Integrand size = 15, antiderivative size = 242 \[ \int \frac {x^4}{\sqrt [4]{-2-3 x^2}} \, dx=\frac {8}{135} x \left (-2-3 x^2\right )^{3/4}-\frac {2}{27} x^3 \left (-2-3 x^2\right )^{3/4}+\frac {32 x \sqrt [4]{-2-3 x^2}}{135 \left (\sqrt {2}+\sqrt {-2-3 x^2}\right )}+\frac {32 \sqrt [4]{2} \sqrt {-\frac {x^2}{\left (\sqrt {2}+\sqrt {-2-3 x^2}\right )^2}} \left (\sqrt {2}+\sqrt {-2-3 x^2}\right ) E\left (2 \arctan \left (\frac {\sqrt [4]{-2-3 x^2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{135 \sqrt {3} x}-\frac {16 \sqrt [4]{2} \sqrt {-\frac {x^2}{\left (\sqrt {2}+\sqrt {-2-3 x^2}\right )^2}} \left (\sqrt {2}+\sqrt {-2-3 x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{-2-3 x^2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{135 \sqrt {3} x} \]
8/135*x*(-3*x^2-2)^(3/4)-2/27*x^3*(-3*x^2-2)^(3/4)+32/135*x*(-3*x^2-2)^(1/ 4)/(2^(1/2)+(-3*x^2-2)^(1/2))+32/405*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)))*Ellip ticE(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)-16/405* 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 = 5.06 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.26 \[ \int \frac {x^4}{\sqrt [4]{-2-3 x^2}} \, dx=\frac {2 x \left (-8-2 x^2+15 x^4+4\ 2^{3/4} \sqrt [4]{2+3 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {3 x^2}{2}\right )\right )}{135 \sqrt [4]{-2-3 x^2}} \]
(2*x*(-8 - 2*x^2 + 15*x^4 + 4*2^(3/4)*(2 + 3*x^2)^(1/4)*Hypergeometric2F1[ 1/4, 1/2, 3/2, (-3*x^2)/2]))/(135*(-2 - 3*x^2)^(1/4))
Time = 0.30 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.14, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {262, 262, 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.
\(\displaystyle \int \frac {x^4}{\sqrt [4]{-3 x^2-2}} \, dx\) |
\(\Big \downarrow \) 262 |
\(\displaystyle -\frac {4}{9} \int \frac {x^2}{\sqrt [4]{-3 x^2-2}}dx-\frac {2}{27} \left (-3 x^2-2\right )^{3/4} x^3\) |
\(\Big \downarrow \) 262 |
\(\displaystyle -\frac {4}{9} \left (-\frac {4}{15} \int \frac {1}{\sqrt [4]{-3 x^2-2}}dx-\frac {2}{15} \left (-3 x^2-2\right )^{3/4} x\right )-\frac {2}{27} \left (-3 x^2-2\right )^{3/4} x^3\) |
\(\Big \downarrow \) 228 |
\(\displaystyle -\frac {4}{9} \left (\frac {4 \sqrt {\frac {2}{3}} \sqrt {-x^2} \int \frac {\sqrt {\frac {2}{3}} \sqrt {-3 x^2-2}}{\sqrt {-x^2}}d\sqrt [4]{-3 x^2-2}}{15 x}-\frac {2}{15} x \left (-3 x^2-2\right )^{3/4}\right )-\frac {2}{27} \left (-3 x^2-2\right )^{3/4} x^3\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {4}{9} \left (\frac {8 \sqrt {-x^2} \int \frac {\sqrt {-3 x^2-2}}{\sqrt {3} \sqrt {-x^2}}d\sqrt [4]{-3 x^2-2}}{15 \sqrt {3} x}-\frac {2}{15} x \left (-3 x^2-2\right )^{3/4}\right )-\frac {2}{27} \left (-3 x^2-2\right )^{3/4} x^3\) |
\(\Big \downarrow \) 834 |
\(\displaystyle -\frac {4}{9} \left (\frac {8 \sqrt {-x^2} \left (\sqrt {2} \int \frac {1}{\sqrt {3} \sqrt {-x^2}}d\sqrt [4]{-3 x^2-2}-\sqrt {2} \int \frac {\sqrt {2}-\sqrt {-3 x^2-2}}{\sqrt {6} \sqrt {-x^2}}d\sqrt [4]{-3 x^2-2}\right )}{15 \sqrt {3} x}-\frac {2}{15} x \left (-3 x^2-2\right )^{3/4}\right )-\frac {2}{27} \left (-3 x^2-2\right )^{3/4} x^3\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {4}{9} \left (\frac {8 \sqrt {-x^2} \left (\sqrt {2} \int \frac {1}{\sqrt {3} \sqrt {-x^2}}d\sqrt [4]{-3 x^2-2}-\int \frac {\sqrt {2}-\sqrt {-3 x^2-2}}{\sqrt {3} \sqrt {-x^2}}d\sqrt [4]{-3 x^2-2}\right )}{15 \sqrt {3} x}-\frac {2}{15} x \left (-3 x^2-2\right )^{3/4}\right )-\frac {2}{27} \left (-3 x^2-2\right )^{3/4} x^3\) |
