Integrand size = 24, antiderivative size = 78 \[ \int \frac {x^7}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\frac {14}{243} \left (-1+3 x^2\right )^{3/4}+\frac {8}{567} \left (-1+3 x^2\right )^{7/4}+\frac {2}{891} \left (-1+3 x^2\right )^{11/4}+\frac {8}{81} \arctan \left (\sqrt [4]{-1+3 x^2}\right )-\frac {8}{81} \text {arctanh}\left (\sqrt [4]{-1+3 x^2}\right ) \]
14/243*(3*x^2-1)^(3/4)+8/567*(3*x^2-1)^(7/4)+2/891*(3*x^2-1)^(11/4)+8/81*a rctan((3*x^2-1)^(1/4))-8/81*arctanh((3*x^2-1)^(1/4))
Time = 0.05 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.73 \[ \int \frac {x^7}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\frac {2 \left (\left (-1+3 x^2\right )^{3/4} \left (428+270 x^2+189 x^4\right )+924 \arctan \left (\sqrt [4]{-1+3 x^2}\right )-924 \text {arctanh}\left (\sqrt [4]{-1+3 x^2}\right )\right )}{18711} \]
(2*((-1 + 3*x^2)^(3/4)*(428 + 270*x^2 + 189*x^4) + 924*ArcTan[(-1 + 3*x^2) ^(1/4)] - 924*ArcTanh[(-1 + 3*x^2)^(1/4)]))/18711
Time = 0.21 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {354, 25, 99, 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.
\(\displaystyle \int \frac {x^7}{\left (3 x^2-2\right ) \sqrt [4]{3 x^2-1}} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int -\frac {x^6}{\left (2-3 x^2\right ) \sqrt [4]{3 x^2-1}}dx^2\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{2} \int \frac {x^6}{\left (2-3 x^2\right ) \sqrt [4]{3 x^2-1}}dx^2\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {1}{2} \int \left (-\frac {1}{27} \left (3 x^2-1\right )^{7/4}-\frac {4}{27} \left (3 x^2-1\right )^{3/4}+\frac {8}{27 \left (2-3 x^2\right ) \sqrt [4]{3 x^2-1}}-\frac {7}{27 \sqrt [4]{3 x^2-1}}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {16}{81} \arctan \left (\sqrt [4]{3 x^2-1}\right )-\frac {16}{81} \text {arctanh}\left (\sqrt [4]{3 x^2-1}\right )+\frac {4}{891} \left (3 x^2-1\right )^{11/4}+\frac {16}{567} \left (3 x^2-1\right )^{7/4}+\frac {28}{243} \left (3 x^2-1\right )^{3/4}\right )\) |
((28*(-1 + 3*x^2)^(3/4))/243 + (16*(-1 + 3*x^2)^(7/4))/567 + (4*(-1 + 3*x^ 2)^(11/4))/891 + (16*ArcTan[(-1 + 3*x^2)^(1/4)])/81 - (16*ArcTanh[(-1 + 3* x^2)^(1/4)])/81)/2
3.11.42.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Time = 4.01 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.83
method | result | size |
pseudoelliptic | \(\frac {4 \ln \left (-1+\left (3 x^{2}-1\right )^{\frac {1}{4}}\right )}{81}-\frac {4 \ln \left (1+\left (3 x^{2}-1\right )^{\frac {1}{4}}\right )}{81}+\frac {\left (378 x^{4}+540 x^{2}+856\right ) \left (3 x^{2}-1\right )^{\frac {3}{4}}}{18711}+\frac {8 \arctan \left (\left (3 x^{2}-1\right )^{\frac {1}{4}}\right )}{81}\) | \(65\) |
trager | \(\left (\frac {2}{99} x^{4}+\frac {20}{693} x^{2}+\frac {856}{18711}\right ) \left (3 x^{2}-1\right )^{\frac {3}{4}}+\frac {4 \ln \left (\frac {2 \left (3 x^{2}-1\right )^{\frac {3}{4}}-2 \sqrt {3 x^{2}-1}-3 x^{2}+2 \left (3 x^{2}-1\right )^{\frac {1}{4}}}{3 x^{2}-2}\right )}{81}-\frac {4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (3 x^{2}-1\right )^{\frac {3}{4}}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (3 x^{2}-1\right )^{\frac {1}{4}}+2 \sqrt {3 x^{2}-1}-3 x^{2}}{3 x^{2}-2}\right )}{81}\) | \(146\) |
risch | \(\frac {2 \left (189 x^{4}+270 x^{2}+428\right ) \left (3 x^{2}-1\right )^{\frac {3}{4}}}{18711}+\frac {4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (3 x^{2}-1\right )^{\frac {3}{4}}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (3 x^{2}-1\right )^{\frac {1}{4}}-2 \sqrt {3 x^{2}-1}+3 x^{2}}{3 x^{2}-2}\right )}{81}+\frac {4 \ln \left (\frac {2 \left (3 x^{2}-1\right )^{\frac {3}{4}}-2 \sqrt {3 x^{2}-1}-3 x^{2}+2 \left (3 x^{2}-1\right )^{\frac {1}{4}}}{3 x^{2}-2}\right )}{81}\) | \(148\) |
