Integrand size = 73, antiderivative size = 21 \[ \int \frac {e \left (192 x+96 x^2+12 x^3\right )+\left (e^5 (-96-24 x)+e \left (192 x+48 x^2\right )\right ) \log \left (e^4-2 x\right )}{\left (e^4 x^3-2 x^4\right ) \log ^3\left (e^4-2 x\right )} \, dx=\frac {3 e (4+x)^2}{x^2 \log ^2\left (e^4-2 x\right )} \] Output:
3/x^2*exp(1)/ln(exp(4)-2*x)^2*(4+x)^2
Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {e \left (192 x+96 x^2+12 x^3\right )+\left (e^5 (-96-24 x)+e \left (192 x+48 x^2\right )\right ) \log \left (e^4-2 x\right )}{\left (e^4 x^3-2 x^4\right ) \log ^3\left (e^4-2 x\right )} \, dx=\frac {3 e (4+x)^2}{x^2 \log ^2\left (e^4-2 x\right )} \] Input:
Integrate[(E*(192*x + 96*x^2 + 12*x^3) + (E^5*(-96 - 24*x) + E*(192*x + 48 *x^2))*Log[E^4 - 2*x])/((E^4*x^3 - 2*x^4)*Log[E^4 - 2*x]^3),x]
Output:
(3*E*(4 + x)^2)/(x^2*Log[E^4 - 2*x]^2)
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.
\(\displaystyle \int \frac {\left (e \left (48 x^2+192 x\right )+e^5 (-24 x-96)\right ) \log \left (e^4-2 x\right )+e \left (12 x^3+96 x^2+192 x\right )}{\left (e^4 x^3-2 x^4\right ) \log ^3\left (e^4-2 x\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (e \left (48 x^2+192 x\right )+e^5 (-24 x-96)\right ) \log \left (e^4-2 x\right )+e \left (12 x^3+96 x^2+192 x\right )}{\left (e^4-2 x\right ) x^3 \log ^3\left (e^4-2 x\right )}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {12 e (x+4) \left (x^2+4 x+4 x \log \left (e^4-2 x\right )-2 e^4 \log \left (e^4-2 x\right )\right )}{\left (e^4-2 x\right ) x^3 \log ^3\left (e^4-2 x\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 12 e \int \frac {(x+4) \left (x^2+4 \log \left (e^4-2 x\right ) x+4 x-2 e^4 \log \left (e^4-2 x\right )\right )}{\left (e^4-2 x\right ) x^3 \log ^3\left (e^4-2 x\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 12 e \int \left (\frac {(x+4)^2}{\left (e^4-2 x\right ) x^2 \log ^3\left (e^4-2 x\right )}-\frac {2 (x+4)}{x^3 \log ^2\left (e^4-2 x\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 12 e \left (-8 \int \frac {1}{x^3 \log ^2\left (e^4-2 x\right )}dx+\frac {16 \int \frac {1}{x^2 \log ^3\left (e^4-2 x\right )}dx}{e^4}-2 \int \frac {1}{x^2 \log ^2\left (e^4-2 x\right )}dx+\frac {8 \left (4+e^4\right ) \int \frac {1}{x \log ^3\left (e^4-2 x\right )}dx}{e^8}+\frac {\left (8+e^4\right )^2}{4 e^8 \log ^2\left (e^4-2 x\right )}\right )\) |
Input:
Int[(E*(192*x + 96*x^2 + 12*x^3) + (E^5*(-96 - 24*x) + E*(192*x + 48*x^2)) *Log[E^4 - 2*x])/((E^4*x^3 - 2*x^4)*Log[E^4 - 2*x]^3),x]
Output:
$Aborted
Time = 1.38 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19
method | result | size |
risch | \(\frac {3 \,{\mathrm e} \left (x^{2}+8 x +16\right )}{x^{2} \ln \left ({\mathrm e}^{4}-2 x \right )^{2}}\) | \(25\) |
norman | \(\frac {24 x \,{\mathrm e}+3 x^{2} {\mathrm e}+48 \,{\mathrm e}}{\ln \left ({\mathrm e}^{4}-2 x \right )^{2} x^{2}}\) | \(31\) |
parallelrisch | \(\frac {12 x^{2} {\mathrm e}+96 x \,{\mathrm e}+192 \,{\mathrm e}}{4 \ln \left ({\mathrm e}^{4}-2 x \right )^{2} x^{2}}\) | \(32\) |
default | \(\frac {\left (-6 \,{\mathrm e} \,{\mathrm e}^{4}-48 \,{\mathrm e}\right ) \left ({\mathrm e}^{4}-2 x \right )+3 \,{\mathrm e} \left ({\mathrm e}^{4}-2 x \right )^{2}+3 \,{\mathrm e} \,{\mathrm e}^{8}+48 \,{\mathrm e} \,{\mathrm e}^{4}+192 \,{\mathrm e}}{4 x^{2} \ln \left ({\mathrm e}^{4}-2 x \right )^{2}}\) | \(64\) |
Input:
int((((-24*x-96)*exp(1)*exp(4)+(48*x^2+192*x)*exp(1))*ln(exp(4)-2*x)+(12*x ^3+96*x^2+192*x)*exp(1))/(x^3*exp(4)-2*x^4)/ln(exp(4)-2*x)^3,x,method=_RET URNVERBOSE)
Output:
