Integrand size = 32, antiderivative size = 101 \[ \int \frac {1+x^2+2 x^4}{\sqrt [4]{1+x^2} \left (2+3 x^2+x^4\right )} \, dx=\frac {4 x}{\sqrt [4]{1+x^2}}+\frac {7}{4} \arctan \left (\frac {-x+\sqrt [4]{1+x^2}}{\sqrt [4]{1+x^2}}\right )-\frac {7}{4} \arctan \left (\frac {x+\sqrt [4]{1+x^2}}{\sqrt [4]{1+x^2}}\right )-\frac {7}{4} \text {arctanh}\left (\frac {2 x \sqrt [4]{1+x^2}}{x^2+2 \sqrt {1+x^2}}\right ) \]
4*x/(x^2+1)^(1/4)+7/4*arctan((-x+(x^2+1)^(1/4))/(x^2+1)^(1/4))-7/4*arctan( (x+(x^2+1)^(1/4))/(x^2+1)^(1/4))-7/4*arctanh(2*x*(x^2+1)^(1/4)/(x^2+2*(x^2 +1)^(1/2)))
Time = 0.31 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.81 \[ \int \frac {1+x^2+2 x^4}{\sqrt [4]{1+x^2} \left (2+3 x^2+x^4\right )} \, dx=\frac {1}{4} \left (\frac {16 x}{\sqrt [4]{1+x^2}}+7 \arctan \left (1-\frac {x}{\sqrt [4]{1+x^2}}\right )-7 \arctan \left (1+\frac {x}{\sqrt [4]{1+x^2}}\right )-7 \text {arctanh}\left (\frac {2 x \sqrt [4]{1+x^2}}{x^2+2 \sqrt {1+x^2}}\right )\right ) \]
((16*x)/(1 + x^2)^(1/4) + 7*ArcTan[1 - x/(1 + x^2)^(1/4)] - 7*ArcTan[1 + x /(1 + x^2)^(1/4)] - 7*ArcTanh[(2*x*(1 + x^2)^(1/4))/(x^2 + 2*Sqrt[1 + x^2] )])/4
Time = 0.39 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.72, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {1387, 7276, 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.
| \(\displaystyle \int \frac {2 x^4+x^2+1}{\sqrt [4]{x^2+1} \left (x^4+3 x^2+2\right )} \, dx\) |
| \(\Big \downarrow \) 1387 |
| \(\displaystyle \int \frac {2 x^4+x^2+1}{\left (x^2+1\right )^{5/4} \left (x^2+2\right )}dx\) |
| \(\Big \downarrow \) 7276 |
| \(\displaystyle \int \left (\frac {2 x^2}{\left (x^2+1\right )^{5/4}}+\frac {7}{\left (x^2+1\right )^{5/4} \left (x^2+2\right )}-\frac {3}{\left (x^2+1\right )^{5/4}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {7}{2} \arctan \left (\frac {\sqrt {x^2+1}+1}{x \sqrt [4]{x^2+1}}\right )+\frac {7}{2} \text {arctanh}\left (\frac {1-\sqrt {x^2+1}}{x \sqrt [4]{x^2+1}}\right )+\frac {4 x}{\sqrt [4]{x^2+1}}\) |
(4*x)/(1 + x^2)^(1/4) + (7*ArcTan[(1 + Sqrt[1 + x^2])/(x*(1 + x^2)^(1/4))] )/2 + (7*ArcTanh[(1 - Sqrt[1 + x^2])/(x*(1 + x^2)^(1/4))])/2
3.15.20.3.1 Defintions of rubi rules used
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + (e_.)* (x_)^(n_))^(q_.), x_Symbol] :> Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^ p, x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && LtQ[c, 0]))
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.70 (sec) , antiderivative size = 291, normalized size of antiderivative = 2.88
| method | result | size |
