Integrand size = 27, antiderivative size = 75 \[ \int \frac {\left (-1+x^4\right ) \sqrt {1+x^4}}{1+3 x^4+x^8} \, dx=-\frac {\arctan \left (\frac {\sqrt {2} x \sqrt {1+x^4}}{1-x^2+x^4}\right )}{2 \sqrt {2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} x \sqrt {1+x^4}}{1+x^2+x^4}\right )}{2 \sqrt {2}} \]
-1/4*arctan(2^(1/2)*x*(x^4+1)^(1/2)/(x^4-x^2+1))*2^(1/2)-1/4*arctanh(2^(1/ 2)*x*(x^4+1)^(1/2)/(x^4+x^2+1))*2^(1/2)
Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.88 \[ \int \frac {\left (-1+x^4\right ) \sqrt {1+x^4}}{1+3 x^4+x^8} \, dx=-\frac {\arctan \left (\frac {\sqrt {2} x \sqrt {1+x^4}}{1-x^2+x^4}\right )+\text {arctanh}\left (\frac {\sqrt {2} x \sqrt {1+x^4}}{1+x^2+x^4}\right )}{2 \sqrt {2}} \]
-1/2*(ArcTan[(Sqrt[2]*x*Sqrt[1 + x^4])/(1 - x^2 + x^4)] + ArcTanh[(Sqrt[2] *x*Sqrt[1 + x^4])/(1 + x^2 + x^4)])/Sqrt[2]
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 2.20 (sec) , antiderivative size = 1130, normalized size of antiderivative = 15.07, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {7279, 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.
| \(\displaystyle \int \frac {\left (x^4-1\right ) \sqrt {x^4+1}}{x^8+3 x^4+1} \, dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (\frac {\left (1-\sqrt {5}\right ) \sqrt {x^4+1}}{2 x^4-\sqrt {5}+3}+\frac {\left (1+\sqrt {5}\right ) \sqrt {x^4+1}}{2 x^4+\sqrt {5}+3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\sqrt [4]{-1} \left (2 i-\sqrt {6-2 \sqrt {5}}\right ) \left (2 i-\sqrt {2 \left (3+\sqrt {5}\right )}\right ) \arctan \left (\frac {\sqrt [4]{-1} x}{\sqrt {x^4+1}}\right )}{16 \sqrt {5}}-\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) \arctan \left (\frac {\sqrt [4]{-1} x}{\sqrt {x^4+1}}\right )}{\sqrt {2}}+\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) \left (2 i+\sqrt {6-2 \sqrt {5}}\right ) \left (2 i+\sqrt {2 \left (3+\sqrt {5}\right )}\right ) \text {arctanh}\left (\frac {\sqrt [4]{-1} x}{\sqrt {x^4+1}}\right )}{\sqrt {10}}-\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) \text {arctanh}\left (\frac {\sqrt [4]{-1} x}{\sqrt {x^4+1}}\right )}{\sqrt {2}}-\frac {\left (3-\sqrt {5}\right ) \left (2+i \sqrt {2 \left (3+\sqrt {5}\right )}\right ) \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 )}{8 \left (5-\sqrt {5}\right ) \sqrt {x^4+1}}-\frac {\left (3-\sqrt {5}\right ) \left (2-i \sqrt {2 \left (3+\sqrt {5}\right )}\right ) \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 )}{8 \left (5-\sqrt {5}\right ) \sqrt {x^4+1}}-\frac {\left (3+\sqrt {5}\right ) \left (1+i \sqrt {\frac {2}{3+\sqrt {5}}}\right ) \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 \left (5+\sqrt {5}\right ) \sqrt {x^4+1}}-\frac {\left (3+\sqrt {5}\right ) \left (1-i \sqrt {\frac {2}{3+\sqrt {5}}}\right ) \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 \left (5+\sqrt {5}\right ) \sqrt {x^4+1}}+\frac {\left (1+\sqrt {5}\right ) \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 {\left (1-\sqrt {5}\right ) \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 {\left (5-2 \sqrt {5}\right ) \left (3+\sqrt {5}\right ) \left (1+\sqrt {5}-2 i \sqrt {2 \left (3+\sqrt {5}\right )}\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticPi}\left (\frac {1}{4} i \sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} \left (i-\sqrt {\frac {2}{3+\sqrt {5}}}\right )^2,2 \arctan (x),\frac {1}{2}\right )}{160 \sqrt {x^4+1}}+\frac {\left (5-2 \sqrt {5}\right ) \left (3+\sqrt {5}\right ) \left (1+\sqrt {5}+2 i \sqrt {2 \left (3+\sqrt {5}\right )}\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticPi}\left (-\frac {1}{4} i \sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} \left (i+\sqrt {\frac {2}{3+\sqrt {5}}}\right )^2,2 \arctan (x),\frac {1}{2}\right )}{160 \sqrt {x^4+1}}-\frac {\left (2 i-\sqrt {6-2 \sqrt {5}}\right ) \left (2-i \sqrt {2 \left (3+\sqrt {5}\right )}\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticPi}\left (\frac {1}{16} i \sqrt {\frac {1}{2} \left (3-\sqrt {5}\right )} \left (2 i-\sqrt {2 \left (3+\sqrt {5}\right )}\right )^2,2 \arctan (x),\frac {1}{2}\right )}{32 \sqrt {5} \sqrt {x^4+1}}+\frac {\left (2-i \sqrt {6-2 \sqrt {5}}\right ) \left (2 i-\sqrt {2 \left (3+\sqrt {5}\right )}\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticPi}\left (-\frac {1}{16} i \sqrt {\frac {1}{2} \left (3-\sqrt {5}\right )} \left (2 i+\sqrt {2 \left (3+\sqrt {5}\right )}\right )^2,2 \arctan (x),\frac {1}{2}\right )}{32 \sqrt {5} \sqrt {x^4+1}}\) |
((-1/4 + I/4)*ArcTan[((-1)^(1/4)*x)/Sqrt[1 + x^4]])/Sqrt[2] - ((-1)^(1/4)* (2*I - Sqrt[6 - 2*Sqrt[5]])*(2*I - Sqrt[2*(3 + Sqrt[5])])*ArcTan[((-1)^(1/ 4)*x)/Sqrt[1 + x^4]])/(16*Sqrt[5]) - ((1/4 - I/4)*ArcTanh[((-1)^(1/4)*x)/S qrt[1 + x^4]])/Sqrt[2] + ((1/16 + I/16)*(2*I + Sqrt[6 - 2*Sqrt[5]])*(2*I + Sqrt[2*(3 + Sqrt[5])])*ArcTanh[((-1)^(1/4)*x)/Sqrt[1 + x^4]])/Sqrt[10] + ((1 - Sqrt[5])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x] , 1/2])/(4*Sqrt[1 + x^4]) + ((1 + Sqrt[5])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x ^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(4*Sqrt[1 + x^4]) - ((3 + Sqrt[5])*(1 - I*Sqrt[2/(3 + Sqrt[5])])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF [2*ArcTan[x], 1/2])/(4*(5 + Sqrt[5])*Sqrt[1 + x^4]) - ((3 + Sqrt[5])*(1 + I*Sqrt[2/(3 + Sqrt[5])])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2 *ArcTan[x], 1/2])/(4*(5 + Sqrt[5])*Sqrt[1 + x^4]) - ((3 - Sqrt[5])*(2 - I* Sqrt[2*(3 + Sqrt[5])])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*A rcTan[x], 1/2])/(8*(5 - Sqrt[5])*Sqrt[1 + x^4]) - ((3 - Sqrt[5])*(2 + I*Sq rt[2*(3 + Sqrt[5])])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*Arc Tan[x], 1/2])/(8*(5 - Sqrt[5])*Sqrt[1 + x^4]) + ((5 - 2*Sqrt[5])*(3 + Sqrt [5])*(1 + Sqrt[5] - (2*I)*Sqrt[2*(3 + Sqrt[5])])*(1 + x^2)*Sqrt[(1 + x^4)/ (1 + x^2)^2]*EllipticPi[(I/4)*Sqrt[(3 + Sqrt[5])/2]*(I - Sqrt[2/(3 + Sqrt[ 5])])^2, 2*ArcTan[x], 1/2])/(160*Sqrt[1 + x^4]) + ((5 - 2*Sqrt[5])*(3 + Sq rt[5])*(1 + Sqrt[5] + (2*I)*Sqrt[2*(3 + Sqrt[5])])*(1 + x^2)*Sqrt[(1 + ...
