Integrand size = 27, antiderivative size = 225 \[ \int \frac {x^4}{\sqrt [4]{1+x^4} \left (1+2 x^4+2 x^8\right )} \, dx=\frac {1}{8} \sqrt {2+\sqrt {2}} \arctan \left (\frac {\sqrt {2-\sqrt {2}} x \sqrt [4]{1+x^4}}{-x^2+\sqrt {1+x^4}}\right )-\frac {1}{8} \sqrt {2-\sqrt {2}} \arctan \left (\frac {\sqrt {2+\sqrt {2}} x \sqrt [4]{1+x^4}}{-x^2+\sqrt {1+x^4}}\right )+\frac {1}{8} \sqrt {2+\sqrt {2}} \text {arctanh}\left (\frac {\sqrt {2-\sqrt {2}} x \sqrt [4]{1+x^4}}{x^2+\sqrt {1+x^4}}\right )-\frac {1}{8} \sqrt {2-\sqrt {2}} \text {arctanh}\left (\frac {\sqrt {2+\sqrt {2}} x \sqrt [4]{1+x^4}}{x^2+\sqrt {1+x^4}}\right ) \]
1/8*(2+2^(1/2))^(1/2)*arctan((2-2^(1/2))^(1/2)*x*(x^4+1)^(1/4)/(-x^2+(x^4+ 1)^(1/2)))-1/8*(2-2^(1/2))^(1/2)*arctan((2+2^(1/2))^(1/2)*x*(x^4+1)^(1/4)/ (-x^2+(x^4+1)^(1/2)))+1/8*(2+2^(1/2))^(1/2)*arctanh((2-2^(1/2))^(1/2)*x*(x ^4+1)^(1/4)/(x^2+(x^4+1)^(1/2)))-1/8*(2-2^(1/2))^(1/2)*arctanh((2+2^(1/2)) ^(1/2)*x*(x^4+1)^(1/4)/(x^2+(x^4+1)^(1/2)))
Time = 0.57 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.97 \[ \int \frac {x^4}{\sqrt [4]{1+x^4} \left (1+2 x^4+2 x^8\right )} \, dx=\frac {1}{8} \left (\sqrt {2-\sqrt {2}} \arctan \left (\frac {\sqrt {2+\sqrt {2}} x \sqrt [4]{1+x^4}}{x^2-\sqrt {1+x^4}}\right )+\sqrt {2+\sqrt {2}} \arctan \left (\frac {\sqrt {2-\sqrt {2}} x \sqrt [4]{1+x^4}}{-x^2+\sqrt {1+x^4}}\right )+\sqrt {2+\sqrt {2}} \text {arctanh}\left (\frac {\sqrt {2-\sqrt {2}} x \sqrt [4]{1+x^4}}{x^2+\sqrt {1+x^4}}\right )-\sqrt {2-\sqrt {2}} \text {arctanh}\left (\frac {\sqrt {2+\sqrt {2}} x \sqrt [4]{1+x^4}}{x^2+\sqrt {1+x^4}}\right )\right ) \]
(Sqrt[2 - Sqrt[2]]*ArcTan[(Sqrt[2 + Sqrt[2]]*x*(1 + x^4)^(1/4))/(x^2 - Sqr t[1 + x^4])] + Sqrt[2 + Sqrt[2]]*ArcTan[(Sqrt[2 - Sqrt[2]]*x*(1 + x^4)^(1/ 4))/(-x^2 + Sqrt[1 + x^4])] + Sqrt[2 + Sqrt[2]]*ArcTanh[(Sqrt[2 - Sqrt[2]] *x*(1 + x^4)^(1/4))/(x^2 + Sqrt[1 + x^4])] - Sqrt[2 - Sqrt[2]]*ArcTanh[(Sq rt[2 + Sqrt[2]]*x*(1 + x^4)^(1/4))/(x^2 + Sqrt[1 + x^4])])/8
Time = 0.38 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.08, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {1852, 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.
