\(\int \frac {x^3 (1+x)^2}{\sqrt {1-x^2}} \, dx\) [54]

Optimal result

Integrand size = 20, antiderivative size = 81 \[ \int \frac {x^3 (1+x)^2}{\sqrt {1-x^2}} \, dx=-\frac {3}{5} x^2 \sqrt {1-x^2}-\frac {1}{2} x^3 \sqrt {1-x^2}-\frac {1}{5} x^4 \sqrt {1-x^2}-\frac {3}{20} (8+5 x) \sqrt {1-x^2}+\frac {3 \arcsin (x)}{4} \]

[Out]

3/4*arcsin(x)-3/5*x^2*(-x^2+1)^(1/2)-1/2*x^3*(-x^2+1)^(1/2)-1/5*x^4*(-x^2+1)^(1/2)-3/20*(8+5*x)*(-x^2+1)^(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1823, 847, 794, 222} \[ \int \frac {x^3 (1+x)^2}{\sqrt {1-x^2}} \, dx=\frac {3 \arcsin (x)}{4}-\frac {3}{5} \sqrt {1-x^2} x^2-\frac {3}{20} (5 x+8) \sqrt {1-x^2}-\frac {1}{5} \sqrt {1-x^2} x^4-\frac {1}{2} \sqrt {1-x^2} x^3 \]

[In]

Int[(x^3*(1 + x)^2)/Sqrt[1 - x^2],x]

[Out]

(-3*x^2*Sqrt[1 - x^2])/5 - (x^3*Sqrt[1 - x^2])/2 - (x^4*Sqrt[1 - x^2])/5 - (3*(8 + 5*x)*Sqrt[1 - x^2])/20 + (3
*ArcSin[x])/4

Rule 222

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 794

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 847

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 1823

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])

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{5} x^4 \sqrt {1-x^2}-\frac {1}{5} \int \frac {(-9-10 x) x^3}{\sqrt {1-x^2}} \, dx \\ & = -\frac {1}{2} x^3 \sqrt {1-x^2}-\frac {1}{5} x^4 \sqrt {1-x^2}+\frac {1}{20} \int \frac {x^2 (30+36 x)}{\sqrt {1-x^2}} \, dx \\ & = -\frac {3}{5} x^2 \sqrt {1-x^2}-\frac {1}{2} x^3 \sqrt {1-x^2}-\frac {1}{5} x^4 \sqrt {1-x^2}-\frac {1}{60} \int \frac {(-72-90 x) x}{\sqrt {1-x^2}} \, dx \\ & = -\frac {3}{5} x^2 \sqrt {1-x^2}-\frac {1}{2} x^3 \sqrt {1-x^2}-\frac {1}{5} x^4 \sqrt {1-x^2}-\frac {3}{20} (8+5 x) \sqrt {1-x^2}+\frac {3}{4} \int \frac {1}{\sqrt {1-x^2}} \, dx \\ & = -\frac {3}{5} x^2 \sqrt {1-x^2}-\frac {1}{2} x^3 \sqrt {1-x^2}-\frac {1}{5} x^4 \sqrt {1-x^2}-\frac {3}{20} (8+5 x) \sqrt {1-x^2}+\frac {3}{4} \sin ^{-1}(x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.72 \[ \int \frac {x^3 (1+x)^2}{\sqrt {1-x^2}} \, dx=\frac {1}{20} \sqrt {1-x^2} \left (-24-15 x-12 x^2-10 x^3-4 x^4\right )+\frac {3}{2} \arctan \left (\frac {x}{-1+\sqrt {1-x^2}}\right ) \]

[In]

Integrate[(x^3*(1 + x)^2)/Sqrt[1 - x^2],x]

[Out]

(Sqrt[1 - x^2]*(-24 - 15*x - 12*x^2 - 10*x^3 - 4*x^4))/20 + (3*ArcTan[x/(-1 + Sqrt[1 - x^2])])/2

