Integrand size = 15, antiderivative size = 223 \[ \int \frac {1}{x^2 \sqrt [4]{-2-3 x^2}} \, dx=\frac {\left (-2-3 x^2\right )^{3/4}}{2 x}+\frac {3 x \sqrt [4]{-2-3 x^2}}{2 \left (\sqrt {2}+\sqrt {-2-3 x^2}\right )}+\frac {\sqrt {3} \sqrt {-\frac {x^2}{\left (\sqrt {2}+\sqrt {-2-3 x^2}\right )^2}} \left (\sqrt {2}+\sqrt {-2-3 x^2}\right ) E\left (2 \arctan \left (\frac {\sqrt [4]{-2-3 x^2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{2^{3/4} x}-\frac {\sqrt {3} \sqrt {-\frac {x^2}{\left (\sqrt {2}+\sqrt {-2-3 x^2}\right )^2}} \left (\sqrt {2}+\sqrt {-2-3 x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{-2-3 x^2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{2\ 2^{3/4} x} \] Output:
1/2*(-3*x^2-2)^(3/4)/x+3*x*(-3*x^2-2)^(1/4)/(2*2^(1/2)+2*(-3*x^2-2)^(1/2)) +1/2*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-1/4*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 = 4.77 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.21 \[ \int \frac {1}{x^2 \sqrt [4]{-2-3 x^2}} \, dx=-\frac {\sqrt [4]{1+\frac {3 x^2}{2}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {1}{2},-\frac {3 x^2}{2}\right )}{x \sqrt [4]{-2-3 x^2}} \] Input:
Integrate[1/(x^2*(-2 - 3*x^2)^(1/4)),x]
Output:
-(((1 + (3*x^2)/2)^(1/4)*Hypergeometric2F1[-1/2, 1/4, 1/2, (-3*x^2)/2])/(x *(-2 - 3*x^2)^(1/4)))
Time = 0.29 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.14, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {264, 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.
\(\displaystyle \int \frac {1}{x^2 \sqrt [4]{-3 x^2-2}} \, dx\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {3}{4} \int \frac {1}{\sqrt [4]{-3 x^2-2}}dx+\frac {\left (-3 x^2-2\right )^{3/4}}{2 x}\) |
\(\Big \downarrow \) 228 |
\(\displaystyle \frac {\left (-3 x^2-2\right )^{3/4}}{2 x}-\frac {\sqrt {\frac {3}{2}} \sqrt {-x^2} \int \frac {\sqrt {\frac {2}{3}} \sqrt {-3 x^2-2}}{\sqrt {-x^2}}d\sqrt [4]{-3 x^2-2}}{2 x}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (-3 x^2-2\right )^{3/4}}{2 x}-\frac {\sqrt {3} \sqrt {-x^2} \int \frac {\sqrt {-3 x^2-2}}{\sqrt {3} \sqrt {-x^2}}d\sqrt [4]{-3 x^2-2}}{2 x}\) |
\(\Big \downarrow \) 834 |
\(\displaystyle \frac {\left (-3 x^2-2\right )^{3/4}}{2 x}-\frac {\sqrt {3} \sqrt {-x^2} \left (\sqrt {2} \int \frac {1}{\sqrt {3} \sqrt {-x^2}}d\sqrt [4]{-3 x^2-2}-\sqrt {2} \int \frac {\sqrt {2}-\sqrt {-3 x^2-2}}{\sqrt {6} \sqrt {-x^2}}d\sqrt [4]{-3 x^2-2}\right )}{2 x}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (-3 x^2-2\right )^{3/4}}{2 x}-\frac {\sqrt {3} \sqrt {-x^2} \left (\sqrt {2} \int \frac {1}{\sqrt {3} \sqrt {-x^2}}d\sqrt [4]{-3 x^2-2}-\int \frac {\sqrt {2}-\sqrt {-3 x^2-2}}{\sqrt {3} \sqrt {-x^2}}d\sqrt [4]{-3 x^2-2}\right )}{2 x}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {\left (-3 x^2-2\right )^{3/4}}{2 x}-\frac {\sqrt {3} \sqrt {-x^2} \left (\frac {\sqrt {-\frac {x^2}{\left (\sqrt {-3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {-3 x^2-2}+\sqrt {2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{-3 x^2-2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{2^{3/4} \sqrt {-x^2}}-\int \frac {\sqrt {2}-\sqrt {-3 x^2-2}}{\sqrt {3} \sqrt {-x^2}}d\sqrt [4]{-3 x^2-2}\right )}{2 x}\) |
\(\Big \downarrow \) 1510 |
