\(\int \frac {(1-x)^{5/2}}{(1+x)^{5/2}} \, dx\) [1128]

Optimal result

Integrand size = 17, antiderivative size = 63 \[ \int \frac {(1-x)^{5/2}}{(1+x)^{5/2}} \, dx=-\frac {2 (1-x)^{5/2}}{3 (1+x)^{3/2}}+\frac {10 (1-x)^{3/2}}{3 \sqrt {1+x}}+5 \sqrt {1-x} \sqrt {1+x}+5 \arcsin (x) \]

[Out]

-2/3*(1-x)^(5/2)/(1+x)^(3/2)+5*arcsin(x)+10/3*(1-x)^(3/2)/(1+x)^(1/2)+5*(1-x)^(1/2)*(1+x)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 63, 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} \[ \int \frac {(1-x)^{5/2}}{(1+x)^{5/2}} \, dx=5 \arcsin (x)-\frac {2 (1-x)^{5/2}}{3 (x+1)^{3/2}}+\frac {10 (1-x)^{3/2}}{3 \sqrt {x+1}}+5 \sqrt {x+1} \sqrt {1-x} \]

[In]

Int[(1 - x)^(5/2)/(1 + x)^(5/2),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.81 \[ \int \frac {(1-x)^{5/2}}{(1+x)^{5/2}} \, dx=\frac {\sqrt {1-x} \left (23+34 x+3 x^2\right )}{3 (1+x)^{3/2}}-10 \arctan \left (\frac {\sqrt {1-x^2}}{-1+x}\right ) \]

[In]

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

[Out]

(Sqrt[1 - x]*(23 + 34*x + 3*x^2))/(3*(1 + x)^(3/2)) - 10*ArcTan[Sqrt[1 - x^2]/(-1 + x)]

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.25

method	result	size
risch	\(-\frac {\left (3 x^{3}+31 x^{2}-11 x -23\right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{3 \left (1+x \right )^{\frac {3}{2}} \sqrt {-\left (-1+x \right ) \left (1+x \right )}\, \sqrt {1-x}}+\frac {5 \sqrt {\left (1+x \right ) \left (1-x \right )}\, \arcsin \left (x \right )}{\sqrt {1+x}\, \sqrt {1-x}}\)	\(79\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.19 \[ \int \frac {(1-x)^{5/2}}{(1+x)^{5/2}} \, dx=\frac {23 \, x^{2} + {\left (3 \, x^{2} + 34 \, x + 23\right )} \sqrt {x + 1} \sqrt {-x + 1} - 30 \, {\left (x^{2} + 2 \, x + 1\right )} \arctan \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) + 46 \, x + 23}{3 \, {\left (x^{2} + 2 \, x + 1\right )}} \]

[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)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 7.04 (sec) , antiderivative size = 162, normalized size of antiderivative = 2.57 \[ \int \frac {(1-x)^{5/2}}{(1+x)^{5/2}} \, dx=\begin {cases} \sqrt {-1 + \frac {2}{x + 1}} \left (x + 1\right ) + \frac {28 \sqrt {-1 + \frac {2}{x + 1}}}{3} - \frac {8 \sqrt {-1 + \frac {2}{x + 1}}}{3 \left (x + 1\right )} + 5 i \log {\left (\frac {1}{x + 1} \right )} + 5 i \log {\left (x + 1 \right )} + 10 \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {x + 1}}{2} \right )} & \text {for}\: \frac {1}{\left |{x + 1}\right |} > \frac {1}{2} \\i \sqrt {1 - \frac {2}{x + 1}} \left (x + 1\right ) + \frac {28 i \sqrt {1 - \frac {2}{x + 1}}}{3} - \frac {8 i \sqrt {1 - \frac {2}{x + 1}}}{3 \left (x + 1\right )} + 5 i \log {\left (\frac {1}{x + 1} \right )} - 10 i \log {\left (\sqrt {1 - \frac {2}{x + 1}} + 1 \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((sqrt(-1 + 2/(x + 1))*(x + 1) + 28*sqrt(-1 + 2/(x + 1))/3 - 8*sqrt(-1 + 2/(x + 1))/(3*(x + 1)) + 5*I
*log(1/(x + 1)) + 5*I*log(x + 1) + 10*asin(sqrt(2)*sqrt(x + 1)/2), 1/Abs(x + 1) > 1/2), (I*sqrt(1 - 2/(x + 1))
*(x + 1) + 28*I*sqrt(1 - 2/(x + 1))/3 - 8*I*sqrt(1 - 2/(x + 1))/(3*(x + 1)) + 5*I*log(1/(x + 1)) - 10*I*log(sq
rt(1 - 2/(x + 1)) + 1), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 98 vs. \(2 (47) = 94\).

Time = 0.28 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.56 \[ \int \frac {(1-x)^{5/2}}{(1+x)^{5/2}} \, dx=\frac {{\left (-x^{2} + 1\right )}^{\frac {5}{2}}}{x^{4} + 4 \, x^{3} + 6 \, x^{2} + 4 \, x + 1} - \frac {5 \, {\left (-x^{2} + 1\right )}^{\frac {3}{2}}}{3 \, {\left (x^{3} + 3 \, x^{2} + 3 \, x + 1\right )}} - \frac {10 \, \sqrt {-x^{2} + 1}}{3 \, {\left (x^{2} + 2 \, x + 1\right )}} + \frac {35 \, \sqrt {-x^{2} + 1}}{3 \, {\left (x + 1\right )}} + 5 \, \arcsin \left (x\right ) \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 115 vs. \(2 (47) = 94\).

Time = 0.32 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.83 \[ \int \frac {(1-x)^{5/2}}{(1+x)^{5/2}} \, dx=\frac {{\left (\sqrt {2} - \sqrt {-x + 1}\right )}^{3}}{6 \, {\left (x + 1\right )}^{\frac {3}{2}}} + \sqrt {x + 1} \sqrt {-x + 1} - \frac {9 \, {\left (\sqrt {2} - \sqrt {-x + 1}\right )}}{2 \, \sqrt {x + 1}} + \frac {{\left (x + 1\right )}^{\frac {3}{2}} {\left (\frac {27 \, {\left (\sqrt {2} - \sqrt {-x + 1}\right )}^{2}}{x + 1} - 1\right )}}{6 \, {\left (\sqrt {2} - \sqrt {-x + 1}\right )}^{3}} + 10 \, \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {x + 1}\right ) \]

[In]

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

[Out]

1/6*(sqrt(2) - sqrt(-x + 1))^3/(x + 1)^(3/2) + sqrt(x + 1)*sqrt(-x + 1) - 9/2*(sqrt(2) - sqrt(-x + 1))/sqrt(x
+ 1) + 1/6*(x + 1)^(3/2)*(27*(sqrt(2) - sqrt(-x + 1))^2/(x + 1) - 1)/(sqrt(2) - sqrt(-x + 1))^3 + 10*arcsin(1/
2*sqrt(2)*sqrt(x + 1))

Mupad [F(-1)]

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

[In]

int((1 - x)^(5/2)/(x + 1)^(5/2),x)

[Out]

int((1 - x)^(5/2)/(x + 1)^(5/2), x)