Integrand size = 102, antiderivative size = 30 \[ \int \frac {e^2 \left (-20 x^4-5 x^5\right ) \log (4)+125 e x \log ^2(4)+\left (e^2 \left (-20 x^4-10 x^5\right ) \log (4)-250 e x \log ^2(4)\right ) \log (x)}{\left (e^2 \left (16 x^6+8 x^7+x^8\right )+e \left (-200 x^3-50 x^4\right ) \log (4)+625 \log ^2(4)\right ) \log ^2(x)} \, dx=\frac {x^2}{\left (-\frac {5}{e}+\frac {x^3 (4+x)}{5 \log (4)}\right ) \log (x)} \] Output:
x^2/(1/10*(4+x)*x^3/ln(2)-5/exp(1))/ln(x)
Time = 0.21 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {e^2 \left (-20 x^4-5 x^5\right ) \log (4)+125 e x \log ^2(4)+\left (e^2 \left (-20 x^4-10 x^5\right ) \log (4)-250 e x \log ^2(4)\right ) \log (x)}{\left (e^2 \left (16 x^6+8 x^7+x^8\right )+e \left (-200 x^3-50 x^4\right ) \log (4)+625 \log ^2(4)\right ) \log ^2(x)} \, dx=\frac {5 e x^2 \log (4)}{\left (e x^3 (4+x)-25 \log (4)\right ) \log (x)} \] Input:
Integrate[(E^2*(-20*x^4 - 5*x^5)*Log[4] + 125*E*x*Log[4]^2 + (E^2*(-20*x^4 - 10*x^5)*Log[4] - 250*E*x*Log[4]^2)*Log[x])/((E^2*(16*x^6 + 8*x^7 + x^8) + E*(-200*x^3 - 50*x^4)*Log[4] + 625*Log[4]^2)*Log[x]^2),x]
Output:
(5*E*x^2*Log[4])/((E*x^3*(4 + x) - 25*Log[4])*Log[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.
\(\displaystyle \int \frac {\left (e^2 \left (-10 x^5-20 x^4\right ) \log (4)-250 e x \log ^2(4)\right ) \log (x)+e^2 \left (-5 x^5-20 x^4\right ) \log (4)+125 e x \log ^2(4)}{\left (e \left (-50 x^4-200 x^3\right ) \log (4)+e^2 \left (x^8+8 x^7+16 x^6\right )+625 \log ^2(4)\right ) \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \frac {\left (e^2 \left (-10 x^5-20 x^4\right ) \log (4)-250 e x \log ^2(4)\right ) \log (x)+e^2 \left (-5 x^5-20 x^4\right ) \log (4)+125 e x \log ^2(4)}{\left (e x^4+4 e x^3-25 \log (4)\right )^2 \log ^2(x)}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {5 e x \log (4) \left (-e x^4-2 e x^4 \log (x)-4 e x^3-4 e x^3 \log (x)-50 \log (4) \log (x)+25 \log (4)\right )}{\left (e x^4+4 e x^3-25 \log (4)\right )^2 \log ^2(x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 5 e \log (4) \int -\frac {x \left (2 e \log (x) x^4+e x^4+4 e \log (x) x^3+4 e x^3+50 \log (4) \log (x)-25 \log (4)\right )}{\left (e x^4+4 e x^3-25 \log (4)\right )^2 \log ^2(x)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -5 e \log (4) \int \frac {x \left (2 e \log (x) x^4+e x^4+4 e \log (x) x^3+4 e x^3+50 \log (4) \log (x)-25 \log (4)\right )}{\left (e x^4+4 e x^3-25 \log (4)\right )^2 \log ^2(x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -5 e \log (4) \int \left (\frac {2 \left (e x^4+2 e x^3+25 \log (4)\right ) x}{\left (e x^4+4 e x^3-25 \log (4)\right )^2 \log (x)}+\frac {x}{\left (e x^4+4 e x^3-25 \log (4)\right ) \log ^2(x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -5 e \log (4) \left (\int \frac {x}{\left (e x^4+4 e x^3-25 \log (4)\right ) \log ^2(x)}dx+2 \int \frac {x \left (e x^4+2 e x^3+25 \log (4)\right )}{\left (e x^4+4 e x^3-25 \log (4)\right )^2 \log (x)}dx\right )\) |
