Integrand size = 20, antiderivative size = 61 \[ \int \frac {x^2 \sqrt {1+x}}{(1-x)^{5/2}} \, dx=-3 \sqrt {1-x} \sqrt {1+x}+\frac {(1+x)^{3/2}}{3 (1-x)^{3/2}}-\frac {2 (1+x)^{3/2}}{\sqrt {1-x}}+3 \arcsin (x) \] Output:
-3*(1-x)^(1/2)*(1+x)^(1/2)+1/3*(1+x)^(3/2)/(1-x)^(3/2)-2*(1+x)^(3/2)/(1-x) ^(1/2)+3*arcsin(x)
Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.84 \[ \int \frac {x^2 \sqrt {1+x}}{(1-x)^{5/2}} \, dx=-\frac {\sqrt {1+x} \left (14-19 x+3 x^2\right )}{3 (1-x)^{3/2}}+6 \arctan \left (\frac {\sqrt {1+x}}{\sqrt {1-x}}\right ) \] Input:
Integrate[(x^2*Sqrt[1 + x])/(1 - x)^(5/2),x]
Output:
-1/3*(Sqrt[1 + x]*(14 - 19*x + 3*x^2))/(1 - x)^(3/2) + 6*ArcTan[Sqrt[1 + x ]/Sqrt[1 - x]]
Time = 0.16 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {100, 27, 57, 50, 223}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^2 \sqrt {x+1}}{(1-x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 100 |
\(\displaystyle \frac {(x+1)^{3/2}}{3 (1-x)^{3/2}}-\frac {1}{3} \int \frac {3 (x+1)^{3/2}}{(1-x)^{3/2}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(x+1)^{3/2}}{3 (1-x)^{3/2}}-\int \frac {(x+1)^{3/2}}{(1-x)^{3/2}}dx\) |
\(\Big \downarrow \) 57 |
\(\displaystyle 3 \int \frac {\sqrt {x+1}}{\sqrt {1-x}}dx-\frac {2 (x+1)^{3/2}}{\sqrt {1-x}}+\frac {(x+1)^{3/2}}{3 (1-x)^{3/2}}\) |
\(\Big \downarrow \) 50 |
\(\displaystyle 3 \left (\int \frac {1}{\sqrt {1-x^2}}dx-\sqrt {1-x^2}\right )-\frac {2 (x+1)^{3/2}}{\sqrt {1-x}}+\frac {(x+1)^{3/2}}{3 (1-x)^{3/2}}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle 3 \left (\arcsin (x)-\sqrt {1-x^2}\right )-\frac {2 (x+1)^{3/2}}{\sqrt {1-x}}+\frac {(x+1)^{3/2}}{3 (1-x)^{3/2}}\) |
Input:
Int[(x^2*Sqrt[1 + x])/(1 - x)^(5/2),x]
Output:
(1 + x)^(3/2)/(3*(1 - x)^(3/2)) - (2*(1 + x)^(3/2))/Sqrt[1 - x] + 3*(-Sqrt [1 - x^2] + ArcSin[x])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a *c + b*d*x^2)^m/(2*d*m), x] + Simp[a Int[(a*c + b*d*x^2)^n, x], x] /; Fre eQ[{a, b, c, d, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[m - n, 1] && GtQ[m, 0 ] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))
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] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & & 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]
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d *e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1)) Int[(c + d*x)^ (n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x , x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] || !SumSimplerQ[p, 1])))
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]
Time = 0.16 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.28
method | result | size |
default | \(\frac {\left (9 \arcsin \left (x \right ) x^{2}-3 \sqrt {-x^{2}+1}\, x^{2}-18 \arcsin \left (x \right ) x +19 x \sqrt {-x^{2}+1}+9 \arcsin \left (x \right )-14 \sqrt {-x^{2}+1}\right ) \sqrt {1+x}}{3 \left (1-x \right )^{\frac {3}{2}} \sqrt {-x^{2}+1}}\) | \(78\) |
