\(\int \sqrt {1-x} \sqrt {1+x} \, dx\) [1067]

Optimal result

Integrand size = 17, antiderivative size = 28 \[ \int \sqrt {1-x} \sqrt {1+x} \, dx=\frac {1}{2} \sqrt {1-x} x \sqrt {1+x}+\frac {\arcsin (x)}{2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {38, 41, 222} \[ \int \sqrt {1-x} \sqrt {1+x} \, dx=\frac {\arcsin (x)}{2}+\frac {1}{2} \sqrt {1-x} \sqrt {x+1} x \]

[In]

Int[Sqrt[1 - x]*Sqrt[1 + x],x]

[Out]

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

Rule 38

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[x*(a + b*x)^m*((c + d*x)^m/(2*m + 1))
, x] + Dist[2*a*c*(m/(2*m + 1)), Int[(a + b*x)^(m - 1)*(c + d*x)^(m - 1), x], x] /; FreeQ[{a, b, c, d}, x] &&
EqQ[b*c + a*d, 0] && IGtQ[m + 1/2, 0]

Rule 41

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(a*c + b*d*x^2)^m, x] /; FreeQ[{a, b
, c, d, m}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))

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]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \sqrt {1-x} \sqrt {1+x} \, dx=\frac {1}{2} x \sqrt {1-x^2}-\arctan \left (\frac {\sqrt {1-x^2}}{1+x}\right ) \]

[In]

Integrate[Sqrt[1 - x]*Sqrt[1 + x],x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(56\) vs. \(2(20)=40\).

Time = 0.31 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.04

method	result	size
default	\(\frac {\sqrt {1-x}\, \left (1+x \right )^{\frac {3}{2}}}{2}-\frac {\sqrt {1-x}\, \sqrt {1+x}}{2}+\frac {\sqrt {\left (1+x \right ) \left (1-x \right )}\, \arcsin \left (x \right )}{2 \sqrt {1+x}\, \sqrt {1-x}}\)	\(57\)
risch	\(-\frac {x \left (-1+x \right ) \sqrt {1+x}\, \sqrt {\left (1+x \right ) \left (1-x \right )}}{2 \sqrt {-\left (-1+x \right ) \left (1+x \right )}\, \sqrt {1-x}}+\frac {\sqrt {\left (1+x \right ) \left (1-x \right )}\, \arcsin \left (x \right )}{2 \sqrt {1+x}\, \sqrt {1-x}}\)	\(68\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \sqrt {1-x} \sqrt {1+x} \, dx=\frac {1}{2} \, \sqrt {x + 1} x \sqrt {-x + 1} - \arctan \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) \]

[In]

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

[Out]

1/2*sqrt(x + 1)*x*sqrt(-x + 1) - arctan((sqrt(x + 1)*sqrt(-x + 1) - 1)/x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.74 (sec) , antiderivative size = 131, normalized size of antiderivative = 4.68 \[ \int \sqrt {1-x} \sqrt {1+x} \, dx=\begin {cases} - i \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {x + 1}}{2} \right )} + \frac {i \left (x + 1\right )^{\frac {5}{2}}}{2 \sqrt {x - 1}} - \frac {3 i \left (x + 1\right )^{\frac {3}{2}}}{2 \sqrt {x - 1}} + \frac {i \sqrt {x + 1}}{\sqrt {x - 1}} & \text {for}\: \left |{x + 1}\right | > 2 \\\operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {x + 1}}{2} \right )} - \frac {\left (x + 1\right )^{\frac {5}{2}}}{2 \sqrt {1 - x}} + \frac {3 \left (x + 1\right )^{\frac {3}{2}}}{2 \sqrt {1 - x}} - \frac {\sqrt {x + 1}}{\sqrt {1 - x}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-I*acosh(sqrt(2)*sqrt(x + 1)/2) + I*(x + 1)**(5/2)/(2*sqrt(x - 1)) - 3*I*(x + 1)**(3/2)/(2*sqrt(x -
1)) + I*sqrt(x + 1)/sqrt(x - 1), Abs(x + 1) > 2), (asin(sqrt(2)*sqrt(x + 1)/2) - (x + 1)**(5/2)/(2*sqrt(1 - x
)) + 3*(x + 1)**(3/2)/(2*sqrt(1 - x)) - sqrt(x + 1)/sqrt(1 - x), True))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61 \[ \int \sqrt {1-x} \sqrt {1+x} \, dx=\frac {1}{2} \, \sqrt {-x^{2} + 1} x + \frac {1}{2} \, \arcsin \left (x\right ) \]

[In]

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

[Out]

1/2*sqrt(-x^2 + 1)*x + 1/2*arcsin(x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (20) = 40\).

Time = 0.30 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \sqrt {1-x} \sqrt {1+x} \, dx=\frac {1}{2} \, \sqrt {x + 1} {\left (x - 2\right )} \sqrt {-x + 1} + \sqrt {x + 1} \sqrt {-x + 1} + \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {x + 1}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \sqrt {1-x} \sqrt {1+x} \, dx=\frac {x\,\sqrt {1-x}\,\sqrt {x+1}}{2}-\frac {\ln \left (x-\sqrt {1-x}\,\sqrt {x+1}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2} \]

[In]

int((1 - x)^(1/2)*(x + 1)^(1/2),x)

[Out]

(x*(1 - x)^(1/2)*(x + 1)^(1/2))/2 - (log(x - (1 - x)^(1/2)*(x + 1)^(1/2)*1i)*1i)/2