Input:
Int[(E^2*(-20*x^4 - 5*x^5)*Log[4] + 125*E*x*Log[4]^2 + (E^2*(-20*x^4 - 10* x^5)*Log[4] - 250*E*x*Log[4]^2)*Log[x])/((E^2*(16*x^6 + 8*x^7 + x^8) + E*( -200*x^3 - 50*x^4)*Log[4] + 625*Log[4]^2)*Log[x]^2),x]
Output:
$Aborted
Time = 2.09 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13
method | result | size |
default | \(\frac {10 \,{\mathrm e} \ln \left (2\right ) x^{2}}{\left (x^{4} {\mathrm e}+4 x^{3} {\mathrm e}-50 \ln \left (2\right )\right ) \ln \left (x \right )}\) | \(34\) |
risch | \(\frac {10 \,{\mathrm e} \ln \left (2\right ) x^{2}}{\left (x^{4} {\mathrm e}+4 x^{3} {\mathrm e}-50 \ln \left (2\right )\right ) \ln \left (x \right )}\) | \(34\) |
parallelrisch | \(\frac {10 \,{\mathrm e} \ln \left (2\right ) x^{2}}{\left (x^{4} {\mathrm e}+4 x^{3} {\mathrm e}-50 \ln \left (2\right )\right ) \ln \left (x \right )}\) | \(34\) |
Input:
int(((-1000*ln(2)^2*exp(1)*x+2*(-10*x^5-20*x^4)*exp(1)^2*ln(2))*ln(x)+500* ln(2)^2*exp(1)*x+2*(-5*x^5-20*x^4)*exp(1)^2*ln(2))/(2500*ln(2)^2+2*(-50*x^ 4-200*x^3)*exp(1)*ln(2)+(x^8+8*x^7+16*x^6)*exp(1)^2)/ln(x)^2,x,method=_RET URNVERBOSE)
Output:
10*exp(1)*ln(2)*x^2/(x^4*exp(1)+4*x^3*exp(1)-50*ln(2))/ln(x)
Time = 0.10 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {e^2 \left (-20 x^4-5 x^5\right ) \log (4)+125 e x \log ^2(4)+\left (e^2 \left (-20 x^4-10 x^5\right ) \log (4)-250 e x \log ^2(4)\right ) \log (x)}{\left (e^2 \left (16 x^6+8 x^7+x^8\right )+e \left (-200 x^3-50 x^4\right ) \log (4)+625 \log ^2(4)\right ) \log ^2(x)} \, dx=\frac {10 \, x^{2} e \log \left (2\right )}{{\left ({\left (x^{4} + 4 \, x^{3}\right )} e - 50 \, \log \left (2\right )\right )} \log \left (x\right )} \] Input:
integrate(((-1000*log(2)^2*exp(1)*x+2*(-10*x^5-20*x^4)*exp(1)^2*log(2))*lo g(x)+500*log(2)^2*exp(1)*x+2*(-5*x^5-20*x^4)*exp(1)^2*log(2))/(2500*log(2) ^2+2*(-50*x^4-200*x^3)*exp(1)*log(2)+(x^8+8*x^7+16*x^6)*exp(1)^2)/log(x)^2 ,x, algorithm="fricas")
Output:
10*x^2*e*log(2)/(((x^4 + 4*x^3)*e - 50*log(2))*log(x))
Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {e^2 \left (-20 x^4-5 x^5\right ) \log (4)+125 e x \log ^2(4)+\left (e^2 \left (-20 x^4-10 x^5\right ) \log (4)-250 e x \log ^2(4)\right ) \log (x)}{\left (e^2 \left (16 x^6+8 x^7+x^8\right )+e \left (-200 x^3-50 x^4\right ) \log (4)+625 \log ^2(4)\right ) \log ^2(x)} \, dx=\frac {10 e x^{2} \log {\left (2 \right )}}{\left (e x^{4} + 4 e x^{3} - 50 \log {\left (2 \right )}\right ) \log {\left (x \right )}} \] Input:
integrate(((-1000*ln(2)**2*exp(1)*x+2*(-10*x**5-20*x**4)*exp(1)**2*ln(2))* ln(x)+500*ln(2)**2*exp(1)*x+2*(-5*x**5-20*x**4)*exp(1)**2*ln(2))/(2500*ln( 2)**2+2*(-50*x**4-200*x**3)*exp(1)*ln(2)+(x**8+8*x**7+16*x**6)*exp(1)**2)/ ln(x)**2,x)
Output:
10*E*x**2*log(2)/((E*x**4 + 4*E*x**3 - 50*log(2))*log(x))
