Integrand size = 17, antiderivative size = 141 \[ \int \frac {1}{\left (1+x^4\right ) \sqrt [4]{2+x^4}} \, dx=-\frac {\arctan \left (1-\frac {\sqrt {2} x}{\sqrt [4]{2+x^4}}\right )}{2 \sqrt {2}}+\frac {\arctan \left (1+\frac {\sqrt {2} x}{\sqrt [4]{2+x^4}}\right )}{2 \sqrt {2}}-\frac {\log \left (1+\frac {x^2}{\sqrt {2+x^4}}-\frac {\sqrt {2} x}{\sqrt [4]{2+x^4}}\right )}{4 \sqrt {2}}+\frac {\log \left (1+\frac {x^2}{\sqrt {2+x^4}}+\frac {\sqrt {2} x}{\sqrt [4]{2+x^4}}\right )}{4 \sqrt {2}} \]
1/4*arctan(-1+x*2^(1/2)/(x^4+2)^(1/4))*2^(1/2)+1/4*arctan(1+x*2^(1/2)/(x^4 +2)^(1/4))*2^(1/2)-1/8*ln(1-x*2^(1/2)/(x^4+2)^(1/4)+x^2/(x^4+2)^(1/2))*2^( 1/2)+1/8*ln(1+x*2^(1/2)/(x^4+2)^(1/4)+x^2/(x^4+2)^(1/2))*2^(1/2)
Time = 0.30 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.54 \[ \int \frac {1}{\left (1+x^4\right ) \sqrt [4]{2+x^4}} \, dx=\frac {\arctan \left (\frac {\sqrt {2} x \sqrt [4]{2+x^4}}{-x^2+\sqrt {2+x^4}}\right )+\text {arctanh}\left (\frac {\sqrt {2} x \sqrt [4]{2+x^4}}{x^2+\sqrt {2+x^4}}\right )}{2 \sqrt {2}} \]
(ArcTan[(Sqrt[2]*x*(2 + x^4)^(1/4))/(-x^2 + Sqrt[2 + x^4])] + ArcTanh[(Sqr t[2]*x*(2 + x^4)^(1/4))/(x^2 + Sqrt[2 + x^4])])/(2*Sqrt[2])
Time = 0.27 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {902, 755, 1476, 1082, 217, 1479, 25, 27, 1103}
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.
| \(\displaystyle \int \frac {1}{\left (x^4+1\right ) \sqrt [4]{x^4+2}} \, dx\) |
| \(\Big \downarrow \) 902 |
| \(\displaystyle \int \frac {1}{\frac {x^4}{x^4+2}+1}d\frac {x}{\sqrt [4]{x^4+2}}\) |
| \(\Big \downarrow \) 755 |
| \(\displaystyle \frac {1}{2} \int \frac {1-\frac {x^2}{\sqrt {x^4+2}}}{\frac {x^4}{x^4+2}+1}d\frac {x}{\sqrt [4]{x^4+2}}+\frac {1}{2} \int \frac {\frac {x^2}{\sqrt {x^4+2}}+1}{\frac {x^4}{x^4+2}+1}d\frac {x}{\sqrt [4]{x^4+2}}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {1}{2} \int \frac {1-\frac {x^2}{\sqrt {x^4+2}}}{\frac {x^4}{x^4+2}+1}d\frac {x}{\sqrt [4]{x^4+2}}+\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{\frac {x^2}{\sqrt {x^4+2}}-\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}+1}d\frac {x}{\sqrt [4]{x^4+2}}+\frac {1}{2} \int \frac {1}{\frac {x^2}{\sqrt {x^4+2}}+\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}+1}d\frac {x}{\sqrt [4]{x^4+2}}\right )\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {1}{2} \left (\frac {\int \frac {1}{-\frac {x^2}{\sqrt {x^4+2}}-1}d\left (1-\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\frac {x^2}{\sqrt {x^4+2}}-1}d\left (\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}+1\right )}{\sqrt {2}}\right )+\frac {1}{2} \int \frac {1-\frac {x^2}{\sqrt {x^4+2}}}{\frac {x^4}{x^4+2}+1}d\frac {x}{\sqrt [4]{x^4+2}}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{2} \int \frac {1-\frac {x^2}{\sqrt {x^4+2}}}{\frac {x^4}{x^4+2}+1}d\frac {x}{\sqrt [4]{x^4+2}}+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}\right )}{\sqrt {2}}\right )\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\int -\frac {\sqrt {2}-\frac {2 x}{\sqrt [4]{x^4+2}}}{\frac {x^2}{\sqrt {x^4+2}}-\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}+1}d\frac {x}{\sqrt [4]{x^4+2}}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}+1\right )}{\frac {x^2}{\sqrt {x^4+2}}+\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}+1}d\frac {x}{\sqrt [4]{x^4+2}}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}\right )}{\sqrt {2}}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-\frac {2 x}{\sqrt [4]{x^4+2}}}{\frac {x^2}{\sqrt {x^4+2}}-\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}+1}d\frac {x}{\sqrt [4]{x^4+2}}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}+1\right )}{\frac {x^2}{\sqrt {x^4+2}}+\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}+1}d\frac {x}{\sqrt [4]{x^4+2}}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}\right )}{\sqrt {2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-\frac {2 x}{\sqrt [4]{x^4+2}}}{\frac {x^2}{\sqrt {x^4+2}}-\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}+1}d\frac {x}{\sqrt [4]{x^4+2}}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}+1}{\frac {x^2}{\sqrt {x^4+2}}+\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}+1}d\frac {x}{\sqrt [4]{x^4+2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}\right )}{\sqrt {2}}\right )\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}+\frac {x^2}{\sqrt {x^4+2}}+1\right )}{2 \sqrt {2}}-\frac {\log \left (-\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}+\frac {x^2}{\sqrt {x^4+2}}+1\right )}{2 \sqrt {2}}\right )\) |
(-(ArcTan[1 - (Sqrt[2]*x)/(2 + x^4)^(1/4)]/Sqrt[2]) + ArcTan[1 + (Sqrt[2]* x)/(2 + x^4)^(1/4)]/Sqrt[2])/2 + (-1/2*Log[1 + x^2/Sqrt[2 + x^4] - (Sqrt[2 ]*x)/(2 + x^4)^(1/4)]/Sqrt[2] + Log[1 + x^2/Sqrt[2 + x^4] + (Sqrt[2]*x)/(2 + x^4)^(1/4)]/(2*Sqrt[2]))/2
3.4.15.3.1 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[(-(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[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] ], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r) Int[(r - s*x^2)/(a + b*x^4) , x], x] + Simp[1/(2*r) Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] & & AtomQ[SplitProduct[SumBaseQ, b]]))
Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Su bst[Int[1/(c - (b*c - a*d)*x^n), x], x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b , c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]
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]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Time = 1.81 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.71
| method | result | size |
| pseudoelliptic | \(-\frac {\sqrt {2}\, \left (\ln \left (\frac {-\left (x^{4}+2\right )^{\frac {1}{4}} \sqrt {2}\, x +x^{2}+\sqrt {x^{4}+2}}{\left (x^{4}+2\right )^{\frac {1}{4}} \sqrt {2}\, x +x^{2}+\sqrt {x^{4}+2}}\right )+2 \arctan \left (\frac {\left (x^{4}+2\right )^{\frac {1}{4}} \sqrt {2}+x}{x}\right )+2 \arctan \left (\frac {\left (x^{4}+2\right )^{\frac {1}{4}} \sqrt {2}-x}{x}\right )\right )}{8}\) | \(100\) |
| trager | \(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (\frac {\left (x^{4}+2\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} x^{3}-\sqrt {x^{4}+2}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) x^{2}+\left (x^{4}+2\right )^{\frac {3}{4}} x +\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3}}{x^{4}+1}\right )}{4}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \ln \left (\frac {-\sqrt {x^{4}+2}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} x^{2}-\left (x^{4}+2\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} x^{3}+\left (x^{4}+2\right )^{\frac {3}{4}} x +\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{x^{4}+1}\right )}{4}\) | \(149\) |
-1/8*2^(1/2)*(ln((-(x^4+2)^(1/4)*2^(1/2)*x+x^2+(x^4+2)^(1/2))/((x^4+2)^(1/ 4)*2^(1/2)*x+x^2+(x^4+2)^(1/2)))+2*arctan(((x^4+2)^(1/4)*2^(1/2)+x)/x)+2*a rctan(((x^4+2)^(1/4)*2^(1/2)-x)/x))
Result contains complex when optimal does not.
