Integrand size = 24, antiderivative size = 48 \[ \int \frac {x^3}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\frac {2}{27} \left (-1+3 x^2\right )^{3/4}+\frac {2}{9} \arctan \left (\sqrt [4]{-1+3 x^2}\right )-\frac {2}{9} \text {arctanh}\left (\sqrt [4]{-1+3 x^2}\right ) \]
Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.92 \[ \int \frac {x^3}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\frac {2}{27} \left (\left (-1+3 x^2\right )^{3/4}+3 \arctan \left (\sqrt [4]{-1+3 x^2}\right )-3 \text {arctanh}\left (\sqrt [4]{-1+3 x^2}\right )\right ) \]
Time = 0.18 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.19, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {354, 25, 90, 73, 827, 216, 219}
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.
\(\displaystyle \int \frac {x^3}{\left (3 x^2-2\right ) \sqrt [4]{3 x^2-1}} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int -\frac {x^2}{\left (2-3 x^2\right ) \sqrt [4]{3 x^2-1}}dx^2\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{2} \int \frac {x^2}{\left (2-3 x^2\right ) \sqrt [4]{3 x^2-1}}dx^2\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {1}{2} \left (\frac {4}{27} \left (3 x^2-1\right )^{3/4}-\frac {2}{3} \int \frac {1}{\left (2-3 x^2\right ) \sqrt [4]{3 x^2-1}}dx^2\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (\frac {4}{27} \left (3 x^2-1\right )^{3/4}-\frac {8}{9} \int \frac {x^4}{1-x^8}d\sqrt [4]{3 x^2-1}\right )\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {1}{2} \left (\frac {4}{27} \left (3 x^2-1\right )^{3/4}-\frac {8}{9} \left (\frac {1}{2} \int \frac {1}{1-x^4}d\sqrt [4]{3 x^2-1}-\frac {1}{2} \int \frac {1}{x^4+1}d\sqrt [4]{3 x^2-1}\right )\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{2} \left (\frac {4}{27} \left (3 x^2-1\right )^{3/4}-\frac {8}{9} \left (\frac {1}{2} \int \frac {1}{1-x^4}d\sqrt [4]{3 x^2-1}-\frac {1}{2} \arctan \left (\sqrt [4]{3 x^2-1}\right )\right )\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {4}{27} \left (3 x^2-1\right )^{3/4}-\frac {8}{9} \left (\frac {1}{2} \text {arctanh}\left (\sqrt [4]{3 x^2-1}\right )-\frac {1}{2} \arctan \left (\sqrt [4]{3 x^2-1}\right )\right )\right )\) |
((4*(-1 + 3*x^2)^(3/4))/27 - (8*(-1/2*ArcTan[(-1 + 3*x^2)^(1/4)] + ArcTanh [(-1 + 3*x^2)^(1/4)]/2))/9)/2
3.11.44.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
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]
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
Time = 3.88 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.10
method | result | size |
pseudoelliptic | \(\frac {2 \left (3 x^{2}-1\right )^{\frac {3}{4}}}{27}+\frac {\ln \left (-1+\left (3 x^{2}-1\right )^{\frac {1}{4}}\right )}{9}-\frac {\ln \left (1+\left (3 x^{2}-1\right )^{\frac {1}{4}}\right )}{9}+\frac {2 \arctan \left (\left (3 x^{2}-1\right )^{\frac {1}{4}}\right )}{9}\) | \(53\) |
trager | \(\frac {2 \left (3 x^{2}-1\right )^{\frac {3}{4}}}{27}-\frac {\ln \left (-\frac {2 \left (3 x^{2}-1\right )^{\frac {3}{4}}+2 \sqrt {3 x^{2}-1}+3 x^{2}+2 \left (3 x^{2}-1\right )^{\frac {1}{4}}}{3 x^{2}-2}\right )}{9}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (3 x^{2}-1\right )^{\frac {3}{4}}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (3 x^{2}-1\right )^{\frac {1}{4}}+2 \sqrt {3 x^{2}-1}-3 x^{2}}{3 x^{2}-2}\right )}{9}\) | \(136\) |
