3.55 \(\int \frac{x^2 (1+x)^2}{\sqrt{1-x^2}} \, dx\)

Optimal. Leaf size=63 \[ -\frac{1}{4} \sqrt{1-x^2} x^3-\frac{2}{3} \sqrt{1-x^2} x^2-\frac{1}{24} (21 x+32) \sqrt{1-x^2}+\frac{7}{8} \sin ^{-1}(x) \]

[Out]

(-2*x^2*Sqrt[1 - x^2])/3 - (x^3*Sqrt[1 - x^2])/4 - ((32 + 21*x)*Sqrt[1 - x^2])/24 + (7*ArcSin[x])/8

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Rubi [A]  time = 0.080565, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1809, 833, 780, 216} \[ -\frac{1}{4} \sqrt{1-x^2} x^3-\frac{2}{3} \sqrt{1-x^2} x^2-\frac{1}{24} (21 x+32) \sqrt{1-x^2}+\frac{7}{8} \sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(1 + x)^2)/Sqrt[1 - x^2],x]

[Out]

(-2*x^2*Sqrt[1 - x^2])/3 - (x^3*Sqrt[1 - x^2])/4 - ((32 + 21*x)*Sqrt[1 - x^2])/24 + (7*ArcSin[x])/8

Rule 1809

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

Rule 833

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(g*(d + e*x)
^m*(a + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p*
Simp[c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, p
}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p]) &&  !(IGtQ[m, 0] && EqQ[f, 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 \frac{x^2 (1+x)^2}{\sqrt{1-x^2}} \, dx &=-\frac{1}{4} x^3 \sqrt{1-x^2}-\frac{1}{4} \int \frac{(-7-8 x) x^2}{\sqrt{1-x^2}} \, dx\\ &=-\frac{2}{3} x^2 \sqrt{1-x^2}-\frac{1}{4} x^3 \sqrt{1-x^2}+\frac{1}{12} \int \frac{x (16+21 x)}{\sqrt{1-x^2}} \, dx\\ &=-\frac{2}{3} x^2 \sqrt{1-x^2}-\frac{1}{4} x^3 \sqrt{1-x^2}-\frac{1}{24} (32+21 x) \sqrt{1-x^2}+\frac{7}{8} \int \frac{1}{\sqrt{1-x^2}} \, dx\\ &=-\frac{2}{3} x^2 \sqrt{1-x^2}-\frac{1}{4} x^3 \sqrt{1-x^2}-\frac{1}{24} (32+21 x) \sqrt{1-x^2}+\frac{7}{8} \sin ^{-1}(x)\\ \end{align*}

Mathematica [A]  time = 0.0302614, size = 37, normalized size = 0.59 \[ \frac{7}{8} \sin ^{-1}(x)-\frac{1}{24} \sqrt{1-x^2} \left (6 x^3+16 x^2+21 x+32\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[1 - x^2]*(32 + 21*x + 16*x^2 + 6*x^3))/24 + (7*ArcSin[x])/8

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Maple [A]  time = 0.048, size = 57, normalized size = 0.9 \begin{align*} -{\frac{{x}^{3}}{4}\sqrt{-{x}^{2}+1}}-{\frac{7\,x}{8}\sqrt{-{x}^{2}+1}}+{\frac{7\,\arcsin \left ( x \right ) }{8}}-{\frac{2\,{x}^{2}}{3}\sqrt{-{x}^{2}+1}}-{\frac{4}{3}\sqrt{-{x}^{2}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*x^3*(-x^2+1)^(1/2)-7/8*x*(-x^2+1)^(1/2)+7/8*arcsin(x)-2/3*x^2*(-x^2+1)^(1/2)-4/3*(-x^2+1)^(1/2)

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Maxima [A]  time = 1.48661, size = 76, normalized size = 1.21 \begin{align*} -\frac{1}{4} \, \sqrt{-x^{2} + 1} x^{3} - \frac{2}{3} \, \sqrt{-x^{2} + 1} x^{2} - \frac{7}{8} \, \sqrt{-x^{2} + 1} x - \frac{4}{3} \, \sqrt{-x^{2} + 1} + \frac{7}{8} \, \arcsin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*sqrt(-x^2 + 1)*x^3 - 2/3*sqrt(-x^2 + 1)*x^2 - 7/8*sqrt(-x^2 + 1)*x - 4/3*sqrt(-x^2 + 1) + 7/8*arcsin(x)

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Fricas [A]  time = 1.85394, size = 119, normalized size = 1.89 \begin{align*} -\frac{1}{24} \,{\left (6 \, x^{3} + 16 \, x^{2} + 21 \, x + 32\right )} \sqrt{-x^{2} + 1} - \frac{7}{4} \, \arctan \left (\frac{\sqrt{-x^{2} + 1} - 1}{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(6*x^3 + 16*x^2 + 21*x + 32)*sqrt(-x^2 + 1) - 7/4*arctan((sqrt(-x^2 + 1) - 1)/x)

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Sympy [A]  time = 0.762138, size = 60, normalized size = 0.95 \begin{align*} - \frac{x^{3} \sqrt{1 - x^{2}}}{4} - \frac{2 x^{2} \sqrt{1 - x^{2}}}{3} - \frac{7 x \sqrt{1 - x^{2}}}{8} - \frac{4 \sqrt{1 - x^{2}}}{3} + \frac{7 \operatorname{asin}{\left (x \right )}}{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x**3*sqrt(1 - x**2)/4 - 2*x**2*sqrt(1 - x**2)/3 - 7*x*sqrt(1 - x**2)/8 - 4*sqrt(1 - x**2)/3 + 7*asin(x)/8

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Giac [A]  time = 1.11217, size = 41, normalized size = 0.65 \begin{align*} -\frac{1}{24} \,{\left ({\left (2 \,{\left (3 \, x + 8\right )} x + 21\right )} x + 32\right )} \sqrt{-x^{2} + 1} + \frac{7}{8} \, \arcsin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*((2*(3*x + 8)*x + 21)*x + 32)*sqrt(-x^2 + 1) + 7/8*arcsin(x)