\(\int \frac {(1-x)^{5/2}}{(1+x)^{3/2}} \, dx\) [1117]

Optimal result

Integrand size = 17, antiderivative size = 65 \[ \int \frac {(1-x)^{5/2}}{(1+x)^{3/2}} \, dx=-\frac {2 (1-x)^{5/2}}{\sqrt {1+x}}-\frac {15}{2} \sqrt {1-x} \sqrt {1+x}-\frac {5}{2} (1-x)^{3/2} \sqrt {1+x}-\frac {15 \arcsin (x)}{2} \]

[Out]

-15/2*arcsin(x)-2*(1-x)^(5/2)/(1+x)^(1/2)-5/2*(1-x)^(3/2)*(1+x)^(1/2)-15/2*(1-x)^(1/2)*(1+x)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {49, 52, 41, 222} \[ \int \frac {(1-x)^{5/2}}{(1+x)^{3/2}} \, dx=-\frac {15 \arcsin (x)}{2}-\frac {2 (1-x)^{5/2}}{\sqrt {x+1}}-\frac {5}{2} \sqrt {x+1} (1-x)^{3/2}-\frac {15}{2} \sqrt {x+1} \sqrt {1-x} \]

[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 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 49

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 52

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 222

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*} \text {integral}& = -\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 [A] (verified)

Time = 0.06 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.75 \[ \int \frac {(1-x)^{5/2}}{(1+x)^{3/2}} \, dx=\frac {\sqrt {1-x} \left (-24-7 x+x^2\right )}{2 \sqrt {1+x}}+15 \arctan \left (\frac {\sqrt {1-x^2}}{-1+x}\right ) \]

[In]

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

[Out]

(Sqrt[1 - x]*(-24 - 7*x + x^2))/(2*Sqrt[1 + x]) + 15*ArcTan[Sqrt[1 - x^2]/(-1 + x)]

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.18

method	result	size
risch	\(-\frac {\left (x^{3}-8 x^{2}-17 x +24\right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{2 \sqrt {-\left (-1+x \right ) \left (1+x \right )}\, \sqrt {1-x}\, \sqrt {1+x}}-\frac {15 \sqrt {\left (1+x \right ) \left (1-x \right )}\, \arcsin \left (x \right )}{2 \sqrt {1+x}\, \sqrt {1-x}}\)	\(77\)

[In]

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

[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)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.89 \[ \int \frac {(1-x)^{5/2}}{(1+x)^{3/2}} \, dx=\frac {{\left (x^{2} - 7 \, x - 24\right )} \sqrt {x + 1} \sqrt {-x + 1} + 30 \, {\left (x + 1\right )} \arctan \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) - 24 \, x - 24}{2 \, {\left (x + 1\right )}} \]

[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)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 6.23 (sec) , antiderivative size = 167, normalized size of antiderivative = 2.57 \[ \int \frac {(1-x)^{5/2}}{(1+x)^{3/2}} \, dx=\begin {cases} 15 i \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {x + 1}}{2} \right )} + \frac {i \left (x + 1\right )^{\frac {5}{2}}}{2 \sqrt {x - 1}} - \frac {11 i \left (x + 1\right )^{\frac {3}{2}}}{2 \sqrt {x - 1}} + \frac {i \sqrt {x + 1}}{\sqrt {x - 1}} + \frac {16 i}{\sqrt {x - 1} \sqrt {x + 1}} & \text {for}\: \left |{x + 1}\right | > 2 \\- 15 \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {x + 1}}{2} \right )} - \frac {\left (x + 1\right )^{\frac {5}{2}}}{2 \sqrt {1 - x}} + \frac {11 \left (x + 1\right )^{\frac {3}{2}}}{2 \sqrt {1 - x}} - \frac {\sqrt {x + 1}}{\sqrt {1 - x}} - \frac {16}{\sqrt {1 - x} \sqrt {x + 1}} & \text {otherwise} \end {cases} \]

[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), (-15*asin(sqrt(2)*sqrt(
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) - 16/
(sqrt(1 - x)*sqrt(x + 1)), True))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.86 \[ \int \frac {(1-x)^{5/2}}{(1+x)^{3/2}} \, dx=-\frac {x^{3}}{2 \, \sqrt {-x^{2} + 1}} + \frac {4 \, x^{2}}{\sqrt {-x^{2} + 1}} + \frac {17 \, x}{2 \, \sqrt {-x^{2} + 1}} - \frac {12}{\sqrt {-x^{2} + 1}} - \frac {15}{2} \, \arcsin \left (x\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.12 \[ \int \frac {(1-x)^{5/2}}{(1+x)^{3/2}} \, dx=\frac {1}{2} \, \sqrt {x + 1} {\left (x - 8\right )} \sqrt {-x + 1} + \frac {4 \, {\left (\sqrt {2} - \sqrt {-x + 1}\right )}}{\sqrt {x + 1}} - \frac {4 \, \sqrt {x + 1}}{\sqrt {2} - \sqrt {-x + 1}} - 15 \, \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {x + 1}\right ) \]

[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))

Mupad [F(-1)]

Timed out. \[ \int \frac {(1-x)^{5/2}}{(1+x)^{3/2}} \, dx=\int \frac {{\left (1-x\right )}^{5/2}}{{\left (x+1\right )}^{3/2}} \,d x \]

[In]

int((1 - x)^(5/2)/(x + 1)^(3/2),x)

[Out]

int((1 - x)^(5/2)/(x + 1)^(3/2), x)