Integrand size = 117, antiderivative size = 22 \[ \int \frac {-480+60 x^2+80 x^4+\left (-60+10 x^2+20 x^4\right ) \log \left (\frac {1}{2} \left (6-x^2-2 x^4\right )\right )}{-384+64 x^2+128 x^4+\left (-96+16 x^2+32 x^4\right ) \log \left (\frac {1}{2} \left (6-x^2-2 x^4\right )\right )+\left (-6+x^2+2 x^4\right ) \log ^2\left (\frac {1}{2} \left (6-x^2-2 x^4\right )\right )} \, dx=\frac {10 x}{8+\log \left (3-\frac {x^2}{2}-x^4\right )} \]
Time = 0.43 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-480+60 x^2+80 x^4+\left (-60+10 x^2+20 x^4\right ) \log \left (\frac {1}{2} \left (6-x^2-2 x^4\right )\right )}{-384+64 x^2+128 x^4+\left (-96+16 x^2+32 x^4\right ) \log \left (\frac {1}{2} \left (6-x^2-2 x^4\right )\right )+\left (-6+x^2+2 x^4\right ) \log ^2\left (\frac {1}{2} \left (6-x^2-2 x^4\right )\right )} \, dx=\frac {10 x}{8+\log \left (3-\frac {x^2}{2}-x^4\right )} \]
Integrate[(-480 + 60*x^2 + 80*x^4 + (-60 + 10*x^2 + 20*x^4)*Log[(6 - x^2 - 2*x^4)/2])/(-384 + 64*x^2 + 128*x^4 + (-96 + 16*x^2 + 32*x^4)*Log[(6 - x^ 2 - 2*x^4)/2] + (-6 + x^2 + 2*x^4)*Log[(6 - x^2 - 2*x^4)/2]^2),x]
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.
\(\displaystyle \int \frac {80 x^4+60 x^2+\left (20 x^4+10 x^2-60\right ) \log \left (\frac {1}{2} \left (-2 x^4-x^2+6\right )\right )-480}{128 x^4+64 x^2+\left (2 x^4+x^2-6\right ) \log ^2\left (\frac {1}{2} \left (-2 x^4-x^2+6\right )\right )+\left (32 x^4+16 x^2-96\right ) \log \left (\frac {1}{2} \left (-2 x^4-x^2+6\right )\right )-384} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-80 x^4-60 x^2-\left (20 x^4+10 x^2-60\right ) \log \left (\frac {1}{2} \left (-2 x^4-x^2+6\right )\right )+480}{\left (-2 x^4-x^2+6\right ) \left (\log \left (-x^4-\frac {x^2}{2}+3\right )+8\right )^2}dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {20 \left (8 x^4+6 x^2+2 x^4 \log \left (-x^4-\frac {x^2}{2}+3\right )+x^2 \log \left (-x^4-\frac {x^2}{2}+3\right )-6 \log \left (-x^4-\frac {x^2}{2}+3\right )-48\right )}{7 \left (2 x^2-3\right ) \left (\log \left (-x^4-\frac {x^2}{2}+3\right )+8\right )^2}-\frac {10 \left (8 x^4+6 x^2+2 x^4 \log \left (-x^4-\frac {x^2}{2}+3\right )+x^2 \log \left (-x^4-\frac {x^2}{2}+3\right )-6 \log \left (-x^4-\frac {x^2}{2}+3\right )-48\right )}{7 \left (x^2+2\right ) \left (\log \left (-x^4-\frac {x^2}{2}+3\right )+8\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -40 \int \frac {1}{\left (\log \left (-x^4-\frac {x^2}{2}+3\right )+8\right )^2}dx+10 i \sqrt {2} \int \frac {1}{\left (i \sqrt {2}-x\right ) \left (\log \left (-x^4-\frac {x^2}{2}+3\right )+8\right )^2}dx+10 i \sqrt {2} \int \frac {1}{\left (x+i \sqrt {2}\right ) \left (\log \left (-x^4-\frac {x^2}{2}+3\right )+8\right )^2}dx+10 \sqrt {3} \int \frac {1}{\left (\sqrt {3}-\sqrt {2} x\right ) \left (\log \left (-x^4-\frac {x^2}{2}+3\right )+8\right )^2}dx+10 \sqrt {3} \int \frac {1}{\left (\sqrt {2} x+\sqrt {3}\right ) \left (\log \left (-x^4-\frac {x^2}{2}+3\right )+8\right )^2}dx+10 \int \frac {1}{\log \left (-x^4-\frac {x^2}{2}+3\right )+8}dx\) |
Int[(-480 + 60*x^2 + 80*x^4 + (-60 + 10*x^2 + 20*x^4)*Log[(6 - x^2 - 2*x^4 )/2])/(-384 + 64*x^2 + 128*x^4 + (-96 + 16*x^2 + 32*x^4)*Log[(6 - x^2 - 2* x^4)/2] + (-6 + x^2 + 2*x^4)*Log[(6 - x^2 - 2*x^4)/2]^2),x]
3.23.15.3.1 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 = 0.76 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95
method | result | size |
