Integrand size = 96, antiderivative size = 26 \[ \int \frac {-4 x^5 \log (x)+\left (-3 x-x^5+\left (-6 x-2 x^5\right ) \log (x)\right ) \log \left (\frac {e^5}{3+x^4}\right )+\left (24+24 x+8 x^4+8 x^5\right ) \log ^2\left (\frac {e^5}{3+x^4}\right )}{\left (12+4 x^4\right ) \log ^2\left (\frac {e^5}{3+x^4}\right )} \, dx=x \left (2+x-\frac {x \log (x)}{4 \log \left (\frac {e^5}{3+x^4}\right )}\right ) \]
Time = 1.50 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {-4 x^5 \log (x)+\left (-3 x-x^5+\left (-6 x-2 x^5\right ) \log (x)\right ) \log \left (\frac {e^5}{3+x^4}\right )+\left (24+24 x+8 x^4+8 x^5\right ) \log ^2\left (\frac {e^5}{3+x^4}\right )}{\left (12+4 x^4\right ) \log ^2\left (\frac {e^5}{3+x^4}\right )} \, dx=\frac {1}{4} x \left (8+4 x-\frac {x \log (x)}{5+\log \left (\frac {1}{3+x^4}\right )}\right ) \]
Integrate[(-4*x^5*Log[x] + (-3*x - x^5 + (-6*x - 2*x^5)*Log[x])*Log[E^5/(3 + x^4)] + (24 + 24*x + 8*x^4 + 8*x^5)*Log[E^5/(3 + x^4)]^2)/((12 + 4*x^4) *Log[E^5/(3 + x^4)]^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.
\(\displaystyle \int \frac {-4 x^5 \log (x)+\left (8 x^5+8 x^4+24 x+24\right ) \log ^2\left (\frac {e^5}{x^4+3}\right )+\left (-x^5+\left (-2 x^5-6 x\right ) \log (x)-3 x\right ) \log \left (\frac {e^5}{x^4+3}\right )}{\left (4 x^4+12\right ) \log ^2\left (\frac {e^5}{x^4+3}\right )} \, dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (-\frac {x (2 \log (x)+1)}{4 \left (\log \left (\frac {1}{x^4+3}\right )+5\right )}-\frac {x^5 \log (x)}{\left (x^4+3\right ) \left (\log \left (\frac {1}{x^4+3}\right )+5\right )^2}+2 (x+1)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} i \sqrt {3} \int \frac {\log (x)}{\left (\sqrt [4]{-3}-x\right ) \left (\log \left (\frac {1}{x^4+3}\right )+5\right )^2}dx-\frac {1}{4} i \sqrt {3} \int \frac {\log (x)}{\left (-x-(-1)^{3/4} \sqrt [4]{3}\right ) \left (\log \left (\frac {1}{x^4+3}\right )+5\right )^2}dx-\int \frac {x \log (x)}{\left (\log \left (\frac {1}{x^4+3}\right )+5\right )^2}dx-\frac {1}{4} i \sqrt {3} \int \frac {\log (x)}{\left (x+\sqrt [4]{-3}\right ) \left (\log \left (\frac {1}{x^4+3}\right )+5\right )^2}dx+\frac {1}{4} i \sqrt {3} \int \frac {\log (x)}{\left (x-(-1)^{3/4} \sqrt [4]{3}\right ) \left (\log \left (\frac {1}{x^4+3}\right )+5\right )^2}dx-\frac {1}{4} \int \frac {x}{\log \left (\frac {1}{x^4+3}\right )+5}dx-\frac {1}{2} \int \frac {x \log (x)}{\log \left (\frac {1}{x^4+3}\right )+5}dx+(x+1)^2\) |
Int[(-4*x^5*Log[x] + (-3*x - x^5 + (-6*x - 2*x^5)*Log[x])*Log[E^5/(3 + x^4 )] + (24 + 24*x + 8*x^4 + 8*x^5)*Log[E^5/(3 + x^4)]^2)/((12 + 4*x^4)*Log[E ^5/(3 + x^4)]^2),x]
3.25.74.3.1 Defintions of rubi rules used
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Result contains complex when optimal does not.
