Integrand size = 108, antiderivative size = 29 \[ \int \frac {4 \log (5)+e^x (16-16 x) \log ^2(5)+\left (1088+64 x^2\right ) \log ^2(5)}{1+\left (544-8 x-32 x^2\right ) \log (5)+16 e^{2 x} \log ^2(5)+\left (73984-2176 x-8688 x^2+128 x^3+256 x^4\right ) \log ^2(5)+e^x \left (8 \log (5)+\left (2176-32 x-128 x^2\right ) \log ^2(5)\right )} \, dx=\frac {x}{4+e^x-x+4 \left (16-x^2\right )+\frac {1}{4 \log (5)}} \] Output:
x/(-4*x^2+68+1/4/ln(5)-x+exp(x))
Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {4 \log (5)+e^x (16-16 x) \log ^2(5)+\left (1088+64 x^2\right ) \log ^2(5)}{1+\left (544-8 x-32 x^2\right ) \log (5)+16 e^{2 x} \log ^2(5)+\left (73984-2176 x-8688 x^2+128 x^3+256 x^4\right ) \log ^2(5)+e^x \left (8 \log (5)+\left (2176-32 x-128 x^2\right ) \log ^2(5)\right )} \, dx=\frac {4 x \log (5)}{1+272 \log (5)+4 e^x \log (5)-4 x \log (5)-16 x^2 \log (5)} \] Input:
Integrate[(4*Log[5] + E^x*(16 - 16*x)*Log[5]^2 + (1088 + 64*x^2)*Log[5]^2) /(1 + (544 - 8*x - 32*x^2)*Log[5] + 16*E^(2*x)*Log[5]^2 + (73984 - 2176*x - 8688*x^2 + 128*x^3 + 256*x^4)*Log[5]^2 + E^x*(8*Log[5] + (2176 - 32*x - 128*x^2)*Log[5]^2)),x]
Output:
(4*x*Log[5])/(1 + 272*Log[5] + 4*E^x*Log[5] - 4*x*Log[5] - 16*x^2*Log[5])
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.
\(\displaystyle \int \frac {\left (64 x^2+1088\right ) \log ^2(5)+e^x (16-16 x) \log ^2(5)+4 \log (5)}{e^x \left (\left (-128 x^2-32 x+2176\right ) \log ^2(5)+8 \log (5)\right )+\left (-32 x^2-8 x+544\right ) \log (5)+\left (256 x^4+128 x^3-8688 x^2-2176 x+73984\right ) \log ^2(5)+16 e^{2 x} \log ^2(5)+1} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {4 \log (5) \left (16 x^2 \log (5)-4 e^x (x-1) \log (5)+1+272 \log (5)\right )}{\left (-16 x^2 \log (5)-4 x \log (5)+4 e^x \log (5)+1+272 \log (5)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \log (5) \int \frac {16 \log (5) x^2+4 e^x (1-x) \log (5)+272 \log (5)+1}{\left (-16 \log (5) x^2-4 \log (5) x+4 e^x \log (5)+272 \log (5)+1\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 4 \log (5) \int \left (\frac {x-1}{16 \log (5) x^2+4 \log (5) x-4 e^x \log (5)-272 \log (5)-1}-\frac {x \left (16 \log (5) x^2-28 \log (5) x-276 \log (5)-1\right )}{\left (16 \log (5) x^2+4 \log (5) x-4 e^x \log (5)-272 \log (5)-1\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 4 \log (5) \left (\int \frac {1}{-16 \log (5) x^2-4 \log (5) x+4 e^x \log (5)+272 \log (5)+1}dx+(1+276 \log (5)) \int \frac {x}{\left (16 \log (5) x^2+4 \log (5) x-4 e^x \log (5)-272 \log (5)-1\right )^2}dx+28 \log (5) \int \frac {x^2}{\left (16 \log (5) x^2+4 \log (5) x-4 e^x \log (5)-272 \log (5)-1\right )^2}dx+\int \frac {x}{16 \log (5) x^2+4 \log (5) x-4 e^x \log (5)-272 \log (5)-1}dx-16 \log (5) \int \frac {x^3}{\left (16 \log (5) x^2+4 \log (5) x-4 e^x \log (5)-272 \log (5)-1\right )^2}dx\right )\) |
Input:
Int[(4*Log[5] + E^x*(16 - 16*x)*Log[5]^2 + (1088 + 64*x^2)*Log[5]^2)/(1 + (544 - 8*x - 32*x^2)*Log[5] + 16*E^(2*x)*Log[5]^2 + (73984 - 2176*x - 8688 *x^2 + 128*x^3 + 256*x^4)*Log[5]^2 + E^x*(8*Log[5] + (2176 - 32*x - 128*x^ 2)*Log[5]^2)),x]
