\(\int \frac {\sqrt {x^2} \sqrt [3]{2-6 x^2}}{\sqrt {2 x^2-3 x^4}} \, dx\) [564]

Optimal result

Integrand size = 34, antiderivative size = 620 \[ \int \frac {\sqrt {x^2} \sqrt [3]{2-6 x^2}}{\sqrt {2 x^2-3 x^4}} \, dx=-\frac {3 \sqrt {x^2} \sqrt {x^2 \left (2-3 x^2\right )}}{x \left (2^{2/3} \left (1+\sqrt {3}\right )-2^{2/3} \left (1-3 x^2\right )^{2/3}\right )}+\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt {x^2} \left (1-\left (1-3 x^2\right )^{2/3}\right ) \sqrt {\frac {\sqrt [3]{2}+\sqrt [3]{2} \left (1-3 x^2\right )^{2/3}+\sqrt [3]{2} \left (1-3 x^2\right )^{4/3}}{\left (2^{2/3} \left (1+\sqrt {3}\right )-2^{2/3} \left (1-3 x^2\right )^{2/3}\right )^2}} E\left (\arcsin \left (\frac {2^{2/3} \left (1-\sqrt {3}\right )-2^{2/3} \left (1-3 x^2\right )^{2/3}}{2^{2/3} \left (1+\sqrt {3}\right )-2^{2/3} \left (1-3 x^2\right )^{2/3}}\right )|-7-4 \sqrt {3}\right )}{2 \sqrt {2} x \sqrt {x^2 \left (2-3 x^2\right )} \sqrt {\frac {2^{2/3}-2^{2/3} \left (1-3 x^2\right )^{2/3}}{\left (2^{2/3} \left (1+\sqrt {3}\right )-2^{2/3} \left (1-3 x^2\right )^{2/3}\right )^2}}}-\frac {\sqrt {x^2} \left (1-\left (1-3 x^2\right )^{2/3}\right ) \sqrt {\frac {\sqrt [3]{2}+\sqrt [3]{2} \left (1-3 x^2\right )^{2/3}+\sqrt [3]{2} \left (1-3 x^2\right )^{4/3}}{\left (2^{2/3} \left (1+\sqrt {3}\right )-2^{2/3} \left (1-3 x^2\right )^{2/3}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {2^{2/3} \left (1-\sqrt {3}\right )-2^{2/3} \left (1-3 x^2\right )^{2/3}}{2^{2/3} \left (1+\sqrt {3}\right )-2^{2/3} \left (1-3 x^2\right )^{2/3}}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} x \sqrt {x^2 \left (2-3 x^2\right )} \sqrt {\frac {2^{2/3}-2^{2/3} \left (1-3 x^2\right )^{2/3}}{\left (2^{2/3} \left (1+\sqrt {3}\right )-2^{2/3} \left (1-3 x^2\right )^{2/3}\right )^2}}} \] Output:

-3*(x^2)^(1/2)*(x^2*(-3*x^2+2))^(1/2)/x/(2^(2/3)*(1+3^(1/2))-2^(2/3)*(-3*x 
^2+1)^(2/3))+1/4*3^(1/4)*(1/2*6^(1/2)-1/2*2^(1/2))*(x^2)^(1/2)*(1-(-3*x^2+ 
1)^(2/3))*((2^(1/3)+2^(1/3)*(-3*x^2+1)^(2/3)+2^(1/3)*(-3*x^2+1)^(4/3))/(2^ 
(2/3)*(1+3^(1/2))-2^(2/3)*(-3*x^2+1)^(2/3))^2)^(1/2)*EllipticE((2^(2/3)*(1 
-3^(1/2))-2^(2/3)*(-3*x^2+1)^(2/3))/(2^(2/3)*(1+3^(1/2))-2^(2/3)*(-3*x^2+1 
)^(2/3)),I*3^(1/2)+2*I)*2^(1/2)/x/(x^2*(-3*x^2+2))^(1/2)/((2^(2/3)-2^(2/3) 
*(-3*x^2+1)^(2/3))/(2^(2/3)*(1+3^(1/2))-2^(2/3)*(-3*x^2+1)^(2/3))^2)^(1/2) 
-1/3*(x^2)^(1/2)*(1-(-3*x^2+1)^(2/3))*((2^(1/3)+2^(1/3)*(-3*x^2+1)^(2/3)+2 
^(1/3)*(-3*x^2+1)^(4/3))/(2^(2/3)*(1+3^(1/2))-2^(2/3)*(-3*x^2+1)^(2/3))^2) 
^(1/2)*EllipticF((2^(2/3)*(1-3^(1/2))-2^(2/3)*(-3*x^2+1)^(2/3))/(2^(2/3)*( 
1+3^(1/2))-2^(2/3)*(-3*x^2+1)^(2/3)),I*3^(1/2)+2*I)*3^(3/4)/x/(x^2*(-3*x^2 
+2))^(1/2)/((2^(2/3)-2^(2/3)*(-3*x^2+1)^(2/3))/(2^(2/3)*(1+3^(1/2))-2^(2/3 
)*(-3*x^2+1)^(2/3))^2)^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 1.15 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.24 \[ \int \frac {\sqrt {x^2} \sqrt [3]{2-6 x^2}}{\sqrt {2 x^2-3 x^4}} \, dx=\frac {6 x \sqrt {x^2} \sqrt [3]{2-6 x^2} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{3},\frac {1}{2},\frac {3}{2},3 x^2,\frac {3 x^2}{2}\right )}{\sqrt {2 x^2-3 x^4} \left (6 \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{3},\frac {1}{2},\frac {3}{2},3 x^2,\frac {3 x^2}{2}\right )+x^2 \left (3 \operatorname {AppellF1}\left (\frac {3}{2},-\frac {1}{3},\frac {3}{2},\frac {5}{2},3 x^2,\frac {3 x^2}{2}\right )-4 \operatorname {AppellF1}\left (\frac {3}{2},\frac {2}{3},\frac {1}{2},\frac {5}{2},3 x^2,\frac {3 x^2}{2}\right )\right )\right )} \] Input:

Integrate[(Sqrt[x^2]*(2 - 6*x^2)^(1/3))/Sqrt[2*x^2 - 3*x^4],x]
 

Output:

(6*x*Sqrt[x^2]*(2 - 6*x^2)^(1/3)*AppellF1[1/2, -1/3, 1/2, 3/2, 3*x^2, (3*x 
^2)/2])/(Sqrt[2*x^2 - 3*x^4]*(6*AppellF1[1/2, -1/3, 1/2, 3/2, 3*x^2, (3*x^ 
2)/2] + x^2*(3*AppellF1[3/2, -1/3, 3/2, 5/2, 3*x^2, (3*x^2)/2] - 4*AppellF 
1[3/2, 2/3, 1/2, 5/2, 3*x^2, (3*x^2)/2])))
 

Rubi [A] (warning: unable to verify)

Time = 0.44 (sec) , antiderivative size = 300, normalized size of antiderivative = 0.48, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {34, 27, 1940, 1118, 27, 266, 807, 832, 759, 2416}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 34

\(\displaystyle \frac {\sqrt {x^2} \int \frac {\sqrt [3]{2} x \sqrt [3]{1-3 x^2}}{\sqrt {2 x^2-3 x^4}}dx}{x}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt [3]{2} \sqrt {x^2} \int \frac {x \sqrt [3]{1-3 x^2}}{\sqrt {2 x^2-3 x^4}}dx}{x}\)

\(\Big \downarrow \) 1940

\(\displaystyle \frac {\sqrt {x^2} \int \frac {\sqrt [3]{1-3 x^2}}{\sqrt {2 x^2-3 x^4}}dx^2}{2^{2/3} x}\)

\(\Big \downarrow \) 1118

\(\displaystyle -\frac {\sqrt {x^2} \int \frac {\sqrt {3} \sqrt [3]{1-3 x^2}}{\sqrt {1-x^4}}d\left (1-3 x^2\right )}{3\ 2^{2/3} x}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\sqrt {x^2} \int \frac {\sqrt [3]{1-3 x^2}}{\sqrt {1-x^4}}d\left (1-3 x^2\right )}{2^{2/3} \sqrt {3} x}\)

\(\Big \downarrow \) 266

\(\displaystyle -\frac {\sqrt {3} \sqrt {x^2} \int \frac {x^6}{\sqrt {1-x^{12}}}d\sqrt [3]{1-3 x^2}}{2^{2/3} x}\)