risch | \(\frac {\left (3 x^{3}-16 x^{2}-5 x +14\right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{3 \left (-1+x \right ) \sqrt {-\left (1+x \right ) \left (-1+x \right )}\, \sqrt {1-x}\, \sqrt {1+x}}+\frac {3 \arcsin \left (x \right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{\sqrt {1-x}\, \sqrt {1+x}}\) | \(84\) |
Input:
int(x^2*(1+x)^(1/2)/(1-x)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/3*(9*arcsin(x)*x^2-3*(-x^2+1)^(1/2)*x^2-18*arcsin(x)*x+19*x*(-x^2+1)^(1/ 2)+9*arcsin(x)-14*(-x^2+1)^(1/2))*(1+x)^(1/2)/(1-x)^(3/2)/(-x^2+1)^(1/2)
Time = 0.10 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.23 \[ \int \frac {x^2 \sqrt {1+x}}{(1-x)^{5/2}} \, dx=-\frac {14 \, x^{2} + {\left (3 \, x^{2} - 19 \, x + 14\right )} \sqrt {x + 1} \sqrt {-x + 1} + 18 \, {\left (x^{2} - 2 \, x + 1\right )} \arctan \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) - 28 \, x + 14}{3 \, {\left (x^{2} - 2 \, x + 1\right )}} \] Input:
integrate(x^2*(1+x)^(1/2)/(1-x)^(5/2),x, algorithm="fricas")
Output:
-1/3*(14*x^2 + (3*x^2 - 19*x + 14)*sqrt(x + 1)*sqrt(-x + 1) + 18*(x^2 - 2* x + 1)*arctan((sqrt(x + 1)*sqrt(-x + 1) - 1)/x) - 28*x + 14)/(x^2 - 2*x + 1)
\[ \int \frac {x^2 \sqrt {1+x}}{(1-x)^{5/2}} \, dx=\int \frac {x^{2} \sqrt {x + 1}}{\left (1 - x\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(x**2*(1+x)**(1/2)/(1-x)**(5/2),x)
Output:
Integral(x**2*sqrt(x + 1)/(1 - x)**(5/2), x)
Timed out. \[ \int \frac {x^2 \sqrt {1+x}}{(1-x)^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(x^2*(1+x)^(1/2)/(1-x)^(5/2),x, algorithm="maxima")
Output:
Timed out
Time = 0.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.72 \[ \int \frac {x^2 \sqrt {1+x}}{(1-x)^{5/2}} \, dx=-\frac {{\left ({\left (3 \, x - 22\right )} {\left (x + 1\right )} + 36\right )} \sqrt {x + 1} \sqrt {-x + 1}}{3 \, {\left (x - 1\right )}^{2}} + 6 \, \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {x + 1}\right ) \] Input:
integrate(x^2*(1+x)^(1/2)/(1-x)^(5/2),x, algorithm="giac")
Output:
-1/3*((3*x - 22)*(x + 1) + 36)*sqrt(x + 1)*sqrt(-x + 1)/(x - 1)^2 + 6*arcs in(1/2*sqrt(2)*sqrt(x + 1))
Timed out. \[ \int \frac {x^2 \sqrt {1+x}}{(1-x)^{5/2}} \, dx=\int \frac {x^2\,\sqrt {x+1}}{{\left (1-x\right )}^{5/2}} \,d x \] Input:
int((x^2*(x + 1)^(1/2))/(1 - x)^(5/2),x)
Output:
int((x^2*(x + 1)^(1/2))/(1 - x)^(5/2), x)
Time = 0.18 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.30 \[ \int \frac {x^2 \sqrt {1+x}}{(1-x)^{5/2}} \, dx=\frac {-18 \sqrt {1-x}\, \mathit {asin} \left (\frac {\sqrt {1-x}}{\sqrt {2}}\right ) x +18 \sqrt {1-x}\, \mathit {asin} \left (\frac {\sqrt {1-x}}{\sqrt {2}}\right )+3 \sqrt {x +1}\, x^{2}-19 \sqrt {x +1}\, x +14 \sqrt {x +1}}{3 \sqrt {1-x}\, \left (x -1\right )} \] Input:
int(x^2*(1+x)^(1/2)/(1-x)^(5/2),x)
Output:
( - 18*sqrt( - x + 1)*asin(sqrt( - x + 1)/sqrt(2))*x + 18*sqrt( - x + 1)*a sin(sqrt( - x + 1)/sqrt(2)) + 3*sqrt(x + 1)*x**2 - 19*sqrt(x + 1)*x + 14*s qrt(x + 1))/(3*sqrt( - x + 1)*(x - 1))