Time = 2.39 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.60 \[ \int \frac {1}{\left (1+x^4\right ) \sqrt [4]{2+x^4}} \, dx=-\left (\frac {1}{16} i + \frac {1}{16}\right ) \, \sqrt {2} \log \left (\frac {\left (i + 1\right ) \, \sqrt {2} {\left (x^{4} + 2\right )}^{\frac {1}{4}} x^{3} - 2 i \, \sqrt {x^{4} + 2} x^{2} + \left (i - 1\right ) \, \sqrt {2} {\left (x^{4} + 2\right )}^{\frac {3}{4}} x + 2}{x^{4} + 1}\right ) + \left (\frac {1}{16} i - \frac {1}{16}\right ) \, \sqrt {2} \log \left (\frac {-\left (i - 1\right ) \, \sqrt {2} {\left (x^{4} + 2\right )}^{\frac {1}{4}} x^{3} + 2 i \, \sqrt {x^{4} + 2} x^{2} - \left (i + 1\right ) \, \sqrt {2} {\left (x^{4} + 2\right )}^{\frac {3}{4}} x + 2}{x^{4} + 1}\right ) - \left (\frac {1}{16} i - \frac {1}{16}\right ) \, \sqrt {2} \log \left (\frac {\left (i - 1\right ) \, \sqrt {2} {\left (x^{4} + 2\right )}^{\frac {1}{4}} x^{3} + 2 i \, \sqrt {x^{4} + 2} x^{2} + \left (i + 1\right ) \, \sqrt {2} {\left (x^{4} + 2\right )}^{\frac {3}{4}} x + 2}{x^{4} + 1}\right ) + \left (\frac {1}{16} i + \frac {1}{16}\right ) \, \sqrt {2} \log \left (\frac {-\left (i + 1\right ) \, \sqrt {2} {\left (x^{4} + 2\right )}^{\frac {1}{4}} x^{3} - 2 i \, \sqrt {x^{4} + 2} x^{2} - \left (i - 1\right ) \, \sqrt {2} {\left (x^{4} + 2\right )}^{\frac {3}{4}} x + 2}{x^{4} + 1}\right ) \]
-(1/16*I + 1/16)*sqrt(2)*log(((I + 1)*sqrt(2)*(x^4 + 2)^(1/4)*x^3 - 2*I*sq rt(x^4 + 2)*x^2 + (I - 1)*sqrt(2)*(x^4 + 2)^(3/4)*x + 2)/(x^4 + 1)) + (1/1 6*I - 1/16)*sqrt(2)*log((-(I - 1)*sqrt(2)*(x^4 + 2)^(1/4)*x^3 + 2*I*sqrt(x ^4 + 2)*x^2 - (I + 1)*sqrt(2)*(x^4 + 2)^(3/4)*x + 2)/(x^4 + 1)) - (1/16*I - 1/16)*sqrt(2)*log(((I - 1)*sqrt(2)*(x^4 + 2)^(1/4)*x^3 + 2*I*sqrt(x^4 + 2)*x^2 + (I + 1)*sqrt(2)*(x^4 + 2)^(3/4)*x + 2)/(x^4 + 1)) + (1/16*I + 1/1 6)*sqrt(2)*log((-(I + 1)*sqrt(2)*(x^4 + 2)^(1/4)*x^3 - 2*I*sqrt(x^4 + 2)*x ^2 - (I - 1)*sqrt(2)*(x^4 + 2)^(3/4)*x + 2)/(x^4 + 1))
\[ \int \frac {1}{\left (1+x^4\right ) \sqrt [4]{2+x^4}} \, dx=\int \frac {1}{\left (x^{4} + 1\right ) \sqrt [4]{x^{4} + 2}}\, dx \]
\[ \int \frac {1}{\left (1+x^4\right ) \sqrt [4]{2+x^4}} \, dx=\int { \frac {1}{{\left (x^{4} + 2\right )}^{\frac {1}{4}} {\left (x^{4} + 1\right )}} \,d x } \]
\[ \int \frac {1}{\left (1+x^4\right ) \sqrt [4]{2+x^4}} \, dx=\int { \frac {1}{{\left (x^{4} + 2\right )}^{\frac {1}{4}} {\left (x^{4} + 1\right )}} \,d x } \]
Timed out. \[ \int \frac {1}{\left (1+x^4\right ) \sqrt [4]{2+x^4}} \, dx=\int \frac {1}{\left (x^4+1\right )\,{\left (x^4+2\right )}^{1/4}} \,d x \]
\[ \int \frac {1}{\left (1+x^4\right ) \sqrt [4]{2+x^4}} \, dx=\int \frac {1}{\left (x^{4}+2\right )^{\frac {1}{4}} x^{4}+\left (x^{4}+2\right )^{\frac {1}{4}}}d x \]