\(\int \sqrt {3 x^2-3 x^4+x^6} \, dx\) [126]

Optimal result

Integrand size = 18, antiderivative size = 86 \[ \int \sqrt {3 x^2-3 x^4+x^6} \, dx=-\frac {\left (3-2 x^2\right ) \sqrt {3 x^2-3 x^4+x^6}}{8 x}-\frac {3 \sqrt {3 x^2-3 x^4+x^6} \text {arcsinh}\left (\frac {3-2 x^2}{\sqrt {3}}\right )}{16 x \sqrt {3-3 x^2+x^4}} \]

[Out]

-1/8*(-2*x^2+3)*(x^6-3*x^4+3*x^2)^(1/2)/x-3/16*arcsinh(1/3*(-2*x^2+3)*3^(1/2))*(x^6-3*x^4+3*x^2)^(1/2)/x/(x^4-
3*x^2+3)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {1917, 1121, 626, 633, 221} \[ \int \sqrt {3 x^2-3 x^4+x^6} \, dx=-\frac {3 \sqrt {x^6-3 x^4+3 x^2} \text {arcsinh}\left (\frac {3-2 x^2}{\sqrt {3}}\right )}{16 x \sqrt {x^4-3 x^2+3}}-\frac {\sqrt {x^6-3 x^4+3 x^2} \left (3-2 x^2\right )}{8 x} \]

[In]

Int[Sqrt[3*x^2 - 3*x^4 + x^6],x]

[Out]

-1/8*((3 - 2*x^2)*Sqrt[3*x^2 - 3*x^4 + x^6])/x - (3*Sqrt[3*x^2 - 3*x^4 + x^6]*ArcSinh[(3 - 2*x^2)/Sqrt[3]])/(1
6*x*Sqrt[3 - 3*x^2 + x^4])

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 626

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x)*((a + b*x + c*x^2)^p/(2*c*(2*p + 1
))), x] - Dist[p*((b^2 - 4*a*c)/(2*c*(2*p + 1))), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 633

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*(-4*(c/(b^2 - 4*a*c)))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 1121

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

Rule 1917

Int[Sqrt[(b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.)], x_Symbol] :> Dist[Sqrt[a*x^q + b*x^n + c*x^(
2*n - q)]/(x^(q/2)*Sqrt[a + b*x^(n - q) + c*x^(2*(n - q))]), Int[x^(q/2)*Sqrt[a + b*x^(n - q) + c*x^(2*(n - q)
)], x], x] /; FreeQ[{a, b, c, n, q}, x] && EqQ[r, 2*n - q] && PosQ[n - q]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {3 x^2-3 x^4+x^6} \int x \sqrt {3-3 x^2+x^4} \, dx}{x \sqrt {3-3 x^2+x^4}} \\ & = \frac {\sqrt {3 x^2-3 x^4+x^6} \text {Subst}\left (\int \sqrt {3-3 x+x^2} \, dx,x,x^2\right )}{2 x \sqrt {3-3 x^2+x^4}} \\ & = -\frac {\left (3-2 x^2\right ) \sqrt {3 x^2-3 x^4+x^6}}{8 x}+\frac {\left (3 \sqrt {3 x^2-3 x^4+x^6}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {3-3 x+x^2}} \, dx,x,x^2\right )}{16 x \sqrt {3-3 x^2+x^4}} \\ & = -\frac {\left (3-2 x^2\right ) \sqrt {3 x^2-3 x^4+x^6}}{8 x}+\frac {\left (\sqrt {3} \sqrt {3 x^2-3 x^4+x^6}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{3}}} \, dx,x,-3+2 x^2\right )}{16 x \sqrt {3-3 x^2+x^4}} \\ & = -\frac {\left (3-2 x^2\right ) \sqrt {3 x^2-3 x^4+x^6}}{8 x}-\frac {3 \sqrt {3 x^2-3 x^4+x^6} \sinh ^{-1}\left (\frac {3-2 x^2}{\sqrt {3}}\right )}{16 x \sqrt {3-3 x^2+x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.93 \[ \int \sqrt {3 x^2-3 x^4+x^6} \, dx=\frac {x \left (-18+30 x^2-18 x^4+4 x^6-3 \sqrt {3-3 x^2+x^4} \log \left (3-2 x^2+2 \sqrt {3-3 x^2+x^4}\right )\right )}{16 \sqrt {x^2 \left (3-3 x^2+x^4\right )}} \]

[In]

Integrate[Sqrt[3*x^2 - 3*x^4 + x^6],x]

[Out]

(x*(-18 + 30*x^2 - 18*x^4 + 4*x^6 - 3*Sqrt[3 - 3*x^2 + x^4]*Log[3 - 2*x^2 + 2*Sqrt[3 - 3*x^2 + x^4]]))/(16*Sqr
t[x^2*(3 - 3*x^2 + x^4)])

