Integrand size = 30, antiderivative size = 105 \[ \int \frac {\sqrt [4]{-1+x^4} \left (2+x^4\right )}{x^2 \left (2+2 x^4+x^8\right )} \, dx=-\frac {\sqrt [4]{-1+x^4}}{x}+\frac {1}{8} \text {RootSum}\left [5-6 \text {$\#$1}^4+2 \text {$\#$1}^8\&,\frac {-5 \log (x)+5 \log \left (\sqrt [4]{-1+x^4}-x \text {$\#$1}\right )+3 \log (x) \text {$\#$1}^4-3 \log \left (\sqrt [4]{-1+x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{-3 \text {$\#$1}^3+2 \text {$\#$1}^7}\&\right ] \] Output:
Unintegrable
Time = 0.00 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt [4]{-1+x^4} \left (2+x^4\right )}{x^2 \left (2+2 x^4+x^8\right )} \, dx=-\frac {\sqrt [4]{-1+x^4}}{x}+\frac {1}{8} \text {RootSum}\left [5-6 \text {$\#$1}^4+2 \text {$\#$1}^8\&,\frac {-5 \log (x)+5 \log \left (\sqrt [4]{-1+x^4}-x \text {$\#$1}\right )+3 \log (x) \text {$\#$1}^4-3 \log \left (\sqrt [4]{-1+x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{-3 \text {$\#$1}^3+2 \text {$\#$1}^7}\&\right ] \] Input:
Integrate[((-1 + x^4)^(1/4)*(2 + x^4))/(x^2*(2 + 2*x^4 + x^8)),x]
Output:
-((-1 + x^4)^(1/4)/x) + RootSum[5 - 6*#1^4 + 2*#1^8 & , (-5*Log[x] + 5*Log [(-1 + x^4)^(1/4) - x*#1] + 3*Log[x]*#1^4 - 3*Log[(-1 + x^4)^(1/4) - x*#1] *#1^4)/(-3*#1^3 + 2*#1^7) & ]/8
Result contains complex when optimal does not.
Time = 1.34 (sec) , antiderivative size = 295, normalized size of antiderivative = 2.81, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, 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.
\(\displaystyle \int \frac {\sqrt [4]{x^4-1} \left (x^4+2\right )}{x^2 \left (x^8+2 x^4+2\right )} \, dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {\sqrt [4]{x^4-1}}{x^2}+\frac {\left (-x^4-1\right ) \sqrt [4]{x^4-1} x^2}{x^8+2 x^4+2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\left (\frac {1}{12}+\frac {i}{12}\right ) \sqrt [4]{x^4-1} x^3 \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{4},1,\frac {7}{4},x^4,\left (-\frac {1}{2}+\frac {i}{2}\right ) x^4\right )}{\sqrt [4]{1-x^4}}-\frac {\left (\frac {1}{12}-\frac {i}{12}\right ) \sqrt [4]{x^4-1} x^3 \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},\left (-\frac {1}{2}-\frac {i}{2}\right ) x^4,x^4\right )}{\sqrt [4]{1-x^4}}-\left (\frac {1}{4}+\frac {i}{2}\right ) \left (\frac {3}{5}-\frac {i}{5}\right )^{3/4} \arctan \left (\frac {x}{\sqrt [4]{\frac {3}{5}-\frac {i}{5}} \sqrt [4]{x^4-1}}\right )-\left (\frac {1}{4}-\frac {i}{2}\right ) \left (\frac {3}{5}+\frac {i}{5}\right )^{3/4} \arctan \left (\frac {x}{\sqrt [4]{\frac {3}{5}+\frac {i}{5}} \sqrt [4]{x^4-1}}\right )+\left (\frac {1}{4}+\frac {i}{2}\right ) \left (\frac {3}{5}-\frac {i}{5}\right )^{3/4} \text {arctanh}\left (\frac {x}{\sqrt [4]{\frac {3}{5}-\frac {i}{5}} \sqrt [4]{x^4-1}}\right )+\left (\frac {1}{4}-\frac {i}{2}\right ) \left (\frac {3}{5}+\frac {i}{5}\right )^{3/4} \text {arctanh}\left (\frac {x}{\sqrt [4]{\frac {3}{5}+\frac {i}{5}} \sqrt [4]{x^4-1}}\right )-\frac {\sqrt [4]{x^4-1}}{x}\) |