\(\Big \downarrow \) 761 |
\(\displaystyle -\frac {4}{9} \left (\frac {8 \sqrt {-x^2} \left (\frac {\sqrt {-\frac {x^2}{\left (\sqrt {-3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {-3 x^2-2}+\sqrt {2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{-3 x^2-2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{2^{3/4} \sqrt {-x^2}}-\int \frac {\sqrt {2}-\sqrt {-3 x^2-2}}{\sqrt {3} \sqrt {-x^2}}d\sqrt [4]{-3 x^2-2}\right )}{15 \sqrt {3} x}-\frac {2}{15} x \left (-3 x^2-2\right )^{3/4}\right )-\frac {2}{27} \left (-3 x^2-2\right )^{3/4} x^3\) |
\(\Big \downarrow \) 1510 |
\(\displaystyle -\frac {4}{9} \left (\frac {8 \sqrt {-x^2} \left (\frac {\sqrt {-\frac {x^2}{\left (\sqrt {-3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {-3 x^2-2}+\sqrt {2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{-3 x^2-2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{2^{3/4} \sqrt {-x^2}}-\frac {\sqrt [4]{2} \sqrt {-\frac {x^2}{\left (\sqrt {-3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {-3 x^2-2}+\sqrt {2}\right ) E\left (2 \arctan \left (\frac {\sqrt [4]{-3 x^2-2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{\sqrt {-x^2}}+\frac {\sqrt {3} \sqrt {-x^2} \sqrt [4]{-3 x^2-2}}{\sqrt {-3 x^2-2}+\sqrt {2}}\right )}{15 \sqrt {3} x}-\frac {2}{15} x \left (-3 x^2-2\right )^{3/4}\right )-\frac {2}{27} \left (-3 x^2-2\right )^{3/4} x^3\) |
(-2*x^3*(-2 - 3*x^2)^(3/4))/27 - (4*((-2*x*(-2 - 3*x^2)^(3/4))/15 + (8*Sqr t[-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])))/(15*Sqrt[3]*x)))/9
3.9.99.3.1 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*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && 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 5 vs. order 4.
Time = 2.17 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.10
method | result | size |
meijerg | \(-\frac {\left (-1\right )^{\frac {3}{4}} 2^{\frac {3}{4}} x^{5} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{4},\frac {5}{2};\frac {7}{2};-\frac {3 x^{2}}{2}\right )}{10}\) | \(23\) |
risch | \(\frac {2 x \left (5 x^{2}-4\right ) \left (3 x^{2}+2\right )}{135 \left (-3 x^{2}-2\right )^{\frac {1}{4}}}-\frac {8 \left (-1\right )^{\frac {3}{4}} 2^{\frac {3}{4}} x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};-\frac {3 x^{2}}{2}\right )}{135}\) | \(48\) |
\[ \int \frac {x^4}{\sqrt [4]{-2-3 x^2}} \, dx=\int { \frac {x^{4}}{{\left (-3 \, x^{2} - 2\right )}^{\frac {1}{4}}} \,d x } \]
1/405*(405*x*integral(-64/405*(-3*x^2 - 2)^(3/4)/(3*x^4 + 2*x^2), x) - 2*( 15*x^4 - 12*x^2 + 16)*(-3*x^2 - 2)^(3/4))/x
Result contains complex when optimal does not.
Time = 0.44 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.14 \[ \int \frac {x^4}{\sqrt [4]{-2-3 x^2}} \, dx=\frac {2^{\frac {3}{4}} x^{5} e^{- \frac {i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {3 x^{2} e^{i \pi }}{2}} \right )}}{10} \]
\[ \int \frac {x^4}{\sqrt [4]{-2-3 x^2}} \, dx=\int { \frac {x^{4}}{{\left (-3 \, x^{2} - 2\right )}^{\frac {1}{4}}} \,d x } \]
\[ \int \frac {x^4}{\sqrt [4]{-2-3 x^2}} \, dx=\int { \frac {x^{4}}{{\left (-3 \, x^{2} - 2\right )}^{\frac {1}{4}}} \,d x } \]
Timed out. \[ \int \frac {x^4}{\sqrt [4]{-2-3 x^2}} \, dx=\int \frac {x^4}{{\left (-3\,x^2-2\right )}^{1/4}} \,d x \]