3.1096 \(\int \frac{(1+x)^{5/2}}{(1-x)^{5/2}} \, dx\)

Optimal. Leaf size=63 \[ \frac{2 (x+1)^{5/2}}{3 (1-x)^{3/2}}-\frac{10 (x+1)^{3/2}}{3 \sqrt{1-x}}-5 \sqrt{1-x} \sqrt{x+1}+5 \sin ^{-1}(x) \]

[Out]

-5*Sqrt[1 - x]*Sqrt[1 + x] - (10*(1 + x)^(3/2))/(3*Sqrt[1 - x]) + (2*(1 + x)^(5/2))/(3*(1 - x)^(3/2)) + 5*ArcS
in[x]

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Rubi [A]  time = 0.0101658, antiderivative size = 63, 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} \[ \frac{2 (x+1)^{5/2}}{3 (1-x)^{3/2}}-\frac{10 (x+1)^{3/2}}{3 \sqrt{1-x}}-5 \sqrt{1-x} \sqrt{x+1}+5 \sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(1 + x)^(5/2)/(1 - x)^(5/2),x]

[Out]

-5*Sqrt[1 - x]*Sqrt[1 + x] - (10*(1 + x)^(3/2))/(3*Sqrt[1 - x]) + (2*(1 + x)^(5/2))/(3*(1 - x)^(3/2)) + 5*ArcS
in[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)^{5/2}}{(1-x)^{5/2}} \, dx &=\frac{2 (1+x)^{5/2}}{3 (1-x)^{3/2}}-\frac{5}{3} \int \frac{(1+x)^{3/2}}{(1-x)^{3/2}} \, dx\\ &=-\frac{10 (1+x)^{3/2}}{3 \sqrt{1-x}}+\frac{2 (1+x)^{5/2}}{3 (1-x)^{3/2}}+5 \int \frac{\sqrt{1+x}}{\sqrt{1-x}} \, dx\\ &=-5 \sqrt{1-x} \sqrt{1+x}-\frac{10 (1+x)^{3/2}}{3 \sqrt{1-x}}+\frac{2 (1+x)^{5/2}}{3 (1-x)^{3/2}}+5 \int \frac{1}{\sqrt{1-x} \sqrt{1+x}} \, dx\\ &=-5 \sqrt{1-x} \sqrt{1+x}-\frac{10 (1+x)^{3/2}}{3 \sqrt{1-x}}+\frac{2 (1+x)^{5/2}}{3 (1-x)^{3/2}}+5 \int \frac{1}{\sqrt{1-x^2}} \, dx\\ &=-5 \sqrt{1-x} \sqrt{1+x}-\frac{10 (1+x)^{3/2}}{3 \sqrt{1-x}}+\frac{2 (1+x)^{5/2}}{3 (1-x)^{3/2}}+5 \sin ^{-1}(x)\\ \end{align*}

Mathematica [C]  time = 0.0080356, size = 37, normalized size = 0.59 \[ \frac{8 \sqrt{2} \, _2F_1\left (-\frac{5}{2},-\frac{3}{2};-\frac{1}{2};\frac{1-x}{2}\right )}{3 (1-x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.017, size = 84, normalized size = 1.3 \begin{align*}{\frac{3\,{x}^{3}-31\,{x}^{2}-11\,x+23}{-3+3\,x}\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}}}}+5\,{\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)^(5/2)/(1-x)^(5/2),x)

[Out]

1/3*(3*x^3-31*x^2-11*x+23)/(-1+x)/(-(1+x)*(-1+x))^(1/2)*((1+x)*(1-x))^(1/2)/(1-x)^(1/2)/(1+x)^(1/2)+5*((1+x)*(
1-x))^(1/2)/(1+x)^(1/2)/(1-x)^(1/2)*arcsin(x)

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Maxima [B]  time = 1.4886, size = 134, normalized size = 2.13 \begin{align*} -\frac{{\left (-x^{2} + 1\right )}^{\frac{5}{2}}}{x^{4} - 4 \, x^{3} + 6 \, x^{2} - 4 \, x + 1} - \frac{5 \,{\left (-x^{2} + 1\right )}^{\frac{3}{2}}}{3 \,{\left (x^{3} - 3 \, x^{2} + 3 \, x - 1\right )}} + \frac{10 \, \sqrt{-x^{2} + 1}}{3 \,{\left (x^{2} - 2 \, x + 1\right )}} + \frac{35 \, \sqrt{-x^{2} + 1}}{3 \,{\left (x - 1\right )}} + 5 \, \arcsin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-x^2 + 1)^(5/2)/(x^4 - 4*x^3 + 6*x^2 - 4*x + 1) - 5/3*(-x^2 + 1)^(3/2)/(x^3 - 3*x^2 + 3*x - 1) + 10/3*sqrt(-
x^2 + 1)/(x^2 - 2*x + 1) + 35/3*sqrt(-x^2 + 1)/(x - 1) + 5*arcsin(x)

