3.211 \(\int \frac{x^2 \sqrt{1-a^2 x^2}}{(1-a x)^5} \, dx\)

Optimal. Leaf size=88 \[ \frac{23 \left (1-a^2 x^2\right )^{3/2}}{105 a^3 (1-a x)^3}-\frac{12 \left (1-a^2 x^2\right )^{3/2}}{35 a^3 (1-a x)^4}+\frac{\left (1-a^2 x^2\right )^{3/2}}{7 a^3 (1-a x)^5} \]

[Out]

(1 - a^2*x^2)^(3/2)/(7*a^3*(1 - a*x)^5) - (12*(1 - a^2*x^2)^(3/2))/(35*a^3*(1 - a*x)^4) + (23*(1 - a^2*x^2)^(3
/2))/(105*a^3*(1 - a*x)^3)

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Rubi [A]  time = 0.123396, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {1639, 793, 659, 651} \[ \frac{23 \left (1-a^2 x^2\right )^{3/2}}{105 a^3 (1-a x)^3}-\frac{12 \left (1-a^2 x^2\right )^{3/2}}{35 a^3 (1-a x)^4}+\frac{\left (1-a^2 x^2\right )^{3/2}}{7 a^3 (1-a x)^5} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*Sqrt[1 - a^2*x^2])/(1 - a*x)^5,x]

[Out]

(1 - a^2*x^2)^(3/2)/(7*a^3*(1 - a*x)^5) - (12*(1 - a^2*x^2)^(3/2))/(35*a^3*(1 - a*x)^4) + (23*(1 - a^2*x^2)^(3
/2))/(105*a^3*(1 - a*x)^3)

Rule 1639

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 - 2*e*f*(m + p + q)*(d + e*x)^(q - 2)*(a*e - c*d*x), x], x], x] /; NeQ[m + q +
2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 793

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

Rubi steps

\begin{align*} \int \frac{x^2 \sqrt{1-a^2 x^2}}{(1-a x)^5} \, dx &=-\frac{\left (1-a^2 x^2\right )^{3/2}}{a^3 (1-a x)^4}+\frac{\int \frac{\left (4 a^2-3 a^3 x\right ) \sqrt{1-a^2 x^2}}{(1-a x)^5} \, dx}{a^4}\\ &=\frac{\left (1-a^2 x^2\right )^{3/2}}{7 a^3 (1-a x)^5}-\frac{\left (1-a^2 x^2\right )^{3/2}}{a^3 (1-a x)^4}+\frac{23 \int \frac{\sqrt{1-a^2 x^2}}{(1-a x)^4} \, dx}{7 a^2}\\ &=\frac{\left (1-a^2 x^2\right )^{3/2}}{7 a^3 (1-a x)^5}-\frac{12 \left (1-a^2 x^2\right )^{3/2}}{35 a^3 (1-a x)^4}+\frac{23 \int \frac{\sqrt{1-a^2 x^2}}{(1-a x)^3} \, dx}{35 a^2}\\ &=\frac{\left (1-a^2 x^2\right )^{3/2}}{7 a^3 (1-a x)^5}-\frac{12 \left (1-a^2 x^2\right )^{3/2}}{35 a^3 (1-a x)^4}+\frac{23 \left (1-a^2 x^2\right )^{3/2}}{105 a^3 (1-a x)^3}\\ \end{align*}

Mathematica [A]  time = 0.0769959, size = 50, normalized size = 0.57 \[ \frac{\sqrt{1-a^2 x^2} \left (23 a^3 x^3+13 a^2 x^2-8 a x+2\right )}{105 a^3 (a x-1)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*Sqrt[1 - a^2*x^2])/(1 - a*x)^5,x]

[Out]

(Sqrt[1 - a^2*x^2]*(2 - 8*a*x + 13*a^2*x^2 + 23*a^3*x^3))/(105*a^3*(-1 + a*x)^4)

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Maple [A]  time = 0.047, size = 44, normalized size = 0.5 \begin{align*}{\frac{ \left ( 23\,{a}^{2}{x}^{2}-10\,ax+2 \right ) \left ( ax+1 \right ) }{105\, \left ( ax-1 \right ) ^{4}{a}^{3}}\sqrt{-{a}^{2}{x}^{2}+1}} \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)^5,x)

[Out]

1/105*(-a^2*x^2+1)^(1/2)*(23*a^2*x^2-10*a*x+2)*(a*x+1)/(a*x-1)^4/a^3

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Maxima [B]  time = 1.12352, size = 207, normalized size = 2.35 \begin{align*} \frac{2 \, \sqrt{-a^{2} x^{2} + 1}}{7 \,{\left (a^{7} x^{4} - 4 \, a^{6} x^{3} + 6 \, a^{5} x^{2} - 4 \, a^{4} x + a^{3}\right )}} + \frac{29 \, \sqrt{-a^{2} x^{2} + 1}}{35 \,{\left (a^{6} x^{3} - 3 \, a^{5} x^{2} + 3 \, a^{4} x - a^{3}\right )}} + \frac{82 \, \sqrt{-a^{2} x^{2} + 1}}{105 \,{\left (a^{5} x^{2} - 2 \, a^{4} x + a^{3}\right )}} + \frac{23 \, \sqrt{-a^{2} x^{2} + 1}}{105 \,{\left (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)^5,x, algorithm="maxima")

