Integrand size = 24, antiderivative size = 65 \[ \int \frac {1+x^2}{\left (-1+x^2\right ) \left (1+2 x^2\right )^{3/2}} \, dx=-\frac {x}{3 \sqrt {1+2 x^2}}-\frac {2 \text {arctanh}\left (\sqrt {\frac {2}{3}}-\sqrt {\frac {2}{3}} x^2+\frac {x \sqrt {1+2 x^2}}{\sqrt {3}}\right )}{3 \sqrt {3}} \]
-1/3*x/(2*x^2+1)^(1/2)+2/9*arctanh(-1/3*6^(1/2)+1/3*x^2*6^(1/2)-1/3*x*(2*x ^2+1)^(1/2)*3^(1/2))*3^(1/2)
Time = 0.14 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.92 \[ \int \frac {1+x^2}{\left (-1+x^2\right ) \left (1+2 x^2\right )^{3/2}} \, dx=\frac {1}{9} \left (-\frac {3 x}{\sqrt {1+2 x^2}}-2 \sqrt {3} \text {arctanh}\left (\frac {1}{3} \left (\sqrt {6}-\sqrt {6} x^2+x \sqrt {3+6 x^2}\right )\right )\right ) \]
((-3*x)/Sqrt[1 + 2*x^2] - 2*Sqrt[3]*ArcTanh[(Sqrt[6] - Sqrt[6]*x^2 + x*Sqr t[3 + 6*x^2])/3])/9
Time = 0.17 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.69, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {402, 27, 291, 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.
\(\displaystyle \int \frac {x^2+1}{\left (x^2-1\right ) \left (2 x^2+1\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 402 |
\(\displaystyle \frac {1}{3} \int -\frac {2}{\left (1-x^2\right ) \sqrt {2 x^2+1}}dx-\frac {x}{3 \sqrt {2 x^2+1}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {2}{3} \int \frac {1}{\left (1-x^2\right ) \sqrt {2 x^2+1}}dx-\frac {x}{3 \sqrt {2 x^2+1}}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle -\frac {2}{3} \int \frac {1}{1-\frac {3 x^2}{2 x^2+1}}d\frac {x}{\sqrt {2 x^2+1}}-\frac {x}{3 \sqrt {2 x^2+1}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {2 \text {arctanh}\left (\frac {\sqrt {3} x}{\sqrt {2 x^2+1}}\right )}{3 \sqrt {3}}-\frac {x}{3 \sqrt {2 x^2+1}}\) |
3.9.52.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
Time = 1.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.71
method | result | size |
pseudoelliptic | \(-\frac {\frac {2 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {3}\, \sqrt {2 x^{2}+1}}{3 x}\right ) \sqrt {2 x^{2}+1}}{3}+x}{3 \sqrt {2 x^{2}+1}}\) | \(46\) |
trager | \(-\frac {x}{3 \sqrt {2 x^{2}+1}}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (-\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) x^{2}+6 \sqrt {2 x^{2}+1}\, x -\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right )}{\left (x -1\right ) \left (1+x \right )}\right )}{9}\) | \(67\) |
risch | \(-\frac {x}{3 \sqrt {2 x^{2}+1}}-\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (4 x +2\right ) \sqrt {3}}{6 \sqrt {2 \left (x -1\right )^{2}+4 x -1}}\right )}{9}+\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (2-4 x \right ) \sqrt {3}}{6 \sqrt {2 \left (1+x \right )^{2}-1-4 x}}\right )}{9}\) | \(74\) |