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Fricas [A]  time = 1.579, size = 205, normalized size = 3.25 \begin{align*} -\frac{23 \, x^{2} +{\left (3 \, x^{2} - 34 \, x + 23\right )} \sqrt{x + 1} \sqrt{-x + 1} + 30 \,{\left (x^{2} - 2 \, x + 1\right )} \arctan \left (\frac{\sqrt{x + 1} \sqrt{-x + 1} - 1}{x}\right ) - 46 \, x + 23}{3 \,{\left (x^{2} - 2 \, x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(23*x^2 + (3*x^2 - 34*x + 23)*sqrt(x + 1)*sqrt(-x + 1) + 30*(x^2 - 2*x + 1)*arctan((sqrt(x + 1)*sqrt(-x +
1) - 1)/x) - 46*x + 23)/(x^2 - 2*x + 1)

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Sympy [B]  time = 14.2402, size = 576, normalized size = 9.14 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)**(5/2)/(1-x)**(5/2),x)

[Out]

Piecewise((-30*I*sqrt(x - 1)*(x + 1)**(27/2)*acosh(sqrt(2)*sqrt(x + 1)/2)/(3*sqrt(x - 1)*(x + 1)**(27/2) - 6*s
qrt(x - 1)*(x + 1)**(25/2)) + 15*pi*sqrt(x - 1)*(x + 1)**(27/2)/(3*sqrt(x - 1)*(x + 1)**(27/2) - 6*sqrt(x - 1)
*(x + 1)**(25/2)) + 60*I*sqrt(x - 1)*(x + 1)**(25/2)*acosh(sqrt(2)*sqrt(x + 1)/2)/(3*sqrt(x - 1)*(x + 1)**(27/
2) - 6*sqrt(x - 1)*(x + 1)**(25/2)) - 30*pi*sqrt(x - 1)*(x + 1)**(25/2)/(3*sqrt(x - 1)*(x + 1)**(27/2) - 6*sqr
t(x - 1)*(x + 1)**(25/2)) - 3*I*(x + 1)**15/(3*sqrt(x - 1)*(x + 1)**(27/2) - 6*sqrt(x - 1)*(x + 1)**(25/2)) +
40*I*(x + 1)**14/(3*sqrt(x - 1)*(x + 1)**(27/2) - 6*sqrt(x - 1)*(x + 1)**(25/2)) - 60*I*(x + 1)**13/(3*sqrt(x
- 1)*(x + 1)**(27/2) - 6*sqrt(x - 1)*(x + 1)**(25/2)), Abs(x + 1)/2 > 1), (30*sqrt(1 - x)*(x + 1)**(27/2)*asin
(sqrt(2)*sqrt(x + 1)/2)/(3*sqrt(1 - x)*(x + 1)**(27/2) - 6*sqrt(1 - x)*(x + 1)**(25/2)) - 60*sqrt(1 - x)*(x +
1)**(25/2)*asin(sqrt(2)*sqrt(x + 1)/2)/(3*sqrt(1 - x)*(x + 1)**(27/2) - 6*sqrt(1 - x)*(x + 1)**(25/2)) + 3*(x
+ 1)**15/(3*sqrt(1 - x)*(x + 1)**(27/2) - 6*sqrt(1 - x)*(x + 1)**(25/2)) - 40*(x + 1)**14/(3*sqrt(1 - x)*(x +
1)**(27/2) - 6*sqrt(1 - x)*(x + 1)**(25/2)) + 60*(x + 1)**13/(3*sqrt(1 - x)*(x + 1)**(27/2) - 6*sqrt(1 - x)*(x
+ 1)**(25/2)), True))

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Giac [A]  time = 1.09, size = 59, normalized size = 0.94 \begin{align*} -\frac{{\left ({\left (3 \, x - 37\right )}{\left (x + 1\right )} + 60\right )} \sqrt{x + 1} \sqrt{-x + 1}}{3 \,{\left (x - 1\right )}^{2}} + 10 \, \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)^(5/2),x, algorithm="giac")

[Out]

-1/3*((3*x - 37)*(x + 1) + 60)*sqrt(x + 1)*sqrt(-x + 1)/(x - 1)^2 + 10*arcsin(1/2*sqrt(2)*sqrt(x + 1))