\(\Big \downarrow \) 807

\(\displaystyle -\frac {\sqrt {3} \sqrt {x^2} \int \frac {x^4}{\sqrt {1-x^6}}dx^4}{2\ 2^{2/3} x}\)

\(\Big \downarrow \) 832

\(\displaystyle -\frac {\sqrt {3} \sqrt {x^2} \left (\left (1-\sqrt {3}\right ) \int \frac {1}{\sqrt {1-x^6}}dx^4-\int \frac {3 x^2-\sqrt {3}}{\sqrt {1-x^6}}dx^4\right )}{2\ 2^{2/3} x}\)

\(\Big \downarrow \) 759

\(\displaystyle -\frac {\sqrt {3} \sqrt {x^2} \left (-\int \frac {3 x^2-\sqrt {3}}{\sqrt {1-x^6}}dx^4-\frac {2 \sqrt [4]{3} \left (1-\sqrt {3}\right ) \sqrt {2+\sqrt {3}} \sqrt {\frac {x^4-3 x^2+2}{\left (3 x^2+\sqrt {3}\right )^2}} x^2 \operatorname {EllipticF}\left (\arcsin \left (\frac {3 x^2-\sqrt {3}}{3 x^2+\sqrt {3}}\right ),-7-4 \sqrt {3}\right )}{\sqrt {\frac {x^2}{\left (3 x^2+\sqrt {3}\right )^2}} \sqrt {1-x^6}}\right )}{2\ 2^{2/3} x}\)

\(\Big \downarrow \) 2416

\(\displaystyle -\frac {\sqrt {3} \sqrt {x^2} \left (-\frac {2 \sqrt [4]{3} \left (1-\sqrt {3}\right ) \sqrt {2+\sqrt {3}} \sqrt {\frac {x^4-3 x^2+2}{\left (3 x^2+\sqrt {3}\right )^2}} x^2 \operatorname {EllipticF}\left (\arcsin \left (\frac {3 x^2-\sqrt {3}}{3 x^2+\sqrt {3}}\right ),-7-4 \sqrt {3}\right )}{\sqrt {\frac {x^2}{\left (3 x^2+\sqrt {3}\right )^2}} \sqrt {1-x^6}}-\frac {3^{3/4} \sqrt {2-\sqrt {3}} \sqrt {\frac {x^4-3 x^2+2}{\left (3 x^2+\sqrt {3}\right )^2}} x^2 E\left (\arcsin \left (\frac {3 x^2-\sqrt {3}}{3 x^2+\sqrt {3}}\right )|-7-4 \sqrt {3}\right )}{\sqrt {\frac {x^2}{\left (3 x^2+\sqrt {3}\right )^2}} \sqrt {1-x^6}}+\frac {2 \sqrt {1-x^6}}{3 x^2+\sqrt {3}}\right )}{2\ 2^{2/3} x}\)

Input:

Int[(Sqrt[x^2]*(2 - 6*x^2)^(1/3))/Sqrt[2*x^2 - 3*x^4],x]
 

Output:

-1/2*(Sqrt[3]*Sqrt[x^2]*((2*Sqrt[1 - x^6])/(Sqrt[3] + 3*x^2) - (3^(3/4)*Sq 
rt[2 - Sqrt[3]]*x^2*Sqrt[(2 - 3*x^2 + x^4)/(Sqrt[3] + 3*x^2)^2]*EllipticE[ 
ArcSin[(-Sqrt[3] + 3*x^2)/(Sqrt[3] + 3*x^2)], -7 - 4*Sqrt[3]])/(Sqrt[x^2/( 
Sqrt[3] + 3*x^2)^2]*Sqrt[1 - x^6]) - (2*3^(1/4)*(1 - Sqrt[3])*Sqrt[2 + Sqr 
t[3]]*x^2*Sqrt[(2 - 3*x^2 + x^4)/(Sqrt[3] + 3*x^2)^2]*EllipticF[ArcSin[(-S 
qrt[3] + 3*x^2)/(Sqrt[3] + 3*x^2)], -7 - 4*Sqrt[3]])/(Sqrt[x^2/(Sqrt[3] + 
3*x^2)^2]*Sqrt[1 - x^6])))/(2^(2/3)*x)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 34

Int[(u_.)*((a_.)*(x_)^(m_))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a*x^m)^F 
racPart[p]/x^(m*FracPart[p]))   Int[u*x^(m*p), x], x] /; FreeQ[{a, m, p}, x 
] &&  !IntegerQ[p]
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 759