Input:
Int[((-1 + x^4)^(1/4)*(2 + x^4))/(x^2*(2 + 2*x^4 + x^8)),x]
Output:
-((-1 + x^4)^(1/4)/x) - ((1/12 + I/12)*x^3*(-1 + x^4)^(1/4)*AppellF1[3/4, -1/4, 1, 7/4, x^4, (-1/2 + I/2)*x^4])/(1 - x^4)^(1/4) - ((1/12 - I/12)*x^3 *(-1 + x^4)^(1/4)*AppellF1[3/4, 1, -1/4, 7/4, (-1/2 - I/2)*x^4, x^4])/(1 - x^4)^(1/4) - (1/4 + I/2)*(3/5 - I/5)^(3/4)*ArcTan[x/((3/5 - I/5)^(1/4)*(- 1 + x^4)^(1/4))] - (1/4 - I/2)*(3/5 + I/5)^(3/4)*ArcTan[x/((3/5 + I/5)^(1/ 4)*(-1 + x^4)^(1/4))] + (1/4 + I/2)*(3/5 - I/5)^(3/4)*ArcTanh[x/((3/5 - I/ 5)^(1/4)*(-1 + x^4)^(1/4))] + (1/4 - I/2)*(3/5 + I/5)^(3/4)*ArcTanh[x/((3/ 5 + I/5)^(1/4)*(-1 + x^4)^(1/4))]
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]
Result contains higher order function than in optimal. Order 9 vs. order 1.
Time = 14.03 (sec) , antiderivative size = 4179, normalized size of antiderivative = 39.80
\[\text {output too large to display}\]
Input:
int((x^4-1)^(1/4)*(x^4+2)/x^2/(x^8+2*x^4+2),x)
Output:
-(x^4-1)^(1/4)/x+(-1024*ln((-4096*x^12*RootOf(33554432*_Z^8-24576*_Z^4+5)^ 4*RootOf(_Z^4+4096*RootOf(33554432*_Z^8-24576*_Z^4+5)^4-3)^2-16384*(x^12-3 *x^8+3*x^4-1)^(1/4)*RootOf(33554432*_Z^8-24576*_Z^4+5)^4*RootOf(_Z^4+4096* RootOf(33554432*_Z^8-24576*_Z^4+5)^4-3)^3*x^9+8192*RootOf(_Z^4+4096*RootOf (33554432*_Z^8-24576*_Z^4+5)^4-3)^2*RootOf(33554432*_Z^8-24576*_Z^4+5)^4*x ^8+4*x^12*RootOf(_Z^4+4096*RootOf(33554432*_Z^8-24576*_Z^4+5)^4-3)^2+32768 *(x^12-3*x^8+3*x^4-1)^(1/4)*RootOf(33554432*_Z^8-24576*_Z^4+5)^4*RootOf(_Z ^4+4096*RootOf(33554432*_Z^8-24576*_Z^4+5)^4-3)^3*x^5+6*(x^12-3*x^8+3*x^4- 1)^(1/4)*RootOf(_Z^4+4096*RootOf(33554432*_Z^8-24576*_Z^4+5)^4-3)^3*x^9+81 92*RootOf(33554432*_Z^8-24576*_Z^4+5)^4*(x^12-3*x^8+3*x^4-1)^(1/2)*x^6+245 76*(x^12-3*x^8+3*x^4-1)^(3/4)*RootOf(33554432*_Z^8-24576*_Z^4+5)^4*RootOf( _Z^4+4096*RootOf(33554432*_Z^8-24576*_Z^4+5)^4-3)*x^3-4096*x^4*RootOf(3355 4432*_Z^8-24576*_Z^4+5)^4*RootOf(_Z^4+4096*RootOf(33554432*_Z^8-24576*_Z^4 +5)^4-3)^2-9*RootOf(_Z^4+4096*RootOf(33554432*_Z^8-24576*_Z^4+5)^4-3)^2*x^ 8-16384*(x^12-3*x^8+3*x^4-1)^(1/4)*RootOf(33554432*_Z^8-24576*_Z^4+5)^4*Ro otOf(_Z^4+4096*RootOf(33554432*_Z^8-24576*_Z^4+5)^4-3)^3*x-12*(x^12-3*x^8+ 3*x^4-1)^(1/4)*RootOf(_Z^4+4096*RootOf(33554432*_Z^8-24576*_Z^4+5)^4-3)^3* x^5-8192*RootOf(33554432*_Z^8-24576*_Z^4+5)^4*(x^12-3*x^8+3*x^4-1)^(1/2)*x ^2-6*(x^12-3*x^8+3*x^4-1)^(1/2)*x^6-8*(x^12-3*x^8+3*x^4-1)^(3/4)*RootOf(_Z ^4+4096*RootOf(33554432*_Z^8-24576*_Z^4+5)^4-3)*x^3+6*x^4*RootOf(_Z^4+4...
