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

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

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Rubi [A]  time = 0.01, antiderivative size = 65, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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) \end {gather*}

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 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 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 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*}

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Mathematica [C]  time = 0.01, size = 35, normalized size = 0.54 \begin {gather*} \frac {8 \sqrt {2} \, _2F_1\left (-\frac {5}{2},-\frac {1}{2};\frac {1}{2};\frac {1-x}{2}\right )}{\sqrt {1-x}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.08, size = 81, normalized size = 1.25 \begin {gather*} \frac {\sqrt {x+1} \left (\frac {15 (1-x)^2}{(x+1)^2}+\frac {25 (1-x)}{x+1}+8\right )}{\sqrt {1-x} \left (\frac {1-x}{x+1}+1\right )^2}+15 \tan ^{-1}\left (\frac {\sqrt {1-x}}{\sqrt {x+1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 + x]*(8 + (15*(1 - x)^2)/(1 + x)^2 + (25*(1 - x))/(1 + x)))/(Sqrt[1 - x]*(1 + (1 - x)/(1 + x))^2) + 15
*ArcTan[Sqrt[1 - x]/Sqrt[1 + x]]

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fricas [A]  time = 1.11, size = 58, normalized size = 0.89 \begin {gather*} \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 {gather*}

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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giac [A]  time = 1.01, size = 42, normalized size = 0.65 \begin {gather*} \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 {gather*}

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))

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maple [A]  time = 0.02, size = 77, normalized size = 1.18 \begin {gather*} -\frac {15 \sqrt {\left (x +1\right ) \left (-x +1\right )}\, \arcsin \relax (x )}{2 \sqrt {x +1}\, \sqrt {-x +1}}-\frac {\left (x^{3}+8 x^{2}-17 x -24\right ) \sqrt {\left (x +1\right ) \left (-x +1\right )}}{2 \sqrt {-\left (x +1\right ) \left (x -1\right )}\, \sqrt {-x +1}\, \sqrt {x +1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.97, size = 56, normalized size = 0.86 \begin {gather*} -\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 \relax (x) \end {gather*}

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (x+1\right )}^{5/2}}{{\left (1-x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 1)^(5/2)/(1 - x)^(3/2),x)

[Out]

int((x + 1)^(5/2)/(1 - x)^(3/2), x)

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sympy [A]  time = 7.76, size = 139, normalized size = 2.14 \begin {gather*} \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 {gather*}

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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