\(\int (1+x) \sqrt {1-x^2} \, dx\) [820]

Optimal result

Integrand size = 15, antiderivative size = 38 \[ \int (1+x) \sqrt {1-x^2} \, dx=\frac {1}{2} x \sqrt {1-x^2}-\frac {1}{3} \left (1-x^2\right )^{3/2}+\frac {\arcsin (x)}{2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {655, 201, 222} \[ \int (1+x) \sqrt {1-x^2} \, dx=\frac {\arcsin (x)}{2}-\frac {1}{3} \left (1-x^2\right )^{3/2}+\frac {1}{2} x \sqrt {1-x^2} \]

[In]

Int[(1 + x)*Sqrt[1 - x^2],x]

[Out]

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

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 655

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[e*((a + c*x^2)^(p + 1)/(2*c*(p + 1))),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{3} \left (1-x^2\right )^{3/2}+\int \sqrt {1-x^2} \, dx \\ & = \frac {1}{2} x \sqrt {1-x^2}-\frac {1}{3} \left (1-x^2\right )^{3/2}+\frac {1}{2} \int \frac {1}{\sqrt {1-x^2}} \, dx \\ & = \frac {1}{2} x \sqrt {1-x^2}-\frac {1}{3} \left (1-x^2\right )^{3/2}+\frac {1}{2} \sin ^{-1}(x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.21 \[ \int (1+x) \sqrt {1-x^2} \, dx=\frac {1}{6} \sqrt {1-x^2} \left (-2+3 x+2 x^2\right )-\arctan \left (\frac {\sqrt {1-x^2}}{1+x}\right ) \]

[In]

Integrate[(1 + x)*Sqrt[1 - x^2],x]

[Out]

(Sqrt[1 - x^2]*(-2 + 3*x + 2*x^2))/6 - ArcTan[Sqrt[1 - x^2]/(1 + x)]

Maple [A] (verified)

Time = 2.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.76

method	result	size
default	\(-\frac {\left (-x^{2}+1\right )^{\frac {3}{2}}}{3}+\frac {\arcsin \left (x \right )}{2}+\frac {x \sqrt {-x^{2}+1}}{2}\)	\(29\)
risch	\(-\frac {\left (2 x^{2}+3 x -2\right ) \left (x^{2}-1\right )}{6 \sqrt {-x^{2}+1}}+\frac {\arcsin \left (x \right )}{2}\)	\(32\)
trager	\(\left (\frac {1}{3} x^{2}+\frac {1}{2} x -\frac {1}{3}\right ) \sqrt {-x^{2}+1}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {-x^{2}+1}+x \right )}{2}\)	\(49\)
meijerg	\(\frac {i \left (-2 i \sqrt {\pi }\, x \sqrt {-x^{2}+1}-2 i \sqrt {\pi }\, \arcsin \left (x \right )\right )}{4 \sqrt {\pi }}+\frac {\frac {4 \sqrt {\pi }}{3}-\frac {2 \sqrt {\pi }\, \left (-2 x^{2}+2\right ) \sqrt {-x^{2}+1}}{3}}{4 \sqrt {\pi }}\)	\(65\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.05 \[ \int (1+x) \sqrt {1-x^2} \, dx=\frac {1}{6} \, {\left (2 \, x^{2} + 3 \, x - 2\right )} \sqrt {-x^{2} + 1} - \arctan \left (\frac {\sqrt {-x^{2} + 1} - 1}{x}\right ) \]

[In]

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

[Out]

1/6*(2*x^2 + 3*x - 2)*sqrt(-x^2 + 1) - arctan((sqrt(-x^2 + 1) - 1)/x)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.03 \[ \int (1+x) \sqrt {1-x^2} \, dx=\frac {x^{2} \sqrt {1 - x^{2}}}{3} + \frac {x \sqrt {1 - x^{2}}}{2} - \frac {\sqrt {1 - x^{2}}}{3} + \frac {\operatorname {asin}{\left (x \right )}}{2} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.74 \[ \int (1+x) \sqrt {1-x^2} \, dx=-\frac {1}{3} \, {\left (-x^{2} + 1\right )}^{\frac {3}{2}} + \frac {1}{2} \, \sqrt {-x^{2} + 1} x + \frac {1}{2} \, \arcsin \left (x\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.66 \[ \int (1+x) \sqrt {1-x^2} \, dx=\frac {1}{6} \, {\left ({\left (2 \, x + 3\right )} x - 2\right )} \sqrt {-x^{2} + 1} + \frac {1}{2} \, \arcsin \left (x\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.66 \[ \int (1+x) \sqrt {1-x^2} \, dx=\frac {\mathrm {asin}\left (x\right )}{2}+\sqrt {1-x^2}\,\left (\frac {x^2}{3}+\frac {x}{2}-\frac {1}{3}\right ) \]

[In]

int((1 - x^2)^(1/2)*(x + 1),x)

[Out]

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