| trager | \(\frac {4 x}{\left (x^{2}+1\right )^{\frac {1}{4}}}+14 \operatorname {RootOf}\left (32 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right ) \ln \left (-\frac {8 \left (x^{2}+1\right )^{\frac {3}{4}} \operatorname {RootOf}\left (32 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right )-x \sqrt {x^{2}+1}+8 \operatorname {RootOf}\left (32 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {1}{4}}-8 \operatorname {RootOf}\left (32 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right ) x -2 \left (x^{2}+1\right )^{\frac {1}{4}}+x}{x^{2}+2}\right )+\frac {7 \ln \left (\frac {8 \left (x^{2}+1\right )^{\frac {3}{4}} \operatorname {RootOf}\left (32 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right )-2 \left (x^{2}+1\right )^{\frac {3}{4}}+x \sqrt {x^{2}+1}+8 \operatorname {RootOf}\left (32 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {1}{4}}-8 \operatorname {RootOf}\left (32 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right ) x +x}{x^{2}+2}\right )}{2}-14 \ln \left (\frac {8 \left (x^{2}+1\right )^{\frac {3}{4}} \operatorname {RootOf}\left (32 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right )-2 \left (x^{2}+1\right )^{\frac {3}{4}}+x \sqrt {x^{2}+1}+8 \operatorname {RootOf}\left (32 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {1}{4}}-8 \operatorname {RootOf}\left (32 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right ) x +x}{x^{2}+2}\right ) \operatorname {RootOf}\left (32 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right )\) | \(291\) |
| risch | \(\frac {4 x}{\left (x^{2}+1\right )^{\frac {1}{4}}}-\frac {7 \ln \left (\frac {4 \left (x^{2}+1\right )^{\frac {3}{4}} \operatorname {RootOf}\left (8 \textit {\_Z}^{2}+4 \textit {\_Z} +1\right )+2 \left (x^{2}+1\right )^{\frac {3}{4}}+x \sqrt {x^{2}+1}+4 \operatorname {RootOf}\left (8 \textit {\_Z}^{2}+4 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {1}{4}}+4 \operatorname {RootOf}\left (8 \textit {\_Z}^{2}+4 \textit {\_Z} +1\right ) x +x}{x^{2}+2}\right )}{2}-7 \ln \left (\frac {4 \left (x^{2}+1\right )^{\frac {3}{4}} \operatorname {RootOf}\left (8 \textit {\_Z}^{2}+4 \textit {\_Z} +1\right )+2 \left (x^{2}+1\right )^{\frac {3}{4}}+x \sqrt {x^{2}+1}+4 \operatorname {RootOf}\left (8 \textit {\_Z}^{2}+4 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {1}{4}}+4 \operatorname {RootOf}\left (8 \textit {\_Z}^{2}+4 \textit {\_Z} +1\right ) x +x}{x^{2}+2}\right ) \operatorname {RootOf}\left (8 \textit {\_Z}^{2}+4 \textit {\_Z} +1\right )+7 \operatorname {RootOf}\left (8 \textit {\_Z}^{2}+4 \textit {\_Z} +1\right ) \ln \left (-\frac {4 \left (x^{2}+1\right )^{\frac {3}{4}} \operatorname {RootOf}\left (8 \textit {\_Z}^{2}+4 \textit {\_Z} +1\right )-x \sqrt {x^{2}+1}+4 \operatorname {RootOf}\left (8 \textit {\_Z}^{2}+4 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {1}{4}}+4 \operatorname {RootOf}\left (8 \textit {\_Z}^{2}+4 \textit {\_Z} +1\right ) x +2 \left (x^{2}+1\right )^{\frac {1}{4}}+x}{x^{2}+2}\right )\) | \(291\) |