3.10.96.3.1 Defintions of rubi rules used
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 9.21 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.25
| method | result | size |
| pseudoelliptic | \(\frac {\sqrt {2}\, \left (\ln \left (\frac {x^{4}-\sqrt {2}\, x \sqrt {x^{4}+1}+x^{2}+1}{x^{4}+\sqrt {2}\, x \sqrt {x^{4}+1}+x^{2}+1}\right )+2 \arctan \left (\frac {\sqrt {x^{4}+1}\, \sqrt {2}+x}{x}\right )+2 \arctan \left (\frac {\sqrt {x^{4}+1}\, \sqrt {2}-x}{x}\right )\right )}{8}\) | \(94\) |
| default | \(\frac {\left (\frac {\ln \left (\frac {x^{4}+1}{x^{2}}-\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{x}+1\right )}{4}+\frac {\arctan \left (-1+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{x}\right )}{2}-\frac {\ln \left (\frac {x^{4}+1}{x^{2}}+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{x}+1\right )}{4}+\frac {\arctan \left (\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{x}+1\right )}{2}\right ) \sqrt {2}}{2}\) | \(102\) |
| elliptic | \(\frac {\left (\frac {\ln \left (\frac {x^{4}+1}{x^{2}}-\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{x}+1\right )}{4}+\frac {\arctan \left (-1+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{x}\right )}{2}-\frac {\ln \left (\frac {x^{4}+1}{x^{2}}+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{x}+1\right )}{4}+\frac {\arctan \left (\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{x}+1\right )}{2}\right ) \sqrt {2}}{2}\) | \(102\) |
| trager | \(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} x^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) x^{4}+2 \sqrt {x^{4}+1}\, x -\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} x^{2}+x^{4}+1}\right )}{4}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{5} x^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} x^{4}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3}-2 \sqrt {x^{4}+1}\, x}{\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} x^{2}-x^{4}-1}\right )}{4}\) | \(150\) |
1/8*2^(1/2)*(ln((x^4-2^(1/2)*x*(x^4+1)^(1/2)+x^2+1)/(x^4+2^(1/2)*x*(x^4+1) ^(1/2)+x^2+1))+2*arctan(((x^4+1)^(1/2)*2^(1/2)+x)/x)+2*arctan(((x^4+1)^(1/ 2)*2^(1/2)-x)/x))
Result contains complex when optimal does not.
