size = 1.8 \begin{align*} -4\,{\frac{ \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{5/2}}{{a}^{5} \left ( x+{a}^{-1} \right ) ^{2}}}-{\frac{11}{3\,{a}^{3}} \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{11\,x}{2\,{a}^{2}}\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}-{\frac{11}{2\,{a}^{2}}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}}-{\frac{1}{{a}^{6} \left ( x+{a}^{-1} \right ) ^{3}} \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/a^5/(x+1/a)^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)-11/3/a^3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(3/2)-11/2/a^2*(-a^2
*(x+1/a)^2+2*a*(x+1/a))^(1/2)*x-11/2/a^2/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2))-
1/a^6/(x+1/a)^3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)

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Maxima [C]  time = 1.49219, size = 239, normalized size = 2.52 \begin{align*} \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{a^{5} x^{2} + 2 \, a^{4} x + a^{3}} - \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{a^{4} x + a^{3}} - \frac{6 \, \sqrt{-a^{2} x^{2} + 1}}{a^{4} x + a^{3}} + \frac{\sqrt{a^{2} x^{2} + 4 \, a x + 3} x}{2 \, a^{2}} + \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{3 \, a^{3}} - \frac{i \, \arcsin \left (a x + 2\right )}{2 \, a^{3}} - \frac{6 \, \arcsin \left (a x\right )}{a^{3}} + \frac{\sqrt{a^{2} x^{2} + 4 \, a x + 3}}{a^{3}} - \frac{3 \, \sqrt{-a^{2} x^{2} + 1}}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-a^2*x^2 + 1)^(3/2)/(a^5*x^2 + 2*a^4*x + a^3) - (-a^2*x^2 + 1)^(3/2)/(a^4*x + a^3) - 6*sqrt(-a^2*x^2 + 1)/(a^
4*x + a^3) + 1/2*sqrt(a^2*x^2 + 4*a*x + 3)*x/a^2 + 1/3*(-a^2*x^2 + 1)^(3/2)/a^3 - 1/2*I*arcsin(a*x + 2)/a^3 -
6*arcsin(a*x)/a^3 + sqrt(a^2*x^2 + 4*a*x + 3)/a^3 - 3*sqrt(-a^2*x^2 + 1)/a^3

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Fricas [A]  time = 1.87613, size = 198, normalized size = 2.08 \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(x^2/(a*x+1)^3*(-a^2*x^2+1)^(3/2),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)*sqr
t(-a^2*x^2 + 1) + 52)/(a^4*x + a^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.24343, size = 117, normalized size = 1.23 \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(x^2/(a*x+1)^3*(-a^2*x^2+1)^(3/2),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))