Integrand size = 81, antiderivative size = 28 \[ \int \frac {e^{-2 x} \left (2 e^{8 x}+\left (-8 e^{5 x}-6 e^{8 x} x\right ) \log \left (\frac {x}{3}\right )+24 e^{5 x} x \log ^2\left (\frac {x}{3}\right )+e^{1+3 x} x \log ^3\left (\frac {x}{3}\right )\right )}{x \log ^3\left (\frac {x}{3}\right )} \, dx=-1+e^{1+x}-\left (4-\frac {e^{3 x}}{\log \left (\frac {x}{3}\right )}\right )^2 \] Output:
exp(1+x)-(4-exp(4*x)/ln(1/3*x)/exp(x))^2-1
Time = 0.76 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {e^{-2 x} \left (2 e^{8 x}+\left (-8 e^{5 x}-6 e^{8 x} x\right ) \log \left (\frac {x}{3}\right )+24 e^{5 x} x \log ^2\left (\frac {x}{3}\right )+e^{1+3 x} x \log ^3\left (\frac {x}{3}\right )\right )}{x \log ^3\left (\frac {x}{3}\right )} \, dx=e^{1+x}-\frac {e^{6 x}}{\log ^2\left (\frac {x}{3}\right )}+\frac {8 e^{3 x}}{\log \left (\frac {x}{3}\right )} \] Input:
Integrate[(2*E^(8*x) + (-8*E^(5*x) - 6*E^(8*x)*x)*Log[x/3] + 24*E^(5*x)*x* Log[x/3]^2 + E^(1 + 3*x)*x*Log[x/3]^3)/(E^(2*x)*x*Log[x/3]^3),x]
Output:
E^(1 + x) - E^(6*x)/Log[x/3]^2 + (8*E^(3*x))/Log[x/3]
Time = 1.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {7293, 2009}
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.
\(\displaystyle \int \frac {e^{-2 x} \left (2 e^{8 x}+e^{3 x+1} x \log ^3\left (\frac {x}{3}\right )+24 e^{5 x} x \log ^2\left (\frac {x}{3}\right )+\left (-6 e^{8 x} x-8 e^{5 x}\right ) \log \left (\frac {x}{3}\right )\right )}{x \log ^3\left (\frac {x}{3}\right )} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (e^{x+1}-\frac {2 e^{6 x} \left (3 x \log \left (\frac {x}{3}\right )-1\right )}{x \log ^3\left (\frac {x}{3}\right )}+\frac {8 e^{3 x} \left (3 x \log \left (\frac {x}{3}\right )-1\right )}{x \log ^2\left (\frac {x}{3}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle e^{x+1}-\frac {e^{6 x}}{\log ^2\left (\frac {x}{3}\right )}+\frac {8 e^{3 x}}{\log \left (\frac {x}{3}\right )}\) |
Input:
Int[(2*E^(8*x) + (-8*E^(5*x) - 6*E^(8*x)*x)*Log[x/3] + 24*E^(5*x)*x*Log[x/ 3]^2 + E^(1 + 3*x)*x*Log[x/3]^3)/(E^(2*x)*x*Log[x/3]^3),x]
Output:
E^(1 + x) - E^(6*x)/Log[x/3]^2 + (8*E^(3*x))/Log[x/3]
Time = 3.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04
method | result | size |
risch | \({\mathrm e}^{1+x}-\frac {{\mathrm e}^{3 x} \left ({\mathrm e}^{3 x}-8 \ln \left (\frac {x}{3}\right )\right )}{\ln \left (\frac {x}{3}\right )^{2}}\) | \(29\) |
parallelrisch | \(\frac {\left ({\mathrm e}^{2 x} {\mathrm e}^{1+x} \ln \left (\frac {x}{3}\right )^{2}+8 \,{\mathrm e}^{x} {\mathrm e}^{4 x} \ln \left (\frac {x}{3}\right )-{\mathrm e}^{8 x}\right ) {\mathrm e}^{-2 x}}{\ln \left (\frac {x}{3}\right )^{2}}\) | \(48\) |
Input:
int((x*exp(x)^2*exp(1+x)*ln(1/3*x)^3+24*x*exp(x)*exp(4*x)*ln(1/3*x)^2+(-6* x*exp(4*x)^2-8*exp(x)*exp(4*x))*ln(1/3*x)+2*exp(4*x)^2)/x/exp(x)^2/ln(1/3* x)^3,x,method=_RETURNVERBOSE)
Output:
exp(1+x)-exp(3*x)*(exp(3*x)-8*ln(1/3*x))/ln(1/3*x)^2
Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {e^{-2 x} \left (2 e^{8 x}+\left (-8 e^{5 x}-6 e^{8 x} x\right ) \log \left (\frac {x}{3}\right )+24 e^{5 x} x \log ^2\left (\frac {x}{3}\right )+e^{1+3 x} x \log ^3\left (\frac {x}{3}\right )\right )}{x \log ^3\left (\frac {x}{3}\right )} \, dx=\frac {{\left (e^{\left (-5 \, x + 1\right )} \log \left (\frac {1}{3} \, x\right )^{2} + 8 \, e^{\left (-3 \, x\right )} \log \left (\frac {1}{3} \, x\right ) - 1\right )} e^{\left (6 \, x\right )}}{\log \left (\frac {1}{3} \, x\right )^{2}} \] Input:
integrate((x*exp(x)^2*exp(1+x)*log(1/3*x)^3+24*x*exp(x)*exp(4*x)*log(1/3*x )^2+(-6*x*exp(4*x)^2-8*exp(x)*exp(4*x))*log(1/3*x)+2*exp(4*x)^2)/x/exp(x)^ 2/log(1/3*x)^3,x, algorithm="fricas")
Output:
(e^(-5*x + 1)*log(1/3*x)^2 + 8*e^(-3*x)*log(1/3*x) - 1)*e^(6*x)/log(1/3*x) ^2
Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (19) = 38\).