3*exp(1)*(x^2+8*x+16)/x^2/ln(exp(4)-2*x)^2
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {e \left (192 x+96 x^2+12 x^3\right )+\left (e^5 (-96-24 x)+e \left (192 x+48 x^2\right )\right ) \log \left (e^4-2 x\right )}{\left (e^4 x^3-2 x^4\right ) \log ^3\left (e^4-2 x\right )} \, dx=\frac {3 \, {\left (x^{2} + 8 \, x + 16\right )} e}{x^{2} \log \left (-2 \, x + e^{4}\right )^{2}} \] Input:
integrate((((-24*x-96)*exp(1)*exp(4)+(48*x^2+192*x)*exp(1))*log(exp(4)-2*x )+(12*x^3+96*x^2+192*x)*exp(1))/(x^3*exp(4)-2*x^4)/log(exp(4)-2*x)^3,x, al gorithm="fricas")
Output:
3*(x^2 + 8*x + 16)*e/(x^2*log(-2*x + e^4)^2)
Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.52 \[ \int \frac {e \left (192 x+96 x^2+12 x^3\right )+\left (e^5 (-96-24 x)+e \left (192 x+48 x^2\right )\right ) \log \left (e^4-2 x\right )}{\left (e^4 x^3-2 x^4\right ) \log ^3\left (e^4-2 x\right )} \, dx=\frac {3 e x^{2} + 24 e x + 48 e}{x^{2} \log {\left (- 2 x + e^{4} \right )}^{2}} \] Input:
integrate((((-24*x-96)*exp(1)*exp(4)+(48*x**2+192*x)*exp(1))*ln(exp(4)-2*x )+(12*x**3+96*x**2+192*x)*exp(1))/(x**3*exp(4)-2*x**4)/ln(exp(4)-2*x)**3,x )
Output:
(3*E*x**2 + 24*E*x + 48*E)/(x**2*log(-2*x + exp(4))**2)
Time = 0.09 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43 \[ \int \frac {e \left (192 x+96 x^2+12 x^3\right )+\left (e^5 (-96-24 x)+e \left (192 x+48 x^2\right )\right ) \log \left (e^4-2 x\right )}{\left (e^4 x^3-2 x^4\right ) \log ^3\left (e^4-2 x\right )} \, dx=\frac {3 \, {\left (x^{2} e + 8 \, x e + 16 \, e\right )}}{x^{2} \log \left (-2 \, x + e^{4}\right )^{2}} \] Input:
integrate((((-24*x-96)*exp(1)*exp(4)+(48*x^2+192*x)*exp(1))*log(exp(4)-2*x )+(12*x^3+96*x^2+192*x)*exp(1))/(x^3*exp(4)-2*x^4)/log(exp(4)-2*x)^3,x, al gorithm="maxima")
Output:
3*(x^2*e + 8*x*e + 16*e)/(x^2*log(-2*x + e^4)^2)
Time = 0.12 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43 \[ \int \frac {e \left (192 x+96 x^2+12 x^3\right )+\left (e^5 (-96-24 x)+e \left (192 x+48 x^2\right )\right ) \log \left (e^4-2 x\right )}{\left (e^4 x^3-2 x^4\right ) \log ^3\left (e^4-2 x\right )} \, dx=\frac {3 \, {\left (x^{2} e + 8 \, x e + 16 \, e\right )}}{x^{2} \log \left (-2 \, x + e^{4}\right )^{2}} \] Input:
integrate((((-24*x-96)*exp(1)*exp(4)+(48*x^2+192*x)*exp(1))*log(exp(4)-2*x )+(12*x^3+96*x^2+192*x)*exp(1))/(x^3*exp(4)-2*x^4)/log(exp(4)-2*x)^3,x, al gorithm="giac")
Output:
3*(x^2*e + 8*x*e + 16*e)/(x^2*log(-2*x + e^4)^2)
Time = 3.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {e \left (192 x+96 x^2+12 x^3\right )+\left (e^5 (-96-24 x)+e \left (192 x+48 x^2\right )\right ) \log \left (e^4-2 x\right )}{\left (e^4 x^3-2 x^4\right ) \log ^3\left (e^4-2 x\right )} \, dx=\frac {3\,\mathrm {e}\,{\left (x+4\right )}^2}{x^2\,{\ln \left ({\mathrm {e}}^4-2\,x\right )}^2} \] Input:
int((exp(1)*(192*x + 96*x^2 + 12*x^3) + log(exp(4) - 2*x)*(exp(1)*(192*x + 48*x^2) - exp(5)*(24*x + 96)))/(log(exp(4) - 2*x)^3*(x^3*exp(4) - 2*x^4)) ,x)
Output:
(3*exp(1)*(x + 4)^2)/(x^2*log(exp(4) - 2*x)^2)
Time = 0.17 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {e \left (192 x+96 x^2+12 x^3\right )+\left (e^5 (-96-24 x)+e \left (192 x+48 x^2\right )\right ) \log \left (e^4-2 x\right )}{\left (e^4 x^3-2 x^4\right ) \log ^3\left (e^4-2 x\right )} \, dx=\frac {3 e \left (x^{2}+8 x +16\right )}{\mathrm {log}\left (e^{4}-2 x \right )^{2} x^{2}} \] Input:
int((((-24*x-96)*exp(1)*exp(4)+(48*x^2+192*x)*exp(1))*log(exp(4)-2*x)+(12* x^3+96*x^2+192*x)*exp(1))/(x^3*exp(4)-2*x^4)/log(exp(4)-2*x)^3,x)
Output:
(3*e*(x**2 + 8*x + 16))/(log(e**4 - 2*x)**2*x**2)