Time = 0.15 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {e^2 \left (-20 x^4-5 x^5\right ) \log (4)+125 e x \log ^2(4)+\left (e^2 \left (-20 x^4-10 x^5\right ) \log (4)-250 e x \log ^2(4)\right ) \log (x)}{\left (e^2 \left (16 x^6+8 x^7+x^8\right )+e \left (-200 x^3-50 x^4\right ) \log (4)+625 \log ^2(4)\right ) \log ^2(x)} \, dx=\frac {10 \, x^{2} e \log \left (2\right )}{{\left (x^{4} e + 4 \, x^{3} e - 50 \, \log \left (2\right )\right )} \log \left (x\right )} \] Input:
integrate(((-1000*log(2)^2*exp(1)*x+2*(-10*x^5-20*x^4)*exp(1)^2*log(2))*lo g(x)+500*log(2)^2*exp(1)*x+2*(-5*x^5-20*x^4)*exp(1)^2*log(2))/(2500*log(2) ^2+2*(-50*x^4-200*x^3)*exp(1)*log(2)+(x^8+8*x^7+16*x^6)*exp(1)^2)/log(x)^2 ,x, algorithm="maxima")
Output:
10*x^2*e*log(2)/((x^4*e + 4*x^3*e - 50*log(2))*log(x))
Timed out. \[ \int \frac {e^2 \left (-20 x^4-5 x^5\right ) \log (4)+125 e x \log ^2(4)+\left (e^2 \left (-20 x^4-10 x^5\right ) \log (4)-250 e x \log ^2(4)\right ) \log (x)}{\left (e^2 \left (16 x^6+8 x^7+x^8\right )+e \left (-200 x^3-50 x^4\right ) \log (4)+625 \log ^2(4)\right ) \log ^2(x)} \, dx=\text {Timed out} \] Input:
integrate(((-1000*log(2)^2*exp(1)*x+2*(-10*x^5-20*x^4)*exp(1)^2*log(2))*lo g(x)+500*log(2)^2*exp(1)*x+2*(-5*x^5-20*x^4)*exp(1)^2*log(2))/(2500*log(2) ^2+2*(-50*x^4-200*x^3)*exp(1)*log(2)+(x^8+8*x^7+16*x^6)*exp(1)^2)/log(x)^2 ,x, algorithm="giac")
Output:
Timed out
Time = 4.00 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {e^2 \left (-20 x^4-5 x^5\right ) \log (4)+125 e x \log ^2(4)+\left (e^2 \left (-20 x^4-10 x^5\right ) \log (4)-250 e x \log ^2(4)\right ) \log (x)}{\left (e^2 \left (16 x^6+8 x^7+x^8\right )+e \left (-200 x^3-50 x^4\right ) \log (4)+625 \log ^2(4)\right ) \log ^2(x)} \, dx=\frac {10\,x^2\,\mathrm {e}\,\ln \left (2\right )}{\ln \left (x\right )\,\left (\mathrm {e}\,x^4+4\,\mathrm {e}\,x^3-50\,\ln \left (2\right )\right )} \] Input:
int(-(log(x)*(1000*x*exp(1)*log(2)^2 + 2*exp(2)*log(2)*(20*x^4 + 10*x^5)) - 500*x*exp(1)*log(2)^2 + 2*exp(2)*log(2)*(20*x^4 + 5*x^5))/(log(x)^2*(exp (2)*(16*x^6 + 8*x^7 + x^8) + 2500*log(2)^2 - 2*exp(1)*log(2)*(200*x^3 + 50 *x^4))),x)
Output:
(10*x^2*exp(1)*log(2))/(log(x)*(4*x^3*exp(1) - 50*log(2) + x^4*exp(1)))
Time = 0.17 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {e^2 \left (-20 x^4-5 x^5\right ) \log (4)+125 e x \log ^2(4)+\left (e^2 \left (-20 x^4-10 x^5\right ) \log (4)-250 e x \log ^2(4)\right ) \log (x)}{\left (e^2 \left (16 x^6+8 x^7+x^8\right )+e \left (-200 x^3-50 x^4\right ) \log (4)+625 \log ^2(4)\right ) \log ^2(x)} \, dx=-\frac {10 \,\mathrm {log}\left (2\right ) e \,x^{2}}{\mathrm {log}\left (x \right ) \left (50 \,\mathrm {log}\left (2\right )-e \,x^{4}-4 e \,x^{3}\right )} \] Input:
int(((-1000*exp(1)*log(2)^2*x+2*(-10*x^5-20*x^4)*exp(1)^2*log(2))*log(x)+5 00*exp(1)*log(2)^2*x+2*(-5*x^5-20*x^4)*exp(1)^2*log(2))/(2500*log(2)^2+2*( -50*x^4-200*x^3)*exp(1)*log(2)+(x^8+8*x^7+16*x^6)*exp(1)^2)/log(x)^2,x)
Output:
( - 10*log(2)*e*x**2)/(log(x)*(50*log(2) - e*x**4 - 4*e*x**3))