default | \(\frac {x}{\sqrt {2 x^{2}+1}}+\frac {1}{3 \sqrt {2 \left (x -1\right )^{2}+4 x -1}}-\frac {2 x}{3 \sqrt {2 \left (x -1\right )^{2}+4 x -1}}-\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (4 x +2\right ) \sqrt {3}}{6 \sqrt {2 \left (x -1\right )^{2}+4 x -1}}\right )}{9}-\frac {1}{3 \sqrt {2 \left (1+x \right )^{2}-1-4 x}}-\frac {2 x}{3 \sqrt {2 \left (1+x \right )^{2}-1-4 x}}+\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (2-4 x \right ) \sqrt {3}}{6 \sqrt {2 \left (1+x \right )^{2}-1-4 x}}\right )}{9}\) | \(139\) |
Time = 0.24 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.02 \[ \int \frac {1+x^2}{\left (-1+x^2\right ) \left (1+2 x^2\right )^{3/2}} \, dx=\frac {\sqrt {3} {\left (2 \, x^{2} + 1\right )} \log \left (\frac {2 \, \sqrt {3} \sqrt {2 \, x^{2} + 1} x - 5 \, x^{2} - 1}{x^{2} - 1}\right ) - 3 \, \sqrt {2 \, x^{2} + 1} x}{9 \, {\left (2 \, x^{2} + 1\right )}} \]
1/9*(sqrt(3)*(2*x^2 + 1)*log((2*sqrt(3)*sqrt(2*x^2 + 1)*x - 5*x^2 - 1)/(x^ 2 - 1)) - 3*sqrt(2*x^2 + 1)*x)/(2*x^2 + 1)
\[ \int \frac {1+x^2}{\left (-1+x^2\right ) \left (1+2 x^2\right )^{3/2}} \, dx=\int \frac {x^{2} + 1}{\left (x - 1\right ) \left (x + 1\right ) \left (2 x^{2} + 1\right )^{\frac {3}{2}}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.23 \[ \int \frac {1+x^2}{\left (-1+x^2\right ) \left (1+2 x^2\right )^{3/2}} \, dx=-\frac {1}{9} \, \sqrt {3} \operatorname {arsinh}\left (\frac {2 \, \sqrt {2} x}{{\left | 2 \, x + 2 \right |}} - \frac {\sqrt {2}}{{\left | 2 \, x + 2 \right |}}\right ) - \frac {1}{9} \, \sqrt {3} \operatorname {arsinh}\left (\frac {2 \, \sqrt {2} x}{{\left | 2 \, x - 2 \right |}} + \frac {\sqrt {2}}{{\left | 2 \, x - 2 \right |}}\right ) - \frac {x}{3 \, \sqrt {2 \, x^{2} + 1}} \]
-1/9*sqrt(3)*arcsinh(2*sqrt(2)*x/abs(2*x + 2) - sqrt(2)/abs(2*x + 2)) - 1/ 9*sqrt(3)*arcsinh(2*sqrt(2)*x/abs(2*x - 2) + sqrt(2)/abs(2*x - 2)) - 1/3*x /sqrt(2*x^2 + 1)
Time = 0.30 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.28 \[ \int \frac {1+x^2}{\left (-1+x^2\right ) \left (1+2 x^2\right )^{3/2}} \, dx=-\frac {1}{18} \, \sqrt {6} \sqrt {2} \log \left (\frac {{\left | 2 \, {\left (\sqrt {2} x - \sqrt {2 \, x^{2} + 1}\right )}^{2} - 4 \, \sqrt {6} - 10 \right |}}{{\left | 2 \, {\left (\sqrt {2} x - \sqrt {2 \, x^{2} + 1}\right )}^{2} + 4 \, \sqrt {6} - 10 \right |}}\right ) - \frac {x}{3 \, \sqrt {2 \, x^{2} + 1}} \]
-1/18*sqrt(6)*sqrt(2)*log(abs(2*(sqrt(2)*x - sqrt(2*x^2 + 1))^2 - 4*sqrt(6 ) - 10)/abs(2*(sqrt(2)*x - sqrt(2*x^2 + 1))^2 + 4*sqrt(6) - 10)) - 1/3*x/s qrt(2*x^2 + 1)
Time = 5.62 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.71 \[ \int \frac {1+x^2}{\left (-1+x^2\right ) \left (1+2 x^2\right )^{3/2}} \, dx=\frac {\sqrt {3}\,\left (\ln \left (x-1\right )-\ln \left (x+\frac {\sqrt {2}\,\sqrt {3}\,\sqrt {x^2+\frac {1}{2}}}{2}+\frac {1}{2}\right )\right )}{9}-\frac {\sqrt {3}\,\left (\ln \left (x+1\right )-\ln \left (x-\frac {\sqrt {2}\,\sqrt {3}\,\sqrt {x^2+\frac {1}{2}}}{2}-\frac {1}{2}\right )\right )}{9}-\frac {\sqrt {2}\,\sqrt {x^2+\frac {1}{2}}}{12\,\left (x-\frac {\sqrt {2}\,1{}\mathrm {i}}{2}\right )}-\frac {\sqrt {2}\,\sqrt {x^2+\frac {1}{2}}}{12\,\left (x+\frac {\sqrt {2}\,1{}\mathrm {i}}{2}\right )} \]