Time = 0.16 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \frac {e^{-2 x} \left (2 e^{8 x}+\left (-8 e^{5 x}-6 e^{8 x} x\right ) \log \left (\frac {x}{3}\right )+24 e^{5 x} x \log ^2\left (\frac {x}{3}\right )+e^{1+3 x} x \log ^3\left (\frac {x}{3}\right )\right )}{x \log ^3\left (\frac {x}{3}\right )} \, dx=\frac {- e^{6 x} \log {\left (\frac {x}{3} \right )} + 8 e^{3 x} \log {\left (\frac {x}{3} \right )}^{2} + e e^{x} \log {\left (\frac {x}{3} \right )}^{3}}{\log {\left (\frac {x}{3} \right )}^{3}} \] Input:
integrate((x*exp(x)**2*exp(1+x)*ln(1/3*x)**3+24*x*exp(x)*exp(4*x)*ln(1/3*x )**2+(-6*x*exp(4*x)**2-8*exp(x)*exp(4*x))*ln(1/3*x)+2*exp(4*x)**2)/x/exp(x )**2/ln(1/3*x)**3,x)
Output:
(-exp(6*x)*log(x/3) + 8*exp(3*x)*log(x/3)**2 + E*exp(x)*log(x/3)**3)/log(x /3)**3
Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (23) = 46\).
Time = 0.15 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.29 \[ \int \frac {e^{-2 x} \left (2 e^{8 x}+\left (-8 e^{5 x}-6 e^{8 x} x\right ) \log \left (\frac {x}{3}\right )+24 e^{5 x} x \log ^2\left (\frac {x}{3}\right )+e^{1+3 x} x \log ^3\left (\frac {x}{3}\right )\right )}{x \log ^3\left (\frac {x}{3}\right )} \, dx=-\frac {8 \, {\left (\log \left (3\right ) - \log \left (x\right )\right )} e^{\left (3 \, x\right )} - {\left (e \log \left (3\right )^{2} - 2 \, e \log \left (3\right ) \log \left (x\right ) + e \log \left (x\right )^{2}\right )} e^{x} + e^{\left (6 \, x\right )}}{\log \left (3\right )^{2} - 2 \, \log \left (3\right ) \log \left (x\right ) + \log \left (x\right )^{2}} \] Input:
integrate((x*exp(x)^2*exp(1+x)*log(1/3*x)^3+24*x*exp(x)*exp(4*x)*log(1/3*x )^2+(-6*x*exp(4*x)^2-8*exp(x)*exp(4*x))*log(1/3*x)+2*exp(4*x)^2)/x/exp(x)^ 2/log(1/3*x)^3,x, algorithm="maxima")
Output:
-(8*(log(3) - log(x))*e^(3*x) - (e*log(3)^2 - 2*e*log(3)*log(x) + e*log(x) ^2)*e^x + e^(6*x))/(log(3)^2 - 2*log(3)*log(x) + log(x)^2)
Time = 0.11 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \frac {e^{-2 x} \left (2 e^{8 x}+\left (-8 e^{5 x}-6 e^{8 x} x\right ) \log \left (\frac {x}{3}\right )+24 e^{5 x} x \log ^2\left (\frac {x}{3}\right )+e^{1+3 x} x \log ^3\left (\frac {x}{3}\right )\right )}{x \log ^3\left (\frac {x}{3}\right )} \, dx=\frac {e^{\left (x + 1\right )} \log \left (\frac {1}{3} \, x\right )^{2} + 8 \, e^{\left (3 \, x\right )} \log \left (\frac {1}{3} \, x\right ) - e^{\left (6 \, x\right )}}{\log \left (\frac {1}{3} \, x\right )^{2}} \] Input:
integrate((x*exp(x)^2*exp(1+x)*log(1/3*x)^3+24*x*exp(x)*exp(4*x)*log(1/3*x )^2+(-6*x*exp(4*x)^2-8*exp(x)*exp(4*x))*log(1/3*x)+2*exp(4*x)^2)/x/exp(x)^ 2/log(1/3*x)^3,x, algorithm="giac")
Output:
(e^(x + 1)*log(1/3*x)^2 + 8*e^(3*x)*log(1/3*x) - e^(6*x))/log(1/3*x)^2
Time = 3.16 (sec) , antiderivative size = 192, normalized size of antiderivative = 6.86 \[ \int \frac {e^{-2 x} \left (2 e^{8 x}+\left (-8 e^{5 x}-6 e^{8 x} x\right ) \log \left (\frac {x}{3}\right )+24 e^{5 x} x \log ^2\left (\frac {x}{3}\right )+e^{1+3 x} x \log ^3\left (\frac {x}{3}\right )\right )}{x \log ^3\left (\frac {x}{3}\right )} \, dx={\mathrm {e}}^{x+1}+24\,x\,{\mathrm {e}}^{3\,x}-\frac {12\,x\,{\mathrm {e}}^{3\,x}\,{\ln \left (\frac {x}{3}\right )}^2-{\mathrm {e}}^{-6}\,\left (4\,{\mathrm {e}}^{3\,x+6}+3\,x\,{\mathrm {e}}^{6\,x+6}\right )\,\ln \left (\frac {x}{3}\right )+{\mathrm {e}}^{6\,x}}{{\ln \left (\frac {x}{3}\right )}^2}+\frac {-12\,x\,{\mathrm {e}}^{-3}\,\left ({\mathrm {e}}^{3\,x+3}+3\,x\,{\mathrm {e}}^{3\,x+3}\right )\,{\ln \left (\frac {x}{3}\right )}^2+3\,x\,{\mathrm {e}}^{-6}\,\left ({\mathrm {e}}^{6\,x+6}-4\,{\mathrm {e}}^{3\,x+6}+6\,x\,{\mathrm {e}}^{6\,x+6}\right )\,\ln \left (\frac {x}{3}\right )+{\mathrm {e}}^{-6}\,\left (4\,{\mathrm {e}}^{3\,x+6}-3\,x\,{\mathrm {e}}^{6\,x+6}\right )}{\ln \left (\frac {x}{3}\right )}-{\mathrm {e}}^{6\,x+6}\,\left (18\,{\mathrm {e}}^{-6}\,x^2+3\,{\mathrm {e}}^{-6}\,x\right )+\ln \left (\frac {x}{3}\right )\,{\mathrm {e}}^{3\,x}\,\left (36\,x^2+12\,x\right ) \] Input:
int((exp(-2*x)*(2*exp(8*x) - log(x/3)*(8*exp(5*x) + 6*x*exp(8*x)) + 24*x*l og(x/3)^2*exp(5*x) + x*log(x/3)^3*exp(2*x)*exp(x + 1)))/(x*log(x/3)^3),x)
Output:
exp(x + 1) + 24*x*exp(3*x) - (exp(6*x) + 12*x*log(x/3)^2*exp(3*x) - log(x/ 3)*exp(-6)*(4*exp(3*x + 6) + 3*x*exp(6*x + 6)))/log(x/3)^2 + (exp(-6)*(4*e xp(3*x + 6) - 3*x*exp(6*x + 6)) - 12*x*log(x/3)^2*exp(-3)*(exp(3*x + 3) + 3*x*exp(3*x + 3)) + 3*x*log(x/3)*exp(-6)*(exp(6*x + 6) - 4*exp(3*x + 6) + 6*x*exp(6*x + 6)))/log(x/3) - exp(6*x + 6)*(3*x*exp(-6) + 18*x^2*exp(-6)) + log(x/3)*exp(3*x)*(12*x + 36*x^2)
Time = 0.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \frac {e^{-2 x} \left (2 e^{8 x}+\left (-8 e^{5 x}-6 e^{8 x} x\right ) \log \left (\frac {x}{3}\right )+24 e^{5 x} x \log ^2\left (\frac {x}{3}\right )+e^{1+3 x} x \log ^3\left (\frac {x}{3}\right )\right )}{x \log ^3\left (\frac {x}{3}\right )} \, dx=\frac {e^{x} \left (-e^{5 x}+8 e^{2 x} \mathrm {log}\left (\frac {x}{3}\right )+\mathrm {log}\left (\frac {x}{3}\right )^{2} e \right )}{\mathrm {log}\left (\frac {x}{3}\right )^{2}} \] Input:
int((x*exp(x)^2*exp(1+x)*log(1/3*x)^3+24*x*exp(x)*exp(4*x)*log(1/3*x)^2+(- 6*x*exp(4*x)^2-8*exp(x)*exp(4*x))*log(1/3*x)+2*exp(4*x)^2)/x/exp(x)^2/log( 1/3*x)^3,x)
Output:
(e**x*( - e**(5*x) + 8*e**(2*x)*log(x/3) + log(x/3)**2*e))/log(x/3)**2