\(\int \frac {x^2 (1+x)^{3/2}}{\sqrt {1-x}} \, dx\) [725]

Optimal result

Integrand size = 20, antiderivative size = 88 \[ \int \frac {x^2 (1+x)^{3/2}}{\sqrt {1-x}} \, dx=-\frac {7}{8} \sqrt {1-x} \sqrt {1+x}-\frac {7}{24} \sqrt {1-x} (1+x)^{3/2}-\frac {1}{6} \sqrt {1-x} (1+x)^{5/2}-\frac {1}{4} \sqrt {1-x} x (1+x)^{5/2}+\frac {7 \arcsin (x)}{8} \]

[Out]

7/8*arcsin(x)-7/24*(1-x)^(1/2)*(1+x)^(3/2)-1/6*(1-x)^(1/2)*(1+x)^(5/2)-1/4*x*(1+x)^(5/2)*(1-x)^(1/2)-7/8*(1-x)
^(1/2)*(1+x)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {92, 81, 52, 41, 222} \[ \int \frac {x^2 (1+x)^{3/2}}{\sqrt {1-x}} \, dx=\frac {7 \arcsin (x)}{8}-\frac {1}{4} \sqrt {1-x} x (x+1)^{5/2}-\frac {1}{6} \sqrt {1-x} (x+1)^{5/2}-\frac {7}{24} \sqrt {1-x} (x+1)^{3/2}-\frac {7}{8} \sqrt {1-x} \sqrt {x+1} \]

[In]

Int[(x^2*(1 + x)^(3/2))/Sqrt[1 - x],x]

[Out]

(-7*Sqrt[1 - x]*Sqrt[1 + x])/8 - (7*Sqrt[1 - x]*(1 + x)^(3/2))/24 - (Sqrt[1 - x]*(1 + x)^(5/2))/6 - (Sqrt[1 -
x]*x*(1 + x)^(5/2))/4 + (7*ArcSin[x])/8

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 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 81

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

Rule 92

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a + b*x
)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 3))), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

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}{4} \sqrt {1-x} x (1+x)^{5/2}-\frac {1}{4} \int \frac {(-1-2 x) (1+x)^{3/2}}{\sqrt {1-x}} \, dx \\ & = -\frac {1}{6} \sqrt {1-x} (1+x)^{5/2}-\frac {1}{4} \sqrt {1-x} x (1+x)^{5/2}+\frac {7}{12} \int \frac {(1+x)^{3/2}}{\sqrt {1-x}} \, dx \\ & = -\frac {7}{24} \sqrt {1-x} (1+x)^{3/2}-\frac {1}{6} \sqrt {1-x} (1+x)^{5/2}-\frac {1}{4} \sqrt {1-x} x (1+x)^{5/2}+\frac {7}{8} \int \frac {\sqrt {1+x}}{\sqrt {1-x}} \, dx \\ & = -\frac {7}{8} \sqrt {1-x} \sqrt {1+x}-\frac {7}{24} \sqrt {1-x} (1+x)^{3/2}-\frac {1}{6} \sqrt {1-x} (1+x)^{5/2}-\frac {1}{4} \sqrt {1-x} x (1+x)^{5/2}+\frac {7}{8} \int \frac {1}{\sqrt {1-x} \sqrt {1+x}} \, dx \\ & = -\frac {7}{8} \sqrt {1-x} \sqrt {1+x}-\frac {7}{24} \sqrt {1-x} (1+x)^{3/2}-\frac {1}{6} \sqrt {1-x} (1+x)^{5/2}-\frac {1}{4} \sqrt {1-x} x (1+x)^{5/2}+\frac {7}{8} \int \frac {1}{\sqrt {1-x^2}} \, dx \\ & = -\frac {7}{8} \sqrt {1-x} \sqrt {1+x}-\frac {7}{24} \sqrt {1-x} (1+x)^{3/2}-\frac {1}{6} \sqrt {1-x} (1+x)^{5/2}-\frac {1}{4} \sqrt {1-x} x (1+x)^{5/2}+\frac {7}{8} \sin ^{-1}(x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.72 \[ \int \frac {x^2 (1+x)^{3/2}}{\sqrt {1-x}} \, dx=-\frac {\sqrt {1-x} \left (32+53 x+37 x^2+22 x^3+6 x^4\right )}{24 \sqrt {1+x}}-\frac {7}{4} \arctan \left (\frac {\sqrt {1-x}}{\sqrt {1+x}}\right ) \]

