Integrand size = 24, antiderivative size = 165 \[ \int \frac {x^6}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\frac {40}{567} x \sqrt [4]{-1+3 x^2}+\frac {2}{63} x^3 \sqrt [4]{-1+3 x^2}+\frac {2}{27} \sqrt {\frac {2}{3}} \arctan \left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {2}{27} \sqrt {\frac {2}{3}} \text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )+\frac {40 \sqrt {\frac {x^2}{\left (1+\sqrt {-1+3 x^2}\right )^2}} \left (1+\sqrt {-1+3 x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{-1+3 x^2}\right ),\frac {1}{2}\right )}{567 \sqrt {3} x} \]
40/567*x*(3*x^2-1)^(1/4)+2/63*x^3*(3*x^2-1)^(1/4)+2/81*arctan(1/2*x*6^(1/2 )/(3*x^2-1)^(1/4))*6^(1/2)-2/81*arctanh(1/2*x*6^(1/2)/(3*x^2-1)^(1/4))*6^( 1/2)+40/1701*(cos(2*arctan((3*x^2-1)^(1/4)))^2)^(1/2)/cos(2*arctan((3*x^2- 1)^(1/4)))*EllipticF(sin(2*arctan((3*x^2-1)^(1/4))),1/2*2^(1/2))*(1+(3*x^2 -1)^(1/2))*(x^2/(1+(3*x^2-1)^(1/2))^2)^(1/2)/x*3^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 6.07 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.12 \[ \int \frac {x^6}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\frac {2 x \left (-20+51 x^2+27 x^4-31 x^2 \left (1-3 x^2\right )^{3/4} \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},1,\frac {5}{2},3 x^2,\frac {3 x^2}{2}\right )-\frac {80 \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{4},1,\frac {3}{2},3 x^2,\frac {3 x^2}{2}\right )}{\left (-2+3 x^2\right ) \left (2 \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{4},1,\frac {3}{2},3 x^2,\frac {3 x^2}{2}\right )+x^2 \left (2 \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},2,\frac {5}{2},3 x^2,\frac {3 x^2}{2}\right )+3 \operatorname {AppellF1}\left (\frac {3}{2},\frac {7}{4},1,\frac {5}{2},3 x^2,\frac {3 x^2}{2}\right )\right )\right )}\right )}{567 \left (-1+3 x^2\right )^{3/4}} \]
(2*x*(-20 + 51*x^2 + 27*x^4 - 31*x^2*(1 - 3*x^2)^(3/4)*AppellF1[3/2, 3/4, 1, 5/2, 3*x^2, (3*x^2)/2] - (80*AppellF1[1/2, 3/4, 1, 3/2, 3*x^2, (3*x^2)/ 2])/((-2 + 3*x^2)*(2*AppellF1[1/2, 3/4, 1, 3/2, 3*x^2, (3*x^2)/2] + x^2*(2 *AppellF1[3/2, 3/4, 2, 5/2, 3*x^2, (3*x^2)/2] + 3*AppellF1[3/2, 7/4, 1, 5/ 2, 3*x^2, (3*x^2)/2])))))/(567*(-1 + 3*x^2)^(3/4))
Time = 0.38 (sec) , antiderivative size = 165, 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.
\(\displaystyle \int \frac {x^6}{\left (3 x^2-2\right ) \left (3 x^2-1\right )^{3/4}} \, dx\) |
\(\Big \downarrow \) 352 |
\(\displaystyle \int \left (\frac {2 x^2}{9 \left (3 x^2-1\right )^{3/4}}+\frac {8}{27 \left (3 x^2-2\right ) \left (3 x^2-1\right )^{3/4}}+\frac {4}{27 \left (3 x^2-1\right )^{3/4}}+\frac {x^4}{3 \left (3 x^2-1\right )^{3/4}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{27} \sqrt {\frac {2}{3}} \arctan \left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )+\frac {40 \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-1}+1\right )^2}} \left (\sqrt {3 x^2-1}+1\right ) \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{3 x^2-1}\right ),\frac {1}{2}\right )}{567 \sqrt {3} x}-\frac {2}{27} \sqrt {\frac {2}{3}} \text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )+\frac {40}{567} \sqrt [4]{3 x^2-1} x+\frac {2}{63} \sqrt [4]{3 x^2-1} x^3\) |
(40*x*(-1 + 3*x^2)^(1/4))/567 + (2*x^3*(-1 + 3*x^2)^(1/4))/63 + (2*Sqrt[2/ 3]*ArcTan[(Sqrt[3/2]*x)/(-1 + 3*x^2)^(1/4)])/27 - (2*Sqrt[2/3]*ArcTanh[(Sq rt[3/2]*x)/(-1 + 3*x^2)^(1/4)])/27 + (40*Sqrt[x^2/(1 + Sqrt[-1 + 3*x^2])^2 ]*(1 + Sqrt[-1 + 3*x^2])*EllipticF[2*ArcTan[(-1 + 3*x^2)^(1/4)], 1/2])/(56 7*Sqrt[3]*x)
3.11.87.3.1 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])
\[\int \frac {x^{6}}{\left (3 x^{2}-2\right ) \left (3 x^{2}-1\right )^{\frac {3}{4}}}d x\]
\[ \int \frac {x^6}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\int { \frac {x^{6}}{{\left (3 \, x^{2} - 1\right )}^{\frac {3}{4}} {\left (3 \, x^{2} - 2\right )}} \,d x } \]
\[ \int \frac {x^6}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\int \frac {x^{6}}{\left (3 x^{2} - 2\right ) \left (3 x^{2} - 1\right )^{\frac {3}{4}}}\, dx \]
\[ \int \frac {x^6}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\int { \frac {x^{6}}{{\left (3 \, x^{2} - 1\right )}^{\frac {3}{4}} {\left (3 \, x^{2} - 2\right )}} \,d x } \]
\[ \int \frac {x^6}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\int { \frac {x^{6}}{{\left (3 \, x^{2} - 1\right )}^{\frac {3}{4}} {\left (3 \, x^{2} - 2\right )}} \,d x } \]
Timed out. \[ \int \frac {x^6}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\int \frac {x^6}{{\left (3\,x^2-1\right )}^{3/4}\,\left (3\,x^2-2\right )} \,d x \]