3.1095 \(\int \frac{(1+x)^{5/2}}{(1-x)^{3/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0103244, 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} \[ \frac{2 (x+1)^{5/2}}{\sqrt{1-x}}+\frac{5}{2} \sqrt{1-x} (x+1)^{3/2}+\frac{15}{2} \sqrt{1-x} \sqrt{x+1}-\frac{15}{2} \sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(1 + x)^(5/2)/(1 - x)^(3/2),x]

[Out]

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

Mathematica [C]  time = 0.0065163, size = 35, normalized size = 0.54 \[ \frac{8 \sqrt{2} \, _2F_1\left (-\frac{5}{2},-\frac{1}{2};\frac{1}{2};\frac{1-x}{2}\right )}{\sqrt{1-x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*Sqrt[2]*Hypergeometric2F1[-5/2, -1/2, 1/2, (1 - x)/2])/Sqrt[1 - x]

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Maple [A]  time = 0.014, 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 = 1.49699, 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 = 1.56274, 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 = 14.3842, size = 139, normalized size = 2.14 \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{5 i \left (x + 1\right )^{\frac{3}{2}}}{2 \sqrt{x - 1}} - \frac{15 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{5 \left (x + 1\right )^{\frac{3}{2}}}{2 \sqrt{1 - x}} + \frac{15 \sqrt{x + 1}}{\sqrt{1 - x}} & \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)) + 5*I*(x + 1)**(3/2)/(2*sqrt(x
- 1)) - 15*I*sqrt(x + 1)/sqrt(x - 1), Abs(x + 1)/2 > 1), (-15*asin(sqrt(2)*sqrt(x + 1)/2) - (x + 1)**(5/2)/(2
*sqrt(1 - x)) - 5*(x + 1)**(3/2)/(2*sqrt(1 - x)) + 15*sqrt(x + 1)/sqrt(1 - x), True))

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Giac [A]  time = 1.08772, size = 57, normalized size = 0.88 \begin{align*} \frac{{\left ({\left (x + 6\right )}{\left (x + 1\right )} - 30\right )} \sqrt{x + 1} \sqrt{-x + 1}}{2 \,{\left (x - 1\right )}} - 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*((x + 6)*(x + 1) - 30)*sqrt(x + 1)*sqrt(-x + 1)/(x - 1) - 15*arcsin(1/2*sqrt(2)*sqrt(x + 1))