Integrand size = 112, antiderivative size = 26 \[ \int \frac {-4 x^3+e^{\frac {2 x+4 \log (3)}{x^2}} \left (30 x-6 x^2+x^3+(120-24 x) \log (3)\right )+\left (-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3\right ) \log \left (-4+e^{\frac {2 x+4 \log (3)}{x^2}}\right )}{-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3} \, dx=(-5+x) \left (\frac {1}{3}+\log \left (-4+e^{\frac {2+\frac {4 \log (3)}{x}}{x}}\right )\right ) \] Output:
(1/3+ln(exp((2+4*ln(3)/x)/x)-4))*(-5+x)
\[ \int \frac {-4 x^3+e^{\frac {2 x+4 \log (3)}{x^2}} \left (30 x-6 x^2+x^3+(120-24 x) \log (3)\right )+\left (-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3\right ) \log \left (-4+e^{\frac {2 x+4 \log (3)}{x^2}}\right )}{-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3} \, dx=\int \frac {-4 x^3+e^{\frac {2 x+4 \log (3)}{x^2}} \left (30 x-6 x^2+x^3+(120-24 x) \log (3)\right )+\left (-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3\right ) \log \left (-4+e^{\frac {2 x+4 \log (3)}{x^2}}\right )}{-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3} \, dx \] Input:
Integrate[(-4*x^3 + E^((2*x + 4*Log[3])/x^2)*(30*x - 6*x^2 + x^3 + (120 - 24*x)*Log[3]) + (-12*x^3 + 3*E^((2*x + 4*Log[3])/x^2)*x^3)*Log[-4 + E^((2* x + 4*Log[3])/x^2)])/(-12*x^3 + 3*E^((2*x + 4*Log[3])/x^2)*x^3),x]
Output:
Integrate[(-4*x^3 + E^((2*x + 4*Log[3])/x^2)*(30*x - 6*x^2 + x^3 + (120 - 24*x)*Log[3]) + (-12*x^3 + 3*E^((2*x + 4*Log[3])/x^2)*x^3)*Log[-4 + E^((2* x + 4*Log[3])/x^2)])/(-12*x^3 + 3*E^((2*x + 4*Log[3])/x^2)*x^3), 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.
| \(\displaystyle \int \frac {-4 x^3+e^{\frac {2 x+4 \log (3)}{x^2}} \left (x^3-6 x^2+30 x+(120-24 x) \log (3)\right )+\left (3 x^3 e^{\frac {2 x+4 \log (3)}{x^2}}-12 x^3\right ) \log \left (e^{\frac {2 x+4 \log (3)}{x^2}}-4\right )}{3 x^3 e^{\frac {2 x+4 \log (3)}{x^2}}-12 x^3} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {4 x^3-e^{\frac {2 x+4 \log (3)}{x^2}} \left (x^3-6 x^2+30 x+(120-24 x) \log (3)\right )-\left (3 x^3 e^{\frac {2 x+4 \log (3)}{x^2}}-12 x^3\right ) \log \left (e^{\frac {2 x+4 \log (3)}{x^2}}-4\right )}{3 \left (4-81^{\frac {1}{x^2}} e^{2/x}\right ) x^3}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \int \frac {4 x^3-3^{\frac {4}{x^2}} e^{2/x} \left (x^3-6 x^2+30 x+24 (5-x) \log (3)\right )+\left (12 x^3-3^{1+\frac {4}{x^2}} e^{2/x} x^3\right ) \log \left (-4+3^{\frac {4}{x^2}} e^{2/x}\right )}{\left (4-81^{\frac {1}{x^2}} e^{2/x}\right ) x^3}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {1}{3} \int \left (-\frac {6 (x-5) (x+\log (81))}{\left (-2+9^{\frac {1}{x^2}} e^{\frac {1}{x}}\right ) x^3}+\frac {6 (x-5) (x+\log (81))}{\left (2+9^{\frac {1}{x^2}} e^{\frac {1}{x}}\right ) x^3}+\frac {3 \log \left (-4+81^{\frac {1}{x^2}} e^{2/x}\right ) x^3+x^3-6 x^2+30 \left (1-\frac {4 \log (3)}{5}\right ) x+120 \log (3)}{x^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{3} \left (-6 (5-\log (81)) \text {Subst}\left (\int \frac {1}{-2+9^{x^2} e^x}dx,x,\frac {1}{x}\right )+6 (5-\log (81)) \text {Subst}\left (\int \frac {1}{2+9^{x^2} e^x}dx,x,\frac {1}{x}\right )-6 \log (81) \text {Subst}\left (\int \frac {e^{\log (81) x^2+2 x}}{-4+81^{x^2} e^{2 x}}dx,x,\frac {1}{x}\right )-6 \int \frac {1}{\left (-2+9^{\frac {1}{x^2}} e^{\frac {1}{x}}\right ) x}dx+6 \int \frac {1}{\left (2+9^{\frac {1}{x^2}} e^{\frac {1}{x}}\right ) x}dx+6 \int \frac {81^{\frac {1}{x^2}} e^{2/x}}{\left (-4+81^{\frac {1}{x^2}} e^{2/x}\right ) x}dx+30 \log (81) \int \frac {1}{\left (-2+9^{\frac {1}{x^2}} e^{\frac {1}{x}}\right ) x^3}dx-30 \log (81) \int \frac {1}{\left (2+9^{\frac {1}{x^2}} e^{\frac {1}{x}}\right ) x^3}dx+3 x \log \left (81^{\frac {1}{x^2}} e^{2/x}-4\right )-\frac {60 \log (3)}{x^2}+x-6 \log (x)-\frac {6 (5-\log (81))}{x}\right )\) |