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.72

method	result	size
pseudoelliptic	\(\frac {4 \sqrt {x^{2} \left (x^{4}-3 x^{2}+3\right )}\, x^{2}+3 \,\operatorname {arcsinh}\left (\frac {\sqrt {3}\, \left (2 x^{2}-3\right )}{3}\right ) x -6 \sqrt {x^{2} \left (x^{4}-3 x^{2}+3\right )}}{16 x}\)	\(62\)
trager	\(\frac {\left (2 x^{2}-3\right ) \sqrt {x^{6}-3 x^{4}+3 x^{2}}}{8 x}-\frac {3 \ln \left (\frac {-2 x^{3}+2 \sqrt {x^{6}-3 x^{4}+3 x^{2}}+3 x}{x}\right )}{16}\)	\(64\)
risch	\(\frac {\left (2 x^{2}-3\right ) \sqrt {x^{2} \left (x^{4}-3 x^{2}+3\right )}}{8 x}+\frac {3 \,\operatorname {arcsinh}\left (\frac {2 \sqrt {3}\, \left (x^{2}-\frac {3}{2}\right )}{3}\right ) \sqrt {x^{2} \left (x^{4}-3 x^{2}+3\right )}}{16 \sqrt {x^{4}-3 x^{2}+3}\, x}\)	\(74\)
default	\(\frac {\sqrt {x^{6}-3 x^{4}+3 x^{2}}\, \left (4 \sqrt {x^{4}-3 x^{2}+3}\, x^{2}+3 \,\operatorname {arcsinh}\left (\frac {\sqrt {3}\, \left (2 x^{2}-3\right )}{3}\right )-6 \sqrt {x^{4}-3 x^{2}+3}\right )}{16 x \sqrt {x^{4}-3 x^{2}+3}}\)	\(81\)

[In]

int((x^6-3*x^4+3*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/16/x*(4*(x^2*(x^4-3*x^2+3))^(1/2)*x^2+3*arcsinh(1/3*3^(1/2)*(2*x^2-3))*x-6*(x^2*(x^4-3*x^2+3))^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.81 \[ \int \sqrt {3 x^2-3 x^4+x^6} \, dx=-\frac {12 \, x \log \left (-\frac {2 \, x^{3} - 3 \, x - 2 \, \sqrt {x^{6} - 3 \, x^{4} + 3 \, x^{2}}}{x}\right ) - 8 \, \sqrt {x^{6} - 3 \, x^{4} + 3 \, x^{2}} {\left (2 \, x^{2} - 3\right )} - 9 \, x}{64 \, x} \]

[In]

integrate((x^6-3*x^4+3*x^2)^(1/2),x, algorithm="fricas")

[Out]

-1/64*(12*x*log(-(2*x^3 - 3*x - 2*sqrt(x^6 - 3*x^4 + 3*x^2))/x) - 8*sqrt(x^6 - 3*x^4 + 3*x^2)*(2*x^2 - 3) - 9*
x)/x

Sympy [F]

\[ \int \sqrt {3 x^2-3 x^4+x^6} \, dx=\int \sqrt {x^{6} - 3 x^{4} + 3 x^{2}}\, dx \]

[In]

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

[Out]

Integral(sqrt(x**6 - 3*x**4 + 3*x**2), x)

Maxima [F]

\[ \int \sqrt {3 x^2-3 x^4+x^6} \, dx=\int { \sqrt {x^{6} - 3 \, x^{4} + 3 \, x^{2}} \,d x } \]

[In]

integrate((x^6-3*x^4+3*x^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(x^6 - 3*x^4 + 3*x^2), x)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.80 \[ \int \sqrt {3 x^2-3 x^4+x^6} \, dx=\frac {1}{16} \, {\left (2 \, \sqrt {x^{4} - 3 \, x^{2} + 3} {\left (2 \, x^{2} - 3\right )} - 3 \, \log \left (-2 \, x^{2} + 2 \, \sqrt {x^{4} - 3 \, x^{2} + 3} + 3\right )\right )} \mathrm {sgn}\left (x\right ) + \frac {3}{16} \, {\left (2 \, \sqrt {3} + \log \left (2 \, \sqrt {3} + 3\right )\right )} \mathrm {sgn}\left (x\right ) \]

[In]

integrate((x^6-3*x^4+3*x^2)^(1/2),x, algorithm="giac")

[Out]

1/16*(2*sqrt(x^4 - 3*x^2 + 3)*(2*x^2 - 3) - 3*log(-2*x^2 + 2*sqrt(x^4 - 3*x^2 + 3) + 3))*sgn(x) + 3/16*(2*sqrt
(3) + log(2*sqrt(3) + 3))*sgn(x)

Mupad [F(-1)]

Timed out. \[ \int \sqrt {3 x^2-3 x^4+x^6} \, dx=\int \sqrt {x^6-3\,x^4+3\,x^2} \,d x \]

[In]

int((3*x^2 - 3*x^4 + x^6)^(1/2),x)

[Out]

int((3*x^2 - 3*x^4 + x^6)^(1/2), x)