Integrand size = 11, antiderivative size = 199 \[ \int \frac {1}{\sqrt [4]{-2+3 x^2}} \, dx=\frac {2 x \sqrt [4]{-2+3 x^2}}{\sqrt {2}+\sqrt {-2+3 x^2}}-\frac {2 \sqrt [4]{2} \sqrt {\frac {x^2}{\left (\sqrt {2}+\sqrt {-2+3 x^2}\right )^2}} \left (\sqrt {2}+\sqrt {-2+3 x^2}\right ) E\left (2 \arctan \left (\frac {\sqrt [4]{-2+3 x^2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{\sqrt {3} x}+\frac {\sqrt [4]{2} \sqrt {\frac {x^2}{\left (\sqrt {2}+\sqrt {-2+3 x^2}\right )^2}} \left (\sqrt {2}+\sqrt {-2+3 x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{-2+3 x^2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{\sqrt {3} x} \] Output:
2*x*(3*x^2-2)^(1/4)/(2^(1/2)+(3*x^2-2)^(1/2))-2/3*2^(1/4)*(x^2/(2^(1/2)+(3 *x^2-2)^(1/2))^2)^(1/2)*(2^(1/2)+(3*x^2-2)^(1/2))*EllipticE(sin(2*arctan(1 /2*(3*x^2-2)^(1/4)*2^(3/4))),1/2*2^(1/2))*3^(1/2)/x+1/3*2^(1/4)*(x^2/(2^(1 /2)+(3*x^2-2)^(1/2))^2)^(1/2)*(2^(1/2)+(3*x^2-2)^(1/2))*InverseJacobiAM(2* arctan(1/2*(3*x^2-2)^(1/4)*2^(3/4)),1/2*2^(1/2))*3^(1/2)/x
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 4.66 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.22 \[ \int \frac {1}{\sqrt [4]{-2+3 x^2}} \, dx=\frac {x \sqrt [4]{1-\frac {3 x^2}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},\frac {3 x^2}{2}\right )}{\sqrt [4]{-2+3 x^2}} \] Input:
Integrate[(-2 + 3*x^2)^(-1/4),x]
Output:
(x*(1 - (3*x^2)/2)^(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2, (3*x^2)/2])/(-2 + 3*x^2)^(1/4)
Time = 0.27 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.13, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {228, 27, 834, 27, 761, 1510}
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.
\(\displaystyle \int \frac {1}{\sqrt [4]{3 x^2-2}} \, dx\) |
\(\Big \downarrow \) 228 |
\(\displaystyle \frac {\sqrt {\frac {2}{3}} \sqrt {x^2} \int \frac {\sqrt {\frac {2}{3}} \sqrt {3 x^2-2}}{\sqrt {x^2}}d\sqrt [4]{3 x^2-2}}{x}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 \sqrt {x^2} \int \frac {\sqrt {3 x^2-2}}{\sqrt {3} \sqrt {x^2}}d\sqrt [4]{3 x^2-2}}{\sqrt {3} x}\) |
\(\Big \downarrow \) 834 |
\(\displaystyle \frac {2 \sqrt {x^2} \left (\sqrt {2} \int \frac {1}{\sqrt {3} \sqrt {x^2}}d\sqrt [4]{3 x^2-2}-\sqrt {2} \int \frac {\sqrt {2}-\sqrt {3 x^2-2}}{\sqrt {6} \sqrt {x^2}}d\sqrt [4]{3 x^2-2}\right )}{\sqrt {3} x}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 \sqrt {x^2} \left (\sqrt {2} \int \frac {1}{\sqrt {3} \sqrt {x^2}}d\sqrt [4]{3 x^2-2}-\int \frac {\sqrt {2}-\sqrt {3 x^2-2}}{\sqrt {3} \sqrt {x^2}}d\sqrt [4]{3 x^2-2}\right )}{\sqrt {3} x}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {2 \sqrt {x^2} \left (\frac {\sqrt {\frac {x^2}{\left (\sqrt {3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {3 x^2-2}+\sqrt {2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{2^{3/4} \sqrt {x^2}}-\int \frac {\sqrt {2}-\sqrt {3 x^2-2}}{\sqrt {3} \sqrt {x^2}}d\sqrt [4]{3 x^2-2}\right )}{\sqrt {3} x}\) |
\(\Big \downarrow \) 1510 |
\(\displaystyle \frac {2 \sqrt {x^2} \left (\frac {\sqrt {\frac {x^2}{\left (\sqrt {3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {3 x^2-2}+\sqrt {2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{2^{3/4} \sqrt {x^2}}-\frac {\sqrt [4]{2} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {3 x^2-2}+\sqrt {2}\right ) E\left (2 \arctan \left (\frac {\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{\sqrt {x^2}}+\frac {\sqrt {3} \sqrt {x^2} \sqrt [4]{3 x^2-2}}{\sqrt {3 x^2-2}+\sqrt {2}}\right )}{\sqrt {3} x}\) |
