Integrand size = 20, antiderivative size = 79 \[ \int \frac {x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {1+\sqrt [3]{2-2 x^2}}{\sqrt {3}}\right )}{2\ 2^{2/3}}-\frac {\log \left (3+x^2\right )}{4\ 2^{2/3}}+\frac {3 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{4\ 2^{2/3}} \]
-1/8*ln(x^2+3)*2^(1/3)+3/8*ln(2^(2/3)-(-x^2+1)^(1/3))*2^(1/3)+1/4*arctan(1 /3*(1+(-2*x^2+2)^(1/3))*3^(1/2))*3^(1/2)*2^(1/3)
Time = 0.12 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.04 \[ \int \frac {x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {1+\sqrt [3]{2-2 x^2}}{\sqrt {3}}\right )+2 \log \left (-2+\sqrt [3]{2-2 x^2}\right )-\log \left (4+2 \sqrt [3]{2-2 x^2}+\left (2-2 x^2\right )^{2/3}\right )}{4\ 2^{2/3}} \]
(2*Sqrt[3]*ArcTan[(1 + (2 - 2*x^2)^(1/3))/Sqrt[3]] + 2*Log[-2 + (2 - 2*x^2 )^(1/3)] - Log[4 + 2*(2 - 2*x^2)^(1/3) + (2 - 2*x^2)^(2/3)])/(4*2^(2/3))
Time = 0.21 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {353, 67, 16, 1082, 217}
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.
\(\displaystyle \int \frac {x}{\sqrt [3]{1-x^2} \left (x^2+3\right )} \, dx\) |
\(\Big \downarrow \) 353 |
\(\displaystyle \frac {1}{2} \int \frac {1}{\sqrt [3]{1-x^2} \left (x^2+3\right )}dx^2\) |
\(\Big \downarrow \) 67 |
\(\displaystyle \frac {1}{2} \left (-\frac {3 \int \frac {1}{2^{2/3}-\sqrt [3]{1-x^2}}d\sqrt [3]{1-x^2}}{2\ 2^{2/3}}+\frac {3}{2} \int \frac {1}{x^4+2^{2/3} \sqrt [3]{1-x^2}+2 \sqrt [3]{2}}d\sqrt [3]{1-x^2}-\frac {\log \left (x^2+3\right )}{2\ 2^{2/3}}\right )\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{2} \int \frac {1}{x^4+2^{2/3} \sqrt [3]{1-x^2}+2 \sqrt [3]{2}}d\sqrt [3]{1-x^2}-\frac {\log \left (x^2+3\right )}{2\ 2^{2/3}}+\frac {3 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{2\ 2^{2/3}}\right )\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {1}{2} \left (-\frac {3 \int \frac {1}{-x^4-3}d\left (\sqrt [3]{2} \sqrt [3]{1-x^2}+1\right )}{2^{2/3}}-\frac {\log \left (x^2+3\right )}{2\ 2^{2/3}}+\frac {3 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{2\ 2^{2/3}}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \left (\frac {\sqrt {3} \arctan \left (\frac {\sqrt [3]{2} \sqrt [3]{1-x^2}+1}{\sqrt {3}}\right )}{2^{2/3}}-\frac {\log \left (x^2+3\right )}{2\ 2^{2/3}}+\frac {3 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{2\ 2^{2/3}}\right )\) |
((Sqrt[3]*ArcTan[(1 + 2^(1/3)*(1 - x^2)^(1/3))/Sqrt[3]])/2^(2/3) - Log[3 + x^2]/(2*2^(2/3)) + (3*Log[2^(2/3) - (1 - x^2)^(1/3)])/(2*2^(2/3)))/2
3.11.10.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[ {q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q), x ] + (Simp[3/(2*b) Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/3)], x] - Simp[3/(2*b*q) Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] / ; FreeQ[{a, b, c, d}, x] && PosQ[(b*c - a*d)/b]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[1/2 Subst[Int[(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]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Time = 3.33 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.04
method | result | size |
pseudoelliptic | \(\frac {2^{\frac {1}{3}} \left (2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (1+2^{\frac {1}{3}} \left (-x^{2}+1\right )^{\frac {1}{3}}\right )}{3}\right )+2 \ln \left (\left (-x^{2}+1\right )^{\frac {1}{3}}-2^{\frac {2}{3}}\right )-\ln \left (\left (-x^{2}+1\right )^{\frac {2}{3}}+2^{\frac {2}{3}} \left (-x^{2}+1\right )^{\frac {1}{3}}+2 \,2^{\frac {1}{3}}\right )\right )}{8}\) | \(82\) |
trager | \(\text {Expression too large to display}\) | \(740\) |
