Integrand size = 20, antiderivative size = 85 \[ \int \frac {1+2 x^2}{x \left (1+x^2\right )^{2/3}} \, dx=3 \sqrt [3]{1+x^2}-\frac {1}{2} \sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1+x^2}}{\sqrt {3}}\right )+\frac {1}{2} \log \left (-1+\sqrt [3]{1+x^2}\right )-\frac {1}{4} \log \left (1+\sqrt [3]{1+x^2}+\left (1+x^2\right )^{2/3}\right ) \]
3*(x^2+1)^(1/3)-1/2*3^(1/2)*arctan(1/3*3^(1/2)+2/3*(x^2+1)^(1/3)*3^(1/2))+ 1/2*ln(-1+(x^2+1)^(1/3))-1/4*ln(1+(x^2+1)^(1/3)+(x^2+1)^(2/3))
Time = 0.05 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.94 \[ \int \frac {1+2 x^2}{x \left (1+x^2\right )^{2/3}} \, dx=\frac {1}{4} \left (12 \sqrt [3]{1+x^2}-2 \sqrt {3} \arctan \left (\frac {1+2 \sqrt [3]{1+x^2}}{\sqrt {3}}\right )+2 \log \left (-1+\sqrt [3]{1+x^2}\right )-\log \left (1+\sqrt [3]{1+x^2}+\left (1+x^2\right )^{2/3}\right )\right ) \]
(12*(1 + x^2)^(1/3) - 2*Sqrt[3]*ArcTan[(1 + 2*(1 + x^2)^(1/3))/Sqrt[3]] + 2*Log[-1 + (1 + x^2)^(1/3)] - Log[1 + (1 + x^2)^(1/3) + (1 + x^2)^(2/3)])/ 4
Time = 0.19 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.81, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {354, 90, 69, 16, 1083, 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.
\(\displaystyle \int \frac {2 x^2+1}{x \left (x^2+1\right )^{2/3}} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {2 x^2+1}{x^2 \left (x^2+1\right )^{2/3}}dx^2\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {1}{2} \left (\int \frac {1}{x^2 \left (x^2+1\right )^{2/3}}dx^2+6 \sqrt [3]{x^2+1}\right )\) |
\(\Big \downarrow \) 69 |
\(\displaystyle \frac {1}{2} \left (-\frac {3}{2} \int \frac {1}{1-\sqrt [3]{x^2+1}}d\sqrt [3]{x^2+1}-\frac {3}{2} \int \frac {1}{x^4+\sqrt [3]{x^2+1}+1}d\sqrt [3]{x^2+1}+6 \sqrt [3]{x^2+1}-\frac {1}{2} \log \left (x^2\right )\right )\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {1}{2} \left (-\frac {3}{2} \int \frac {1}{x^4+\sqrt [3]{x^2+1}+1}d\sqrt [3]{x^2+1}+6 \sqrt [3]{x^2+1}-\frac {1}{2} \log \left (x^2\right )+\frac {3}{2} \log \left (1-\sqrt [3]{x^2+1}\right )\right )\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {1}{2} \left (3 \int \frac {1}{-x^4-3}d\left (2 \sqrt [3]{x^2+1}+1\right )+6 \sqrt [3]{x^2+1}-\frac {1}{2} \log \left (x^2\right )+\frac {3}{2} \log \left (1-\sqrt [3]{x^2+1}\right )\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \left (-\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{x^2+1}+1}{\sqrt {3}}\right )+6 \sqrt [3]{x^2+1}-\frac {\log \left (x^2\right )}{2}+\frac {3}{2} \log \left (1-\sqrt [3]{x^2+1}\right )\right )\) |
(6*(1 + x^2)^(1/3) - Sqrt[3]*ArcTan[(1 + 2*(1 + x^2)^(1/3))/Sqrt[3]] - Log [x^2]/2 + (3*Log[1 - (1 + x^2)^(1/3)])/2)/2
3.12.31.3.1 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_))^(2/3)), x_Symbol] :> With[ {q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Simp[3/(2*b*q) Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1 /3)], x] - Simp[3/(2*b*q^2) 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_))*((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[(-(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_)^(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[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Result contains higher order function than in optimal. Order 5 vs. order 3.