\(\displaystyle \frac {\left (-3 x^2-2\right )^{3/4}}{2 x}-\frac {\sqrt {3} \sqrt {-x^2} \left (\frac {\sqrt {-\frac {x^2}{\left (\sqrt {-3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {-3 x^2-2}+\sqrt {2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{-3 x^2-2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{2^{3/4} \sqrt {-x^2}}-\frac {\sqrt [4]{2} \sqrt {-\frac {x^2}{\left (\sqrt {-3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {-3 x^2-2}+\sqrt {2}\right ) E\left (2 \arctan \left (\frac {\sqrt [4]{-3 x^2-2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{\sqrt {-x^2}}+\frac {\sqrt {3} \sqrt {-x^2} \sqrt [4]{-3 x^2-2}}{\sqrt {-3 x^2-2}+\sqrt {2}}\right )}{2 x}\) |
Input:
Int[1/(x^2*(-2 - 3*x^2)^(1/4)),x]
Output:
(-2 - 3*x^2)^(3/4)/(2*x) - (Sqrt[3]*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] + Sqrt[-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])))/(2*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[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
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 = 0.37 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.10
method | result | size |
meijerg | \(\frac {\left (-1\right )^{\frac {3}{4}} 2^{\frac {3}{4}} \operatorname {hypergeom}\left (\left [-\frac {1}{2}, \frac {1}{4}\right ], \left [\frac {1}{2}\right ], -\frac {3 x^{2}}{2}\right )}{2 x}\) | \(23\) |
risch | \(-\frac {3 x^{2}+2}{2 x \left (-3 x^{2}-2\right )^{\frac {1}{4}}}-\frac {3 \left (-1\right )^{\frac {3}{4}} 2^{\frac {3}{4}} x \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {1}{2}\right ], \left [\frac {3}{2}\right ], -\frac {3 x^{2}}{2}\right )}{8}\) | \(43\) |
Input:
int(1/x^2/(-3*x^2-2)^(1/4),x,method=_RETURNVERBOSE)
Output:
1/2*(-1)^(3/4)*2^(3/4)/x*hypergeom([-1/2,1/4],[1/2],-3/2*x^2)
\[ \int \frac {1}{x^2 \sqrt [4]{-2-3 x^2}} \, dx=\int { \frac {1}{{\left (-3 \, x^{2} - 2\right )}^{\frac {1}{4}} x^{2}} \,d x } \] Input:
integrate(1/x^2/(-3*x^2-2)^(1/4),x, algorithm="fricas")
Output:
1/2*(2*x*integral(-3/4*(-3*x^2 - 2)^(3/4)/(3*x^2 + 2), x) + (-3*x^2 - 2)^( 3/4))/x
Result contains complex when optimal does not.
Time = 0.46 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.16 \[ \int \frac {1}{x^2 \sqrt [4]{-2-3 x^2}} \, dx=\frac {2^{\frac {3}{4}} e^{\frac {3 i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {1}{2} \end {matrix}\middle | {\frac {3 x^{2} e^{i \pi }}{2}} \right )}}{2 x} \] Input:
integrate(1/x**2/(-3*x**2-2)**(1/4),x)
Output:
2**(3/4)*exp(3*I*pi/4)*hyper((-1/2, 1/4), (1/2,), 3*x**2*exp_polar(I*pi)/2 )/(2*x)
\[ \int \frac {1}{x^2 \sqrt [4]{-2-3 x^2}} \, dx=\int { \frac {1}{{\left (-3 \, x^{2} - 2\right )}^{\frac {1}{4}} x^{2}} \,d x } \] Input:
integrate(1/x^2/(-3*x^2-2)^(1/4),x, algorithm="maxima")
Output:
integrate(1/((-3*x^2 - 2)^(1/4)*x^2), x)
\[ \int \frac {1}{x^2 \sqrt [4]{-2-3 x^2}} \, dx=\int { \frac {1}{{\left (-3 \, x^{2} - 2\right )}^{\frac {1}{4}} x^{2}} \,d x } \] Input:
integrate(1/x^2/(-3*x^2-2)^(1/4),x, algorithm="giac")
Output:
integrate(1/((-3*x^2 - 2)^(1/4)*x^2), x)
Time = 0.42 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.16 \[ \int \frac {1}{x^2 \sqrt [4]{-2-3 x^2}} \, dx=-\frac {2\,3^{3/4}\,{\left (\frac {2}{x^2}+3\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {3}{4};\ \frac {7}{4};\ -\frac {2}{3\,x^2}\right )}{9\,x\,{\left (-3\,x^2-2\right )}^{1/4}} \] Input:
int(1/(x^2*(- 3*x^2 - 2)^(1/4)),x)
Output:
-(2*3^(3/4)*(2/x^2 + 3)^(1/4)*hypergeom([1/4, 3/4], 7/4, -2/(3*x^2)))/(9*x *(- 3*x^2 - 2)^(1/4))
\[ \int \frac {1}{x^2 \sqrt [4]{-2-3 x^2}} \, dx=\int \frac {1}{\left (-3 x^{2}-2\right )^{\frac {1}{4}} x^{2}}d x \] Input:
int(1/x^2/(-3*x^2-2)^(1/4),x)
Output:
int(1/(( - 3*x**2 - 2)**(1/4)*x**2),x)