Integrand size = 79, antiderivative size = 24 \[ \int \frac {e^x \left (-3+x+x^2\right )-4 e^{2 x} \log (3)}{18 x^2+12 x^3+2 x^4+e^x \left (144 x+96 x^2+16 x^3\right ) \log (3)+e^{2 x} \left (288+192 x+32 x^2\right ) \log ^2(3)} \, dx=\frac {1}{8 (3+x) \left (\frac {e^{-x} x}{4}+\log (3)\right )} \] Output:
1/(8*ln(3)+2*x/exp(x))/(3+x)
Time = 2.41 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.92 \[ \int \frac {e^x \left (-3+x+x^2\right )-4 e^{2 x} \log (3)}{18 x^2+12 x^3+2 x^4+e^x \left (144 x+96 x^2+16 x^3\right ) \log (3)+e^{2 x} \left (288+192 x+32 x^2\right ) \log ^2(3)} \, dx=\frac {e^x \left (x^2 \log (81)+x \log (6561)-\log (531441)\right )}{2 (-1+x) (3+x)^2 \log (81) \left (x+e^x \log (81)\right )} \] Input:
Integrate[(E^x*(-3 + x + x^2) - 4*E^(2*x)*Log[3])/(18*x^2 + 12*x^3 + 2*x^4 + E^x*(144*x + 96*x^2 + 16*x^3)*Log[3] + E^(2*x)*(288 + 192*x + 32*x^2)*L og[3]^2),x]
Output:
(E^x*(x^2*Log[81] + x*Log[6561] - Log[531441]))/(2*(-1 + x)*(3 + x)^2*Log[ 81]*(x + E^x*Log[81]))
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.
\(\displaystyle \int \frac {e^x \left (x^2+x-3\right )-4 e^{2 x} \log (3)}{2 x^4+12 x^3+18 x^2+e^{2 x} \left (32 x^2+192 x+288\right ) \log ^2(3)+e^x \left (16 x^3+96 x^2+144 x\right ) \log (3)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^x \left (x^2+x-4 e^x \log (3)-3\right )}{2 (x+3)^2 \left (x+e^x \log (81)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int -\frac {e^x \left (-x^2-x+4 e^x \log (3)+3\right )}{(x+3)^2 \left (x+e^x \log (81)\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{2} \int \frac {e^x \left (-x^2-x+4 e^x \log (3)+3\right )}{(x+3)^2 \left (x+e^x \log (81)\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{2} \int \left (\frac {e^x}{(x+3)^2 \left (x+e^x \log (81)\right )}-\frac {e^x \left (\log (81) x^2+2 \log (81) x-\log (531441)\right )}{(x+3)^2 \log (81) \left (x+e^x \log (81)\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {\int \frac {1}{\left (x+e^x \log (81)\right )^2}dx}{\log (81)}-4 \int \frac {e^x}{(x+3) \left (x+e^x \log (81)\right )^2}dx-\int \frac {e^x}{(x+3)^2 \left (x+e^x \log (81)\right )}dx-\frac {1}{\log (81) \left (x+e^x \log (81)\right )}\right )\) |
Input:
Int[(E^x*(-3 + x + x^2) - 4*E^(2*x)*Log[3])/(18*x^2 + 12*x^3 + 2*x^4 + E^x *(144*x + 96*x^2 + 16*x^3)*Log[3] + E^(2*x)*(288 + 192*x + 32*x^2)*Log[3]^ 2),x]
Output:
$Aborted
Time = 0.35 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83
method | result | size |
norman | \(\frac {{\mathrm e}^{x}}{2 \left (3+x \right ) \left (4 \ln \left (3\right ) {\mathrm e}^{x}+x \right )}\) | \(20\) |
parallelrisch | \(\frac {{\mathrm e}^{x}}{8 x \ln \left (3\right ) {\mathrm e}^{x}+24 \ln \left (3\right ) {\mathrm e}^{x}+2 x^{2}+6 x}\) | \(27\) |
risch | \(\frac {1}{8 \ln \left (3\right ) \left (3+x \right )}-\frac {x}{8 \ln \left (3\right ) \left (3+x \right ) \left (4 \ln \left (3\right ) {\mathrm e}^{x}+x \right )}\) | \(35\) |
Input:
int((-4*ln(3)*exp(x)^2+(x^2+x-3)*exp(x))/((32*x^2+192*x+288)*ln(3)^2*exp(x )^2+(16*x^3+96*x^2+144*x)*ln(3)*exp(x)+2*x^4+12*x^3+18*x^2),x,method=_RETU RNVERBOSE)
Output:
1/2*exp(x)/(3+x)/(4*ln(3)*exp(x)+x)
Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {e^x \left (-3+x+x^2\right )-4 e^{2 x} \log (3)}{18 x^2+12 x^3+2 x^4+e^x \left (144 x+96 x^2+16 x^3\right ) \log (3)+e^{2 x} \left (288+192 x+32 x^2\right ) \log ^2(3)} \, dx=\frac {e^{x}}{2 \, {\left (4 \, {\left (x + 3\right )} e^{x} \log \left (3\right ) + x^{2} + 3 \, x\right )}} \] Input:
integrate((-4*log(3)*exp(x)^2+(x^2+x-3)*exp(x))/((32*x^2+192*x+288)*log(3) ^2*exp(x)^2+(16*x^3+96*x^2+144*x)*log(3)*exp(x)+2*x^4+12*x^3+18*x^2),x, al gorithm="fricas")
Output:
1/2*e^x/(4*(x + 3)*e^x*log(3) + x^2 + 3*x)
Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (15) = 30\).