4/81*ln(-1+(3*x^2-1)^(1/4))-4/81*ln(1+(3*x^2-1)^(1/4))+1/18711*(378*x^4+54 0*x^2+856)*(3*x^2-1)^(3/4)+8/81*arctan((3*x^2-1)^(1/4))
Time = 0.27 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.82 \[ \int \frac {x^7}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\frac {2}{18711} \, {\left (189 \, x^{4} + 270 \, x^{2} + 428\right )} {\left (3 \, x^{2} - 1\right )}^{\frac {3}{4}} + \frac {8}{81} \, \arctan \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) - \frac {4}{81} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {4}{81} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - 1\right ) \]
2/18711*(189*x^4 + 270*x^2 + 428)*(3*x^2 - 1)^(3/4) + 8/81*arctan((3*x^2 - 1)^(1/4)) - 4/81*log((3*x^2 - 1)^(1/4) + 1) + 4/81*log((3*x^2 - 1)^(1/4) - 1)
Time = 6.30 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.13 \[ \int \frac {x^7}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\frac {2 \left (3 x^{2} - 1\right )^{\frac {11}{4}}}{891} + \frac {8 \left (3 x^{2} - 1\right )^{\frac {7}{4}}}{567} + \frac {14 \left (3 x^{2} - 1\right )^{\frac {3}{4}}}{243} + \frac {4 \log {\left (\sqrt [4]{3 x^{2} - 1} - 1 \right )}}{81} - \frac {4 \log {\left (\sqrt [4]{3 x^{2} - 1} + 1 \right )}}{81} + \frac {8 \operatorname {atan}{\left (\sqrt [4]{3 x^{2} - 1} \right )}}{81} \]
2*(3*x**2 - 1)**(11/4)/891 + 8*(3*x**2 - 1)**(7/4)/567 + 14*(3*x**2 - 1)** (3/4)/243 + 4*log((3*x**2 - 1)**(1/4) - 1)/81 - 4*log((3*x**2 - 1)**(1/4) + 1)/81 + 8*atan((3*x**2 - 1)**(1/4))/81
Time = 0.28 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.95 \[ \int \frac {x^7}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\frac {2}{891} \, {\left (3 \, x^{2} - 1\right )}^{\frac {11}{4}} + \frac {8}{567} \, {\left (3 \, x^{2} - 1\right )}^{\frac {7}{4}} + \frac {14}{243} \, {\left (3 \, x^{2} - 1\right )}^{\frac {3}{4}} + \frac {8}{81} \, \arctan \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) - \frac {4}{81} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {4}{81} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - 1\right ) \]
2/891*(3*x^2 - 1)^(11/4) + 8/567*(3*x^2 - 1)^(7/4) + 14/243*(3*x^2 - 1)^(3 /4) + 8/81*arctan((3*x^2 - 1)^(1/4)) - 4/81*log((3*x^2 - 1)^(1/4) + 1) + 4 /81*log((3*x^2 - 1)^(1/4) - 1)
Time = 0.36 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.96 \[ \int \frac {x^7}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\frac {2}{891} \, {\left (3 \, x^{2} - 1\right )}^{\frac {11}{4}} + \frac {8}{567} \, {\left (3 \, x^{2} - 1\right )}^{\frac {7}{4}} + \frac {14}{243} \, {\left (3 \, x^{2} - 1\right )}^{\frac {3}{4}} + \frac {8}{81} \, \arctan \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) - \frac {4}{81} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {4}{81} \, \log \left ({\left | {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - 1 \right |}\right ) \]
2/891*(3*x^2 - 1)^(11/4) + 8/567*(3*x^2 - 1)^(7/4) + 14/243*(3*x^2 - 1)^(3 /4) + 8/81*arctan((3*x^2 - 1)^(1/4)) - 4/81*log((3*x^2 - 1)^(1/4) + 1) + 4 /81*log(abs((3*x^2 - 1)^(1/4) - 1))
Time = 0.13 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.79 \[ \int \frac {x^7}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\frac {8\,\mathrm {atan}\left ({\left (3\,x^2-1\right )}^{1/4}\right )}{81}+\frac {14\,{\left (3\,x^2-1\right )}^{3/4}}{243}+\frac {8\,{\left (3\,x^2-1\right )}^{7/4}}{567}+\frac {2\,{\left (3\,x^2-1\right )}^{11/4}}{891}+\frac {\mathrm {atan}\left ({\left (3\,x^2-1\right )}^{1/4}\,1{}\mathrm {i}\right )\,8{}\mathrm {i}}{81} \]