\(\int \frac {x (1+x)^2}{\sqrt {1-x^2}} \, dx\) [56]

Optimal result

Integrand size = 18, antiderivative size = 41 \[ \int \frac {x (1+x)^2}{\sqrt {1-x^2}} \, dx=-\frac {1}{3} x^2 \sqrt {1-x^2}-\frac {1}{3} (5+3 x) \sqrt {1-x^2}+\arcsin (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1823, 794, 222} \[ \int \frac {x (1+x)^2}{\sqrt {1-x^2}} \, dx=\arcsin (x)-\frac {1}{3} \sqrt {1-x^2} x^2-\frac {1}{3} (3 x+5) \sqrt {1-x^2} \]

[In]

Int[(x*(1 + x)^2)/Sqrt[1 - x^2],x]

[Out]

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

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 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}{3} x^2 \sqrt {1-x^2}-\frac {1}{3} \int \frac {(-5-6 x) x}{\sqrt {1-x^2}} \, dx \\ & = -\frac {1}{3} x^2 \sqrt {1-x^2}-\frac {1}{3} (5+3 x) \sqrt {1-x^2}+\int \frac {1}{\sqrt {1-x^2}} \, dx \\ & = -\frac {1}{3} x^2 \sqrt {1-x^2}-\frac {1}{3} (5+3 x) \sqrt {1-x^2}+\sin ^{-1}(x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.12 \[ \int \frac {x (1+x)^2}{\sqrt {1-x^2}} \, dx=\frac {1}{3} \sqrt {1-x^2} \left (-5-3 x-x^2\right )+2 \arctan \left (\frac {x}{-1+\sqrt {1-x^2}}\right ) \]

[In]

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

[Out]

(Sqrt[1 - x^2]*(-5 - 3*x - x^2))/3 + 2*ArcTan[x/(-1 + Sqrt[1 - x^2])]

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.68

method	result	size
risch	\(\frac {\left (x^{2}+3 x +5\right ) \left (x^{2}-1\right )}{3 \sqrt {-x^{2}+1}}+\arcsin \left (x \right )\)	\(28\)
default	\(-\frac {5 \sqrt {-x^{2}+1}}{3}-\frac {x^{2} \sqrt {-x^{2}+1}}{3}-x \sqrt {-x^{2}+1}+\arcsin \left (x \right )\)	\(41\)
trager	\(\left (-\frac {1}{3} x^{2}-x -\frac {5}{3}\right ) \sqrt {-x^{2}+1}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {-x^{2}+1}+x \right )\)	\(48\)
meijerg	\(-\frac {-2 \sqrt {\pi }+2 \sqrt {\pi }\, \sqrt {-x^{2}+1}}{2 \sqrt {\pi }}+\frac {i \left (i \sqrt {\pi }\, x \sqrt {-x^{2}+1}-i \sqrt {\pi }\, \arcsin \left (x \right )\right )}{\sqrt {\pi }}+\frac {\frac {4 \sqrt {\pi }}{3}-\frac {\sqrt {\pi }\, \left (4 x^{2}+8\right ) \sqrt {-x^{2}+1}}{6}}{2 \sqrt {\pi }}\)	\(90\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.93 \[ \int \frac {x (1+x)^2}{\sqrt {1-x^2}} \, dx=-\frac {1}{3} \, {\left (x^{2} + 3 \, x + 5\right )} \sqrt {-x^{2} + 1} - 2 \, \arctan \left (\frac {\sqrt {-x^{2} + 1} - 1}{x}\right ) \]

[In]

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

[Out]

-1/3*(x^2 + 3*x + 5)*sqrt(-x^2 + 1) - 2*arctan((sqrt(-x^2 + 1) - 1)/x)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.90 \[ \int \frac {x (1+x)^2}{\sqrt {1-x^2}} \, dx=- \frac {x^{2} \sqrt {1 - x^{2}}}{3} - x \sqrt {1 - x^{2}} - \frac {5 \sqrt {1 - x^{2}}}{3} + \operatorname {asin}{\left (x \right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.98 \[ \int \frac {x (1+x)^2}{\sqrt {1-x^2}} \, dx=-\frac {1}{3} \, \sqrt {-x^{2} + 1} x^{2} - \sqrt {-x^{2} + 1} x - \frac {5}{3} \, \sqrt {-x^{2} + 1} + \arcsin \left (x\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.51 \[ \int \frac {x (1+x)^2}{\sqrt {1-x^2}} \, dx=-\frac {1}{3} \, {\left ({\left (x + 3\right )} x + 5\right )} \sqrt {-x^{2} + 1} + \arcsin \left (x\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.54 \[ \int \frac {x (1+x)^2}{\sqrt {1-x^2}} \, dx=\mathrm {asin}\left (x\right )-\sqrt {1-x^2}\,\left (\frac {x^2}{3}+x+\frac {5}{3}\right ) \]

[In]

int((x*(x + 1)^2)/(1 - x^2)^(1/2),x)

[Out]

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