Time = 0.11 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.00 \[ \int \frac {e^x \left (-3+x+x^2\right )-4 e^{2 x} \log (3)}{18 x^2+12 x^3+2 x^4+e^x \left (144 x+96 x^2+16 x^3\right ) \log (3)+e^{2 x} \left (288+192 x+32 x^2\right ) \log ^2(3)} \, dx=- \frac {x}{8 x^{2} \log {\left (3 \right )} + 24 x \log {\left (3 \right )} + \left (32 x \log {\left (3 \right )}^{2} + 96 \log {\left (3 \right )}^{2}\right ) e^{x}} + \frac {1}{8 x \log {\left (3 \right )} + 24 \log {\left (3 \right )}} \] Input:
integrate((-4*ln(3)*exp(x)**2+(x**2+x-3)*exp(x))/((32*x**2+192*x+288)*ln(3 )**2*exp(x)**2+(16*x**3+96*x**2+144*x)*ln(3)*exp(x)+2*x**4+12*x**3+18*x**2 ),x)
Output:
-x/(8*x**2*log(3) + 24*x*log(3) + (32*x*log(3)**2 + 96*log(3)**2)*exp(x)) + 1/(8*x*log(3) + 24*log(3))
Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^x \left (-3+x+x^2\right )-4 e^{2 x} \log (3)}{18 x^2+12 x^3+2 x^4+e^x \left (144 x+96 x^2+16 x^3\right ) \log (3)+e^{2 x} \left (288+192 x+32 x^2\right ) \log ^2(3)} \, dx=\frac {e^{x}}{2 \, {\left (x^{2} + 4 \, {\left (x \log \left (3\right ) + 3 \, \log \left (3\right )\right )} e^{x} + 3 \, x\right )}} \] Input:
integrate((-4*log(3)*exp(x)^2+(x^2+x-3)*exp(x))/((32*x^2+192*x+288)*log(3) ^2*exp(x)^2+(16*x^3+96*x^2+144*x)*log(3)*exp(x)+2*x^4+12*x^3+18*x^2),x, al gorithm="maxima")
Output:
1/2*e^x/(x^2 + 4*(x*log(3) + 3*log(3))*e^x + 3*x)
Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^x \left (-3+x+x^2\right )-4 e^{2 x} \log (3)}{18 x^2+12 x^3+2 x^4+e^x \left (144 x+96 x^2+16 x^3\right ) \log (3)+e^{2 x} \left (288+192 x+32 x^2\right ) \log ^2(3)} \, dx=\frac {e^{x}}{2 \, {\left (4 \, x e^{x} \log \left (3\right ) + x^{2} + 12 \, e^{x} \log \left (3\right ) + 3 \, x\right )}} \] Input:
integrate((-4*log(3)*exp(x)^2+(x^2+x-3)*exp(x))/((32*x^2+192*x+288)*log(3) ^2*exp(x)^2+(16*x^3+96*x^2+144*x)*log(3)*exp(x)+2*x^4+12*x^3+18*x^2),x, al gorithm="giac")
Output:
1/2*e^x/(4*x*e^x*log(3) + x^2 + 12*e^x*log(3) + 3*x)
Time = 2.22 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {e^x \left (-3+x+x^2\right )-4 e^{2 x} \log (3)}{18 x^2+12 x^3+2 x^4+e^x \left (144 x+96 x^2+16 x^3\right ) \log (3)+e^{2 x} \left (288+192 x+32 x^2\right ) \log ^2(3)} \, dx=\frac {{\mathrm {e}}^x}{2\,\left (x+4\,{\mathrm {e}}^x\,\ln \left (3\right )\right )\,\left (x+3\right )} \] Input:
int(-(4*exp(2*x)*log(3) - exp(x)*(x + x^2 - 3))/(18*x^2 + 12*x^3 + 2*x^4 + exp(x)*log(3)*(144*x + 96*x^2 + 16*x^3) + exp(2*x)*log(3)^2*(192*x + 32*x ^2 + 288)),x)
Output:
exp(x)/(2*(x + 4*exp(x)*log(3))*(x + 3))
Time = 0.18 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {e^x \left (-3+x+x^2\right )-4 e^{2 x} \log (3)}{18 x^2+12 x^3+2 x^4+e^x \left (144 x+96 x^2+16 x^3\right ) \log (3)+e^{2 x} \left (288+192 x+32 x^2\right ) \log ^2(3)} \, dx=\frac {e^{x}}{8 e^{x} \mathrm {log}\left (3\right ) x +24 e^{x} \mathrm {log}\left (3\right )+2 x^{2}+6 x} \] Input:
int((-4*log(3)*exp(x)^2+(x^2+x-3)*exp(x))/((32*x^2+192*x+288)*log(3)^2*exp (x)^2+(16*x^3+96*x^2+144*x)*log(3)*exp(x)+2*x^4+12*x^3+18*x^2),x)
Output:
e**x/(2*(4*e**x*log(3)*x + 12*e**x*log(3) + x**2 + 3*x))