Integrand size = 20, antiderivative size = 53 \[ \int \frac {x^2}{\left (-1+x^4\right ) \sqrt {1+x^4}} \, dx=\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{4 \sqrt {2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{4 \sqrt {2}} \]
Time = 0.22 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.87 \[ \int \frac {x^2}{\left (-1+x^4\right ) \sqrt {1+x^4}} \, dx=\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )-\text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{4 \sqrt {2}} \]
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.37 (sec) , antiderivative size = 149, normalized size of antiderivative = 2.81, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {993, 1535, 761, 2213, 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.
| \(\displaystyle \int \frac {x^2}{\left (x^4-1\right ) \sqrt {x^4+1}} \, dx\) |
| \(\Big \downarrow \) 993 |
| \(\displaystyle \frac {1}{2} \int \frac {1}{\left (x^2+1\right ) \sqrt {x^4+1}}dx-\frac {1}{2} \int \frac {1}{\left (1-x^2\right ) \sqrt {x^4+1}}dx\) |
| \(\Big \downarrow \) 1535 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1}{\sqrt {x^4+1}}dx+\frac {1}{2} \int \frac {1-x^2}{\left (x^2+1\right ) \sqrt {x^4+1}}dx\right )+\frac {1}{2} \left (-\frac {1}{2} \int \frac {1}{\sqrt {x^4+1}}dx-\frac {1}{2} \int \frac {x^2+1}{\left (1-x^2\right ) \sqrt {x^4+1}}dx\right )\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1-x^2}{\left (x^2+1\right ) \sqrt {x^4+1}}dx+\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{4 \sqrt {x^4+1}}\right )+\frac {1}{2} \left (-\frac {1}{2} \int \frac {x^2+1}{\left (1-x^2\right ) \sqrt {x^4+1}}dx-\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{4 \sqrt {x^4+1}}\right )\) |
| \(\Big \downarrow \) 2213 |
| \(\displaystyle \frac {1}{2} \left (-\frac {1}{2} \int \frac {1}{1-\frac {2 x^2}{x^4+1}}d\frac {x}{\sqrt {x^4+1}}-\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{4 \sqrt {x^4+1}}\right )+\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{\frac {2 x^2}{x^4+1}+1}d\frac {x}{\sqrt {x^4+1}}+\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{4 \sqrt {x^4+1}}\right )\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {1}{2} \left (-\frac {1}{2} \int \frac {1}{1-\frac {2 x^2}{x^4+1}}d\frac {x}{\sqrt {x^4+1}}-\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{4 \sqrt {x^4+1}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{2 \sqrt {2}}+\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{4 \sqrt {x^4+1}}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{4 \sqrt {x^4+1}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{2 \sqrt {2}}+\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{4 \sqrt {x^4+1}}\right )\) |
(-1/2*ArcTanh[(Sqrt[2]*x)/Sqrt[1 + x^4]]/Sqrt[2] - ((1 + x^2)*Sqrt[(1 + x^ 4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(4*Sqrt[1 + x^4]))/2 + (ArcTa n[(Sqrt[2]*x)/Sqrt[1 + x^4]]/(2*Sqrt[2]) + ((1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(4*Sqrt[1 + x^4]))/2
3.7.67.3.1 Defintions of rubi rules used
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[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(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)*Sqrt[c + d*x^4]), x], x] - Simp[s/(2*b) Int[1/((r - s*x^2)*Sqrt[c + d*x^4]), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Simp[ 1/(2*d) Int[1/Sqrt[a + c*x^4], x], x] + Simp[1/(2*d) Int[(d - e*x^2)/(( d + e*x^2)*Sqrt[a + c*x^4]), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[c*d^2 - a*e^2, 0]
Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]) , x_Symbol] :> Simp[A Subst[Int[1/(d + 2*a*e*x^2), x], x, x/Sqrt[a + c*x^ 4]], x] /; FreeQ[{a, c, d, e, A, B}, x] && EqQ[c*d^2 - a*e^2, 0] && EqQ[B*d + A*e, 0]
Time = 2.23 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.81
| method | result | size |
| elliptic | \(\frac {\left (-\frac {\operatorname {arctanh}\left (\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{4}-\frac {\arctan \left (\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{4}\right ) \sqrt {2}}{2}\) | \(43\) |
| default | \(\frac {\sqrt {2}\, \left (2 \arctan \left (\frac {\sqrt {2}\, x}{\sqrt {x^{4}+1}}\right )+\operatorname {arctanh}\left (\frac {\left (x^{2}+x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right )-\operatorname {arctanh}\left (\frac {\left (x^{2}-x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right )\right )}{16}\) | \(62\) |
| pseudoelliptic | \(\frac {\sqrt {2}\, \left (2 \arctan \left (\frac {\sqrt {2}\, x}{\sqrt {x^{4}+1}}\right )+\operatorname {arctanh}\left (\frac {\left (x^{2}+x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right )-\operatorname {arctanh}\left (\frac {\left (x^{2}-x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right )\right )}{16}\) | \(62\) |
| trager | \(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x +\sqrt {x^{4}+1}}{\left (x -1\right ) \left (1+x \right )}\right )}{8}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \ln \left (\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x +\sqrt {x^{4}+1}}{x^{2}+1}\right )}{8}\) | \(73\) |
1/2*(-1/4*arctanh(1/2*2^(1/2)/x*(x^4+1)^(1/2))-1/4*arctan(1/2*2^(1/2)/x*(x ^4+1)^(1/2)))*2^(1/2)
Time = 0.29 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.15 \[ \int \frac {x^2}{\left (-1+x^4\right ) \sqrt {1+x^4}} \, dx=\frac {1}{8} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} x}{\sqrt {x^{4} + 1}}\right ) + \frac {1}{16} \, \sqrt {2} \log \left (\frac {x^{4} - 2 \, \sqrt {2} \sqrt {x^{4} + 1} x + 2 \, x^{2} + 1}{x^{4} - 2 \, x^{2} + 1}\right ) \]
1/8*sqrt(2)*arctan(sqrt(2)*x/sqrt(x^4 + 1)) + 1/16*sqrt(2)*log((x^4 - 2*sq rt(2)*sqrt(x^4 + 1)*x + 2*x^2 + 1)/(x^4 - 2*x^2 + 1))
\[ \int \frac {x^2}{\left (-1+x^4\right ) \sqrt {1+x^4}} \, dx=\int \frac {x^{2}}{\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right ) \sqrt {x^{4} + 1}}\, dx \]
\[ \int \frac {x^2}{\left (-1+x^4\right ) \sqrt {1+x^4}} \, dx=\int { \frac {x^{2}}{\sqrt {x^{4} + 1} {\left (x^{4} - 1\right )}} \,d x } \]
\[ \int \frac {x^2}{\left (-1+x^4\right ) \sqrt {1+x^4}} \, dx=\int { \frac {x^{2}}{\sqrt {x^{4} + 1} {\left (x^{4} - 1\right )}} \,d x } \]
Timed out. \[ \int \frac {x^2}{\left (-1+x^4\right ) \sqrt {1+x^4}} \, dx=\int \frac {x^2}{\left (x^4-1\right )\,\sqrt {x^4+1}} \,d x \]