Time = 1.88 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.73
method | result | size |
meijerg | \(x^{2} \operatorname {hypergeom}\left (\left [\frac {2}{3}, 1\right ], \left [2\right ], -x^{2}\right )+\frac {-\frac {2 \Gamma \left (\frac {2}{3}\right ) x^{2} \operatorname {hypergeom}\left (\left [1, 1, \frac {5}{3}\right ], \left [2, 2\right ], -x^{2}\right )}{3}+\left (\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \left (3\right )}{2}+2 \ln \left (x \right )\right ) \Gamma \left (\frac {2}{3}\right )}{2 \Gamma \left (\frac {2}{3}\right )}\) | \(62\) |
pseudoelliptic | \(3 \left (x^{2}+1\right )^{\frac {1}{3}}+\frac {\ln \left (-1+\left (x^{2}+1\right )^{\frac {1}{3}}\right )}{2}-\frac {\ln \left (1+\left (x^{2}+1\right )^{\frac {1}{3}}+\left (x^{2}+1\right )^{\frac {2}{3}}\right )}{4}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (1+2 \left (x^{2}+1\right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )}{2}\) | \(64\) |
trager | \(3 \left (x^{2}+1\right )^{\frac {1}{3}}+3 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \ln \left (-\frac {144 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2} x^{2}+90 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {2}{3}}+102 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) x^{2}+24 \left (x^{2}+1\right )^{\frac {2}{3}}+90 \left (x^{2}+1\right )^{\frac {1}{3}} \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )-144 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2}+15 x^{2}+24 \left (x^{2}+1\right )^{\frac {1}{3}}+66 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )+20}{x^{2}}\right )-\frac {\ln \left (\frac {-144 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2} x^{2}+90 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {2}{3}}+54 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) x^{2}-9 \left (x^{2}+1\right )^{\frac {2}{3}}+90 \left (x^{2}+1\right )^{\frac {1}{3}} \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )+144 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2}-2 x^{2}-9 \left (x^{2}+1\right )^{\frac {1}{3}}+114 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )-5}{x^{2}}\right )}{2}-3 \ln \left (\frac {-144 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2} x^{2}+90 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {2}{3}}+54 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) x^{2}-9 \left (x^{2}+1\right )^{\frac {2}{3}}+90 \left (x^{2}+1\right )^{\frac {1}{3}} \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )+144 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2}-2 x^{2}-9 \left (x^{2}+1\right )^{\frac {1}{3}}+114 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )-5}{x^{2}}\right ) \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )\) | \(436\) |
x^2*hypergeom([2/3,1],[2],-x^2)+1/2/GAMMA(2/3)*(-2/3*GAMMA(2/3)*x^2*hyperg eom([1,1,5/3],[2,2],-x^2)+(1/6*Pi*3^(1/2)-3/2*ln(3)+2*ln(x))*GAMMA(2/3))
Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.76 \[ \int \frac {1+2 x^2}{x \left (1+x^2\right )^{2/3}} \, dx=-\frac {1}{2} \, \sqrt {3} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (x^{2} + 1\right )}^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) + 3 \, {\left (x^{2} + 1\right )}^{\frac {1}{3}} - \frac {1}{4} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {2}{3}} + {\left (x^{2} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{2} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{3}} - 1\right ) \]
-1/2*sqrt(3)*arctan(2/3*sqrt(3)*(x^2 + 1)^(1/3) + 1/3*sqrt(3)) + 3*(x^2 + 1)^(1/3) - 1/4*log((x^2 + 1)^(2/3) + (x^2 + 1)^(1/3) + 1) + 1/2*log((x^2 + 1)^(1/3) - 1)
Time = 4.59 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.86 \[ \int \frac {1+2 x^2}{x \left (1+x^2\right )^{2/3}} \, dx=3 \sqrt [3]{x^{2} + 1} + \frac {\log {\left (\sqrt [3]{x^{2} + 1} - 1 \right )}}{2} - \frac {\log {\left (\left (x^{2} + 1\right )^{\frac {2}{3}} + \sqrt [3]{x^{2} + 1} + 1 \right )}}{4} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} \left (\sqrt [3]{x^{2} + 1} + \frac {1}{2}\right )}{3} \right )}}{2} \]
3*(x**2 + 1)**(1/3) + log((x**2 + 1)**(1/3) - 1)/2 - log((x**2 + 1)**(2/3) + (x**2 + 1)**(1/3) + 1)/4 - sqrt(3)*atan(2*sqrt(3)*((x**2 + 1)**(1/3) + 1/2)/3)/2
Time = 0.27 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.74 \[ \int \frac {1+2 x^2}{x \left (1+x^2\right )^{2/3}} \, dx=-\frac {1}{2} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{2} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + 3 \, {\left (x^{2} + 1\right )}^{\frac {1}{3}} - \frac {1}{4} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {2}{3}} + {\left (x^{2} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{2} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{3}} - 1\right ) \]
-1/2*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^2 + 1)^(1/3) + 1)) + 3*(x^2 + 1)^(1/ 3) - 1/4*log((x^2 + 1)^(2/3) + (x^2 + 1)^(1/3) + 1) + 1/2*log((x^2 + 1)^(1 /3) - 1)
Time = 0.28 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.74 \[ \int \frac {1+2 x^2}{x \left (1+x^2\right )^{2/3}} \, dx=-\frac {1}{2} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{2} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + 3 \, {\left (x^{2} + 1\right )}^{\frac {1}{3}} - \frac {1}{4} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {2}{3}} + {\left (x^{2} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{2} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{3}} - 1\right ) \]
-1/2*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^2 + 1)^(1/3) + 1)) + 3*(x^2 + 1)^(1/ 3) - 1/4*log((x^2 + 1)^(2/3) + (x^2 + 1)^(1/3) + 1) + 1/2*log((x^2 + 1)^(1 /3) - 1)
Time = 5.96 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.93 \[ \int \frac {1+2 x^2}{x \left (1+x^2\right )^{2/3}} \, dx=\frac {\ln \left (\frac {9\,{\left (x^2+1\right )}^{1/3}}{4}-\frac {9}{4}\right )}{2}+3\,{\left (x^2+1\right )}^{1/3}+\ln \left (\frac {9\,{\left (x^2+1\right )}^{1/3}}{2}+\frac {9}{4}-\frac {\sqrt {3}\,9{}\mathrm {i}}{4}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )-\ln \left (\frac {9\,{\left (x^2+1\right )}^{1/3}}{2}+\frac {9}{4}+\frac {\sqrt {3}\,9{}\mathrm {i}}{4}\right )\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right ) \]