risch | \(\frac {2 \left (3 x^{2}-1\right )^{\frac {3}{4}}}{27}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (3 x^{2}-1\right )^{\frac {3}{4}}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (3 x^{2}-1\right )^{\frac {1}{4}}-2 \sqrt {3 x^{2}-1}+3 x^{2}}{3 x^{2}-2}\right )}{9}-\frac {\ln \left (-\frac {2 \left (3 x^{2}-1\right )^{\frac {3}{4}}+2 \sqrt {3 x^{2}-1}+3 x^{2}+2 \left (3 x^{2}-1\right )^{\frac {1}{4}}}{3 x^{2}-2}\right )}{9}\) | \(137\) |
2/27*(3*x^2-1)^(3/4)+1/9*ln(-1+(3*x^2-1)^(1/4))-1/9*ln(1+(3*x^2-1)^(1/4))+ 2/9*arctan((3*x^2-1)^(1/4))
Time = 0.28 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.08 \[ \int \frac {x^3}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\frac {2}{27} \, {\left (3 \, x^{2} - 1\right )}^{\frac {3}{4}} + \frac {2}{9} \, \arctan \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{9} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{9} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - 1\right ) \]
2/27*(3*x^2 - 1)^(3/4) + 2/9*arctan((3*x^2 - 1)^(1/4)) - 1/9*log((3*x^2 - 1)^(1/4) + 1) + 1/9*log((3*x^2 - 1)^(1/4) - 1)
Time = 4.05 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.21 \[ \int \frac {x^3}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\frac {2 \left (3 x^{2} - 1\right )^{\frac {3}{4}}}{27} + \frac {\log {\left (\sqrt [4]{3 x^{2} - 1} - 1 \right )}}{9} - \frac {\log {\left (\sqrt [4]{3 x^{2} - 1} + 1 \right )}}{9} + \frac {2 \operatorname {atan}{\left (\sqrt [4]{3 x^{2} - 1} \right )}}{9} \]
2*(3*x**2 - 1)**(3/4)/27 + log((3*x**2 - 1)**(1/4) - 1)/9 - log((3*x**2 - 1)**(1/4) + 1)/9 + 2*atan((3*x**2 - 1)**(1/4))/9
Time = 0.28 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.08 \[ \int \frac {x^3}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\frac {2}{27} \, {\left (3 \, x^{2} - 1\right )}^{\frac {3}{4}} + \frac {2}{9} \, \arctan \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{9} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{9} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - 1\right ) \]
2/27*(3*x^2 - 1)^(3/4) + 2/9*arctan((3*x^2 - 1)^(1/4)) - 1/9*log((3*x^2 - 1)^(1/4) + 1) + 1/9*log((3*x^2 - 1)^(1/4) - 1)
Time = 0.38 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.10 \[ \int \frac {x^3}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\frac {2}{27} \, {\left (3 \, x^{2} - 1\right )}^{\frac {3}{4}} + \frac {2}{9} \, \arctan \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{9} \, \log \left ({\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{9} \, \log \left ({\left | {\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} - 1 \right |}\right ) \]
2/27*(3*x^2 - 1)^(3/4) + 2/9*arctan((3*x^2 - 1)^(1/4)) - 1/9*log((3*x^2 - 1)^(1/4) + 1) + 1/9*log(abs((3*x^2 - 1)^(1/4) - 1))
Time = 5.39 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.75 \[ \int \frac {x^3}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\frac {2\,\mathrm {atan}\left ({\left (3\,x^2-1\right )}^{1/4}\right )}{9}-\frac {2\,\mathrm {atanh}\left ({\left (3\,x^2-1\right )}^{1/4}\right )}{9}+\frac {2\,{\left (3\,x^2-1\right )}^{3/4}}{27} \]