Integrand size = 56, antiderivative size = 18 \[ \int \frac {-16 x^4+8 x^4 \log (x)+\left (-8+2 x^4+\left (8-2 x^4\right ) \log (x)\right ) \log \left (-4+x^4\right )}{e \left (-4+x^4\right ) \log ^2\left (-4+x^4\right )} \, dx=-\frac {2 x (-2+\log (x))}{e \log \left (-4+x^4\right )} \] Output:
-2*x/exp(1)/ln(x^4-4)*(ln(x)-2)
Time = 0.31 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {-16 x^4+8 x^4 \log (x)+\left (-8+2 x^4+\left (8-2 x^4\right ) \log (x)\right ) \log \left (-4+x^4\right )}{e \left (-4+x^4\right ) \log ^2\left (-4+x^4\right )} \, dx=-\frac {2 x (-2+\log (x))}{e \log \left (-4+x^4\right )} \] Input:
Integrate[(-16*x^4 + 8*x^4*Log[x] + (-8 + 2*x^4 + (8 - 2*x^4)*Log[x])*Log[ -4 + x^4])/(E*(-4 + x^4)*Log[-4 + x^4]^2),x]
Output:
(-2*x*(-2 + Log[x]))/(E*Log[-4 + x^4])
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.
\(\displaystyle \int \frac {-16 x^4+8 x^4 \log (x)+\left (2 x^4+\left (8-2 x^4\right ) \log (x)-8\right ) \log \left (x^4-4\right )}{e \left (x^4-4\right ) \log ^2\left (x^4-4\right )} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {2 \left (-4 \log (x) x^4+8 x^4+\left (-x^4-\left (4-x^4\right ) \log (x)+4\right ) \log \left (x^4-4\right )\right )}{\left (4-x^4\right ) \log ^2\left (x^4-4\right )}dx}{e}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 \int \frac {-4 \log (x) x^4+8 x^4+\left (-x^4-\left (4-x^4\right ) \log (x)+4\right ) \log \left (x^4-4\right )}{\left (4-x^4\right ) \log ^2\left (x^4-4\right )}dx}{e}\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \frac {2 \int \left (\frac {4 (\log (x)-2) x^4}{\left (x^4-4\right ) \log ^2\left (x^4-4\right )}+\frac {1-\log (x)}{\log \left (x^4-4\right )}\right )dx}{e}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \left (-8 \int \frac {1}{\log ^2\left (x^4-4\right )}dx+4 \int \frac {\log (x)}{\log ^2\left (x^4-4\right )}dx-i \sqrt {2} \int \frac {\log (x)}{\left (i \sqrt {2}-x\right ) \log ^2\left (x^4-4\right )}dx-\sqrt {2} \int \frac {\log (x)}{\left (\sqrt {2}-x\right ) \log ^2\left (x^4-4\right )}dx-i \sqrt {2} \int \frac {\log (x)}{\left (x+i \sqrt {2}\right ) \log ^2\left (x^4-4\right )}dx-\sqrt {2} \int \frac {\log (x)}{\left (x+\sqrt {2}\right ) \log ^2\left (x^4-4\right )}dx+\int \frac {1}{\log \left (x^4-4\right )}dx-\int \frac {\log (x)}{\log \left (x^4-4\right )}dx-8 \int \frac {1}{\left (x^2-2\right ) \log ^2\left (x^4-4\right )}dx+8 \int \frac {1}{\left (x^2+2\right ) \log ^2\left (x^4-4\right )}dx\right )}{e}\) |
Input:
Int[(-16*x^4 + 8*x^4*Log[x] + (-8 + 2*x^4 + (8 - 2*x^4)*Log[x])*Log[-4 + x ^4])/(E*(-4 + x^4)*Log[-4 + x^4]^2),x]
Output:
$Aborted
Time = 1.28 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00
method | result | size |
risch | \(-\frac {2 x \,{\mathrm e}^{-1} \left (\ln \left (x \right )-2\right )}{\ln \left (x^{4}-4\right )}\) | \(18\) |
parallelrisch | \(\frac {{\mathrm e}^{-1} \left (-16 x \ln \left (x \right )+32 x \right )}{8 \ln \left (x^{4}-4\right )}\) | \(24\) |
Input:
int((((-2*x^4+8)*ln(x)+2*x^4-8)*ln(x^4-4)+8*x^4*ln(x)-16*x^4)/(x^4-4)/exp( 1)/ln(x^4-4)^2,x,method=_RETURNVERBOSE)
Output:
-2*x*exp(-1)/ln(x^4-4)*(ln(x)-2)
Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {-16 x^4+8 x^4 \log (x)+\left (-8+2 x^4+\left (8-2 x^4\right ) \log (x)\right ) \log \left (-4+x^4\right )}{e \left (-4+x^4\right ) \log ^2\left (-4+x^4\right )} \, dx=-\frac {2 \, {\left (x \log \left (x\right ) - 2 \, x\right )} e^{\left (-1\right )}}{\log \left (x^{4} - 4\right )} \] Input:
integrate((((-2*x^4+8)*log(x)+2*x^4-8)*log(x^4-4)+8*x^4*log(x)-16*x^4)/(x^ 4-4)/exp(1)/log(x^4-4)^2,x, algorithm="fricas")
Output:
-2*(x*log(x) - 2*x)*e^(-1)/log(x^4 - 4)
Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {-16 x^4+8 x^4 \log (x)+\left (-8+2 x^4+\left (8-2 x^4\right ) \log (x)\right ) \log \left (-4+x^4\right )}{e \left (-4+x^4\right ) \log ^2\left (-4+x^4\right )} \, dx=\frac {- 2 x \log {\left (x \right )} + 4 x}{e \log {\left (x^{4} - 4 \right )}} \] Input:
integrate((((-2*x**4+8)*ln(x)+2*x**4-8)*ln(x**4-4)+8*x**4*ln(x)-16*x**4)/( x**4-4)/exp(1)/ln(x**4-4)**2,x)
Output:
(-2*x*log(x) + 4*x)*exp(-1)/log(x**4 - 4)
Time = 0.16 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.50 \[ \int \frac {-16 x^4+8 x^4 \log (x)+\left (-8+2 x^4+\left (8-2 x^4\right ) \log (x)\right ) \log \left (-4+x^4\right )}{e \left (-4+x^4\right ) \log ^2\left (-4+x^4\right )} \, dx=-\frac {2 \, {\left (x \log \left (x\right ) - 2 \, x\right )} e^{\left (-1\right )}}{\log \left (x^{2} + 2\right ) + \log \left (x^{2} - 2\right )} \] Input:
integrate((((-2*x^4+8)*log(x)+2*x^4-8)*log(x^4-4)+8*x^4*log(x)-16*x^4)/(x^ 4-4)/exp(1)/log(x^4-4)^2,x, algorithm="maxima")
Output:
-2*(x*log(x) - 2*x)*e^(-1)/(log(x^2 + 2) + log(x^2 - 2))
Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {-16 x^4+8 x^4 \log (x)+\left (-8+2 x^4+\left (8-2 x^4\right ) \log (x)\right ) \log \left (-4+x^4\right )}{e \left (-4+x^4\right ) \log ^2\left (-4+x^4\right )} \, dx=-\frac {2 \, {\left (x \log \left (x\right ) - 2 \, x\right )} e^{\left (-1\right )}}{\log \left (x^{4} - 4\right )} \] Input:
integrate((((-2*x^4+8)*log(x)+2*x^4-8)*log(x^4-4)+8*x^4*log(x)-16*x^4)/(x^ 4-4)/exp(1)/log(x^4-4)^2,x, algorithm="giac")
Output:
-2*(x*log(x) - 2*x)*e^(-1)/log(x^4 - 4)
Time = 7.57 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {-16 x^4+8 x^4 \log (x)+\left (-8+2 x^4+\left (8-2 x^4\right ) \log (x)\right ) \log \left (-4+x^4\right )}{e \left (-4+x^4\right ) \log ^2\left (-4+x^4\right )} \, dx=-\frac {2\,x\,{\mathrm {e}}^{-1}\,\left (\ln \left (x\right )-2\right )}{\ln \left (x^4-4\right )} \] Input:
int(-(exp(-1)*(log(x^4 - 4)*(log(x)*(2*x^4 - 8) - 2*x^4 + 8) - 8*x^4*log(x ) + 16*x^4))/(log(x^4 - 4)^2*(x^4 - 4)),x)
Output:
-(2*x*exp(-1)*(log(x) - 2))/log(x^4 - 4)
Time = 0.16 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {-16 x^4+8 x^4 \log (x)+\left (-8+2 x^4+\left (8-2 x^4\right ) \log (x)\right ) \log \left (-4+x^4\right )}{e \left (-4+x^4\right ) \log ^2\left (-4+x^4\right )} \, dx=\frac {2 x \left (-\mathrm {log}\left (x \right )+2\right )}{\mathrm {log}\left (x^{4}-4\right ) e} \] Input:
int((((-2*x^4+8)*log(x)+2*x^4-8)*log(x^4-4)+8*x^4*log(x)-16*x^4)/(x^4-4)/e xp(1)/log(x^4-4)^2,x)
Output:
(2*x*( - log(x) + 2))/(log(x**4 - 4)*e)