3.1116 \(\int \frac{(1-x)^{7/2}}{(1+x)^{3/2}} \, dx\)

Optimal. Leaf size=85 \[ -\frac{2 (1-x)^{7/2}}{\sqrt{x+1}}-\frac{7}{3} \sqrt{x+1} (1-x)^{5/2}-\frac{35}{6} \sqrt{x+1} (1-x)^{3/2}-\frac{35}{2} \sqrt{x+1} \sqrt{1-x}-\frac{35}{2} \sin ^{-1}(x) \]

[Out]

(-2*(1 - x)^(7/2))/Sqrt[1 + x] - (35*Sqrt[1 - x]*Sqrt[1 + x])/2 - (35*(1 - x)^(3/2)*Sqrt[1 + x])/6 - (7*(1 - x
)^(5/2)*Sqrt[1 + x])/3 - (35*ArcSin[x])/2

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Rubi [A]  time = 0.0162829, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {47, 50, 41, 216} \[ -\frac{2 (1-x)^{7/2}}{\sqrt{x+1}}-\frac{7}{3} \sqrt{x+1} (1-x)^{5/2}-\frac{35}{6} \sqrt{x+1} (1-x)^{3/2}-\frac{35}{2} \sqrt{x+1} \sqrt{1-x}-\frac{35}{2} \sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(1 - x)^(7/2)/(1 + x)^(3/2),x]

[Out]

(-2*(1 - x)^(7/2))/Sqrt[1 + x] - (35*Sqrt[1 - x]*Sqrt[1 + x])/2 - (35*(1 - x)^(3/2)*Sqrt[1 + x])/6 - (7*(1 - x
)^(5/2)*Sqrt[1 + x])/3 - (35*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)^{7/2}}{(1+x)^{3/2}} \, dx &=-\frac{2 (1-x)^{7/2}}{\sqrt{1+x}}-7 \int \frac{(1-x)^{5/2}}{\sqrt{1+x}} \, dx\\ &=-\frac{2 (1-x)^{7/2}}{\sqrt{1+x}}-\frac{7}{3} (1-x)^{5/2} \sqrt{1+x}-\frac{35}{3} \int \frac{(1-x)^{3/2}}{\sqrt{1+x}} \, dx\\ &=-\frac{2 (1-x)^{7/2}}{\sqrt{1+x}}-\frac{35}{6} (1-x)^{3/2} \sqrt{1+x}-\frac{7}{3} (1-x)^{5/2} \sqrt{1+x}-\frac{35}{2} \int \frac{\sqrt{1-x}}{\sqrt{1+x}} \, dx\\ &=-\frac{2 (1-x)^{7/2}}{\sqrt{1+x}}-\frac{35}{2} \sqrt{1-x} \sqrt{1+x}-\frac{35}{6} (1-x)^{3/2} \sqrt{1+x}-\frac{7}{3} (1-x)^{5/2} \sqrt{1+x}-\frac{35}{2} \int \frac{1}{\sqrt{1-x} \sqrt{1+x}} \, dx\\ &=-\frac{2 (1-x)^{7/2}}{\sqrt{1+x}}-\frac{35}{2} \sqrt{1-x} \sqrt{1+x}-\frac{35}{6} (1-x)^{3/2} \sqrt{1+x}-\frac{7}{3} (1-x)^{5/2} \sqrt{1+x}-\frac{35}{2} \int \frac{1}{\sqrt{1-x^2}} \, dx\\ &=-\frac{2 (1-x)^{7/2}}{\sqrt{1+x}}-\frac{35}{2} \sqrt{1-x} \sqrt{1+x}-\frac{35}{6} (1-x)^{3/2} \sqrt{1+x}-\frac{7}{3} (1-x)^{5/2} \sqrt{1+x}-\frac{35}{2} \sin ^{-1}(x)\\ \end{align*}

Mathematica [C]  time = 0.0121307, size = 37, normalized size = 0.44 \[ -\frac{(1-x)^{9/2} \, _2F_1\left (\frac{3}{2},\frac{9}{2};\frac{11}{2};\frac{1-x}{2}\right )}{9 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - x)^(7/2)/(1 + x)^(3/2),x]

[Out]

-((1 - x)^(9/2)*Hypergeometric2F1[3/2, 9/2, 11/2, (1 - x)/2])/(9*Sqrt[2])

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Maple [A]  time = 0.015, size = 84, normalized size = 1. \begin{align*}{\frac{2\,{x}^{4}-15\,{x}^{3}+68\,{x}^{2}+111\,x-166}{6}\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{35\,\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)^(7/2)/(1+x)^(3/2),x)

[Out]

1/6*(2*x^4-15*x^3+68*x^2+111*x-166)/(-(1+x)*(-1+x))^(1/2)*((1+x)*(1-x))^(1/2)/(1-x)^(1/2)/(1+x)^(1/2)-35/2*((1
+x)*(1-x))^(1/2)/(1+x)^(1/2)/(1-x)^(1/2)*arcsin(x)

