\(\int \frac {x^2}{(1+x)^{3/2} (1-x+x^2)^{3/2}} \, dx\) [511]

Optimal result

Integrand size = 23, antiderivative size = 23 \[ \int \frac {x^2}{(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx=-\frac {2}{3 \sqrt {1+x} \sqrt {1-x+x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {927} \[ \int \frac {x^2}{(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx=-\frac {2}{3 \sqrt {x+1} \sqrt {x^2-x+1}} \]

[In]

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

[Out]

-2/(3*Sqrt[1 + x]*Sqrt[1 - x + x^2])

Rule 927

Int[(x_)^2*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(
m + 1)*((a + b*x + c*x^2)^(p + 1)/(c*e*(m + 2*p + 3))), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b*e*(m +
p + 2) + 2*c*d*(p + 1), 0] && EqQ[b*d*(p + 1) + a*e*(m + 1), 0] && NeQ[m + 2*p + 3, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {2}{3 \sqrt {1+x} \sqrt {1-x+x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 10.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {x^2}{(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx=-\frac {2}{3 \sqrt {1+x} \sqrt {1-x+x^2}} \]

[In]

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

[Out]

-2/(3*Sqrt[1 + x]*Sqrt[1 - x + x^2])

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78

method	result	size
gosper	\(-\frac {2}{3 \sqrt {1+x}\, \sqrt {x^{2}-x +1}}\)	\(18\)
risch	\(-\frac {2}{3 \sqrt {1+x}\, \sqrt {x^{2}-x +1}}\)	\(18\)
default	\(-\frac {2 \sqrt {1+x}\, \sqrt {x^{2}-x +1}}{3 \left (x^{3}+1\right )}\)	\(25\)
elliptic	\(-\frac {2 \sqrt {\left (1+x \right ) \left (x^{2}-x +1\right )}}{3 \sqrt {1+x}\, \sqrt {x^{2}-x +1}\, \sqrt {x^{3}+1}}\)	\(39\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {x^2}{(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx=-\frac {2 \, \sqrt {x^{2} - x + 1} \sqrt {x + 1}}{3 \, {\left (x^{3} + 1\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {x^2}{(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx=-\frac {2}{3 \, \sqrt {x^{2} - x + 1} \sqrt {x + 1}} \]

[In]

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

[Out]

-2/3/(sqrt(x^2 - x + 1)*sqrt(x + 1))

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.33 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {x^2}{(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx=-\frac {2}{3\,\sqrt {x+1}\,\sqrt {x^2-x+1}} \]

[In]

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

[Out]

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