3.1118 \(\int \frac{(1-x)^{3/2}}{(1+x)^{3/2}} \, dx\)

Optimal. Leaf size=41 \[ -\frac{2 (1-x)^{3/2}}{\sqrt{x+1}}-3 \sqrt{x+1} \sqrt{1-x}-3 \sin ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.0065795, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {47, 50, 41, 216} \[ -\frac{2 (1-x)^{3/2}}{\sqrt{x+1}}-3 \sqrt{x+1} \sqrt{1-x}-3 \sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(1 - x)^(3/2)/(1 + x)^(3/2),x]

[Out]

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

Rule 47

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]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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 216

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

Mathematica [C]  time = 0.0085057, size = 37, normalized size = 0.9 \[ -\frac{(1-x)^{5/2} \, _2F_1\left (\frac{3}{2},\frac{5}{2};\frac{7}{2};\frac{1-x}{2}\right )}{5 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - x)^(3/2)/(1 + x)^(3/2),x]

[Out]

-((1 - x)^(5/2)*Hypergeometric2F1[3/2, 5/2, 7/2, (1 - x)/2])/(5*Sqrt[2])

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Maple [B]  time = 0.013, size = 71, normalized size = 1.7 \begin{align*}{({x}^{2}+4\,x-5)\sqrt{ \left ( 1+x \right ) \left ( 1-x \right ) }{\frac{1}{\sqrt{- \left ( 1+x \right ) \left ( -1+x \right ) }}}{\frac{1}{\sqrt{1-x}}}{\frac{1}{\sqrt{1+x}}}}-3\,{\frac{\sqrt{ \left ( 1+x \right ) \left ( 1-x \right ) }\arcsin \left ( x \right ) }{\sqrt{1-x}\sqrt{1+x}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-x)^(3/2)/(1+x)^(3/2),x)

[Out]

(x^2+4*x-5)/(-(1+x)*(-1+x))^(1/2)*((1+x)*(1-x))^(1/2)/(1-x)^(1/2)/(1+x)^(1/2)-3*((1+x)*(1-x))^(1/2)/(1+x)^(1/2
)/(1-x)^(1/2)*arcsin(x)

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Maxima [A]  time = 1.57728, size = 55, normalized size = 1.34 \begin{align*} \frac{{\left (-x^{2} + 1\right )}^{\frac{3}{2}}}{x^{2} + 2 \, x + 1} - \frac{6 \, \sqrt{-x^{2} + 1}}{x + 1} - 3 \, \arcsin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-x)^(3/2)/(1+x)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.85204, size = 146, normalized size = 3.56 \begin{align*} -\frac{{\left (x + 5\right )} \sqrt{x + 1} \sqrt{-x + 1} - 6 \,{\left (x + 1\right )} \arctan \left (\frac{\sqrt{x + 1} \sqrt{-x + 1} - 1}{x}\right ) + 5 \, x + 5}{x + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-x)^(3/2)/(1+x)^(3/2),x, algorithm="fricas")

[Out]

-((x + 5)*sqrt(x + 1)*sqrt(-x + 1) - 6*(x + 1)*arctan((sqrt(x + 1)*sqrt(-x + 1) - 1)/x) + 5*x + 5)/(x + 1)

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Sympy [A]  time = 2.81974, size = 133, normalized size = 3.24 \begin{align*} \begin{cases} 6 i \operatorname{acosh}{\left (\frac{\sqrt{2} \sqrt{x + 1}}{2} \right )} - \frac{i \left (x + 1\right )^{\frac{3}{2}}}{\sqrt{x - 1}} - \frac{2 i \sqrt{x + 1}}{\sqrt{x - 1}} + \frac{8 i}{\sqrt{x - 1} \sqrt{x + 1}} & \text{for}\: \frac{\left |{x + 1}\right |}{2} > 1 \\- 6 \operatorname{asin}{\left (\frac{\sqrt{2} \sqrt{x + 1}}{2} \right )} + \frac{\left (x + 1\right )^{\frac{3}{2}}}{\sqrt{1 - x}} + \frac{2 \sqrt{x + 1}}{\sqrt{1 - x}} - \frac{8}{\sqrt{1 - x} \sqrt{x + 1}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-x)**(3/2)/(1+x)**(3/2),x)

[Out]

Piecewise((6*I*acosh(sqrt(2)*sqrt(x + 1)/2) - I*(x + 1)**(3/2)/sqrt(x - 1) - 2*I*sqrt(x + 1)/sqrt(x - 1) + 8*I
/(sqrt(x - 1)*sqrt(x + 1)), Abs(x + 1)/2 > 1), (-6*asin(sqrt(2)*sqrt(x + 1)/2) + (x + 1)**(3/2)/sqrt(1 - x) +
2*sqrt(x + 1)/sqrt(1 - x) - 8/(sqrt(1 - x)*sqrt(x + 1)), True))

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Giac [B]  time = 1.0769, size = 95, normalized size = 2.32 \begin{align*} -\sqrt{x + 1} \sqrt{-x + 1} + \frac{2 \,{\left (\sqrt{2} - \sqrt{-x + 1}\right )}}{\sqrt{x + 1}} - \frac{2 \, \sqrt{x + 1}}{\sqrt{2} - \sqrt{-x + 1}} - 6 \, \arcsin \left (\frac{1}{2} \, \sqrt{2} \sqrt{x + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-x)^(3/2)/(1+x)^(3/2),x, algorithm="giac")

[Out]

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