Input:
Int[(-4*x^3 + E^((2*x + 4*Log[3])/x^2)*(30*x - 6*x^2 + x^3 + (120 - 24*x)* Log[3]) + (-12*x^3 + 3*E^((2*x + 4*Log[3])/x^2)*x^3)*Log[-4 + E^((2*x + 4* Log[3])/x^2)])/(-12*x^3 + 3*E^((2*x + 4*Log[3])/x^2)*x^3),x]
Output:
$Aborted
Time = 1.51 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50
| method | result | size |
| risch | \(x \ln \left (81^{\frac {1}{x^{2}}} {\mathrm e}^{\frac {2}{x}}-4\right )+\frac {x}{3}-5 \ln \left (81^{\frac {1}{x^{2}}} {\mathrm e}^{\frac {2}{x}}-4\right )\) | \(39\) |
| parallelrisch | \(x \ln \left ({\mathrm e}^{\frac {4 \ln \left (3\right )+2 x}{x^{2}}}-4\right )+\frac {x}{3}-5 \ln \left ({\mathrm e}^{\frac {4 \ln \left (3\right )+2 x}{x^{2}}}-4\right )\) | \(39\) |
| norman | \(\frac {x^{3} \ln \left ({\mathrm e}^{\frac {4 \ln \left (3\right )+2 x}{x^{2}}}-4\right )-5 x^{2} \ln \left ({\mathrm e}^{\frac {4 \ln \left (3\right )+2 x}{x^{2}}}-4\right )+\frac {x^{3}}{3}}{x^{2}}\) | \(52\) |
Input:
int(((3*x^3*exp((4*ln(3)+2*x)/x^2)-12*x^3)*ln(exp((4*ln(3)+2*x)/x^2)-4)+(( -24*x+120)*ln(3)+x^3-6*x^2+30*x)*exp((4*ln(3)+2*x)/x^2)-4*x^3)/(3*x^3*exp( (4*ln(3)+2*x)/x^2)-12*x^3),x,method=_RETURNVERBOSE)
Output:
x*ln(81^(1/x^2)*exp(2/x)-4)+1/3*x-5*ln(81^(1/x^2)*exp(2/x)-4)
Time = 0.08 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {-4 x^3+e^{\frac {2 x+4 \log (3)}{x^2}} \left (30 x-6 x^2+x^3+(120-24 x) \log (3)\right )+\left (-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3\right ) \log \left (-4+e^{\frac {2 x+4 \log (3)}{x^2}}\right )}{-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3} \, dx={\left (x - 5\right )} \log \left (e^{\left (\frac {2 \, {\left (x + 2 \, \log \left (3\right )\right )}}{x^{2}}\right )} - 4\right ) + \frac {1}{3} \, x \] Input:
integrate(((3*x^3*exp((4*log(3)+2*x)/x^2)-12*x^3)*log(exp((4*log(3)+2*x)/x ^2)-4)+((-24*x+120)*log(3)+x^3-6*x^2+30*x)*exp((4*log(3)+2*x)/x^2)-4*x^3)/ (3*x^3*exp((4*log(3)+2*x)/x^2)-12*x^3),x, algorithm="fricas")
Output:
(x - 5)*log(e^(2*(x + 2*log(3))/x^2) - 4) + 1/3*x
Time = 0.21 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \[ \int \frac {-4 x^3+e^{\frac {2 x+4 \log (3)}{x^2}} \left (30 x-6 x^2+x^3+(120-24 x) \log (3)\right )+\left (-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3\right ) \log \left (-4+e^{\frac {2 x+4 \log (3)}{x^2}}\right )}{-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3} \, dx=x \log {\left (e^{\frac {2 x + 4 \log {\left (3 \right )}}{x^{2}}} - 4 \right )} + \frac {x}{3} - 5 \log {\left (e^{\frac {2 x + 4 \log {\left (3 \right )}}{x^{2}}} - 4 \right )} \] Input:
integrate(((3*x**3*exp((4*ln(3)+2*x)/x**2)-12*x**3)*ln(exp((4*ln(3)+2*x)/x **2)-4)+((-24*x+120)*ln(3)+x**3-6*x**2+30*x)*exp((4*ln(3)+2*x)/x**2)-4*x** 3)/(3*x**3*exp((4*ln(3)+2*x)/x**2)-12*x**3),x)
Output:
x*log(exp((2*x + 4*log(3))/x**2) - 4) + x/3 - 5*log(exp((2*x + 4*log(3))/x **2) - 4)
Leaf count of result is larger than twice the leaf count of optimal. 99 vs. \(2 (27) = 54\).