1/8*2^(1/3)*(2*3^(1/2)*arctan(1/3*3^(1/2)*(1+2^(1/3)*(-x^2+1)^(1/3)))+2*ln ((-x^2+1)^(1/3)-2^(2/3))-ln((-x^2+1)^(2/3)+2^(2/3)*(-x^2+1)^(1/3)+2*2^(1/3 )))
Time = 0.26 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.09 \[ \int \frac {x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\frac {1}{4} \cdot 4^{\frac {1}{6}} \sqrt {3} \arctan \left (\frac {1}{6} \cdot 4^{\frac {1}{6}} \sqrt {3} {\left (4^{\frac {1}{3}} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right )}\right ) - \frac {1}{16} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{8} \cdot 4^{\frac {2}{3}} \log \left (-4^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) \]
1/4*4^(1/6)*sqrt(3)*arctan(1/6*4^(1/6)*sqrt(3)*(4^(1/3) + 2*(-x^2 + 1)^(1/ 3))) - 1/16*4^(2/3)*log(4^(2/3) + 4^(1/3)*(-x^2 + 1)^(1/3) + (-x^2 + 1)^(2 /3)) + 1/8*4^(2/3)*log(-4^(1/3) + (-x^2 + 1)^(1/3))
Time = 54.01 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.99 \[ \int \frac {x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\begin {cases} \sqrt [3]{2} \left (\frac {\log {\left (\sqrt [3]{2 - 2 x^{2}} - 2 \right )}}{4} - \frac {\log {\left (\left (2 - 2 x^{2}\right )^{\frac {2}{3}} + 2 \sqrt [3]{2 - 2 x^{2}} + 4 \right )}}{8} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {\sqrt {3} \left (\sqrt [3]{2 - 2 x^{2}} + 1\right )}{3} \right )}}{4}\right ) & \text {for}\: x > -1 \wedge x < 1 \end {cases} \]
Piecewise((2**(1/3)*(log((2 - 2*x**2)**(1/3) - 2)/4 - log((2 - 2*x**2)**(2 /3) + 2*(2 - 2*x**2)**(1/3) + 4)/8 + sqrt(3)*atan(sqrt(3)*((2 - 2*x**2)**( 1/3) + 1)/3)/4), (x > -1) & (x < 1)))
Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.09 \[ \int \frac {x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\frac {1}{8} \cdot 4^{\frac {2}{3}} \sqrt {3} \arctan \left (\frac {1}{12} \cdot 4^{\frac {2}{3}} \sqrt {3} {\left (4^{\frac {1}{3}} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right )}\right ) - \frac {1}{16} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{8} \cdot 4^{\frac {2}{3}} \log \left (-4^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) \]
1/8*4^(2/3)*sqrt(3)*arctan(1/12*4^(2/3)*sqrt(3)*(4^(1/3) + 2*(-x^2 + 1)^(1 /3))) - 1/16*4^(2/3)*log(4^(2/3) + 4^(1/3)*(-x^2 + 1)^(1/3) + (-x^2 + 1)^( 2/3)) + 1/8*4^(2/3)*log(-4^(1/3) + (-x^2 + 1)^(1/3))
Time = 0.31 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.09 \[ \int \frac {x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\frac {1}{8} \cdot 4^{\frac {2}{3}} \sqrt {3} \arctan \left (\frac {1}{12} \cdot 4^{\frac {2}{3}} \sqrt {3} {\left (4^{\frac {1}{3}} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right )}\right ) - \frac {1}{16} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{8} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {1}{3}} - {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) \]
1/8*4^(2/3)*sqrt(3)*arctan(1/12*4^(2/3)*sqrt(3)*(4^(1/3) + 2*(-x^2 + 1)^(1 /3))) - 1/16*4^(2/3)*log(4^(2/3) + 4^(1/3)*(-x^2 + 1)^(1/3) + (-x^2 + 1)^( 2/3)) + 1/8*4^(2/3)*log(4^(1/3) - (-x^2 + 1)^(1/3))
Time = 5.35 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.34 \[ \int \frac {x}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\frac {2^{1/3}\,\ln \left (\frac {9\,{\left (1-x^2\right )}^{1/3}}{4}-\frac {9\,2^{2/3}}{4}\right )}{4}+\frac {2^{1/3}\,\ln \left (\frac {9\,{\left (1-x^2\right )}^{1/3}}{4}-\frac {9\,2^{2/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{16}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{8}-\frac {2^{1/3}\,\ln \left (\frac {9\,{\left (1-x^2\right )}^{1/3}}{4}-\frac {9\,2^{2/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{16}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{8} \]