Output:
$Aborted
Time = 0.31 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10
method | result | size |
risch | \(-\frac {4 \ln \left (5\right ) x}{16 x^{2} \ln \left (5\right )+4 x \ln \left (5\right )-4 \,{\mathrm e}^{x} \ln \left (5\right )-272 \ln \left (5\right )-1}\) | \(32\) |
parallelrisch | \(-\frac {4 \ln \left (5\right ) x}{16 x^{2} \ln \left (5\right )+4 x \ln \left (5\right )-4 \,{\mathrm e}^{x} \ln \left (5\right )-272 \ln \left (5\right )-1}\) | \(32\) |
norman | \(\frac {-4 \,{\mathrm e}^{x} \ln \left (5\right )+16 x^{2} \ln \left (5\right )-1-272 \ln \left (5\right )}{16 x^{2} \ln \left (5\right )+4 x \ln \left (5\right )-4 \,{\mathrm e}^{x} \ln \left (5\right )-272 \ln \left (5\right )-1}\) | \(47\) |
Input:
int(((-16*x+16)*ln(5)^2*exp(x)+(64*x^2+1088)*ln(5)^2+4*ln(5))/(16*ln(5)^2* exp(x)^2+((-128*x^2-32*x+2176)*ln(5)^2+8*ln(5))*exp(x)+(256*x^4+128*x^3-86 88*x^2-2176*x+73984)*ln(5)^2+(-32*x^2-8*x+544)*ln(5)+1),x,method=_RETURNVE RBOSE)
Output:
-4*ln(5)*x/(16*x^2*ln(5)+4*x*ln(5)-4*exp(x)*ln(5)-272*ln(5)-1)
Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {4 \log (5)+e^x (16-16 x) \log ^2(5)+\left (1088+64 x^2\right ) \log ^2(5)}{1+\left (544-8 x-32 x^2\right ) \log (5)+16 e^{2 x} \log ^2(5)+\left (73984-2176 x-8688 x^2+128 x^3+256 x^4\right ) \log ^2(5)+e^x \left (8 \log (5)+\left (2176-32 x-128 x^2\right ) \log ^2(5)\right )} \, dx=-\frac {4 \, x \log \left (5\right )}{4 \, {\left (4 \, x^{2} + x - 68\right )} \log \left (5\right ) - 4 \, e^{x} \log \left (5\right ) - 1} \] Input:
integrate(((-16*x+16)*log(5)^2*exp(x)+(64*x^2+1088)*log(5)^2+4*log(5))/(16 *exp(x)^2*log(5)^2+((-128*x^2-32*x+2176)*log(5)^2+8*log(5))*exp(x)+(256*x^ 4+128*x^3-8688*x^2-2176*x+73984)*log(5)^2+(-32*x^2-8*x+544)*log(5)+1),x, a lgorithm="fricas")
Output:
-4*x*log(5)/(4*(4*x^2 + x - 68)*log(5) - 4*e^x*log(5) - 1)
Time = 0.11 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {4 \log (5)+e^x (16-16 x) \log ^2(5)+\left (1088+64 x^2\right ) \log ^2(5)}{1+\left (544-8 x-32 x^2\right ) \log (5)+16 e^{2 x} \log ^2(5)+\left (73984-2176 x-8688 x^2+128 x^3+256 x^4\right ) \log ^2(5)+e^x \left (8 \log (5)+\left (2176-32 x-128 x^2\right ) \log ^2(5)\right )} \, dx=\frac {4 x \log {\left (5 \right )}}{- 16 x^{2} \log {\left (5 \right )} - 4 x \log {\left (5 \right )} + 4 e^{x} \log {\left (5 \right )} + 1 + 272 \log {\left (5 \right )}} \] Input:
integrate(((-16*x+16)*ln(5)**2*exp(x)+(64*x**2+1088)*ln(5)**2+4*ln(5))/(16 *exp(x)**2*ln(5)**2+((-128*x**2-32*x+2176)*ln(5)**2+8*ln(5))*exp(x)+(256*x **4+128*x**3-8688*x**2-2176*x+73984)*ln(5)**2+(-32*x**2-8*x+544)*ln(5)+1), x)
Output:
4*x*log(5)/(-16*x**2*log(5) - 4*x*log(5) + 4*exp(x)*log(5) + 1 + 272*log(5 ))
Time = 0.15 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {4 \log (5)+e^x (16-16 x) \log ^2(5)+\left (1088+64 x^2\right ) \log ^2(5)}{1+\left (544-8 x-32 x^2\right ) \log (5)+16 e^{2 x} \log ^2(5)+\left (73984-2176 x-8688 x^2+128 x^3+256 x^4\right ) \log ^2(5)+e^x \left (8 \log (5)+\left (2176-32 x-128 x^2\right ) \log ^2(5)\right )} \, dx=-\frac {4 \, x \log \left (5\right )}{16 \, x^{2} \log \left (5\right ) + 4 \, x \log \left (5\right ) - 4 \, e^{x} \log \left (5\right ) - 272 \, \log \left (5\right ) - 1} \] Input:
integrate(((-16*x+16)*log(5)^2*exp(x)+(64*x^2+1088)*log(5)^2+4*log(5))/(16 *exp(x)^2*log(5)^2+((-128*x^2-32*x+2176)*log(5)^2+8*log(5))*exp(x)+(256*x^ 4+128*x^3-8688*x^2-2176*x+73984)*log(5)^2+(-32*x^2-8*x+544)*log(5)+1),x, a lgorithm="maxima")
Output:
-4*x*log(5)/(16*x^2*log(5) + 4*x*log(5) - 4*e^x*log(5) - 272*log(5) - 1)
Time = 0.13 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {4 \log (5)+e^x (16-16 x) \log ^2(5)+\left (1088+64 x^2\right ) \log ^2(5)}{1+\left (544-8 x-32 x^2\right ) \log (5)+16 e^{2 x} \log ^2(5)+\left (73984-2176 x-8688 x^2+128 x^3+256 x^4\right ) \log ^2(5)+e^x \left (8 \log (5)+\left (2176-32 x-128 x^2\right ) \log ^2(5)\right )} \, dx=-\frac {4 \, x \log \left (5\right )}{16 \, x^{2} \log \left (5\right ) + 4 \, x \log \left (5\right ) - 4 \, e^{x} \log \left (5\right ) - 272 \, \log \left (5\right ) - 1} \] Input:
integrate(((-16*x+16)*log(5)^2*exp(x)+(64*x^2+1088)*log(5)^2+4*log(5))/(16 *exp(x)^2*log(5)^2+((-128*x^2-32*x+2176)*log(5)^2+8*log(5))*exp(x)+(256*x^ 4+128*x^3-8688*x^2-2176*x+73984)*log(5)^2+(-32*x^2-8*x+544)*log(5)+1),x, a lgorithm="giac")
Output:
-4*x*log(5)/(16*x^2*log(5) + 4*x*log(5) - 4*e^x*log(5) - 272*log(5) - 1)
Time = 0.48 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {4 \log (5)+e^x (16-16 x) \log ^2(5)+\left (1088+64 x^2\right ) \log ^2(5)}{1+\left (544-8 x-32 x^2\right ) \log (5)+16 e^{2 x} \log ^2(5)+\left (73984-2176 x-8688 x^2+128 x^3+256 x^4\right ) \log ^2(5)+e^x \left (8 \log (5)+\left (2176-32 x-128 x^2\right ) \log ^2(5)\right )} \, dx=\frac {4\,x\,\ln \left (5\right )}{272\,\ln \left (5\right )-4\,x\,\ln \left (5\right )-16\,x^2\,\ln \left (5\right )+4\,{\mathrm {e}}^x\,\ln \left (5\right )+1} \] Input:
int((4*log(5) + log(5)^2*(64*x^2 + 1088) - exp(x)*log(5)^2*(16*x - 16))/(e xp(x)*(8*log(5) - log(5)^2*(32*x + 128*x^2 - 2176)) + log(5)^2*(128*x^3 - 8688*x^2 - 2176*x + 256*x^4 + 73984) - log(5)*(8*x + 32*x^2 - 544) + 16*ex p(2*x)*log(5)^2 + 1),x)
Output:
(4*x*log(5))/(272*log(5) - 4*x*log(5) - 16*x^2*log(5) + 4*exp(x)*log(5) + 1)
Time = 0.18 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {4 \log (5)+e^x (16-16 x) \log ^2(5)+\left (1088+64 x^2\right ) \log ^2(5)}{1+\left (544-8 x-32 x^2\right ) \log (5)+16 e^{2 x} \log ^2(5)+\left (73984-2176 x-8688 x^2+128 x^3+256 x^4\right ) \log ^2(5)+e^x \left (8 \log (5)+\left (2176-32 x-128 x^2\right ) \log ^2(5)\right )} \, dx=\frac {4 \,\mathrm {log}\left (5\right ) x}{4 e^{x} \mathrm {log}\left (5\right )-16 \,\mathrm {log}\left (5\right ) x^{2}-4 \,\mathrm {log}\left (5\right ) x +272 \,\mathrm {log}\left (5\right )+1} \] Input:
int(((-16*x+16)*log(5)^2*exp(x)+(64*x^2+1088)*log(5)^2+4*log(5))/(16*exp(x )^2*log(5)^2+((-128*x^2-32*x+2176)*log(5)^2+8*log(5))*exp(x)+(256*x^4+128* x^3-8688*x^2-2176*x+73984)*log(5)^2+(-32*x^2-8*x+544)*log(5)+1),x)
Output:
(4*log(5)*x)/(4*e**x*log(5) - 16*log(5)*x**2 - 4*log(5)*x + 272*log(5) + 1 )