3.1117 \(\int \frac{(1-x)^{5/2}}{(1+x)^{3/2}} \, dx\)

Optimal. Leaf size=65 \[ -\frac{2 (1-x)^{5/2}}{\sqrt{x+1}}-\frac{5}{2} \sqrt{x+1} (1-x)^{3/2}-\frac{15}{2} \sqrt{x+1} \sqrt{1-x}-\frac{15}{2} \sin ^{-1}(x) \]

[Out]

(-2*(1 - x)^(5/2))/Sqrt[1 + x] - (15*Sqrt[1 - x]*Sqrt[1 + x])/2 - (5*(1 - x)^(3/2)*Sqrt[1 + x])/2 - (15*ArcSin
[x])/2

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Rubi [A]  time = 0.0108989, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {47, 50, 41, 216} \[ -\frac{2 (1-x)^{5/2}}{\sqrt{x+1}}-\frac{5}{2} \sqrt{x+1} (1-x)^{3/2}-\frac{15}{2} \sqrt{x+1} \sqrt{1-x}-\frac{15}{2} \sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(1 - x)^(5/2)/(1 + x)^(3/2),x]

[Out]

(-2*(1 - x)^(5/2))/Sqrt[1 + x] - (15*Sqrt[1 - x]*Sqrt[1 + x])/2 - (5*(1 - x)^(3/2)*Sqrt[1 + x])/2 - (15*ArcSin
[x])/2

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)^{5/2}}{(1+x)^{3/2}} \, dx &=-\frac{2 (1-x)^{5/2}}{\sqrt{1+x}}-5 \int \frac{(1-x)^{3/2}}{\sqrt{1+x}} \, dx\\ &=-\frac{2 (1-x)^{5/2}}{\sqrt{1+x}}-\frac{5}{2} (1-x)^{3/2} \sqrt{1+x}-\frac{15}{2} \int \frac{\sqrt{1-x}}{\sqrt{1+x}} \, dx\\ &=-\frac{2 (1-x)^{5/2}}{\sqrt{1+x}}-\frac{15}{2} \sqrt{1-x} \sqrt{1+x}-\frac{5}{2} (1-x)^{3/2} \sqrt{1+x}-\frac{15}{2} \int \frac{1}{\sqrt{1-x} \sqrt{1+x}} \, dx\\ &=-\frac{2 (1-x)^{5/2}}{\sqrt{1+x}}-\frac{15}{2} \sqrt{1-x} \sqrt{1+x}-\frac{5}{2} (1-x)^{3/2} \sqrt{1+x}-\frac{15}{2} \int \frac{1}{\sqrt{1-x^2}} \, dx\\ &=-\frac{2 (1-x)^{5/2}}{\sqrt{1+x}}-\frac{15}{2} \sqrt{1-x} \sqrt{1+x}-\frac{5}{2} (1-x)^{3/2} \sqrt{1+x}-\frac{15}{2} \sin ^{-1}(x)\\ \end{align*}

Mathematica [C]  time = 0.0102438, size = 37, normalized size = 0.57 \[ -\frac{(1-x)^{7/2} \, _2F_1\left (\frac{3}{2},\frac{7}{2};\frac{9}{2};\frac{1-x}{2}\right )}{7 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - x)^(5/2)/(1 + x)^(3/2),x]

[Out]

-((1 - x)^(7/2)*Hypergeometric2F1[3/2, 7/2, 9/2, (1 - x)/2])/(7*Sqrt[2])

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Maple [A]  time = 0.012, size = 77, normalized size = 1.2 \begin{align*} -{\frac{{x}^{3}-8\,{x}^{2}-17\,x+24}{2}\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}}}}-{\frac{15\,\arcsin \left ( x \right ) }{2}\sqrt{ \left ( 1+x \right ) \left ( 1-x \right ) }{\frac{1}{\sqrt{1-x}}}{\frac{1}{\sqrt{1+x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-x)^(5/2)/(1+x)^(3/2),x)

[Out]

-1/2*(x^3-8*x^2-17*x+24)/(-(1+x)*(-1+x))^(1/2)*((1+x)*(1-x))^(1/2)/(1-x)^(1/2)/(1+x)^(1/2)-15/2*((1+x)*(1-x))^
(1/2)/(1+x)^(1/2)/(1-x)^(1/2)*arcsin(x)