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 6.32 (sec) , antiderivative size = 883, normalized size of antiderivative = 8.41 \[ \int \frac {\sqrt [4]{-1+x^4} \left (2+x^4\right )}{x^2 \left (2+2 x^4+x^8\right )} \, dx=\text {Too large to display} \] Input:
integrate((x^4-1)^(1/4)*(x^4+2)/x^2/(x^8+2*x^4+2),x, algorithm="fricas")
Output:
-1/16*(x*sqrt(-sqrt(-1/2*I + 3/2))*log(-(2*sqrt(-1/2*I + 3/2)*((63*I + 16) *x^7 + (47*I + 79)*x^3)*(x^4 - 1)^(1/4) + 2*(-(63*I + 16)*x^5 - (47*I + 79 )*x)*(x^4 - 1)^(3/4) - (sqrt(-1/2*I + 3/2)*((104*I + 13)*x^8 + (50*I + 120 )*x^4 - 44*I - 38) - 2*((63*I + 16)*x^6 + (47*I + 79)*x^2)*sqrt(x^4 - 1))* sqrt(-sqrt(-1/2*I + 3/2)))/(x^8 + 2*x^4 + 2)) - x*sqrt(-sqrt(-1/2*I + 3/2) )*log(-(2*sqrt(-1/2*I + 3/2)*((63*I + 16)*x^7 + (47*I + 79)*x^3)*(x^4 - 1) ^(1/4) + 2*(-(63*I + 16)*x^5 - (47*I + 79)*x)*(x^4 - 1)^(3/4) - (sqrt(-1/2 *I + 3/2)*(-(104*I + 13)*x^8 - (50*I + 120)*x^4 + 44*I + 38) - 2*(-(63*I + 16)*x^6 - (47*I + 79)*x^2)*sqrt(x^4 - 1))*sqrt(-sqrt(-1/2*I + 3/2)))/(x^8 + 2*x^4 + 2)) + x*sqrt(-sqrt(1/2*I + 3/2))*log(-(2*sqrt(1/2*I + 3/2)*(-(6 3*I - 16)*x^7 - (47*I - 79)*x^3)*(x^4 - 1)^(1/4) + 2*((63*I - 16)*x^5 + (4 7*I - 79)*x)*(x^4 - 1)^(3/4) - (sqrt(1/2*I + 3/2)*(-(104*I - 13)*x^8 - (50 *I - 120)*x^4 + 44*I - 38) - 2*(-(63*I - 16)*x^6 - (47*I - 79)*x^2)*sqrt(x ^4 - 1))*sqrt(-sqrt(1/2*I + 3/2)))/(x^8 + 2*x^4 + 2)) - x*sqrt(-sqrt(1/2*I + 3/2))*log(-(2*sqrt(1/2*I + 3/2)*(-(63*I - 16)*x^7 - (47*I - 79)*x^3)*(x ^4 - 1)^(1/4) + 2*((63*I - 16)*x^5 + (47*I - 79)*x)*(x^4 - 1)^(3/4) - (sqr t(1/2*I + 3/2)*((104*I - 13)*x^8 + (50*I - 120)*x^4 - 44*I + 38) - 2*((63* I - 16)*x^6 + (47*I - 79)*x^2)*sqrt(x^4 - 1))*sqrt(-sqrt(1/2*I + 3/2)))/(x ^8 + 2*x^4 + 2)) - (1/2*I + 3/2)^(1/4)*x*log(-(2*sqrt(1/2*I + 3/2)*((63*I - 16)*x^7 + (47*I - 79)*x^3)*(x^4 - 1)^(1/4) + 2*((63*I - 16)*x^5 + (47...
