\(\int \frac {1+x}{\sqrt {4-x^2}} \, dx\) [586]

Optimal result

Integrand size = 15, antiderivative size = 20 \[ \int \frac {1+x}{\sqrt {4-x^2}} \, dx=-\sqrt {4-x^2}+\arcsin \left (\frac {x}{2}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {655, 222} \[ \int \frac {1+x}{\sqrt {4-x^2}} \, dx=\arcsin \left (\frac {x}{2}\right )-\sqrt {4-x^2} \]

[In]

Int[(1 + x)/Sqrt[4 - x^2],x]

[Out]

-Sqrt[4 - x^2] + ArcSin[x/2]

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}& = -\sqrt {4-x^2}+\int \frac {1}{\sqrt {4-x^2}} \, dx \\ & = -\sqrt {4-x^2}+\sin ^{-1}\left (\frac {x}{2}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.70 \[ \int \frac {1+x}{\sqrt {4-x^2}} \, dx=-\sqrt {4-x^2}-2 \arctan \left (\frac {\sqrt {4-x^2}}{2+x}\right ) \]

[In]

Integrate[(1 + x)/Sqrt[4 - x^2],x]

[Out]

-Sqrt[4 - x^2] - 2*ArcTan[Sqrt[4 - x^2]/(2 + x)]

Maple [A] (verified)

Time = 2.13 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85

method	result	size
default	\(\arcsin \left (\frac {x}{2}\right )-\sqrt {-x^{2}+4}\)	\(17\)
risch	\(\frac {x^{2}-4}{\sqrt {-x^{2}+4}}+\arcsin \left (\frac {x}{2}\right )\)	\(21\)
meijerg	\(\arcsin \left (\frac {x}{2}\right )-\frac {-2 \sqrt {\pi }+2 \sqrt {\pi }\, \sqrt {-\frac {x^{2}}{4}+1}}{\sqrt {\pi }}\)	\(31\)
trager	\(-\sqrt {-x^{2}+4}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {-x^{2}+4}+x \right )\)	\(39\)

[In]

int((1+x)/(-x^2+4)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.50 \[ \int \frac {1+x}{\sqrt {4-x^2}} \, dx=-\sqrt {-x^{2} + 4} - 2 \, \arctan \left (\frac {\sqrt {-x^{2} + 4} - 2}{x}\right ) \]

[In]

integrate((1+x)/(-x^2+4)^(1/2),x, algorithm="fricas")

[Out]

-sqrt(-x^2 + 4) - 2*arctan((sqrt(-x^2 + 4) - 2)/x)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.60 \[ \int \frac {1+x}{\sqrt {4-x^2}} \, dx=- \sqrt {4 - x^{2}} + \operatorname {asin}{\left (\frac {x}{2} \right )} \]

[In]

integrate((1+x)/(-x**2+4)**(1/2),x)

[Out]

-sqrt(4 - x**2) + asin(x/2)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {1+x}{\sqrt {4-x^2}} \, dx=-\sqrt {-x^{2} + 4} + \arcsin \left (\frac {1}{2} \, x\right ) \]

[In]

integrate((1+x)/(-x^2+4)^(1/2),x, algorithm="maxima")

[Out]

-sqrt(-x^2 + 4) + arcsin(1/2*x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {1+x}{\sqrt {4-x^2}} \, dx=-\sqrt {-x^{2} + 4} + \arcsin \left (\frac {1}{2} \, x\right ) \]

[In]

integrate((1+x)/(-x^2+4)^(1/2),x, algorithm="giac")

[Out]

-sqrt(-x^2 + 4) + arcsin(1/2*x)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {1+x}{\sqrt {4-x^2}} \, dx=\mathrm {asin}\left (\frac {x}{2}\right )-\sqrt {4-x^2} \]

[In]

int((x + 1)/(4 - x^2)^(1/2),x)

[Out]

asin(x/2) - (4 - x^2)^(1/2)