Integrand size = 24, antiderivative size = 182 \[ \int \frac {x^6}{\left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=\frac {80}{567} x \sqrt [4]{2-3 x^2}+\frac {2}{63} x^3 \sqrt [4]{2-3 x^2}+\frac {8\ 2^{3/4} \arctan \left (\frac {2^{3/4}-\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{27 \sqrt {3}}-\frac {8\ 2^{3/4} \text {arctanh}\left (\frac {2^{3/4}+\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{27 \sqrt {3}}-\frac {160\ 2^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\sqrt {\frac {3}{2}} x\right ),2\right )}{567 \sqrt {3}} \]
80/567*x*(-3*x^2+2)^(1/4)+2/63*x^3*(-3*x^2+2)^(1/4)+8/81*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)-8 /81*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)-160/1701*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 = 6.39 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.04 \[ \int \frac {x^6}{\left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=\frac {2}{567} x \left (31 \sqrt [4]{2} x^2 \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},1,\frac {5}{2},\frac {3 x^2}{2},\frac {3 x^2}{4}\right )+\frac {80-102 x^2-27 x^4+\frac {1280 \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{4},1,\frac {3}{2},\frac {3 x^2}{2},\frac {3 x^2}{4}\right )}{\left (-4+3 x^2\right ) \left (4 \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{4},1,\frac {3}{2},\frac {3 x^2}{2},\frac {3 x^2}{4}\right )+x^2 \left (2 \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},2,\frac {5}{2},\frac {3 x^2}{2},\frac {3 x^2}{4}\right )+3 \operatorname {AppellF1}\left (\frac {3}{2},\frac {7}{4},1,\frac {5}{2},\frac {3 x^2}{2},\frac {3 x^2}{4}\right )\right )\right )}}{\left (2-3 x^2\right )^{3/4}}\right ) \]
(2*x*(31*2^(1/4)*x^2*AppellF1[3/2, 3/4, 1, 5/2, (3*x^2)/2, (3*x^2)/4] + (8 0 - 102*x^2 - 27*x^4 + (1280*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)))/567
Time = 0.28 (sec) , antiderivative size = 182, 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.
\(\displaystyle \int \frac {x^6}{\left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx\) |
\(\Big \downarrow \) 352 |
\(\displaystyle \int \left (-\frac {4 x^2}{9 \left (2-3 x^2\right )^{3/4}}-\frac {16}{27 \left (2-3 x^2\right )^{3/4}}+\frac {64}{27 \left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )}-\frac {x^4}{3 \left (2-3 x^2\right )^{3/4}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {160\ 2^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\sqrt {\frac {3}{2}} x\right ),2\right )}{567 \sqrt {3}}+\frac {8\ 2^{3/4} \arctan \left (\frac {2^{3/4}-\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{27 \sqrt {3}}-\frac {8\ 2^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt {2-3 x^2}+2^{3/4}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{27 \sqrt {3}}+\frac {80}{567} \sqrt [4]{2-3 x^2} x+\frac {2}{63} \sqrt [4]{2-3 x^2} x^3\) |
(80*x*(2 - 3*x^2)^(1/4))/567 + (2*x^3*(2 - 3*x^2)^(1/4))/63 + (8*2^(3/4)*A rcTan[(2^(3/4) - 2^(1/4)*Sqrt[2 - 3*x^2])/(Sqrt[3]*x*(2 - 3*x^2)^(1/4))])/ (27*Sqrt[3]) - (8*2^(3/4)*ArcTanh[(2^(3/4) + 2^(1/4)*Sqrt[2 - 3*x^2])/(Sqr t[3]*x*(2 - 3*x^2)^(1/4))])/(27*Sqrt[3]) - (160*2^(3/4)*EllipticF[ArcSin[S qrt[3/2]*x]/2, 2])/(567*Sqrt[3])
3.11.67.3.1 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])
\[\int \frac {x^{6}}{\left (-3 x^{2}+2\right )^{\frac {3}{4}} \left (-3 x^{2}+4\right )}d x\]
\[ \int \frac {x^6}{\left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=\int { -\frac {x^{6}}{{\left (3 \, x^{2} - 4\right )} {\left (-3 \, x^{2} + 2\right )}^{\frac {3}{4}}} \,d x } \]
\[ \int \frac {x^6}{\left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=- \int \frac {x^{6}}{3 x^{2} \left (2 - 3 x^{2}\right )^{\frac {3}{4}} - 4 \left (2 - 3 x^{2}\right )^{\frac {3}{4}}}\, dx \]
\[ \int \frac {x^6}{\left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=\int { -\frac {x^{6}}{{\left (3 \, x^{2} - 4\right )} {\left (-3 \, x^{2} + 2\right )}^{\frac {3}{4}}} \,d x } \]
\[ \int \frac {x^6}{\left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=\int { -\frac {x^{6}}{{\left (3 \, x^{2} - 4\right )} {\left (-3 \, x^{2} + 2\right )}^{\frac {3}{4}}} \,d x } \]
Timed out. \[ \int \frac {x^6}{\left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=-\int \frac {x^6}{{\left (2-3\,x^2\right )}^{3/4}\,\left (3\,x^2-4\right )} \,d x \]