Input:
Int[(-2 + 3*x^2)^(-1/4),x]
Output:
(2*Sqrt[x^2]*((Sqrt[3]*Sqrt[x^2]*(-2 + 3*x^2)^(1/4))/(Sqrt[2] + Sqrt[-2 + 3*x^2]) - (2^(1/4)*Sqrt[x^2/(Sqrt[2] + Sqrt[-2 + 3*x^2])^2]*(Sqrt[2] + Sqr t[-2 + 3*x^2])*EllipticE[2*ArcTan[(-2 + 3*x^2)^(1/4)/2^(1/4)], 1/2])/Sqrt[ x^2] + (Sqrt[x^2/(Sqrt[2] + Sqrt[-2 + 3*x^2])^2]*(Sqrt[2] + Sqrt[-2 + 3*x^ 2])*EllipticF[2*ArcTan[(-2 + 3*x^2)^(1/4)/2^(1/4)], 1/2])/(2^(3/4)*Sqrt[x^ 2])))/(Sqrt[3]*x)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[2*(Sqrt[(-b)*(x^2/a)]/( b*x)) Subst[Int[x^2/Sqrt[1 - x^4/a], x], x, (a + b*x^2)^(1/4)], x] /; Fre eQ[{a, b}, x] && NegQ[a]
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S imp[1/q Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* (1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e }, x] && PosQ[c/a]
Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.20
method | result | size |
meijerg | \(\frac {2^{\frac {3}{4}} {\left (-\operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )\right )}^{\frac {1}{4}} x \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {1}{2}\right ], \left [\frac {3}{2}\right ], \frac {3 x^{2}}{2}\right )}{2 \operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )^{\frac {1}{4}}}\) | \(40\) |
Input:
int(1/(3*x^2-2)^(1/4),x,method=_RETURNVERBOSE)
Output:
1/2*2^(3/4)/signum(-1+3/2*x^2)^(1/4)*(-signum(-1+3/2*x^2))^(1/4)*x*hyperge om([1/4,1/2],[3/2],3/2*x^2)
\[ \int \frac {1}{\sqrt [4]{-2+3 x^2}} \, dx=\int { \frac {1}{{\left (3 \, x^{2} - 2\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate(1/(3*x^2-2)^(1/4),x, algorithm="fricas")
Output:
integral((3*x^2 - 2)^(-1/4), x)
Result contains complex when optimal does not.
Time = 0.39 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.14 \[ \int \frac {1}{\sqrt [4]{-2+3 x^2}} \, dx=\frac {2^{\frac {3}{4}} x e^{- \frac {i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {3 x^{2}}{2}} \right )}}{2} \] Input:
integrate(1/(3*x**2-2)**(1/4),x)
Output:
2**(3/4)*x*exp(-I*pi/4)*hyper((1/4, 1/2), (3/2,), 3*x**2/2)/2
\[ \int \frac {1}{\sqrt [4]{-2+3 x^2}} \, dx=\int { \frac {1}{{\left (3 \, x^{2} - 2\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate(1/(3*x^2-2)^(1/4),x, algorithm="maxima")
Output:
integrate((3*x^2 - 2)^(-1/4), x)
\[ \int \frac {1}{\sqrt [4]{-2+3 x^2}} \, dx=\int { \frac {1}{{\left (3 \, x^{2} - 2\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate(1/(3*x^2-2)^(1/4),x, algorithm="giac")
Output:
integrate((3*x^2 - 2)^(-1/4), x)
Time = 0.21 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.17 \[ \int \frac {1}{\sqrt [4]{-2+3 x^2}} \, dx=\frac {2^{3/4}\,x\,{\left (2-3\,x^2\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {1}{2};\ \frac {3}{2};\ \frac {3\,x^2}{2}\right )}{2\,{\left (3\,x^2-2\right )}^{1/4}} \] Input:
int(1/(3*x^2 - 2)^(1/4),x)
Output:
(2^(3/4)*x*(2 - 3*x^2)^(1/4)*hypergeom([1/4, 1/2], 3/2, (3*x^2)/2))/(2*(3* x^2 - 2)^(1/4))
\[ \int \frac {1}{\sqrt [4]{-2+3 x^2}} \, dx=\int \frac {1}{\left (3 x^{2}-2\right )^{\frac {1}{4}}}d x \] Input:
int(1/(3*x^2-2)^(1/4),x)
Output:
int(1/(3*x**2 - 2)**(1/4),x)