\(\int x (1+2 x^2) \sqrt {-1+2 x^2+2 x^4} \, dx\) [179]

Optimal result

Integrand size = 25, antiderivative size = 20 \[ \int x \left (1+2 x^2\right ) \sqrt {-1+2 x^2+2 x^4} \, dx=\frac {1}{6} \left (-1+2 x^2+2 x^4\right )^{3/2} \]

[Out]

1/6*(2*x^4+2*x^2-1)^(3/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {1261, 643} \[ \int x \left (1+2 x^2\right ) \sqrt {-1+2 x^2+2 x^4} \, dx=\frac {1}{6} \left (2 x^4+2 x^2-1\right )^{3/2} \]

[In]

Int[x*(1 + 2*x^2)*Sqrt[-1 + 2*x^2 + 2*x^4],x]

[Out]

(-1 + 2*x^2 + 2*x^4)^(3/2)/6

Rule 643

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

Rule 1261

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int (1+2 x) \sqrt {-1+2 x+2 x^2} \, dx,x,x^2\right ) \\ & = \frac {1}{6} \left (-1+2 x^2+2 x^4\right )^{3/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int x \left (1+2 x^2\right ) \sqrt {-1+2 x^2+2 x^4} \, dx=\frac {1}{6} \left (-1+2 x^2+2 x^4\right )^{3/2} \]

[In]

Integrate[x*(1 + 2*x^2)*Sqrt[-1 + 2*x^2 + 2*x^4],x]

[Out]

(-1 + 2*x^2 + 2*x^4)^(3/2)/6

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85

method	result	size
gosper	\(\frac {\left (2 x^{4}+2 x^{2}-1\right )^{\frac {3}{2}}}{6}\)	\(17\)
default	\(\frac {\left (2 x^{4}+2 x^{2}-1\right )^{\frac {3}{2}}}{6}\)	\(17\)
risch	\(\frac {\left (2 x^{4}+2 x^{2}-1\right )^{\frac {3}{2}}}{6}\)	\(17\)
pseudoelliptic	\(\frac {\left (2 x^{4}+2 x^{2}-1\right )^{\frac {3}{2}}}{6}\)	\(17\)
trager	\(\left (\frac {1}{3} x^{4}+\frac {1}{3} x^{2}-\frac {1}{6}\right ) \sqrt {2 x^{4}+2 x^{2}-1}\)	\(28\)

[In]

int(x*(2*x^2+1)*(2*x^4+2*x^2-1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/6*(2*x^4+2*x^2-1)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int x \left (1+2 x^2\right ) \sqrt {-1+2 x^2+2 x^4} \, dx=\frac {1}{6} \, {\left (2 \, x^{4} + 2 \, x^{2} - 1\right )}^{\frac {3}{2}} \]

[In]

integrate(x*(2*x^2+1)*(2*x^4+2*x^2-1)^(1/2),x, algorithm="fricas")

[Out]

1/6*(2*x^4 + 2*x^2 - 1)^(3/2)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (15) = 30\).

Time = 0.09 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.80 \[ \int x \left (1+2 x^2\right ) \sqrt {-1+2 x^2+2 x^4} \, dx=\frac {x^{4} \sqrt {2 x^{4} + 2 x^{2} - 1}}{3} + \frac {x^{2} \sqrt {2 x^{4} + 2 x^{2} - 1}}{3} - \frac {\sqrt {2 x^{4} + 2 x^{2} - 1}}{6} \]

[In]

integrate(x*(2*x**2+1)*(2*x**4+2*x**2-1)**(1/2),x)

[Out]

x**4*sqrt(2*x**4 + 2*x**2 - 1)/3 + x**2*sqrt(2*x**4 + 2*x**2 - 1)/3 - sqrt(2*x**4 + 2*x**2 - 1)/6

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int x \left (1+2 x^2\right ) \sqrt {-1+2 x^2+2 x^4} \, dx=\frac {1}{6} \, {\left (2 \, x^{4} + 2 \, x^{2} - 1\right )}^{\frac {3}{2}} \]

[In]

integrate(x*(2*x^2+1)*(2*x^4+2*x^2-1)^(1/2),x, algorithm="maxima")

[Out]

1/6*(2*x^4 + 2*x^2 - 1)^(3/2)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int x \left (1+2 x^2\right ) \sqrt {-1+2 x^2+2 x^4} \, dx=\frac {1}{6} \, {\left (2 \, x^{4} + 2 \, x^{2} - 1\right )}^{\frac {3}{2}} \]

[In]

integrate(x*(2*x^2+1)*(2*x^4+2*x^2-1)^(1/2),x, algorithm="giac")

[Out]

1/6*(2*x^4 + 2*x^2 - 1)^(3/2)

Mupad [B] (verification not implemented)

Time = 5.08 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int x \left (1+2 x^2\right ) \sqrt {-1+2 x^2+2 x^4} \, dx=\frac {{\left (2\,x^4+2\,x^2-1\right )}^{3/2}}{6} \]

[In]

int(x*(2*x^2 + 1)*(2*x^2 + 2*x^4 - 1)^(1/2),x)

[Out]

(2*x^2 + 2*x^4 - 1)^(3/2)/6