Time = 0.35 (sec) , antiderivative size = 261, normalized size of antiderivative = 3.48 \[ \int \frac {\left (-1+x^4\right ) \sqrt {1+x^4}}{1+3 x^4+x^8} \, dx=-\left (\frac {1}{16} i + \frac {1}{16}\right ) \, \sqrt {2} \log \left (\frac {\sqrt {2} {\left (\left (i + 1\right ) \, x^{8} - \left (2 i - 2\right ) \, x^{6} + \left (i + 1\right ) \, x^{4} - \left (2 i - 2\right ) \, x^{2} + i + 1\right )} + 4 \, {\left (x^{5} - i \, x^{3} + x\right )} \sqrt {x^{4} + 1}}{x^{8} + 3 \, x^{4} + 1}\right ) + \left (\frac {1}{16} i - \frac {1}{16}\right ) \, \sqrt {2} \log \left (\frac {\sqrt {2} {\left (-\left (i - 1\right ) \, x^{8} + \left (2 i + 2\right ) \, x^{6} - \left (i - 1\right ) \, x^{4} + \left (2 i + 2\right ) \, x^{2} - i + 1\right )} + 4 \, {\left (x^{5} + i \, x^{3} + x\right )} \sqrt {x^{4} + 1}}{x^{8} + 3 \, x^{4} + 1}\right ) - \left (\frac {1}{16} i - \frac {1}{16}\right ) \, \sqrt {2} \log \left (\frac {\sqrt {2} {\left (\left (i - 1\right ) \, x^{8} - \left (2 i + 2\right ) \, x^{6} + \left (i - 1\right ) \, x^{4} - \left (2 i + 2\right ) \, x^{2} + i - 1\right )} + 4 \, {\left (x^{5} + i \, x^{3} + x\right )} \sqrt {x^{4} + 1}}{x^{8} + 3 \, x^{4} + 1}\right ) + \left (\frac {1}{16} i + \frac {1}{16}\right ) \, \sqrt {2} \log \left (\frac {\sqrt {2} {\left (-\left (i + 1\right ) \, x^{8} + \left (2 i - 2\right ) \, x^{6} - \left (i + 1\right ) \, x^{4} + \left (2 i - 2\right ) \, x^{2} - i - 1\right )} + 4 \, {\left (x^{5} - i \, x^{3} + x\right )} \sqrt {x^{4} + 1}}{x^{8} + 3 \, x^{4} + 1}\right ) \]
-(1/16*I + 1/16)*sqrt(2)*log((sqrt(2)*((I + 1)*x^8 - (2*I - 2)*x^6 + (I + 1)*x^4 - (2*I - 2)*x^2 + I + 1) + 4*(x^5 - I*x^3 + x)*sqrt(x^4 + 1))/(x^8 + 3*x^4 + 1)) + (1/16*I - 1/16)*sqrt(2)*log((sqrt(2)*(-(I - 1)*x^8 + (2*I + 2)*x^6 - (I - 1)*x^4 + (2*I + 2)*x^2 - I + 1) + 4*(x^5 + I*x^3 + x)*sqrt (x^4 + 1))/(x^8 + 3*x^4 + 1)) - (1/16*I - 1/16)*sqrt(2)*log((sqrt(2)*((I - 1)*x^8 - (2*I + 2)*x^6 + (I - 1)*x^4 - (2*I + 2)*x^2 + I - 1) + 4*(x^5 + I*x^3 + x)*sqrt(x^4 + 1))/(x^8 + 3*x^4 + 1)) + (1/16*I + 1/16)*sqrt(2)*log ((sqrt(2)*(-(I + 1)*x^8 + (2*I - 2)*x^6 - (I + 1)*x^4 + (2*I - 2)*x^2 - I - 1) + 4*(x^5 - I*x^3 + x)*sqrt(x^4 + 1))/(x^8 + 3*x^4 + 1))
\[ \int \frac {\left (-1+x^4\right ) \sqrt {1+x^4}}{1+3 x^4+x^8} \, dx=\int \frac {\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right ) \sqrt {x^{4} + 1}}{x^{8} + 3 x^{4} + 1}\, dx \]
\[ \int \frac {\left (-1+x^4\right ) \sqrt {1+x^4}}{1+3 x^4+x^8} \, dx=\int { \frac {\sqrt {x^{4} + 1} {\left (x^{4} - 1\right )}}{x^{8} + 3 \, x^{4} + 1} \,d x } \]
\[ \int \frac {\left (-1+x^4\right ) \sqrt {1+x^4}}{1+3 x^4+x^8} \, dx=\int { \frac {\sqrt {x^{4} + 1} {\left (x^{4} - 1\right )}}{x^{8} + 3 \, x^{4} + 1} \,d x } \]
Timed out. \[ \int \frac {\left (-1+x^4\right ) \sqrt {1+x^4}}{1+3 x^4+x^8} \, dx=\int \frac {\left (x^4-1\right )\,\sqrt {x^4+1}}{x^8+3\,x^4+1} \,d x \]