| \(\displaystyle \int \frac {x^4}{\sqrt [4]{x^4+1} \left (2 x^8+2 x^4+1\right )} \, dx\) |
| \(\Big \downarrow \) 1852 |
| \(\displaystyle \int \left (\frac {1+i}{\left (4 x^4+(2-2 i)\right ) \sqrt [4]{x^4+1}}+\frac {1-i}{\left (4 x^4+(2+2 i)\right ) \sqrt [4]{x^4+1}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {1}{4} (-1)^{5/8} \arctan \left (\frac {(-1)^{7/8} x}{\sqrt [4]{x^4+1}}\right )-\frac {(-1)^{5/8} \arctan \left (1-\frac {(-1)^{7/8} \sqrt {2} x}{\sqrt [4]{x^4+1}}\right )}{4 \sqrt {2}}+\frac {(-1)^{5/8} \arctan \left (\frac {(-1)^{7/8} \sqrt {2} x}{\sqrt [4]{x^4+1}}+1\right )}{4 \sqrt {2}}-\frac {1}{4} (-1)^{5/8} \text {arctanh}\left (\frac {(-1)^{7/8} x}{\sqrt [4]{x^4+1}}\right )-\frac {(-1)^{5/8} \log \left (\frac {\sqrt [8]{-1} \sqrt {2} x}{\sqrt [4]{x^4+1}}+\frac {x^2}{\sqrt {x^4+1}}+\sqrt [4]{-1}\right )}{8 \sqrt {2}}+\frac {(-1)^{5/8} \log \left (\frac {(-1)^{7/8} \sqrt {2} x}{\sqrt [4]{x^4+1}}-\frac {(-1)^{3/4} x^2}{\sqrt {x^4+1}}+1\right )}{8 \sqrt {2}}\) |
-1/4*((-1)^(5/8)*ArcTan[((-1)^(7/8)*x)/(1 + x^4)^(1/4)]) - ((-1)^(5/8)*Arc Tan[1 - ((-1)^(7/8)*Sqrt[2]*x)/(1 + x^4)^(1/4)])/(4*Sqrt[2]) + ((-1)^(5/8) *ArcTan[1 + ((-1)^(7/8)*Sqrt[2]*x)/(1 + x^4)^(1/4)])/(4*Sqrt[2]) - ((-1)^( 5/8)*ArcTanh[((-1)^(7/8)*x)/(1 + x^4)^(1/4)])/4 - ((-1)^(5/8)*Log[(-1)^(1/ 4) + x^2/Sqrt[1 + x^4] + ((-1)^(1/8)*Sqrt[2]*x)/(1 + x^4)^(1/4)])/(8*Sqrt[ 2]) + ((-1)^(5/8)*Log[1 - ((-1)^(3/4)*x^2)/Sqrt[1 + x^4] + ((-1)^(7/8)*Sqr t[2]*x)/(1 + x^4)^(1/4)])/(8*Sqrt[2])
3.26.99.3.1 Defintions of rubi rules used
Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))^(q_))/((a_) + (c_.)*(x_)^( n2_.) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^n)^q, ( f*x)^m/(a + b*x^n + c*x^(2*n)), x], x] /; FreeQ[{a, b, c, d, e, f, q, n}, x ] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && !IntegerQ[q] && IntegerQ[m]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 3.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.15
| method | result | size |
| pseudoelliptic | \(-\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )}{\sum }\frac {\ln \left (\frac {-\textit {\_R} x +\left (x^{4}+1\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}^{5}}\right )}{8}\) | \(33\) |
| trager | \(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{7} \ln \left (\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{11} x^{4}-2 \left (x^{4}+1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{6} x^{3}-2 \sqrt {x^{4}+1}\, \operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{5} x^{2}+2 \left (x^{4}+1\right )^{\frac {3}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{4} x -\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{7}+\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{3} x^{4}+2 \left (x^{4}+1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{2} x^{3}-2 \sqrt {x^{4}+1}\, \operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right ) x^{2}-2 \left (x^{4}+1\right )^{\frac {3}{4}} x +\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{3}}{\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{4} x^{4}-x^{4}-1}\right )}{8}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{3} \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{11} x^{4}+2 \operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{7} x^{4}+2 \left (x^{4}+1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{6} x^{3}+2 \sqrt {x^{4}+1}\, \operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{5} x^{2}+2 \left (x^{4}+1\right )^{\frac {3}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{4} x +\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{7}+\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{3} x^{4}-2 \left (x^{4}+1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{2} x^{3}-2 \sqrt {x^{4}+1}\, \operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right ) x^{2}-2 \left (x^{4}+1\right )^{\frac {3}{4}} x +\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{3}}{\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{4} x^{4}-x^{4}-1}\right )}{8}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{5} \ln \left (-\frac {-2 \sqrt {x^{4}+1}\, \operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{7} x^{2}-2 \left (x^{4}+1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{6} x^{3}-\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{5} x^{4}+\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right ) x^{4}+2 \left (x^{4}+1\right )^{\frac {3}{4}} x +\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )}{\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{4} x^{4}+x^{4}+1}\right )}{8}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{9} x^{4}+2 \left (x^{4}+1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{6} x^{3}-\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{5} x^{4}+2 \sqrt {x^{4}+1}\, \operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{3} x^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{5}+2 \left (x^{4}+1\right )^{\frac {3}{4}} x}{\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )^{4} x^{4}+x^{4}+1}\right )}{8}\) | \(606\) |
Result contains complex when optimal does not.
