∫ √ ___ 1 − x (1+x)3∕2 dx [1119]

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Rubi [A]
time = 0.00, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {49, 41, 222} \begin {gather*} -\text {ArcSin}(x)-\frac {2 \sqrt {1-x}}{\sqrt {x+1}} \end {gather*}

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]))

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(

m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},

x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] && !(IntegerQ[n] && !IntegerQ[m]) && !(ILeQ[m + n + 2, 0

] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}

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

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Mathematica [A]
time = 0.04, size = 39, normalized size = 1.70 \begin {gather*} -\frac {2 \sqrt {1-x}}{\sqrt {1+x}}+2 \tan ^{-1}\left (\frac {\sqrt {1-x}}{\sqrt {1+x}}\right ) \end {gather*}

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(66\) vs. \(2(19)=38\).
time = 0.14, size = 67, normalized size = 2.91


method	result	size

risch	\(\frac {2 \left (-1+x \right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{\sqrt {-\left (1+x \right ) \left (-1+x \right )}\, \sqrt {1-x}\, \sqrt {1+x}}-\frac {\sqrt {\left (1+x \right ) \left (1-x \right )}\, \arcsin \left (x \right )}{\sqrt {1+x}\, \sqrt {1-x}}\)	\(67\)

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

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Maxima [A]
time = 0.48, size = 21, normalized size = 0.91 \begin {gather*} -\frac {2 \, \sqrt {-x^{2} + 1}}{x + 1} - \arcsin \left (x\right ) \end {gather*}

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (19) = 38\).
time = 1.34, size = 50, normalized size = 2.17 \begin {gather*} \frac {2 \, {\left ({\left (x + 1\right )} \arctan \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) - x - \sqrt {x + 1} \sqrt {-x + 1} - 1\right )}}{x + 1} \end {gather*}

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

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Sympy [C] Result contains complex when optimal does not.
time = 0.73, size = 102, normalized size = 4.43 \begin {gather*} \begin {cases} 2 i \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {x + 1}}{2} \right )} - \frac {2 i \sqrt {x + 1}}{\sqrt {x - 1}} + \frac {4 i}{\sqrt {x - 1} \sqrt {x + 1}} & \text {for}\: \left |{x + 1}\right | > 2 \\- 2 \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {x + 1}}{2} \right )} + \frac {2 \sqrt {x + 1}}{\sqrt {1 - x}} - \frac {4}{\sqrt {1 - x} \sqrt {x + 1}} & \text {otherwise} \end {cases} \end {gather*}

Piecewise((2*I*acosh(sqrt(2)*sqrt(x + 1)/2) - 2*I*sqrt(x + 1)/sqrt(x - 1) + 4*I/(sqrt(x - 1)*sqrt(x + 1)), Abs

(x + 1) > 2), (-2*asin(sqrt(2)*sqrt(x + 1)/2) + 2*sqrt(x + 1)/sqrt(1 - x) - 4/(sqrt(1 - x)*sqrt(x + 1)), True)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (19) = 38\).
time = 1.26, size = 55, normalized size = 2.39 \begin {gather*} \frac {\sqrt {2} - \sqrt {-x + 1}}{\sqrt {x + 1}} - \frac {\sqrt {x + 1}}{\sqrt {2} - \sqrt {-x + 1}} - 2 \, \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {x + 1}\right ) \end {gather*}

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {\sqrt {1-x}}{{\left (x+1\right )}^{3/2}} \,d x \end {gather*}

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3.12.19 \(\int \frac {\sqrt {1-x}}{(1+x)^{3/2}} \, dx\) [1119]