Time = 3.12 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15
method | result | size |
risch | \(x^{2}+2 x -\frac {i x^{2} \ln \left (x \right )}{2 \left (-2 i \ln \left (x^{4}+3\right )+10 i\right )}\) | \(30\) |
parallelrisch | \(-\frac {6 x^{2} \ln \left (x \right )-24 x^{2} \ln \left (\frac {{\mathrm e}^{5}}{x^{4}+3}\right )-48 \ln \left (\frac {{\mathrm e}^{5}}{x^{4}+3}\right ) x}{24 \ln \left (\frac {{\mathrm e}^{5}}{x^{4}+3}\right )}\) | \(54\) |
int(((8*x^5+8*x^4+24*x+24)*ln(exp(5)/(x^4+3))^2+((-2*x^5-6*x)*ln(x)-x^5-3* x)*ln(exp(5)/(x^4+3))-4*x^5*ln(x))/(4*x^4+12)/ln(exp(5)/(x^4+3))^2,x,metho d=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {-4 x^5 \log (x)+\left (-3 x-x^5+\left (-6 x-2 x^5\right ) \log (x)\right ) \log \left (\frac {e^5}{3+x^4}\right )+\left (24+24 x+8 x^4+8 x^5\right ) \log ^2\left (\frac {e^5}{3+x^4}\right )}{\left (12+4 x^4\right ) \log ^2\left (\frac {e^5}{3+x^4}\right )} \, dx=-\frac {x^{2} \log \left (x\right ) - 4 \, {\left (x^{2} + 2 \, x\right )} \log \left (\frac {e^{5}}{x^{4} + 3}\right )}{4 \, \log \left (\frac {e^{5}}{x^{4} + 3}\right )} \]
integrate(((8*x^5+8*x^4+24*x+24)*log(exp(5)/(x^4+3))^2+((-2*x^5-6*x)*log(x )-x^5-3*x)*log(exp(5)/(x^4+3))-4*x^5*log(x))/(4*x^4+12)/log(exp(5)/(x^4+3) )^2,x, algorithm=\
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {-4 x^5 \log (x)+\left (-3 x-x^5+\left (-6 x-2 x^5\right ) \log (x)\right ) \log \left (\frac {e^5}{3+x^4}\right )+\left (24+24 x+8 x^4+8 x^5\right ) \log ^2\left (\frac {e^5}{3+x^4}\right )}{\left (12+4 x^4\right ) \log ^2\left (\frac {e^5}{3+x^4}\right )} \, dx=- \frac {x^{2} \log {\left (x \right )}}{4 \log {\left (\frac {e^{5}}{x^{4} + 3} \right )}} + x^{2} + 2 x \]
integrate(((8*x**5+8*x**4+24*x+24)*ln(exp(5)/(x**4+3))**2+((-2*x**5-6*x)*l n(x)-x**5-3*x)*ln(exp(5)/(x**4+3))-4*x**5*ln(x))/(4*x**4+12)/ln(exp(5)/(x* *4+3))**2,x)
Time = 0.32 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {-4 x^5 \log (x)+\left (-3 x-x^5+\left (-6 x-2 x^5\right ) \log (x)\right ) \log \left (\frac {e^5}{3+x^4}\right )+\left (24+24 x+8 x^4+8 x^5\right ) \log ^2\left (\frac {e^5}{3+x^4}\right )}{\left (12+4 x^4\right ) \log ^2\left (\frac {e^5}{3+x^4}\right )} \, dx=\frac {x^{2} \log \left (x\right ) - 20 \, x^{2} + 4 \, {\left (x^{2} + 2 \, x\right )} \log \left (x^{4} + 3\right ) - 40 \, x}{4 \, {\left (\log \left (x^{4} + 3\right ) - 5\right )}} \]
integrate(((8*x^5+8*x^4+24*x+24)*log(exp(5)/(x^4+3))^2+((-2*x^5-6*x)*log(x )-x^5-3*x)*log(exp(5)/(x^4+3))-4*x^5*log(x))/(4*x^4+12)/log(exp(5)/(x^4+3) )^2,x, algorithm=\
Time = 0.43 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {-4 x^5 \log (x)+\left (-3 x-x^5+\left (-6 x-2 x^5\right ) \log (x)\right ) \log \left (\frac {e^5}{3+x^4}\right )+\left (24+24 x+8 x^4+8 x^5\right ) \log ^2\left (\frac {e^5}{3+x^4}\right )}{\left (12+4 x^4\right ) \log ^2\left (\frac {e^5}{3+x^4}\right )} \, dx=x^{2} + \frac {x^{2} \log \left (x\right )}{4 \, {\left (\log \left (x^{4} + 3\right ) - 5\right )}} + 2 \, x \]
integrate(((8*x^5+8*x^4+24*x+24)*log(exp(5)/(x^4+3))^2+((-2*x^5-6*x)*log(x )-x^5-3*x)*log(exp(5)/(x^4+3))-4*x^5*log(x))/(4*x^4+12)/log(exp(5)/(x^4+3) )^2,x, algorithm=\
Time = 10.05 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.96 \[ \int \frac {-4 x^5 \log (x)+\left (-3 x-x^5+\left (-6 x-2 x^5\right ) \log (x)\right ) \log \left (\frac {e^5}{3+x^4}\right )+\left (24+24 x+8 x^4+8 x^5\right ) \log ^2\left (\frac {e^5}{3+x^4}\right )}{\left (12+4 x^4\right ) \log ^2\left (\frac {e^5}{3+x^4}\right )} \, dx=2\,x+\frac {3}{16\,x^2}+\frac {17\,x^2}{16}-\frac {\frac {x^2\,\ln \left (x\right )}{4}+\frac {\ln \left (\frac {{\mathrm {e}}^5}{x^4+3}\right )\,\left (x^4+3\right )\,\left (2\,\ln \left (x\right )+1\right )}{16\,x^2}}{\ln \left (\frac {{\mathrm {e}}^5}{x^4+3}\right )}+\frac {\ln \left (x\right )\,\left (\frac {x^4}{8}+\frac {3}{8}\right )}{x^2} \]
int(-(4*x^5*log(x) - log(exp(5)/(x^4 + 3))^2*(24*x + 8*x^4 + 8*x^5 + 24) + log(exp(5)/(x^4 + 3))*(3*x + log(x)*(6*x + 2*x^5) + x^5))/(log(exp(5)/(x^ 4 + 3))^2*(4*x^4 + 12)),x)