Integrand size = 68, antiderivative size = 21 \[ \int \frac {e^{9 x} \left (8-72 x+24 x^2-72 x^3+10 x^4-18 x^5\right ) \log (27)+\left (16 x^3+8 x^5\right ) \log (27)}{e^{18 x}+2 e^{9 x} x+x^2} \, dx=\frac {2 x \left (2+x^2\right )^2 \log (27)}{e^{9 x}+x} \] Output:
6*ln(3)*(x^2+2)^2/(x+exp(9*x))*x
Time = 0.94 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {e^{9 x} \left (8-72 x+24 x^2-72 x^3+10 x^4-18 x^5\right ) \log (27)+\left (16 x^3+8 x^5\right ) \log (27)}{e^{18 x}+2 e^{9 x} x+x^2} \, dx=\frac {2 x \left (2+x^2\right )^2 \log (27)}{e^{9 x}+x} \] Input:
Integrate[(E^(9*x)*(8 - 72*x + 24*x^2 - 72*x^3 + 10*x^4 - 18*x^5)*Log[27] + (16*x^3 + 8*x^5)*Log[27])/(E^(18*x) + 2*E^(9*x)*x + x^2),x]
Output:
(2*x*(2 + x^2)^2*Log[27])/(E^(9*x) + 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.
\(\displaystyle \int \frac {\left (8 x^5+16 x^3\right ) \log (27)+e^{9 x} \left (-18 x^5+10 x^4-72 x^3+24 x^2-72 x+8\right ) \log (27)}{x^2+2 e^{9 x} x+e^{18 x}} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {2 \left (x^2+2\right ) \left (-9 e^{9 x} x^3+4 x^3+5 e^{9 x} x^2-18 e^{9 x} x+2 e^{9 x}\right ) \log (27)}{\left (x+e^{9 x}\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \log (27) \int \frac {\left (x^2+2\right ) \left (-9 e^{9 x} x^3+4 x^3+5 e^{9 x} x^2-18 e^{9 x} x+2 e^{9 x}\right )}{\left (x+e^{9 x}\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 2 \log (27) \int \left (\frac {x (9 x-1) \left (x^2+2\right )^2}{\left (x+e^{9 x}\right )^2}-\frac {9 x^5-5 x^4+36 x^3-12 x^2+36 x-4}{x+e^{9 x}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \log (27) \left (9 \int \frac {x^6}{\left (x+e^{9 x}\right )^2}dx-\int \frac {x^5}{\left (x+e^{9 x}\right )^2}dx-9 \int \frac {x^5}{x+e^{9 x}}dx+36 \int \frac {x^4}{\left (x+e^{9 x}\right )^2}dx+5 \int \frac {x^4}{x+e^{9 x}}dx-4 \int \frac {x^3}{\left (x+e^{9 x}\right )^2}dx-36 \int \frac {x^3}{x+e^{9 x}}dx+36 \int \frac {x^2}{\left (x+e^{9 x}\right )^2}dx+12 \int \frac {x^2}{x+e^{9 x}}dx-4 \int \frac {x}{\left (x+e^{9 x}\right )^2}dx+4 \int \frac {1}{x+e^{9 x}}dx-36 \int \frac {x}{x+e^{9 x}}dx\right )\) |
Input:
Int[(E^(9*x)*(8 - 72*x + 24*x^2 - 72*x^3 + 10*x^4 - 18*x^5)*Log[27] + (16* x^3 + 8*x^5)*Log[27])/(E^(18*x) + 2*E^(9*x)*x + x^2),x]
Output:
$Aborted
Time = 0.23 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14
method | result | size |
risch | \(\frac {6 \left (x^{4}+4 x^{2}+4\right ) \ln \left (3\right ) x}{x +{\mathrm e}^{9 x}}\) | \(24\) |
parallelrisch | \(\frac {6 x^{5} \ln \left (3\right )+24 x^{3} \ln \left (3\right )+24 x \ln \left (3\right )}{x +{\mathrm e}^{9 x}}\) | \(30\) |
norman | \(\frac {-24 \ln \left (3\right ) {\mathrm e}^{9 x}+24 x^{3} \ln \left (3\right )+6 x^{5} \ln \left (3\right )}{x +{\mathrm e}^{9 x}}\) | \(33\) |
Input:
int((3*(-18*x^5+10*x^4-72*x^3+24*x^2-72*x+8)*ln(3)*exp(9*x)+3*(8*x^5+16*x^ 3)*ln(3))/(exp(9*x)^2+2*x*exp(9*x)+x^2),x,method=_RETURNVERBOSE)
Output:
6*(x^4+4*x^2+4)*ln(3)*x/(x+exp(9*x))
Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {e^{9 x} \left (8-72 x+24 x^2-72 x^3+10 x^4-18 x^5\right ) \log (27)+\left (16 x^3+8 x^5\right ) \log (27)}{e^{18 x}+2 e^{9 x} x+x^2} \, dx=\frac {6 \, {\left (x^{5} + 4 \, x^{3} + 4 \, x\right )} \log \left (3\right )}{x + e^{\left (9 \, x\right )}} \] Input:
integrate((3*(-18*x^5+10*x^4-72*x^3+24*x^2-72*x+8)*log(3)*exp(9*x)+3*(8*x^ 5+16*x^3)*log(3))/(exp(9*x)^2+2*x*exp(9*x)+x^2),x, algorithm="fricas")
Output:
6*(x^5 + 4*x^3 + 4*x)*log(3)/(x + e^(9*x))
Time = 0.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38 \[ \int \frac {e^{9 x} \left (8-72 x+24 x^2-72 x^3+10 x^4-18 x^5\right ) \log (27)+\left (16 x^3+8 x^5\right ) \log (27)}{e^{18 x}+2 e^{9 x} x+x^2} \, dx=\frac {6 x^{5} \log {\left (3 \right )} + 24 x^{3} \log {\left (3 \right )} + 24 x \log {\left (3 \right )}}{x + e^{9 x}} \] Input:
integrate((3*(-18*x**5+10*x**4-72*x**3+24*x**2-72*x+8)*ln(3)*exp(9*x)+3*(8 *x**5+16*x**3)*ln(3))/(exp(9*x)**2+2*x*exp(9*x)+x**2),x)
Output:
(6*x**5*log(3) + 24*x**3*log(3) + 24*x*log(3))/(x + exp(9*x))
Time = 0.14 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38 \[ \int \frac {e^{9 x} \left (8-72 x+24 x^2-72 x^3+10 x^4-18 x^5\right ) \log (27)+\left (16 x^3+8 x^5\right ) \log (27)}{e^{18 x}+2 e^{9 x} x+x^2} \, dx=\frac {6 \, {\left (x^{5} \log \left (3\right ) + 4 \, x^{3} \log \left (3\right ) + 4 \, x \log \left (3\right )\right )}}{x + e^{\left (9 \, x\right )}} \] Input:
integrate((3*(-18*x^5+10*x^4-72*x^3+24*x^2-72*x+8)*log(3)*exp(9*x)+3*(8*x^ 5+16*x^3)*log(3))/(exp(9*x)^2+2*x*exp(9*x)+x^2),x, algorithm="maxima")
Output:
6*(x^5*log(3) + 4*x^3*log(3) + 4*x*log(3))/(x + e^(9*x))
Time = 0.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38 \[ \int \frac {e^{9 x} \left (8-72 x+24 x^2-72 x^3+10 x^4-18 x^5\right ) \log (27)+\left (16 x^3+8 x^5\right ) \log (27)}{e^{18 x}+2 e^{9 x} x+x^2} \, dx=\frac {6 \, {\left (x^{5} \log \left (3\right ) + 4 \, x^{3} \log \left (3\right ) + 4 \, x \log \left (3\right )\right )}}{x + e^{\left (9 \, x\right )}} \] Input:
integrate((3*(-18*x^5+10*x^4-72*x^3+24*x^2-72*x+8)*log(3)*exp(9*x)+3*(8*x^ 5+16*x^3)*log(3))/(exp(9*x)^2+2*x*exp(9*x)+x^2),x, algorithm="giac")
Output:
6*(x^5*log(3) + 4*x^3*log(3) + 4*x*log(3))/(x + e^(9*x))
Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {e^{9 x} \left (8-72 x+24 x^2-72 x^3+10 x^4-18 x^5\right ) \log (27)+\left (16 x^3+8 x^5\right ) \log (27)}{e^{18 x}+2 e^{9 x} x+x^2} \, dx=\frac {6\,x\,\ln \left (3\right )\,{\left (x^2+2\right )}^2}{x+{\mathrm {e}}^{9\,x}} \] Input:
int((3*log(3)*(16*x^3 + 8*x^5) - 3*exp(9*x)*log(3)*(72*x - 24*x^2 + 72*x^3 - 10*x^4 + 18*x^5 - 8))/(exp(18*x) + 2*x*exp(9*x) + x^2),x)
Output:
(6*x*log(3)*(x^2 + 2)^2)/(x + exp(9*x))
Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38 \[ \int \frac {e^{9 x} \left (8-72 x+24 x^2-72 x^3+10 x^4-18 x^5\right ) \log (27)+\left (16 x^3+8 x^5\right ) \log (27)}{e^{18 x}+2 e^{9 x} x+x^2} \, dx=\frac {6 \,\mathrm {log}\left (3\right ) \left (-4 e^{9 x}+x^{5}+4 x^{3}\right )}{e^{9 x}+x} \] Input:
int((3*(-18*x^5+10*x^4-72*x^3+24*x^2-72*x+8)*log(3)*exp(9*x)+3*(8*x^5+16*x ^3)*log(3))/(exp(9*x)^2+2*x*exp(9*x)+x^2),x)
Output:
(6*log(3)*( - 4*e**(9*x) + x**5 + 4*x**3))/(e**(9*x) + x)