Integrand size = 24, antiderivative size = 164 \[ \int \frac {x^4}{\left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=\frac {2}{27} x \sqrt [4]{2-3 x^2}+\frac {2\ 2^{3/4} \arctan \left (\frac {2^{3/4}-\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{9 \sqrt {3}}-\frac {2\ 2^{3/4} \text {arctanh}\left (\frac {2^{3/4}+\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{9 \sqrt {3}}-\frac {4\ 2^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\sqrt {\frac {3}{2}} x\right ),2\right )}{27 \sqrt {3}} \]
2/27*x*(-3*x^2+2)^(1/4)+2/27*2^(3/4)*arctan(1/3*(2^(3/4)-2^(1/4)*(-3*x^2+2 )^(1/2))/x/(-3*x^2+2)^(1/4)*3^(1/2))*3^(1/2)-2/27*2^(3/4)*arctanh(1/3*(2^( 3/4)+2^(1/4)*(-3*x^2+2)^(1/2))/x/(-3*x^2+2)^(1/4)*3^(1/2))*3^(1/2)-4/81*2^ (3/4)*(cos(1/2*arcsin(1/2*x*6^(1/2)))^2)^(1/2)/cos(1/2*arcsin(1/2*x*6^(1/2 )))*EllipticF(sin(1/2*arcsin(1/2*x*6^(1/2))),2^(1/2))*3^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 5.94 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.12 \[ \int \frac {x^4}{\left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=\frac {2}{27} x \left (\sqrt [4]{2} x^2 \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},1,\frac {5}{2},\frac {3 x^2}{2},\frac {3 x^2}{4}\right )+\frac {2-3 x^2+\frac {32 \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{4},1,\frac {3}{2},\frac {3 x^2}{2},\frac {3 x^2}{4}\right )}{\left (-4+3 x^2\right ) \left (4 \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{4},1,\frac {3}{2},\frac {3 x^2}{2},\frac {3 x^2}{4}\right )+x^2 \left (2 \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},2,\frac {5}{2},\frac {3 x^2}{2},\frac {3 x^2}{4}\right )+3 \operatorname {AppellF1}\left (\frac {3}{2},\frac {7}{4},1,\frac {5}{2},\frac {3 x^2}{2},\frac {3 x^2}{4}\right )\right )\right )}}{\left (2-3 x^2\right )^{3/4}}\right ) \]
(2*x*(2^(1/4)*x^2*AppellF1[3/2, 3/4, 1, 5/2, (3*x^2)/2, (3*x^2)/4] + (2 - 3*x^2 + (32*AppellF1[1/2, 3/4, 1, 3/2, (3*x^2)/2, (3*x^2)/4])/((-4 + 3*x^2 )*(4*AppellF1[1/2, 3/4, 1, 3/2, (3*x^2)/2, (3*x^2)/4] + x^2*(2*AppellF1[3/ 2, 3/4, 2, 5/2, (3*x^2)/2, (3*x^2)/4] + 3*AppellF1[3/2, 7/4, 1, 5/2, (3*x^ 2)/2, (3*x^2)/4]))))/(2 - 3*x^2)^(3/4)))/27
Time = 0.27 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {352, 2009}
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.
\(\displaystyle \int \frac {x^4}{\left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx\) |
\(\Big \downarrow \) 352 |
\(\displaystyle \int \left (-\frac {x^2}{3 \left (2-3 x^2\right )^{3/4}}-\frac {4}{9 \left (2-3 x^2\right )^{3/4}}+\frac {16}{9 \left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {4\ 2^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\sqrt {\frac {3}{2}} x\right ),2\right )}{27 \sqrt {3}}+\frac {2\ 2^{3/4} \arctan \left (\frac {2^{3/4}-\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{9 \sqrt {3}}-\frac {2\ 2^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt {2-3 x^2}+2^{3/4}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{9 \sqrt {3}}+\frac {2}{27} \sqrt [4]{2-3 x^2} x\) |
(2*x*(2 - 3*x^2)^(1/4))/27 + (2*2^(3/4)*ArcTan[(2^(3/4) - 2^(1/4)*Sqrt[2 - 3*x^2])/(Sqrt[3]*x*(2 - 3*x^2)^(1/4))])/(9*Sqrt[3]) - (2*2^(3/4)*ArcTanh[ (2^(3/4) + 2^(1/4)*Sqrt[2 - 3*x^2])/(Sqrt[3]*x*(2 - 3*x^2)^(1/4))])/(9*Sqr t[3]) - (4*2^(3/4)*EllipticF[ArcSin[Sqrt[3/2]*x]/2, 2])/(27*Sqrt[3])
3.11.68.3.1 Defintions of rubi rules used
Int[(x_)^(m_)/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol ] :> Int[ExpandIntegrand[x^m/((a + b*x^2)^(3/4)*(c + d*x^2)), x], x] /; Fre eQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d, 0] && IntegerQ[m] && (PosQ[a] || In tegerQ[m/2])
\[\int \frac {x^{4}}{\left (-3 x^{2}+2\right )^{\frac {3}{4}} \left (-3 x^{2}+4\right )}d x\]
\[ \int \frac {x^4}{\left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=\int { -\frac {x^{4}}{{\left (3 \, x^{2} - 4\right )} {\left (-3 \, x^{2} + 2\right )}^{\frac {3}{4}}} \,d x } \]
\[ \int \frac {x^4}{\left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=- \int \frac {x^{4}}{3 x^{2} \left (2 - 3 x^{2}\right )^{\frac {3}{4}} - 4 \left (2 - 3 x^{2}\right )^{\frac {3}{4}}}\, dx \]
\[ \int \frac {x^4}{\left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=\int { -\frac {x^{4}}{{\left (3 \, x^{2} - 4\right )} {\left (-3 \, x^{2} + 2\right )}^{\frac {3}{4}}} \,d x } \]
\[ \int \frac {x^4}{\left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=\int { -\frac {x^{4}}{{\left (3 \, x^{2} - 4\right )} {\left (-3 \, x^{2} + 2\right )}^{\frac {3}{4}}} \,d x } \]
Timed out. \[ \int \frac {x^4}{\left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=-\int \frac {x^4}{{\left (2-3\,x^2\right )}^{3/4}\,\left (3\,x^2-4\right )} \,d x \]