3.210 \(\int \frac{x^2 \sqrt{1-a^2 x^2}}{(1-a x)^4} \, dx\)

Optimal. Leaf size=95 \[ -\frac{3 \left (1-a^2 x^2\right )^{3/2}}{5 a^3 (1-a x)^3}+\frac{\left (1-a^2 x^2\right )^{3/2}}{5 a^3 (1-a x)^4}+\frac{2 \sqrt{1-a^2 x^2}}{a^3 (1-a x)}-\frac{\sin ^{-1}(a x)}{a^3} \]

[Out]

(2*Sqrt[1 - a^2*x^2])/(a^3*(1 - a*x)) + (1 - a^2*x^2)^(3/2)/(5*a^3*(1 - a*x)^4) - (3*(1 - a^2*x^2)^(3/2))/(5*a
^3*(1 - a*x)^3) - ArcSin[a*x]/a^3

________________________________________________________________________________________

Rubi [A]  time = 0.132497, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {1637, 659, 651, 663, 216} \[ -\frac{3 \left (1-a^2 x^2\right )^{3/2}}{5 a^3 (1-a x)^3}+\frac{\left (1-a^2 x^2\right )^{3/2}}{5 a^3 (1-a x)^4}+\frac{2 \sqrt{1-a^2 x^2}}{a^3 (1-a x)}-\frac{\sin ^{-1}(a x)}{a^3} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*Sqrt[1 - a^2*x^2])/(1 - a*x)^4,x]

[Out]

(2*Sqrt[1 - a^2*x^2])/(a^3*(1 - a*x)) + (1 - a^2*x^2)^(3/2)/(5*a^3*(1 - a*x)^4) - (3*(1 - a^2*x^2)^(3/2))/(5*a
^3*(1 - a*x)^3) - ArcSin[a*x]/a^3

Rule 1637

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

Rule 659

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

Rule 651

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

Rule 663

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

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 \frac{x^2 \sqrt{1-a^2 x^2}}{(1-a x)^4} \, dx &=\int \left (\frac{\sqrt{1-a^2 x^2}}{a^2 (-1+a x)^4}+\frac{2 \sqrt{1-a^2 x^2}}{a^2 (-1+a x)^3}+\frac{\sqrt{1-a^2 x^2}}{a^2 (-1+a x)^2}\right ) \, dx\\ &=\frac{\int \frac{\sqrt{1-a^2 x^2}}{(-1+a x)^4} \, dx}{a^2}+\frac{\int \frac{\sqrt{1-a^2 x^2}}{(-1+a x)^2} \, dx}{a^2}+\frac{2 \int \frac{\sqrt{1-a^2 x^2}}{(-1+a x)^3} \, dx}{a^2}\\ &=\frac{2 \sqrt{1-a^2 x^2}}{a^3 (1-a x)}+\frac{\left (1-a^2 x^2\right )^{3/2}}{5 a^3 (1-a x)^4}-\frac{2 \left (1-a^2 x^2\right )^{3/2}}{3 a^3 (1-a x)^3}-\frac{\int \frac{\sqrt{1-a^2 x^2}}{(-1+a x)^3} \, dx}{5 a^2}-\frac{\int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{a^2}\\ &=\frac{2 \sqrt{1-a^2 x^2}}{a^3 (1-a x)}+\frac{\left (1-a^2 x^2\right )^{3/2}}{5 a^3 (1-a x)^4}-\frac{3 \left (1-a^2 x^2\right )^{3/2}}{5 a^3 (1-a x)^3}-\frac{\sin ^{-1}(a x)}{a^3}\\ \end{align*}

Mathematica [A]  time = 0.126385, size = 50, normalized size = 0.53 \[ \frac{\frac{\left (-13 a^2 x^2+19 a x-8\right ) \sqrt{1-a^2 x^2}}{(a x-1)^3}-5 \sin ^{-1}(a x)}{5 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*Sqrt[1 - a^2*x^2])/(1 - a*x)^4,x]

[Out]

(((-8 + 19*a*x - 13*a^2*x^2)*Sqrt[1 - a^2*x^2])/(-1 + a*x)^3 - 5*ArcSin[a*x])/(5*a^3)

