Integrand size = 77, antiderivative size = 25 \[ \int \frac {-16 e^{2 x} \log (5)+e^x \left (-48 x^2+12 x^3\right ) \log (5)-32 e^{2 x} \log (5) \log (6 x)}{9 x^7+24 e^x x^5 \log (6 x)+16 e^{2 x} x^3 \log ^2(6 x)} \, dx=\frac {\log (5)}{x^2 \left (\frac {3}{4} e^{-x} x^2+\log (6 x)\right )} \] Output:
ln(5)/x^2/(3/4*x^2/exp(x)+ln(6*x))
Time = 0.61 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {-16 e^{2 x} \log (5)+e^x \left (-48 x^2+12 x^3\right ) \log (5)-32 e^{2 x} \log (5) \log (6 x)}{9 x^7+24 e^x x^5 \log (6 x)+16 e^{2 x} x^3 \log ^2(6 x)} \, dx=\frac {4 e^x \log (5)}{x^2 \left (3 x^2+4 e^x \log (6 x)\right )} \] Input:
Integrate[(-16*E^(2*x)*Log[5] + E^x*(-48*x^2 + 12*x^3)*Log[5] - 32*E^(2*x) *Log[5]*Log[6*x])/(9*x^7 + 24*E^x*x^5*Log[6*x] + 16*E^(2*x)*x^3*Log[6*x]^2 ),x]
Output:
(4*E^x*Log[5])/(x^2*(3*x^2 + 4*E^x*Log[6*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.
\(\displaystyle \int \frac {e^x \left (12 x^3-48 x^2\right ) \log (5)-32 e^{2 x} \log (5) \log (6 x)-16 e^{2 x} \log (5)}{9 x^7+24 e^x x^5 \log (6 x)+16 e^{2 x} x^3 \log ^2(6 x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {4 e^x \log (5) \left (3 x^3-12 x^2-4 e^x-8 e^x \log (6 x)\right )}{x^3 \left (3 x^2+4 e^x \log (6 x)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \log (5) \int -\frac {e^x \left (-3 x^3+12 x^2+4 e^x+8 e^x \log (6 x)\right )}{x^3 \left (3 x^2+4 e^x \log (6 x)\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -4 \log (5) \int \frac {e^x \left (-3 x^3+12 x^2+4 e^x+8 e^x \log (6 x)\right )}{x^3 \left (3 x^2+4 e^x \log (6 x)\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -4 \log (5) \int \left (\frac {e^x (2 \log (6 x)+1)}{x^3 \log (6 x) \left (3 x^2+4 e^x \log (6 x)\right )}-\frac {3 e^x (x \log (6 x)-2 \log (6 x)+1)}{x \log (6 x) \left (3 x^2+4 e^x \log (6 x)\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \log (5) \left (-3 \int \frac {e^x}{\left (3 x^2+4 e^x \log (6 x)\right )^2}dx+6 \int \frac {e^x}{x \left (3 x^2+4 e^x \log (6 x)\right )^2}dx-3 \int \frac {e^x}{x \log (6 x) \left (3 x^2+4 e^x \log (6 x)\right )^2}dx+2 \int \frac {e^x}{x^3 \left (3 x^2+4 e^x \log (6 x)\right )}dx+\int \frac {e^x}{x^3 \log (6 x) \left (3 x^2+4 e^x \log (6 x)\right )}dx\right )\) |
Input:
Int[(-16*E^(2*x)*Log[5] + E^x*(-48*x^2 + 12*x^3)*Log[5] - 32*E^(2*x)*Log[5 ]*Log[6*x])/(9*x^7 + 24*E^x*x^5*Log[6*x] + 16*E^(2*x)*x^3*Log[6*x]^2),x]
Output:
$Aborted
Time = 0.91 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04
method | result | size |
risch | \(\frac {4 \ln \left (5\right ) {\mathrm e}^{x}}{x^{2} \left (4 \,{\mathrm e}^{x} \ln \left (6 x \right )+3 x^{2}\right )}\) | \(26\) |
parallelrisch | \(\frac {4 \ln \left (5\right ) {\mathrm e}^{x}}{x^{2} \left (4 \,{\mathrm e}^{x} \ln \left (6 x \right )+3 x^{2}\right )}\) | \(26\) |
Input:
int((-32*ln(5)*exp(x)^2*ln(6*x)-16*ln(5)*exp(x)^2+(12*x^3-48*x^2)*ln(5)*ex p(x))/(16*x^3*exp(x)^2*ln(6*x)^2+24*x^5*exp(x)*ln(6*x)+9*x^7),x,method=_RE TURNVERBOSE)
Output:
4/x^2*ln(5)*exp(x)/(4*exp(x)*ln(6*x)+3*x^2)
Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {-16 e^{2 x} \log (5)+e^x \left (-48 x^2+12 x^3\right ) \log (5)-32 e^{2 x} \log (5) \log (6 x)}{9 x^7+24 e^x x^5 \log (6 x)+16 e^{2 x} x^3 \log ^2(6 x)} \, dx=\frac {4 \, e^{x} \log \left (5\right )}{3 \, x^{4} + 4 \, x^{2} e^{x} \log \left (6 \, x\right )} \] Input:
integrate((-32*log(5)*exp(x)^2*log(6*x)-16*exp(x)^2*log(5)+(12*x^3-48*x^2) *log(5)*exp(x))/(16*x^3*exp(x)^2*log(6*x)^2+24*x^5*exp(x)*log(6*x)+9*x^7), x, algorithm="fricas")
Output:
4*e^x*log(5)/(3*x^4 + 4*x^2*e^x*log(6*x))
Time = 0.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {-16 e^{2 x} \log (5)+e^x \left (-48 x^2+12 x^3\right ) \log (5)-32 e^{2 x} \log (5) \log (6 x)}{9 x^7+24 e^x x^5 \log (6 x)+16 e^{2 x} x^3 \log ^2(6 x)} \, dx=- \frac {3 \log {\left (5 \right )}}{3 x^{2} \log {\left (6 x \right )} + 4 e^{x} \log {\left (6 x \right )}^{2}} + \frac {\log {\left (5 \right )}}{x^{2} \log {\left (6 x \right )}} \] Input:
integrate((-32*ln(5)*exp(x)**2*ln(6*x)-16*exp(x)**2*ln(5)+(12*x**3-48*x**2 )*ln(5)*exp(x))/(16*x**3*exp(x)**2*ln(6*x)**2+24*x**5*exp(x)*ln(6*x)+9*x** 7),x)
Output:
-3*log(5)/(3*x**2*log(6*x) + 4*exp(x)*log(6*x)**2) + log(5)/(x**2*log(6*x) )
Time = 0.21 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int \frac {-16 e^{2 x} \log (5)+e^x \left (-48 x^2+12 x^3\right ) \log (5)-32 e^{2 x} \log (5) \log (6 x)}{9 x^7+24 e^x x^5 \log (6 x)+16 e^{2 x} x^3 \log ^2(6 x)} \, dx=\frac {4 \, e^{x} \log \left (5\right )}{3 \, x^{4} + 4 \, {\left (x^{2} {\left (\log \left (3\right ) + \log \left (2\right )\right )} + x^{2} \log \left (x\right )\right )} e^{x}} \] Input:
integrate((-32*log(5)*exp(x)^2*log(6*x)-16*exp(x)^2*log(5)+(12*x^3-48*x^2) *log(5)*exp(x))/(16*x^3*exp(x)^2*log(6*x)^2+24*x^5*exp(x)*log(6*x)+9*x^7), x, algorithm="maxima")
Output:
4*e^x*log(5)/(3*x^4 + 4*(x^2*(log(3) + log(2)) + x^2*log(x))*e^x)
Time = 0.14 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {-16 e^{2 x} \log (5)+e^x \left (-48 x^2+12 x^3\right ) \log (5)-32 e^{2 x} \log (5) \log (6 x)}{9 x^7+24 e^x x^5 \log (6 x)+16 e^{2 x} x^3 \log ^2(6 x)} \, dx=\frac {4 \, e^{x} \log \left (5\right )}{3 \, x^{4} + 4 \, x^{2} e^{x} \log \left (6 \, x\right )} \] Input:
integrate((-32*log(5)*exp(x)^2*log(6*x)-16*exp(x)^2*log(5)+(12*x^3-48*x^2) *log(5)*exp(x))/(16*x^3*exp(x)^2*log(6*x)^2+24*x^5*exp(x)*log(6*x)+9*x^7), x, algorithm="giac")
Output:
4*e^x*log(5)/(3*x^4 + 4*x^2*e^x*log(6*x))
Time = 3.35 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.64 \[ \int \frac {-16 e^{2 x} \log (5)+e^x \left (-48 x^2+12 x^3\right ) \log (5)-32 e^{2 x} \log (5) \log (6 x)}{9 x^7+24 e^x x^5 \log (6 x)+16 e^{2 x} x^3 \log ^2(6 x)} \, dx=\frac {4\,{\mathrm {e}}^{2\,x}\,\ln \left (5\right )\,\left (4\,{\mathrm {e}}^x+6\,x^2-3\,x^3\right )}{x\,\left (4\,\ln \left (6\,x\right )\,{\mathrm {e}}^x+3\,x^2\right )\,\left (4\,x\,{\mathrm {e}}^{2\,x}+6\,x^3\,{\mathrm {e}}^x-3\,x^4\,{\mathrm {e}}^x\right )} \] Input:
int(-(16*exp(2*x)*log(5) + 32*log(6*x)*exp(2*x)*log(5) + exp(x)*log(5)*(48 *x^2 - 12*x^3))/(9*x^7 + 16*x^3*log(6*x)^2*exp(2*x) + 24*x^5*log(6*x)*exp( x)),x)
Output:
(4*exp(2*x)*log(5)*(4*exp(x) + 6*x^2 - 3*x^3))/(x*(4*log(6*x)*exp(x) + 3*x ^2)*(4*x*exp(2*x) + 6*x^3*exp(x) - 3*x^4*exp(x)))
Time = 0.15 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {-16 e^{2 x} \log (5)+e^x \left (-48 x^2+12 x^3\right ) \log (5)-32 e^{2 x} \log (5) \log (6 x)}{9 x^7+24 e^x x^5 \log (6 x)+16 e^{2 x} x^3 \log ^2(6 x)} \, dx=\frac {4 e^{x} \mathrm {log}\left (5\right )}{x^{2} \left (4 e^{x} \mathrm {log}\left (6 x \right )+3 x^{2}\right )} \] Input:
int((-32*log(5)*exp(x)^2*log(6*x)-16*exp(x)^2*log(5)+(12*x^3-48*x^2)*log(5 )*exp(x))/(16*x^3*exp(x)^2*log(6*x)^2+24*x^5*exp(x)*log(6*x)+9*x^7),x)
Output:
(4*e**x*log(5))/(x**2*(4*e**x*log(6*x) + 3*x**2))