Timed out. \[ \int \frac {\sqrt [4]{-1+x^4} \left (2+x^4\right )}{x^2 \left (2+2 x^4+x^8\right )} \, dx=\text {Timed out} \] Input:
integrate((x**4-1)**(1/4)*(x**4+2)/x**2/(x**8+2*x**4+2),x)
Output:
Timed out
Not integrable
Time = 0.17 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.29 \[ \int \frac {\sqrt [4]{-1+x^4} \left (2+x^4\right )}{x^2 \left (2+2 x^4+x^8\right )} \, dx=\int { \frac {{\left (x^{4} + 2\right )} {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{{\left (x^{8} + 2 \, x^{4} + 2\right )} x^{2}} \,d x } \] Input:
integrate((x^4-1)^(1/4)*(x^4+2)/x^2/(x^8+2*x^4+2),x, algorithm="maxima")
Output:
integrate((x^4 + 2)*(x^4 - 1)^(1/4)/((x^8 + 2*x^4 + 2)*x^2), x)
Exception generated. \[ \int \frac {\sqrt [4]{-1+x^4} \left (2+x^4\right )}{x^2 \left (2+2 x^4+x^8\right )} \, dx=\text {Exception raised: NotImplementedError} \] Input:
integrate((x^4-1)^(1/4)*(x^4+2)/x^2/(x^8+2*x^4+2),x, algorithm="giac")
Output:
Exception raised: NotImplementedError >> unable to parse Giac output: %%{p oly1[11728124029819/13521606404067752414158038073344*sin(1/4*atan(1/3))*at an((4123168604160*(sageVARx^4-1)^(1/4)/sageVARx-sin(1/4*atan(13/9))*exp(1/ 4*ln(14450882888647
Not integrable
Time = 0.00 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.29 \[ \int \frac {\sqrt [4]{-1+x^4} \left (2+x^4\right )}{x^2 \left (2+2 x^4+x^8\right )} \, dx=\int \frac {{\left (x^4-1\right )}^{1/4}\,\left (x^4+2\right )}{x^2\,\left (x^8+2\,x^4+2\right )} \,d x \] Input:
int(((x^4 - 1)^(1/4)*(x^4 + 2))/(x^2*(2*x^4 + x^8 + 2)),x)
Output:
int(((x^4 - 1)^(1/4)*(x^4 + 2))/(x^2*(2*x^4 + x^8 + 2)), x)
Not integrable
Time = 0.55 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.62 \[ \int \frac {\sqrt [4]{-1+x^4} \left (2+x^4\right )}{x^2 \left (2+2 x^4+x^8\right )} \, dx=\frac {\left (x^{4}-1\right )^{\frac {1}{4}}-6 \left (\int \frac {\left (x^{4}-1\right )^{\frac {1}{4}}}{x^{14}+x^{10}-2 x^{2}}d x \right ) x +\left (\int \frac {\left (x^{4}-1\right )^{\frac {1}{4}} x^{6}}{x^{12}+x^{8}-2}d x \right ) x}{2 x} \] Input:
int((x^4-1)^(1/4)*(x^4+2)/x^2/(x^8+2*x^4+2),x)
Output:
((x**4 - 1)**(1/4) - 6*int((x**4 - 1)**(1/4)/(x**14 + x**10 - 2*x**2),x)*x + int(((x**4 - 1)**(1/4)*x**6)/(x**12 + x**8 - 2),x)*x)/(2*x)