norman | \(\frac {10 x}{\ln \left (-x^{4}-\frac {1}{2} x^{2}+3\right )+8}\) | \(21\) |
risch | \(\frac {10 x}{\ln \left (-x^{4}-\frac {1}{2} x^{2}+3\right )+8}\) | \(21\) |
parallelrisch | \(\frac {10 x}{\ln \left (-x^{4}-\frac {1}{2} x^{2}+3\right )+8}\) | \(21\) |
int(((20*x^4+10*x^2-60)*ln(-x^4-1/2*x^2+3)+80*x^4+60*x^2-480)/((2*x^4+x^2- 6)*ln(-x^4-1/2*x^2+3)^2+(32*x^4+16*x^2-96)*ln(-x^4-1/2*x^2+3)+128*x^4+64*x ^2-384),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {-480+60 x^2+80 x^4+\left (-60+10 x^2+20 x^4\right ) \log \left (\frac {1}{2} \left (6-x^2-2 x^4\right )\right )}{-384+64 x^2+128 x^4+\left (-96+16 x^2+32 x^4\right ) \log \left (\frac {1}{2} \left (6-x^2-2 x^4\right )\right )+\left (-6+x^2+2 x^4\right ) \log ^2\left (\frac {1}{2} \left (6-x^2-2 x^4\right )\right )} \, dx=\frac {10 \, x}{\log \left (-x^{4} - \frac {1}{2} \, x^{2} + 3\right ) + 8} \]
integrate(((20*x^4+10*x^2-60)*log(-x^4-1/2*x^2+3)+80*x^4+60*x^2-480)/((2*x ^4+x^2-6)*log(-x^4-1/2*x^2+3)^2+(32*x^4+16*x^2-96)*log(-x^4-1/2*x^2+3)+128 *x^4+64*x^2-384),x, algorithm=\
Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {-480+60 x^2+80 x^4+\left (-60+10 x^2+20 x^4\right ) \log \left (\frac {1}{2} \left (6-x^2-2 x^4\right )\right )}{-384+64 x^2+128 x^4+\left (-96+16 x^2+32 x^4\right ) \log \left (\frac {1}{2} \left (6-x^2-2 x^4\right )\right )+\left (-6+x^2+2 x^4\right ) \log ^2\left (\frac {1}{2} \left (6-x^2-2 x^4\right )\right )} \, dx=\frac {10 x}{\log {\left (- x^{4} - \frac {x^{2}}{2} + 3 \right )} + 8} \]
integrate(((20*x**4+10*x**2-60)*ln(-x**4-1/2*x**2+3)+80*x**4+60*x**2-480)/ ((2*x**4+x**2-6)*ln(-x**4-1/2*x**2+3)**2+(32*x**4+16*x**2-96)*ln(-x**4-1/2 *x**2+3)+128*x**4+64*x**2-384),x)
Time = 0.31 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {-480+60 x^2+80 x^4+\left (-60+10 x^2+20 x^4\right ) \log \left (\frac {1}{2} \left (6-x^2-2 x^4\right )\right )}{-384+64 x^2+128 x^4+\left (-96+16 x^2+32 x^4\right ) \log \left (\frac {1}{2} \left (6-x^2-2 x^4\right )\right )+\left (-6+x^2+2 x^4\right ) \log ^2\left (\frac {1}{2} \left (6-x^2-2 x^4\right )\right )} \, dx=-\frac {10 \, x}{\log \left (2\right ) - \log \left (x^{2} + 2\right ) - \log \left (-2 \, x^{2} + 3\right ) - 8} \]
integrate(((20*x^4+10*x^2-60)*log(-x^4-1/2*x^2+3)+80*x^4+60*x^2-480)/((2*x ^4+x^2-6)*log(-x^4-1/2*x^2+3)^2+(32*x^4+16*x^2-96)*log(-x^4-1/2*x^2+3)+128 *x^4+64*x^2-384),x, algorithm=\
Time = 0.37 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {-480+60 x^2+80 x^4+\left (-60+10 x^2+20 x^4\right ) \log \left (\frac {1}{2} \left (6-x^2-2 x^4\right )\right )}{-384+64 x^2+128 x^4+\left (-96+16 x^2+32 x^4\right ) \log \left (\frac {1}{2} \left (6-x^2-2 x^4\right )\right )+\left (-6+x^2+2 x^4\right ) \log ^2\left (\frac {1}{2} \left (6-x^2-2 x^4\right )\right )} \, dx=\frac {10 \, x}{\log \left (-x^{4} - \frac {1}{2} \, x^{2} + 3\right ) + 8} \]
integrate(((20*x^4+10*x^2-60)*log(-x^4-1/2*x^2+3)+80*x^4+60*x^2-480)/((2*x ^4+x^2-6)*log(-x^4-1/2*x^2+3)^2+(32*x^4+16*x^2-96)*log(-x^4-1/2*x^2+3)+128 *x^4+64*x^2-384),x, algorithm=\
Time = 0.69 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {-480+60 x^2+80 x^4+\left (-60+10 x^2+20 x^4\right ) \log \left (\frac {1}{2} \left (6-x^2-2 x^4\right )\right )}{-384+64 x^2+128 x^4+\left (-96+16 x^2+32 x^4\right ) \log \left (\frac {1}{2} \left (6-x^2-2 x^4\right )\right )+\left (-6+x^2+2 x^4\right ) \log ^2\left (\frac {1}{2} \left (6-x^2-2 x^4\right )\right )} \, dx=\frac {10\,x}{\ln \left (-x^4-\frac {x^2}{2}+3\right )+8} \]