\(\int \frac {x \sqrt {1+x}}{(1-x)^{5/2}} \, dx\) [830]

Optimal result

Integrand size = 18, antiderivative size = 41 \[ \int \frac {x \sqrt {1+x}}{(1-x)^{5/2}} \, dx=-\frac {2 \sqrt {1+x}}{\sqrt {1-x}}+\frac {(1+x)^{3/2}}{3 (1-x)^{3/2}}+\arcsin (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {79, 49, 41, 222} \[ \int \frac {x \sqrt {1+x}}{(1-x)^{5/2}} \, dx=\arcsin (x)+\frac {(x+1)^{3/2}}{3 (1-x)^{3/2}}-\frac {2 \sqrt {x+1}}{\sqrt {1-x}} \]

[In]

Int[(x*Sqrt[1 + x])/(1 - x)^(5/2),x]

[Out]

(-2*Sqrt[1 + x])/Sqrt[1 - x] + (1 + x)^(3/2)/(3*(1 - x)^(3/2)) + ArcSin[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 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

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 {(1+x)^{3/2}}{3 (1-x)^{3/2}}-\int \frac {\sqrt {1+x}}{(1-x)^{3/2}} \, dx \\ & = -\frac {2 \sqrt {1+x}}{\sqrt {1-x}}+\frac {(1+x)^{3/2}}{3 (1-x)^{3/2}}+\int \frac {1}{\sqrt {1-x} \sqrt {1+x}} \, dx \\ & = -\frac {2 \sqrt {1+x}}{\sqrt {1-x}}+\frac {(1+x)^{3/2}}{3 (1-x)^{3/2}}+\int \frac {1}{\sqrt {1-x^2}} \, dx \\ & = -\frac {2 \sqrt {1+x}}{\sqrt {1-x}}+\frac {(1+x)^{3/2}}{3 (1-x)^{3/2}}+\sin ^{-1}(x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.12 \[ \int \frac {x \sqrt {1+x}}{(1-x)^{5/2}} \, dx=\frac {\sqrt {1+x} (-5+7 x)}{3 (1-x)^{3/2}}+2 \arctan \left (\frac {\sqrt {1+x}}{\sqrt {1-x}}\right ) \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(68\) vs. \(2(31)=62\).

Time = 0.59 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.68

method	result	size
default	\(\frac {\left (3 \arcsin \left (x \right ) x^{2}-6 \arcsin \left (x \right ) x +7 x \sqrt {-x^{2}+1}+3 \arcsin \left (x \right )-5 \sqrt {-x^{2}+1}\right ) \sqrt {1-x}\, \sqrt {1+x}}{3 \left (-1+x \right )^{2} \sqrt {-x^{2}+1}}\)	\(69\)
risch	\(-\frac {\left (7 x^{2}+2 x -5\right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{3 \left (-1+x \right ) \sqrt {-\left (-1+x \right ) \left (1+x \right )}\, \sqrt {1-x}\, \sqrt {1+x}}+\frac {\arcsin \left (x \right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{\sqrt {1-x}\, \sqrt {1+x}}\)	\(78\)

[In]

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

[Out]

1/3*(3*arcsin(x)*x^2-6*arcsin(x)*x+7*x*(-x^2+1)^(1/2)+3*arcsin(x)-5*(-x^2+1)^(1/2))*(1-x)^(1/2)*(1+x)^(1/2)/(-
1+x)^2/(-x^2+1)^(1/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (31) = 62\).

Time = 0.23 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.73 \[ \int \frac {x \sqrt {1+x}}{(1-x)^{5/2}} \, dx=-\frac {5 \, x^{2} - {\left (7 \, x - 5\right )} \sqrt {x + 1} \sqrt {-x + 1} + 6 \, {\left (x^{2} - 2 \, x + 1\right )} \arctan \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) - 10 \, x + 5}{3 \, {\left (x^{2} - 2 \, x + 1\right )}} \]

[In]

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

[Out]

-1/3*(5*x^2 - (7*x - 5)*sqrt(x + 1)*sqrt(-x + 1) + 6*(x^2 - 2*x + 1)*arctan((sqrt(x + 1)*sqrt(-x + 1) - 1)/x)
- 10*x + 5)/(x^2 - 2*x + 1)

Sympy [F]

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

[In]

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

[Out]

Integral(x*sqrt(x + 1)/(1 - x)**(5/2), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.93 \[ \int \frac {x \sqrt {1+x}}{(1-x)^{5/2}} \, dx=\frac {{\left (7 \, x - 5\right )} \sqrt {x + 1} \sqrt {-x + 1}}{3 \, {\left (x - 1\right )}^{2}} + 2 \, \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {x + 1}\right ) \]

[In]

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

[Out]

1/3*(7*x - 5)*sqrt(x + 1)*sqrt(-x + 1)/(x - 1)^2 + 2*arcsin(1/2*sqrt(2)*sqrt(x + 1))

Mupad [F(-1)]

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

[In]

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

[Out]

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