Time = 5.51 (sec) , antiderivative size = 1113, normalized size of antiderivative = 4.95 \[ \int \frac {x^4}{\sqrt [4]{1+x^4} \left (1+2 x^4+2 x^8\right )} \, dx=\text {Too large to display} \]
-(1/32*I - 1/32)*sqrt(2)*(-1)^(1/8)*log(-(2*sqrt(2)*sqrt(x^4 + 1)*((I + 1) *(-1)^(7/8)*x^2 + (-1)^(3/8)*(-(2*I + 2)*x^6 - (I + 1)*x^2)) + 4*(2*x^5 - (I - 1)*x)*(x^4 + 1)^(3/4) - sqrt(2)*((-1)^(5/8)*((2*I - 2)*x^8 + (4*I - 4 )*x^4 + I - 1) + (-1)^(1/8)*(-(2*I - 2)*x^8 + I - 1)) + 4*(x^4 + 1)^(1/4)* (I*(-1)^(1/4)*x^3 + (-1)^(3/4)*(2*I*x^7 + I*x^3)))/(2*x^8 + 2*x^4 + 1)) + (1/32*I + 1/32)*sqrt(2)*(-1)^(1/8)*log(-(2*sqrt(2)*sqrt(x^4 + 1)*(-(I - 1) *(-1)^(7/8)*x^2 + (-1)^(3/8)*((2*I - 2)*x^6 + (I - 1)*x^2)) + 4*(2*x^5 - ( I - 1)*x)*(x^4 + 1)^(3/4) - sqrt(2)*((-1)^(5/8)*(-(2*I + 2)*x^8 - (4*I + 4 )*x^4 - I - 1) + (-1)^(1/8)*((2*I + 2)*x^8 - I - 1)) + 4*(x^4 + 1)^(1/4)*( -I*(-1)^(1/4)*x^3 + (-1)^(3/4)*(-2*I*x^7 - I*x^3)))/(2*x^8 + 2*x^4 + 1)) - (1/32*I + 1/32)*sqrt(2)*(-1)^(1/8)*log(-(2*sqrt(2)*sqrt(x^4 + 1)*((I - 1) *(-1)^(7/8)*x^2 + (-1)^(3/8)*(-(2*I - 2)*x^6 - (I - 1)*x^2)) + 4*(2*x^5 - (I - 1)*x)*(x^4 + 1)^(3/4) - sqrt(2)*((-1)^(5/8)*((2*I + 2)*x^8 + (4*I + 4 )*x^4 + I + 1) + (-1)^(1/8)*(-(2*I + 2)*x^8 + I + 1)) + 4*(x^4 + 1)^(1/4)* (-I*(-1)^(1/4)*x^3 + (-1)^(3/4)*(-2*I*x^7 - I*x^3)))/(2*x^8 + 2*x^4 + 1)) + (1/32*I - 1/32)*sqrt(2)*(-1)^(1/8)*log(-(2*sqrt(2)*sqrt(x^4 + 1)*(-(I + 1)*(-1)^(7/8)*x^2 + (-1)^(3/8)*((2*I + 2)*x^6 + (I + 1)*x^2)) + 4*(2*x^5 - (I - 1)*x)*(x^4 + 1)^(3/4) - sqrt(2)*((-1)^(5/8)*(-(2*I - 2)*x^8 - (4*I - 4)*x^4 - I + 1) + (-1)^(1/8)*((2*I - 2)*x^8 - I + 1)) + 4*(x^4 + 1)^(1/4) *(I*(-1)^(1/4)*x^3 + (-1)^(3/4)*(2*I*x^7 + I*x^3)))/(2*x^8 + 2*x^4 + 1)...
Timed out. \[ \int \frac {x^4}{\sqrt [4]{1+x^4} \left (1+2 x^4+2 x^8\right )} \, dx=\text {Timed out} \]
\[ \int \frac {x^4}{\sqrt [4]{1+x^4} \left (1+2 x^4+2 x^8\right )} \, dx=\int { \frac {x^{4}}{{\left (2 \, x^{8} + 2 \, x^{4} + 1\right )} {\left (x^{4} + 1\right )}^{\frac {1}{4}}} \,d x } \]
\[ \int \frac {x^4}{\sqrt [4]{1+x^4} \left (1+2 x^4+2 x^8\right )} \, dx=\int { \frac {x^{4}}{{\left (2 \, x^{8} + 2 \, x^{4} + 1\right )} {\left (x^{4} + 1\right )}^{\frac {1}{4}}} \,d x } \]
Timed out. \[ \int \frac {x^4}{\sqrt [4]{1+x^4} \left (1+2 x^4+2 x^8\right )} \, dx=\int \frac {x^4}{{\left (x^4+1\right )}^{1/4}\,\left (2\,x^8+2\,x^4+1\right )} \,d x \]