Time = 0.18 (sec) , antiderivative size = 99, normalized size of antiderivative = 3.81 \[ \int \frac {-4 x^3+e^{\frac {2 x+4 \log (3)}{x^2}} \left (30 x-6 x^2+x^3+(120-24 x) \log (3)\right )+\left (-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3\right ) \log \left (-4+e^{\frac {2 x+4 \log (3)}{x^2}}\right )}{-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3} \, dx=\frac {3 \, x^{2} \log \left (e^{\left (\frac {1}{x} + \frac {2 \, \log \left (3\right )}{x^{2}}\right )} + 2\right ) + 3 \, x^{2} \log \left (e^{\left (\frac {1}{x} + \frac {2 \, \log \left (3\right )}{x^{2}}\right )} - 2\right ) + x^{2} - 30}{3 \, x} - 5 \, \log \left ({\left (e^{\left (\frac {1}{x} + \frac {2 \, \log \left (3\right )}{x^{2}}\right )} + 2\right )} e^{\left (-\frac {1}{x}\right )}\right ) - 5 \, \log \left ({\left (e^{\left (\frac {1}{x} + \frac {2 \, \log \left (3\right )}{x^{2}}\right )} - 2\right )} e^{\left (-\frac {1}{x}\right )}\right ) \] Input:
integrate(((3*x^3*exp((4*log(3)+2*x)/x^2)-12*x^3)*log(exp((4*log(3)+2*x)/x ^2)-4)+((-24*x+120)*log(3)+x^3-6*x^2+30*x)*exp((4*log(3)+2*x)/x^2)-4*x^3)/ (3*x^3*exp((4*log(3)+2*x)/x^2)-12*x^3),x, algorithm="maxima")
Output:
1/3*(3*x^2*log(e^(1/x + 2*log(3)/x^2) + 2) + 3*x^2*log(e^(1/x + 2*log(3)/x ^2) - 2) + x^2 - 30)/x - 5*log((e^(1/x + 2*log(3)/x^2) + 2)*e^(-1/x)) - 5* log((e^(1/x + 2*log(3)/x^2) - 2)*e^(-1/x))
Time = 0.23 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \[ \int \frac {-4 x^3+e^{\frac {2 x+4 \log (3)}{x^2}} \left (30 x-6 x^2+x^3+(120-24 x) \log (3)\right )+\left (-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3\right ) \log \left (-4+e^{\frac {2 x+4 \log (3)}{x^2}}\right )}{-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3} \, dx=x \log \left (e^{\left (\frac {2 \, {\left (x + 2 \, \log \left (3\right )\right )}}{x^{2}}\right )} - 4\right ) + \frac {1}{3} \, x - 5 \, \log \left (e^{\left (\frac {2 \, {\left (x + 2 \, \log \left (3\right )\right )}}{x^{2}}\right )} - 4\right ) \] Input:
integrate(((3*x^3*exp((4*log(3)+2*x)/x^2)-12*x^3)*log(exp((4*log(3)+2*x)/x ^2)-4)+((-24*x+120)*log(3)+x^3-6*x^2+30*x)*exp((4*log(3)+2*x)/x^2)-4*x^3)/ (3*x^3*exp((4*log(3)+2*x)/x^2)-12*x^3),x, algorithm="giac")
Output:
x*log(e^(2*(x + 2*log(3))/x^2) - 4) + 1/3*x - 5*log(e^(2*(x + 2*log(3))/x^ 2) - 4)
Time = 2.41 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {-4 x^3+e^{\frac {2 x+4 \log (3)}{x^2}} \left (30 x-6 x^2+x^3+(120-24 x) \log (3)\right )+\left (-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3\right ) \log \left (-4+e^{\frac {2 x+4 \log (3)}{x^2}}\right )}{-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3} \, dx=\frac {x}{3}-5\,\ln \left (3^{\frac {4}{x^2}}\,{\mathrm {e}}^{2/x}-4\right )+x\,\ln \left (3^{\frac {4}{x^2}}\,{\mathrm {e}}^{2/x}-4\right ) \] Input:
int((log(exp((2*x + 4*log(3))/x^2) - 4)*(3*x^3*exp((2*x + 4*log(3))/x^2) - 12*x^3) + exp((2*x + 4*log(3))/x^2)*(30*x - log(3)*(24*x - 120) - 6*x^2 + x^3) - 4*x^3)/(3*x^3*exp((2*x + 4*log(3))/x^2) - 12*x^3),x)
Output:
x/3 - 5*log(3^(4/x^2)*exp(2/x) - 4) + x*log(3^(4/x^2)*exp(2/x) - 4)
\[ \int \frac {-4 x^3+e^{\frac {2 x+4 \log (3)}{x^2}} \left (30 x-6 x^2+x^3+(120-24 x) \log (3)\right )+\left (-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3\right ) \log \left (-4+e^{\frac {2 x+4 \log (3)}{x^2}}\right )}{-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3} \, dx=\frac {\left (\int \frac {e^{\frac {4 \,\mathrm {log}\left (3\right )+2 x}{x^{2}}}}{e^{\frac {4 \,\mathrm {log}\left (3\right )+2 x}{x^{2}}}-4}d x \right )}{3}+40 \left (\int \frac {e^{\frac {4 \,\mathrm {log}\left (3\right )+2 x}{x^{2}}}}{e^{\frac {4 \,\mathrm {log}\left (3\right )+2 x}{x^{2}}} x^{3}-4 x^{3}}d x \right ) \mathrm {log}\left (3\right )-8 \left (\int \frac {e^{\frac {4 \,\mathrm {log}\left (3\right )+2 x}{x^{2}}}}{e^{\frac {4 \,\mathrm {log}\left (3\right )+2 x}{x^{2}}} x^{2}-4 x^{2}}d x \right ) \mathrm {log}\left (3\right )+10 \left (\int \frac {e^{\frac {4 \,\mathrm {log}\left (3\right )+2 x}{x^{2}}}}{e^{\frac {4 \,\mathrm {log}\left (3\right )+2 x}{x^{2}}} x^{2}-4 x^{2}}d x \right )-2 \left (\int \frac {e^{\frac {4 \,\mathrm {log}\left (3\right )+2 x}{x^{2}}}}{e^{\frac {4 \,\mathrm {log}\left (3\right )+2 x}{x^{2}}} x -4 x}d x \right )-4 \left (\int \frac {\mathrm {log}\left (e^{\frac {4 \,\mathrm {log}\left (3\right )+2 x}{x^{2}}}-4\right )}{e^{\frac {4 \,\mathrm {log}\left (3\right )+2 x}{x^{2}}}-4}d x \right )+\int \frac {e^{\frac {4 \,\mathrm {log}\left (3\right )+2 x}{x^{2}}} \mathrm {log}\left (e^{\frac {4 \,\mathrm {log}\left (3\right )+2 x}{x^{2}}}-4\right )}{e^{\frac {4 \,\mathrm {log}\left (3\right )+2 x}{x^{2}}}-4}d x -\frac {4 \left (\int \frac {1}{e^{\frac {4 \,\mathrm {log}\left (3\right )+2 x}{x^{2}}}-4}d x \right )}{3} \] Input:
int(((3*x^3*exp((4*log(3)+2*x)/x^2)-12*x^3)*log(exp((4*log(3)+2*x)/x^2)-4) +((-24*x+120)*log(3)+x^3-6*x^2+30*x)*exp((4*log(3)+2*x)/x^2)-4*x^3)/(3*x^3 *exp((4*log(3)+2*x)/x^2)-12*x^3),x)
Output:
(int(e**((4*log(3) + 2*x)/x**2)/(e**((4*log(3) + 2*x)/x**2) - 4),x) + 120* int(e**((4*log(3) + 2*x)/x**2)/(e**((4*log(3) + 2*x)/x**2)*x**3 - 4*x**3), x)*log(3) - 24*int(e**((4*log(3) + 2*x)/x**2)/(e**((4*log(3) + 2*x)/x**2)* x**2 - 4*x**2),x)*log(3) + 30*int(e**((4*log(3) + 2*x)/x**2)/(e**((4*log(3 ) + 2*x)/x**2)*x**2 - 4*x**2),x) - 6*int(e**((4*log(3) + 2*x)/x**2)/(e**(( 4*log(3) + 2*x)/x**2)*x - 4*x),x) - 12*int(log(e**((4*log(3) + 2*x)/x**2) - 4)/(e**((4*log(3) + 2*x)/x**2) - 4),x) + 3*int((e**((4*log(3) + 2*x)/x** 2)*log(e**((4*log(3) + 2*x)/x**2) - 4))/(e**((4*log(3) + 2*x)/x**2) - 4),x ) - 4*int(1/(e**((4*log(3) + 2*x)/x**2) - 4),x))/3