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Maxima [A]  time = 1.54404, size = 95, normalized size = 1.12 \begin{align*} \frac{x^{4}}{3 \, \sqrt{-x^{2} + 1}} - \frac{5 \, x^{3}}{2 \, \sqrt{-x^{2} + 1}} + \frac{34 \, x^{2}}{3 \, \sqrt{-x^{2} + 1}} + \frac{37 \, x}{2 \, \sqrt{-x^{2} + 1}} - \frac{83}{3 \, \sqrt{-x^{2} + 1}} - \frac{35}{2} \, \arcsin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-x)^(7/2)/(1+x)^(3/2),x, algorithm="maxima")

[Out]

1/3*x^4/sqrt(-x^2 + 1) - 5/2*x^3/sqrt(-x^2 + 1) + 34/3*x^2/sqrt(-x^2 + 1) + 37/2*x/sqrt(-x^2 + 1) - 83/3/sqrt(
-x^2 + 1) - 35/2*arcsin(x)

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Fricas [A]  time = 1.85903, size = 189, normalized size = 2.22 \begin{align*} -\frac{{\left (2 \, x^{3} - 13 \, x^{2} + 55 \, x + 166\right )} \sqrt{x + 1} \sqrt{-x + 1} - 210 \,{\left (x + 1\right )} \arctan \left (\frac{\sqrt{x + 1} \sqrt{-x + 1} - 1}{x}\right ) + 166 \, x + 166}{6 \,{\left (x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-x)^(7/2)/(1+x)^(3/2),x, algorithm="fricas")

[Out]

-1/6*((2*x^3 - 13*x^2 + 55*x + 166)*sqrt(x + 1)*sqrt(-x + 1) - 210*(x + 1)*arctan((sqrt(x + 1)*sqrt(-x + 1) -
1)/x) + 166*x + 166)/(x + 1)

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Sympy [A]  time = 61.9999, size = 207, normalized size = 2.44 \begin{align*} \begin{cases} 35 i \operatorname{acosh}{\left (\frac{\sqrt{2} \sqrt{x + 1}}{2} \right )} - \frac{i \left (x + 1\right )^{\frac{7}{2}}}{3 \sqrt{x - 1}} + \frac{23 i \left (x + 1\right )^{\frac{5}{2}}}{6 \sqrt{x - 1}} - \frac{125 i \left (x + 1\right )^{\frac{3}{2}}}{6 \sqrt{x - 1}} + \frac{13 i \sqrt{x + 1}}{\sqrt{x - 1}} + \frac{32 i}{\sqrt{x - 1} \sqrt{x + 1}} & \text{for}\: \frac{\left |{x + 1}\right |}{2} > 1 \\- 35 \operatorname{asin}{\left (\frac{\sqrt{2} \sqrt{x + 1}}{2} \right )} + \frac{\left (x + 1\right )^{\frac{7}{2}}}{3 \sqrt{1 - x}} - \frac{23 \left (x + 1\right )^{\frac{5}{2}}}{6 \sqrt{1 - x}} + \frac{125 \left (x + 1\right )^{\frac{3}{2}}}{6 \sqrt{1 - x}} - \frac{13 \sqrt{x + 1}}{\sqrt{1 - x}} - \frac{32}{\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)**(7/2)/(1+x)**(3/2),x)

[Out]

Piecewise((35*I*acosh(sqrt(2)*sqrt(x + 1)/2) - I*(x + 1)**(7/2)/(3*sqrt(x - 1)) + 23*I*(x + 1)**(5/2)/(6*sqrt(
x - 1)) - 125*I*(x + 1)**(3/2)/(6*sqrt(x - 1)) + 13*I*sqrt(x + 1)/sqrt(x - 1) + 32*I/(sqrt(x - 1)*sqrt(x + 1))
, Abs(x + 1)/2 > 1), (-35*asin(sqrt(2)*sqrt(x + 1)/2) + (x + 1)**(7/2)/(3*sqrt(1 - x)) - 23*(x + 1)**(5/2)/(6*
sqrt(1 - x)) + 125*(x + 1)**(3/2)/(6*sqrt(1 - x)) - 13*sqrt(x + 1)/sqrt(1 - x) - 32/(sqrt(1 - x)*sqrt(x + 1)),
True))

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Giac [A]  time = 1.30713, size = 109, normalized size = 1.28 \begin{align*} -\frac{1}{6} \,{\left ({\left (2 \, x - 17\right )}{\left (x + 1\right )} + 87\right )} \sqrt{x + 1} \sqrt{-x + 1} + \frac{8 \,{\left (\sqrt{2} - \sqrt{-x + 1}\right )}}{\sqrt{x + 1}} - \frac{8 \, \sqrt{x + 1}}{\sqrt{2} - \sqrt{-x + 1}} - 35 \, \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)^(7/2)/(1+x)^(3/2),x, algorithm="giac")

[Out]

-1/6*((2*x - 17)*(x + 1) + 87)*sqrt(x + 1)*sqrt(-x + 1) + 8*(sqrt(2) - sqrt(-x + 1))/sqrt(x + 1) - 8*sqrt(x +
1)/(sqrt(2) - sqrt(-x + 1)) - 35*arcsin(1/2*sqrt(2)*sqrt(x + 1))