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Maxima [A]  time = 1.47965, size = 76, normalized size = 1.17 \begin{align*} -\frac{x^{3}}{2 \, \sqrt{-x^{2} + 1}} + \frac{4 \, x^{2}}{\sqrt{-x^{2} + 1}} + \frac{17 \, x}{2 \, \sqrt{-x^{2} + 1}} - \frac{12}{\sqrt{-x^{2} + 1}} - \frac{15}{2} \, \arcsin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-x)^(5/2)/(1+x)^(3/2),x, algorithm="maxima")

[Out]

-1/2*x^3/sqrt(-x^2 + 1) + 4*x^2/sqrt(-x^2 + 1) + 17/2*x/sqrt(-x^2 + 1) - 12/sqrt(-x^2 + 1) - 15/2*arcsin(x)

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Fricas [A]  time = 1.79065, size = 166, normalized size = 2.55 \begin{align*} \frac{{\left (x^{2} - 7 \, x - 24\right )} \sqrt{x + 1} \sqrt{-x + 1} + 30 \,{\left (x + 1\right )} \arctan \left (\frac{\sqrt{x + 1} \sqrt{-x + 1} - 1}{x}\right ) - 24 \, x - 24}{2 \,{\left (x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-x)^(5/2)/(1+x)^(3/2),x, algorithm="fricas")

[Out]

1/2*((x^2 - 7*x - 24)*sqrt(x + 1)*sqrt(-x + 1) + 30*(x + 1)*arctan((sqrt(x + 1)*sqrt(-x + 1) - 1)/x) - 24*x -
24)/(x + 1)

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Sympy [A]  time = 13.6217, size = 168, normalized size = 2.58 \begin{align*} \begin{cases} 15 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{11 i \left (x + 1\right )^{\frac{3}{2}}}{2 \sqrt{x - 1}} + \frac{i \sqrt{x + 1}}{\sqrt{x - 1}} + \frac{16 i}{\sqrt{x - 1} \sqrt{x + 1}} & \text{for}\: \frac{\left |{x + 1}\right |}{2} > 1 \\- 15 \operatorname{asin}{\left (\frac{\sqrt{2} \sqrt{x + 1}}{2} \right )} - \frac{\left (x + 1\right )^{\frac{5}{2}}}{2 \sqrt{1 - x}} + \frac{11 \left (x + 1\right )^{\frac{3}{2}}}{2 \sqrt{1 - x}} - \frac{\sqrt{x + 1}}{\sqrt{1 - x}} - \frac{16}{\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)**(5/2)/(1+x)**(3/2),x)

[Out]

Piecewise((15*I*acosh(sqrt(2)*sqrt(x + 1)/2) + I*(x + 1)**(5/2)/(2*sqrt(x - 1)) - 11*I*(x + 1)**(3/2)/(2*sqrt(
x - 1)) + I*sqrt(x + 1)/sqrt(x - 1) + 16*I/(sqrt(x - 1)*sqrt(x + 1)), Abs(x + 1)/2 > 1), (-15*asin(sqrt(2)*sqr
t(x + 1)/2) - (x + 1)**(5/2)/(2*sqrt(1 - x)) + 11*(x + 1)**(3/2)/(2*sqrt(1 - x)) - sqrt(x + 1)/sqrt(1 - x) - 1
6/(sqrt(1 - x)*sqrt(x + 1)), True))

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Giac [A]  time = 1.10505, size = 99, normalized size = 1.52 \begin{align*} \frac{1}{2} \, \sqrt{x + 1}{\left (x - 8\right )} \sqrt{-x + 1} + \frac{4 \,{\left (\sqrt{2} - \sqrt{-x + 1}\right )}}{\sqrt{x + 1}} - \frac{4 \, \sqrt{x + 1}}{\sqrt{2} - \sqrt{-x + 1}} - 15 \, \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)^(5/2)/(1+x)^(3/2),x, algorithm="giac")

[Out]

1/2*sqrt(x + 1)*(x - 8)*sqrt(-x + 1) + 4*(sqrt(2) - sqrt(-x + 1))/sqrt(x + 1) - 4*sqrt(x + 1)/(sqrt(2) - sqrt(
-x + 1)) - 15*arcsin(1/2*sqrt(2)*sqrt(x + 1))