\(\int \frac {x^2}{\sqrt {1+x} \sqrt {1-x+x^2}} \, dx\) [504]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.01 (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} \[ \int \frac {x^2}{\sqrt {1+x} \sqrt {1-x+x^2}} \, dx=\frac {2}{3} \sqrt {x+1} \sqrt {x^2-x+1} \]

[In]

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

[Out]

(2*Sqrt[1 + x]*Sqrt[1 - x + x^2])/3

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 \[ \int \frac {x^2}{\sqrt {1+x} \sqrt {1-x+x^2}} \, dx=\frac {2}{3} \sqrt {1+x} \sqrt {1-x+x^2} \]

[In]

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

[Out]

(2*Sqrt[1 + x]*Sqrt[1 - x + x^2])/3

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {x^2}{\sqrt {1+x} \sqrt {1-x+x^2}} \, dx=\frac {2}{3} \, \sqrt {x^{2} - x + 1} \sqrt {x + 1} \]

[In]

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

[Out]

2/3*sqrt(x^2 - x + 1)*sqrt(x + 1)

Sympy [F]

\[ \int \frac {x^2}{\sqrt {1+x} \sqrt {1-x+x^2}} \, dx=\int \frac {x^{2}}{\sqrt {x + 1} \sqrt {x^{2} - x + 1}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {x^2}{\sqrt {1+x} \sqrt {1-x+x^2}} \, dx=\frac {2 \, {\left (x^{3} + 1\right )}}{3 \, \sqrt {x^{2} - x + 1} \sqrt {x + 1}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {x^2}{\sqrt {1+x} \sqrt {1-x+x^2}} \, dx=\frac {2}{3} \, \sqrt {{\left (x + 1\right )}^{2} - 3 \, x} \sqrt {x + 1} \]

[In]

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

[Out]

2/3*sqrt((x + 1)^2 - 3*x)*sqrt(x + 1)

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.39 \[ \int \frac {x^2}{\sqrt {1+x} \sqrt {1-x+x^2}} \, dx=\frac {2\,\sqrt {x^3+1}}{3} \]

[In]

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

[Out]

(2*(x^3 + 1)^(1/2))/3