________________________________________________________________________________________

Maple [B]  time = 0.064, size = 200, normalized size = 2.1 \begin{align*}{\frac{1}{{a}^{5}} \left ( - \left ( x-{a}^{-1} \right ) ^{2}{a}^{2}-2\,a \left ( x-{a}^{-1} \right ) \right ) ^{{\frac{3}{2}}} \left ( x-{a}^{-1} \right ) ^{-2}}+{\frac{1}{{a}^{3}}\sqrt{- \left ( x-{a}^{-1} \right ) ^{2}{a}^{2}-2\,a \left ( x-{a}^{-1} \right ) }}-{\frac{1}{{a}^{2}}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{- \left ( x-{a}^{-1} \right ) ^{2}{a}^{2}-2\,a \left ( x-{a}^{-1} \right ) }}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}}+{\frac{3}{5\,{a}^{6}} \left ( - \left ( x-{a}^{-1} \right ) ^{2}{a}^{2}-2\,a \left ( x-{a}^{-1} \right ) \right ) ^{{\frac{3}{2}}} \left ( x-{a}^{-1} \right ) ^{-3}}+{\frac{1}{5\,{a}^{7}} \left ( - \left ( x-{a}^{-1} \right ) ^{2}{a}^{2}-2\,a \left ( x-{a}^{-1} \right ) \right ) ^{{\frac{3}{2}}} \left ( x-{a}^{-1} \right ) ^{-4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^5/(x-1/a)^2*(-(x-1/a)^2*a^2-2*a*(x-1/a))^(3/2)+1/a^3*(-(x-1/a)^2*a^2-2*a*(x-1/a))^(1/2)-1/a^2/(a^2)^(1/2)*
arctan((a^2)^(1/2)*x/(-(x-1/a)^2*a^2-2*a*(x-1/a))^(1/2))+3/5/a^6/(x-1/a)^3*(-(x-1/a)^2*a^2-2*a*(x-1/a))^(3/2)+
1/5/a^7/(x-1/a)^4*(-(x-1/a)^2*a^2-2*a*(x-1/a))^(3/2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-a^2*x^2+1)^(1/2)/(-a*x+1)^4,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.60036, size = 278, normalized size = 2.93 \begin{align*} \frac{8 \, a^{3} x^{3} - 24 \, a^{2} x^{2} + 24 \, a x + 10 \,{\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) -{\left (13 \, a^{2} x^{2} - 19 \, a x + 8\right )} \sqrt{-a^{2} x^{2} + 1} - 8}{5 \,{\left (a^{6} x^{3} - 3 \, a^{5} x^{2} + 3 \, a^{4} x - a^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-a^2*x^2+1)^(1/2)/(-a*x+1)^4,x, algorithm="fricas")

[Out]

1/5*(8*a^3*x^3 - 24*a^2*x^2 + 24*a*x + 10*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a
*x)) - (13*a^2*x^2 - 19*a*x + 8)*sqrt(-a^2*x^2 + 1) - 8)/(a^6*x^3 - 3*a^5*x^2 + 3*a^4*x - a^3)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(-a**2*x**2+1)**(1/2)/(-a*x+1)**4,x)

[Out]

Integral(x**2*sqrt(-(a*x - 1)*(a*x + 1))/(a*x - 1)**4, x)

________________________________________________________________________________________

Giac [A]  time = 1.11031, size = 217, normalized size = 2.28 \begin{align*} -\frac{\arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{a^{2}{\left | a \right |}} - \frac{2 \,{\left (\frac{35 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}}{a^{2} x} - \frac{55 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2}}{a^{4} x^{2}} + \frac{25 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3}}{a^{6} x^{3}} - \frac{5 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{4}}{a^{8} x^{4}} - 8\right )}}{5 \, a^{2}{\left (\frac{\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a}{a^{2} x} - 1\right )}^{5}{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-a^2*x^2+1)^(1/2)/(-a*x+1)^4,x, algorithm="giac")

[Out]

-arcsin(a*x)*sgn(a)/(a^2*abs(a)) - 2/5*(35*(sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 55*(sqrt(-a^2*x^2 + 1)*ab
s(a) + a)^2/(a^4*x^2) + 25*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^3/(a^6*x^3) - 5*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^4/(
a^8*x^4) - 8)/(a^2*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)^5*abs(a))