4*x/(x^2+1)^(1/4)+14*RootOf(32*_Z^2-8*_Z+1)*ln(-(8*(x^2+1)^(3/4)*RootOf(32 *_Z^2-8*_Z+1)-x*(x^2+1)^(1/2)+8*RootOf(32*_Z^2-8*_Z+1)*(x^2+1)^(1/4)-8*Roo tOf(32*_Z^2-8*_Z+1)*x-2*(x^2+1)^(1/4)+x)/(x^2+2))+7/2*ln((8*(x^2+1)^(3/4)* RootOf(32*_Z^2-8*_Z+1)-2*(x^2+1)^(3/4)+x*(x^2+1)^(1/2)+8*RootOf(32*_Z^2-8* _Z+1)*(x^2+1)^(1/4)-8*RootOf(32*_Z^2-8*_Z+1)*x+x)/(x^2+2))-14*ln((8*(x^2+1 )^(3/4)*RootOf(32*_Z^2-8*_Z+1)-2*(x^2+1)^(3/4)+x*(x^2+1)^(1/2)+8*RootOf(32 *_Z^2-8*_Z+1)*(x^2+1)^(1/4)-8*RootOf(32*_Z^2-8*_Z+1)*x+x)/(x^2+2))*RootOf( 32*_Z^2-8*_Z+1)
Time = 0.25 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.36 \[ \int \frac {1+x^2+2 x^4}{\sqrt [4]{1+x^2} \left (2+3 x^2+x^4\right )} \, dx=\frac {14 \, {\left (x^{2} + 1\right )} \arctan \left (\frac {x + 2 \, {\left (x^{2} + 1\right )}^{\frac {1}{4}}}{x}\right ) + 14 \, {\left (x^{2} + 1\right )} \arctan \left (-\frac {x - 2 \, {\left (x^{2} + 1\right )}^{\frac {1}{4}}}{x}\right ) - 7 \, {\left (x^{2} + 1\right )} \log \left (\frac {x^{2} + 2 \, {\left (x^{2} + 1\right )}^{\frac {1}{4}} x + 2 \, \sqrt {x^{2} + 1}}{x^{2}}\right ) + 7 \, {\left (x^{2} + 1\right )} \log \left (\frac {x^{2} - 2 \, {\left (x^{2} + 1\right )}^{\frac {1}{4}} x + 2 \, \sqrt {x^{2} + 1}}{x^{2}}\right ) + 32 \, {\left (x^{2} + 1\right )}^{\frac {3}{4}} x}{8 \, {\left (x^{2} + 1\right )}} \]
1/8*(14*(x^2 + 1)*arctan((x + 2*(x^2 + 1)^(1/4))/x) + 14*(x^2 + 1)*arctan( -(x - 2*(x^2 + 1)^(1/4))/x) - 7*(x^2 + 1)*log((x^2 + 2*(x^2 + 1)^(1/4)*x + 2*sqrt(x^2 + 1))/x^2) + 7*(x^2 + 1)*log((x^2 - 2*(x^2 + 1)^(1/4)*x + 2*sq rt(x^2 + 1))/x^2) + 32*(x^2 + 1)^(3/4)*x)/(x^2 + 1)
\[ \int \frac {1+x^2+2 x^4}{\sqrt [4]{1+x^2} \left (2+3 x^2+x^4\right )} \, dx=\int \frac {2 x^{4} + x^{2} + 1}{\left (x^{2} + 1\right )^{\frac {5}{4}} \left (x^{2} + 2\right )}\, dx \]
\[ \int \frac {1+x^2+2 x^4}{\sqrt [4]{1+x^2} \left (2+3 x^2+x^4\right )} \, dx=\int { \frac {2 \, x^{4} + x^{2} + 1}{{\left (x^{4} + 3 \, x^{2} + 2\right )} {\left (x^{2} + 1\right )}^{\frac {1}{4}}} \,d x } \]
\[ \int \frac {1+x^2+2 x^4}{\sqrt [4]{1+x^2} \left (2+3 x^2+x^4\right )} \, dx=\int { \frac {2 \, x^{4} + x^{2} + 1}{{\left (x^{4} + 3 \, x^{2} + 2\right )} {\left (x^{2} + 1\right )}^{\frac {1}{4}}} \,d x } \]
Timed out. \[ \int \frac {1+x^2+2 x^4}{\sqrt [4]{1+x^2} \left (2+3 x^2+x^4\right )} \, dx=\int \frac {2\,x^4+x^2+1}{{\left (x^2+1\right )}^{1/4}\,\left (x^4+3\,x^2+2\right )} \,d x \]