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

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

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Rubi [A]  time = 0.13, antiderivative size = 95, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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} \end {gather*}

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 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]

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 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 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 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]

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*}

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Mathematica [A]  time = 0.11, size = 50, normalized size = 0.53 \begin {gather*} \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} \end {gather*}

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)

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IntegrateAlgebraic [A]  time = 0.69, size = 85, normalized size = 0.89 \begin {gather*} \frac {\left (-13 a^2 x^2+19 a x-8\right ) \sqrt {1-a^2 x^2}}{5 a^3 (a x-1)^3}-\frac {\sqrt {-a^2} \log \left (\sqrt {1-a^2 x^2}-\sqrt {-a^2} x\right )}{a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

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fricas [A]  time = 0.40, size = 126, normalized size = 1.33 \begin {gather*} \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 {gather*}

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)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [B]  time = 0.02, size = 200, normalized size = 2.11 \begin {gather*} -\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 \left (x -\frac {1}{a}\right ) a}}\right )}{\sqrt {a^{2}}\, a^{2}}+\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 \left (x -\frac {1}{a}\right ) a}}{a^{3}}+\frac {\left (-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 \left (x -\frac {1}{a}\right ) a \right )^{\frac {3}{2}}}{\left (x -\frac {1}{a}\right )^{2} a^{5}}+\frac {3 \left (-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 \left (x -\frac {1}{a}\right ) a \right )^{\frac {3}{2}}}{5 \left (x -\frac {1}{a}\right )^{3} a^{6}}+\frac {\left (-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 \left (x -\frac {1}{a}\right ) a \right )^{\frac {3}{2}}}{5 \left (x -\frac {1}{a}\right )^{4} a^{7}} \end {gather*}

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {-a^{2} x^{2} + 1} x^{2}}{{\left (a x - 1\right )}^{4}}\,{d x} \end {gather*}

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)

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mupad [B]  time = 2.70, size = 220, normalized size = 2.32 \begin {gather*} \frac {4\,a^2\,\sqrt {1-a^2\,x^2}}{15\,\left (a^7\,x^2-2\,a^6\,x+a^5\right )}-\frac {\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{a^2\,\sqrt {-a^2}}-\frac {2\,\sqrt {1-a^2\,x^2}}{5\,\sqrt {-a^2}\,\left (a\,\sqrt {-a^2}-3\,a^2\,x\,\sqrt {-a^2}+3\,a^3\,x^2\,\sqrt {-a^2}-a^4\,x^3\,\sqrt {-a^2}\right )}-\frac {13\,\sqrt {1-a^2\,x^2}}{5\,\left (a\,\sqrt {-a^2}-a^2\,x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}-\frac {5\,\sqrt {1-a^2\,x^2}}{3\,\left (a^5\,x^2-2\,a^4\,x+a^3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*a^2*(1 - a^2*x^2)^(1/2))/(15*(a^5 - 2*a^6*x + a^7*x^2)) - asinh(x*(-a^2)^(1/2))/(a^2*(-a^2)^(1/2)) - (2*(1
- a^2*x^2)^(1/2))/(5*(-a^2)^(1/2)*(a*(-a^2)^(1/2) - 3*a^2*x*(-a^2)^(1/2) + 3*a^3*x^2*(-a^2)^(1/2) - a^4*x^3*(-
a^2)^(1/2))) - (13*(1 - a^2*x^2)^(1/2))/(5*(a*(-a^2)^(1/2) - a^2*x*(-a^2)^(1/2))*(-a^2)^(1/2)) - (5*(1 - a^2*x
^2)^(1/2))/(3*(a^3 - 2*a^4*x + a^5*x^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}{\left (a x - 1\right )^{4}}\, dx \end {gather*}

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)

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