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.52

method	result	size
risch	\(\frac {\left (4 x^{4}+10 x^{3}+12 x^{2}+15 x +24\right ) \left (x^{2}-1\right )}{20 \sqrt {-x^{2}+1}}+\frac {3 \arcsin \left (x \right )}{4}\)	\(42\)
trager	\(\left (-\frac {1}{5} x^{4}-\frac {1}{2} x^{3}-\frac {3}{5} x^{2}-\frac {3}{4} x -\frac {6}{5}\right ) \sqrt {-x^{2}+1}+\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {-x^{2}+1}+x \right )}{4}\)	\(59\)
default	\(-\frac {x^{4} \sqrt {-x^{2}+1}}{5}-\frac {3 x^{2} \sqrt {-x^{2}+1}}{5}-\frac {6 \sqrt {-x^{2}+1}}{5}-\frac {x^{3} \sqrt {-x^{2}+1}}{2}-\frac {3 x \sqrt {-x^{2}+1}}{4}+\frac {3 \arcsin \left (x \right )}{4}\)	\(71\)
meijerg	\(\frac {\frac {4 \sqrt {\pi }}{3}-\frac {\sqrt {\pi }\, \left (4 x^{2}+8\right ) \sqrt {-x^{2}+1}}{6}}{2 \sqrt {\pi }}-\frac {i \left (-\frac {i \sqrt {\pi }\, x \left (10 x^{2}+15\right ) \sqrt {-x^{2}+1}}{20}+\frac {3 i \sqrt {\pi }\, \arcsin \left (x \right )}{4}\right )}{\sqrt {\pi }}-\frac {-\frac {16 \sqrt {\pi }}{15}+\frac {\sqrt {\pi }\, \left (6 x^{4}+8 x^{2}+16\right ) \sqrt {-x^{2}+1}}{15}}{2 \sqrt {\pi }}\)	\(109\)

[In]

int(x^3*(1+x)^2/(-x^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/20*(4*x^4+10*x^3+12*x^2+15*x+24)*(x^2-1)/(-x^2+1)^(1/2)+3/4*arcsin(x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.62 \[ \int \frac {x^3 (1+x)^2}{\sqrt {1-x^2}} \, dx=-\frac {1}{20} \, {\left (4 \, x^{4} + 10 \, x^{3} + 12 \, x^{2} + 15 \, x + 24\right )} \sqrt {-x^{2} + 1} - \frac {3}{2} \, \arctan \left (\frac {\sqrt {-x^{2} + 1} - 1}{x}\right ) \]

[In]

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

[Out]

-1/20*(4*x^4 + 10*x^3 + 12*x^2 + 15*x + 24)*sqrt(-x^2 + 1) - 3/2*arctan((sqrt(-x^2 + 1) - 1)/x)

Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.90 \[ \int \frac {x^3 (1+x)^2}{\sqrt {1-x^2}} \, dx=- \frac {x^{4} \sqrt {1 - x^{2}}}{5} - \frac {x^{3} \sqrt {1 - x^{2}}}{2} - \frac {3 x^{2} \sqrt {1 - x^{2}}}{5} - \frac {3 x \sqrt {1 - x^{2}}}{4} - \frac {6 \sqrt {1 - x^{2}}}{5} + \frac {3 \operatorname {asin}{\left (x \right )}}{4} \]

[In]

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

[Out]

-x**4*sqrt(1 - x**2)/5 - x**3*sqrt(1 - x**2)/2 - 3*x**2*sqrt(1 - x**2)/5 - 3*x*sqrt(1 - x**2)/4 - 6*sqrt(1 - x
**2)/5 + 3*asin(x)/4

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.86 \[ \int \frac {x^3 (1+x)^2}{\sqrt {1-x^2}} \, dx=-\frac {1}{5} \, \sqrt {-x^{2} + 1} x^{4} - \frac {1}{2} \, \sqrt {-x^{2} + 1} x^{3} - \frac {3}{5} \, \sqrt {-x^{2} + 1} x^{2} - \frac {3}{4} \, \sqrt {-x^{2} + 1} x - \frac {6}{5} \, \sqrt {-x^{2} + 1} + \frac {3}{4} \, \arcsin \left (x\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.42 \[ \int \frac {x^3 (1+x)^2}{\sqrt {1-x^2}} \, dx=-\frac {1}{20} \, {\left ({\left (2 \, {\left ({\left (2 \, x + 5\right )} x + 6\right )} x + 15\right )} x + 24\right )} \sqrt {-x^{2} + 1} + \frac {3}{4} \, \arcsin \left (x\right ) \]

[In]

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

[Out]

-1/20*((2*((2*x + 5)*x + 6)*x + 15)*x + 24)*sqrt(-x^2 + 1) + 3/4*arcsin(x)

Mupad [B] (verification not implemented)

Time = 11.17 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.44 \[ \int \frac {x^3 (1+x)^2}{\sqrt {1-x^2}} \, dx=\frac {3\,\mathrm {asin}\left (x\right )}{4}-\sqrt {1-x^2}\,\left (\frac {x^4}{5}+\frac {x^3}{2}+\frac {3\,x^2}{5}+\frac {3\,x}{4}+\frac {6}{5}\right ) \]

[In]

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

[Out]

(3*asin(x))/4 - (1 - x^2)^(1/2)*((3*x)/4 + (3*x^2)/5 + x^3/2 + x^4/5 + 6/5)