Integrand size = 15, antiderivative size = 260 \[ \int \frac {1}{x^6 \sqrt [4]{-2+3 x^2}} \, dx=\frac {\left (-2+3 x^2\right )^{3/4}}{10 x^5}+\frac {7 \left (-2+3 x^2\right )^{3/4}}{40 x^3}+\frac {63 \left (-2+3 x^2\right )^{3/4}}{160 x}-\frac {189 x \sqrt [4]{-2+3 x^2}}{160 \left (\sqrt {2}+\sqrt {-2+3 x^2}\right )}+\frac {63 \sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {2}+\sqrt {-2+3 x^2}\right )^2}} \left (\sqrt {2}+\sqrt {-2+3 x^2}\right ) E\left (2 \arctan \left (\frac {\sqrt [4]{-2+3 x^2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{80\ 2^{3/4} x}-\frac {63 \sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {2}+\sqrt {-2+3 x^2}\right )^2}} \left (\sqrt {2}+\sqrt {-2+3 x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{-2+3 x^2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{160\ 2^{3/4} x} \]
1/10*(3*x^2-2)^(3/4)/x^5+7/40*(3*x^2-2)^(3/4)/x^3+63/160*(3*x^2-2)^(3/4)/x -189/160*x*(3*x^2-2)^(1/4)/(2^(1/2)+(3*x^2-2)^(1/2))+63/160*2^(1/4)*(cos(2 *arctan(1/2*(3*x^2-2)^(1/4)*2^(3/4)))^2)^(1/2)/cos(2*arctan(1/2*(3*x^2-2)^ (1/4)*2^(3/4)))*EllipticE(sin(2*arctan(1/2*(3*x^2-2)^(1/4)*2^(3/4))),1/2*2 ^(1/2))*(2^(1/2)+(3*x^2-2)^(1/2))*(x^2/(2^(1/2)+(3*x^2-2)^(1/2))^2)^(1/2)/ x*3^(1/2)-63/320*2^(1/4)*(cos(2*arctan(1/2*(3*x^2-2)^(1/4)*2^(3/4)))^2)^(1 /2)/cos(2*arctan(1/2*(3*x^2-2)^(1/4)*2^(3/4)))*EllipticF(sin(2*arctan(1/2* (3*x^2-2)^(1/4)*2^(3/4))),1/2*2^(1/2))*(2^(1/2)+(3*x^2-2)^(1/2))*(x^2/(2^( 1/2)+(3*x^2-2)^(1/2))^2)^(1/2)/x*3^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.01 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.18 \[ \int \frac {1}{x^6 \sqrt [4]{-2+3 x^2}} \, dx=-\frac {\sqrt [4]{1-\frac {3 x^2}{2}} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {1}{4},-\frac {3}{2},\frac {3 x^2}{2}\right )}{5 x^5 \sqrt [4]{-2+3 x^2}} \]
-1/5*((1 - (3*x^2)/2)^(1/4)*Hypergeometric2F1[-5/2, 1/4, -3/2, (3*x^2)/2]) /(x^5*(-2 + 3*x^2)^(1/4))
Time = 0.33 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {264, 264, 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.
\(\displaystyle \int \frac {1}{x^6 \sqrt [4]{3 x^2-2}} \, dx\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {21}{20} \int \frac {1}{x^4 \sqrt [4]{3 x^2-2}}dx+\frac {\left (3 x^2-2\right )^{3/4}}{10 x^5}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {21}{20} \left (\frac {3}{4} \int \frac {1}{x^2 \sqrt [4]{3 x^2-2}}dx+\frac {\left (3 x^2-2\right )^{3/4}}{6 x^3}\right )+\frac {\left (3 x^2-2\right )^{3/4}}{10 x^5}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {21}{20} \left (\frac {3}{4} \left (\frac {\left (3 x^2-2\right )^{3/4}}{2 x}-\frac {3}{4} \int \frac {1}{\sqrt [4]{3 x^2-2}}dx\right )+\frac {\left (3 x^2-2\right )^{3/4}}{6 x^3}\right )+\frac {\left (3 x^2-2\right )^{3/4}}{10 x^5}\) |
\(\Big \downarrow \) 228 |
\(\displaystyle \frac {21}{20} \left (\frac {3}{4} \left (\frac {\left (3 x^2-2\right )^{3/4}}{2 x}-\frac {\sqrt {\frac {3}{2}} \sqrt {x^2} \int \frac {\sqrt {\frac {2}{3}} \sqrt {3 x^2-2}}{\sqrt {x^2}}d\sqrt [4]{3 x^2-2}}{2 x}\right )+\frac {\left (3 x^2-2\right )^{3/4}}{6 x^3}\right )+\frac {\left (3 x^2-2\right )^{3/4}}{10 x^5}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {21}{20} \left (\frac {3}{4} \left (\frac {\left (3 x^2-2\right )^{3/4}}{2 x}-\frac {\sqrt {3} \sqrt {x^2} \int \frac {\sqrt {3 x^2-2}}{\sqrt {3} \sqrt {x^2}}d\sqrt [4]{3 x^2-2}}{2 x}\right )+\frac {\left (3 x^2-2\right )^{3/4}}{6 x^3}\right )+\frac {\left (3 x^2-2\right )^{3/4}}{10 x^5}\) |
\(\Big \downarrow \) 834 |