[In]

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

[Out]

-1/24*(Sqrt[1 - x]*(32 + 53*x + 37*x^2 + 22*x^3 + 6*x^4))/Sqrt[1 + x] - (7*ArcTan[Sqrt[1 - x]/Sqrt[1 + x]])/4

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.91

method	result	size
default	\(\frac {\sqrt {1+x}\, \sqrt {1-x}\, \left (-6 x^{3} \sqrt {-x^{2}+1}-16 x^{2} \sqrt {-x^{2}+1}-21 x \sqrt {-x^{2}+1}+21 \arcsin \left (x \right )-32 \sqrt {-x^{2}+1}\right )}{24 \sqrt {-x^{2}+1}}\)	\(80\)
risch	\(\frac {\left (6 x^{3}+16 x^{2}+21 x +32\right ) \left (-1+x \right ) \sqrt {1+x}\, \sqrt {\left (1+x \right ) \left (1-x \right )}}{24 \sqrt {-\left (-1+x \right ) \left (1+x \right )}\, \sqrt {1-x}}+\frac {7 \arcsin \left (x \right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{8 \sqrt {1-x}\, \sqrt {1+x}}\)	\(82\)

[In]

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

[Out]

1/24*(1+x)^(1/2)*(1-x)^(1/2)*(-6*x^3*(-x^2+1)^(1/2)-16*x^2*(-x^2+1)^(1/2)-21*x*(-x^2+1)^(1/2)+21*arcsin(x)-32*
(-x^2+1)^(1/2))/(-x^2+1)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.59 \[ \int \frac {x^2 (1+x)^{3/2}}{\sqrt {1-x}} \, dx=-\frac {1}{24} \, {\left (6 \, x^{3} + 16 \, x^{2} + 21 \, x + 32\right )} \sqrt {x + 1} \sqrt {-x + 1} - \frac {7}{4} \, \arctan \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) \]

[In]

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

[Out]

-1/24*(6*x^3 + 16*x^2 + 21*x + 32)*sqrt(x + 1)*sqrt(-x + 1) - 7/4*arctan((sqrt(x + 1)*sqrt(-x + 1) - 1)/x)

Sympy [F]

\[ \int \frac {x^2 (1+x)^{3/2}}{\sqrt {1-x}} \, dx=\int \frac {x^{2} \left (x + 1\right )^{\frac {3}{2}}}{\sqrt {1 - x}}\, dx \]

[In]

integrate(x**2*(1+x)**(3/2)/(1-x)**(1/2),x)

[Out]

Integral(x**2*(x + 1)**(3/2)/sqrt(1 - x), x)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.64 \[ \int \frac {x^2 (1+x)^{3/2}}{\sqrt {1-x}} \, dx=-\frac {1}{4} \, \sqrt {-x^{2} + 1} x^{3} - \frac {2}{3} \, \sqrt {-x^{2} + 1} x^{2} - \frac {7}{8} \, \sqrt {-x^{2} + 1} x - \frac {4}{3} \, \sqrt {-x^{2} + 1} + \frac {7}{8} \, \arcsin \left (x\right ) \]

[In]

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

[Out]

-1/4*sqrt(-x^2 + 1)*x^3 - 2/3*sqrt(-x^2 + 1)*x^2 - 7/8*sqrt(-x^2 + 1)*x - 4/3*sqrt(-x^2 + 1) + 7/8*arcsin(x)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.52 \[ \int \frac {x^2 (1+x)^{3/2}}{\sqrt {1-x}} \, dx=-\frac {1}{24} \, {\left ({\left (2 \, {\left (3 \, x + 2\right )} {\left (x + 1\right )} + 7\right )} {\left (x + 1\right )} + 21\right )} \sqrt {x + 1} \sqrt {-x + 1} + \frac {7}{4} \, \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {x + 1}\right ) \]

[In]

integrate(x^2*(1+x)^(3/2)/(1-x)^(1/2),x, algorithm="giac")

[Out]

-1/24*((2*(3*x + 2)*(x + 1) + 7)*(x + 1) + 21)*sqrt(x + 1)*sqrt(-x + 1) + 7/4*arcsin(1/2*sqrt(2)*sqrt(x + 1))

Mupad [F(-1)]

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

[In]

int((x^2*(x + 1)^(3/2))/(1 - x)^(1/2),x)

[Out]

int((x^2*(x + 1)^(3/2))/(1 - x)^(1/2), x)