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], 
s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 + Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s 
*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[s* 
((s + r*x)/((1 + Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 - Sqrt[3])*s 
+ r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x] & 
& PosQ[a]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 832

Int[(x_)/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3] 
], s = Denom[Rt[b/a, 3]]}, Simp[(-(1 - Sqrt[3]))*(s/r)   Int[1/Sqrt[a + b*x 
^3], x], x] + Simp[1/r   Int[((1 - Sqrt[3])*s + r*x)/Sqrt[a + b*x^3], x], x 
]] /; FreeQ[{a, b}, x] && PosQ[a]
 

rule 1118

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> Simp[1/e   Subst[Int[x^m*(a - b^2/(4*c) + (c*x^2)/e^2)^p, x], x, 
d + e*x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1940

Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_) 
^(n_))^(q_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1) 
*(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x] /; 
FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] &&  !IntegerQ[p] && NeQ[k, j] && I 
ntegerQ[Simplify[j/n]] && IntegerQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 
 1)/n]] && NeQ[n^2, 1]
 

rule 2416

Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = N 
umer[Simplify[(1 - Sqrt[3])*(d/c)]], s = Denom[Simplify[(1 - Sqrt[3])*(d/c) 
]]}, Simp[2*d*s^3*(Sqrt[a + b*x^3]/(a*r^2*((1 + Sqrt[3])*s + r*x))), x] - S 
imp[3^(1/4)*Sqrt[2 - Sqrt[3]]*d*s*(s + r*x)*(Sqrt[(s^2 - r*s*x + r^2*x^2)/( 
(1 + Sqrt[3])*s + r*x)^2]/(r^2*Sqrt[a + b*x^3]*Sqrt[s*((s + r*x)/((1 + Sqrt 
[3])*s + r*x)^2)]))*EllipticE[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3]) 
*s + r*x)], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b, c, d}, x] && PosQ[a] && Eq 
Q[b*c^3 - 2*(5 - 3*Sqrt[3])*a*d^3, 0]
 

Maple [F]

\[\int \frac {\sqrt {x^{2}}\, \left (-6 x^{2}+2\right )^{\frac {1}{3}}}{\sqrt {-3 x^{4}+2 x^{2}}}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int \frac {\sqrt {x^2} \sqrt [3]{2-6 x^2}}{\sqrt {2 x^2-3 x^4}} \, dx=\int { \frac {\sqrt {x^{2}} {\left (-6 \, x^{2} + 2\right )}^{\frac {1}{3}}}{\sqrt {-3 \, x^{4} + 2 \, x^{2}}} \,d x } \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {\sqrt {x^2} \sqrt [3]{2-6 x^2}}{\sqrt {2 x^2-3 x^4}} \, dx=\sqrt [3]{2} \int \frac {\sqrt [3]{1 - 3 x^{2}} \sqrt {x^{2}}}{\sqrt {- 3 x^{4} + 2 x^{2}}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {\sqrt {x^2} \sqrt [3]{2-6 x^2}}{\sqrt {2 x^2-3 x^4}} \, dx=\int { \frac {\sqrt {x^{2}} {\left (-6 \, x^{2} + 2\right )}^{\frac {1}{3}}}{\sqrt {-3 \, x^{4} + 2 \, x^{2}}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {\sqrt {x^2} \sqrt [3]{2-6 x^2}}{\sqrt {2 x^2-3 x^4}} \, dx=\int { \frac {\sqrt {x^{2}} {\left (-6 \, x^{2} + 2\right )}^{\frac {1}{3}}}{\sqrt {-3 \, x^{4} + 2 \, x^{2}}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {x^2} \sqrt [3]{2-6 x^2}}{\sqrt {2 x^2-3 x^4}} \, dx=\int \frac {{\left (2-6\,x^2\right )}^{1/3}\,\sqrt {x^2}}{\sqrt {2\,x^2-3\,x^4}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt {x^2} \sqrt [3]{2-6 x^2}}{\sqrt {2 x^2-3 x^4}} \, dx=-2^{\frac {1}{3}} \left (\int \frac {\left (-3 x^{2}+1\right )^{\frac {1}{3}} \sqrt {-3 x^{2}+2}}{3 x^{2}-2}d x \right ) \] Input:

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

Output:

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