\(\displaystyle \frac {21}{20} \left (\frac {3}{4} \left (\frac {\left (3 x^2-2\right )^{3/4}}{2 x}-\frac {\sqrt {3} \sqrt {x^2} \left (\sqrt {2} \int \frac {1}{\sqrt {3} \sqrt {x^2}}d\sqrt [4]{3 x^2-2}-\sqrt {2} \int \frac {\sqrt {2}-\sqrt {3 x^2-2}}{\sqrt {6} \sqrt {x^2}}d\sqrt [4]{3 x^2-2}\right )}{2 x}\right )+\frac {\left (3 x^2-2\right )^{3/4}}{6 x^3}\right )+\frac {\left (3 x^2-2\right )^{3/4}}{10 x^5}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {21}{20} \left (\frac {3}{4} \left (\frac {\left (3 x^2-2\right )^{3/4}}{2 x}-\frac {\sqrt {3} \sqrt {x^2} \left (\sqrt {2} \int \frac {1}{\sqrt {3} \sqrt {x^2}}d\sqrt [4]{3 x^2-2}-\int \frac {\sqrt {2}-\sqrt {3 x^2-2}}{\sqrt {3} \sqrt {x^2}}d\sqrt [4]{3 x^2-2}\right )}{2 x}\right )+\frac {\left (3 x^2-2\right )^{3/4}}{6 x^3}\right )+\frac {\left (3 x^2-2\right )^{3/4}}{10 x^5}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {21}{20} \left (\frac {3}{4} \left (\frac {\left (3 x^2-2\right )^{3/4}}{2 x}-\frac {\sqrt {3} \sqrt {x^2} \left (\frac {\sqrt {\frac {x^2}{\left (\sqrt {3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {3 x^2-2}+\sqrt {2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{2^{3/4} \sqrt {x^2}}-\int \frac {\sqrt {2}-\sqrt {3 x^2-2}}{\sqrt {3} \sqrt {x^2}}d\sqrt [4]{3 x^2-2}\right )}{2 x}\right )+\frac {\left (3 x^2-2\right )^{3/4}}{6 x^3}\right )+\frac {\left (3 x^2-2\right )^{3/4}}{10 x^5}\) |
\(\Big \downarrow \) 1510 |
\(\displaystyle \frac {21}{20} \left (\frac {3}{4} \left (\frac {\left (3 x^2-2\right )^{3/4}}{2 x}-\frac {\sqrt {3} \sqrt {x^2} \left (\frac {\sqrt {\frac {x^2}{\left (\sqrt {3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {3 x^2-2}+\sqrt {2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{2^{3/4} \sqrt {x^2}}-\frac {\sqrt [4]{2} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {3 x^2-2}+\sqrt {2}\right ) E\left (2 \arctan \left (\frac {\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{\sqrt {x^2}}+\frac {\sqrt {3} \sqrt {x^2} \sqrt [4]{3 x^2-2}}{\sqrt {3 x^2-2}+\sqrt {2}}\right )}{2 x}\right )+\frac {\left (3 x^2-2\right )^{3/4}}{6 x^3}\right )+\frac {\left (3 x^2-2\right )^{3/4}}{10 x^5}\) |
(-2 + 3*x^2)^(3/4)/(10*x^5) + (21*((-2 + 3*x^2)^(3/4)/(6*x^3) + (3*((-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)))/4))/20
3.9.97.3.1 Defintions of rubi rules used
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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 2.21 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.16
method | result | size |
meijerg | \(-\frac {2^{\frac {3}{4}} {\left (-\operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )\right )}^{\frac {1}{4}} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {5}{2},\frac {1}{4};-\frac {3}{2};\frac {3 x^{2}}{2}\right )}{10 \operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )^{\frac {1}{4}} x^{5}}\) | \(42\) |
risch | \(\frac {189 x^{6}-42 x^{4}-8 x^{2}-32}{160 x^{5} \left (3 x^{2}-2\right )^{\frac {1}{4}}}-\frac {189 \,2^{\frac {3}{4}} {\left (-\operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )\right )}^{\frac {1}{4}} x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};\frac {3 x^{2}}{2}\right )}{640 \operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )^{\frac {1}{4}}}\) | \(72\) |
-1/10*2^(3/4)/signum(-1+3/2*x^2)^(1/4)*(-signum(-1+3/2*x^2))^(1/4)/x^5*hyp ergeom([-5/2,1/4],[-3/2],3/2*x^2)
\[ \int \frac {1}{x^6 \sqrt [4]{-2+3 x^2}} \, dx=\int { \frac {1}{{\left (3 \, x^{2} - 2\right )}^{\frac {1}{4}} x^{6}} \,d x } \]
Result contains complex when optimal does not.
Time = 0.60 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.13 \[ \int \frac {1}{x^6 \sqrt [4]{-2+3 x^2}} \, dx=\frac {2^{\frac {3}{4}} e^{\frac {3 i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{2}, \frac {1}{4} \\ - \frac {3}{2} \end {matrix}\middle | {\frac {3 x^{2}}{2}} \right )}}{10 x^{5}} \]
\[ \int \frac {1}{x^6 \sqrt [4]{-2+3 x^2}} \, dx=\int { \frac {1}{{\left (3 \, x^{2} - 2\right )}^{\frac {1}{4}} x^{6}} \,d x } \]
\[ \int \frac {1}{x^6 \sqrt [4]{-2+3 x^2}} \, dx=\int { \frac {1}{{\left (3 \, x^{2} - 2\right )}^{\frac {1}{4}} x^{6}} \,d x } \]
Timed out. \[ \int \frac {1}{x^6 \sqrt [4]{-2+3 x^2}} \, dx=\int \frac {1}{x^6\,{\left (3\,x^2-2\right )}^{1/4}} \,d x \]