[Out]

2/7*sqrt(-a^2*x^2 + 1)/(a^7*x^4 - 4*a^6*x^3 + 6*a^5*x^2 - 4*a^4*x + a^3) + 29/35*sqrt(-a^2*x^2 + 1)/(a^6*x^3 -
3*a^5*x^2 + 3*a^4*x - a^3) + 82/105*sqrt(-a^2*x^2 + 1)/(a^5*x^2 - 2*a^4*x + a^3) + 23/105*sqrt(-a^2*x^2 + 1)/
(a^4*x - a^3)

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Fricas [A]  time = 1.60369, size = 223, normalized size = 2.53 \begin{align*} \frac{2 \, a^{4} x^{4} - 8 \, a^{3} x^{3} + 12 \, a^{2} x^{2} - 8 \, a x +{\left (23 \, a^{3} x^{3} + 13 \, a^{2} x^{2} - 8 \, a x + 2\right )} \sqrt{-a^{2} x^{2} + 1} + 2}{105 \,{\left (a^{7} x^{4} - 4 \, a^{6} x^{3} + 6 \, a^{5} x^{2} - 4 \, 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)^5,x, algorithm="fricas")

[Out]

1/105*(2*a^4*x^4 - 8*a^3*x^3 + 12*a^2*x^2 - 8*a*x + (23*a^3*x^3 + 13*a^2*x^2 - 8*a*x + 2)*sqrt(-a^2*x^2 + 1) +
2)/(a^7*x^4 - 4*a^6*x^3 + 6*a^5*x^2 - 4*a^4*x + a^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{x^{2} \sqrt{- a^{2} x^{2} + 1}}{a^{5} x^{5} - 5 a^{4} x^{4} + 10 a^{3} x^{3} - 10 a^{2} x^{2} + 5 a x - 1}\, 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)**5,x)

[Out]

-Integral(x**2*sqrt(-a**2*x**2 + 1)/(a**5*x**5 - 5*a**4*x**4 + 10*a**3*x**3 - 10*a**2*x**2 + 5*a*x - 1), x)

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Giac [C]  time = 1.10011, size = 258, normalized size = 2.93 \begin{align*} \frac{1}{420} \,{\left (-\frac{92 i \, \mathrm{sgn}\left (\frac{1}{a x - 1}\right ) \mathrm{sgn}\left (a\right )}{a^{4}} - \frac{140 \,{\left (-\frac{2}{a x - 1} - 1\right )}^{\frac{3}{2}} \mathrm{sgn}\left (\frac{1}{a x - 1}\right ) \mathrm{sgn}\left (a\right ) -{\left (15 \,{\left (\frac{2}{a x - 1} + 1\right )}^{3} \sqrt{-\frac{2}{a x - 1} - 1} - 42 \,{\left (\frac{2}{a x - 1} + 1\right )}^{2} \sqrt{-\frac{2}{a x - 1} - 1} - 35 \,{\left (-\frac{2}{a x - 1} - 1\right )}^{\frac{3}{2}}\right )} \mathrm{sgn}\left (\frac{1}{a x - 1}\right ) \mathrm{sgn}\left (a\right ) - 28 \,{\left (3 \,{\left (\frac{2}{a x - 1} + 1\right )}^{2} \sqrt{-\frac{2}{a x - 1} - 1} + 5 \,{\left (-\frac{2}{a x - 1} - 1\right )}^{\frac{3}{2}}\right )} \mathrm{sgn}\left (\frac{1}{a x - 1}\right ) \mathrm{sgn}\left (a\right )}{a^{4}}\right )}{\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)^5,x, algorithm="giac")

[Out]

1/420*(-92*I*sgn(1/(a*x - 1))*sgn(a)/a^4 - (140*(-2/(a*x - 1) - 1)^(3/2)*sgn(1/(a*x - 1))*sgn(a) - (15*(2/(a*x
- 1) + 1)^3*sqrt(-2/(a*x - 1) - 1) - 42*(2/(a*x - 1) + 1)^2*sqrt(-2/(a*x - 1) - 1) - 35*(-2/(a*x - 1) - 1)^(3
/2))*sgn(1/(a*x - 1))*sgn(a) - 28*(3*(2/(a*x - 1) + 1)^2*sqrt(-2/(a*x - 1) - 1) + 5*(-2/(a*x - 1) - 1)^(3/2))*
sgn(1/(a*x - 1))*sgn(a))/a^4)*abs(a)