Integrand size = 112, antiderivative size = 26 \[ \int \frac {-4 x^3+e^{\frac {2 x+4 \log (3)}{x^2}} \left (30 x-6 x^2+x^3+(120-24 x) \log (3)\right )+\left (-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3\right ) \log \left (-4+e^{\frac {2 x+4 \log (3)}{x^2}}\right )}{-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3} \, dx=(-5+x) \left (\frac {1}{3}+\log \left (-4+e^{\frac {2+\frac {4 \log (3)}{x}}{x}}\right )\right ) \]
\[ \int \frac {-4 x^3+e^{\frac {2 x+4 \log (3)}{x^2}} \left (30 x-6 x^2+x^3+(120-24 x) \log (3)\right )+\left (-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3\right ) \log \left (-4+e^{\frac {2 x+4 \log (3)}{x^2}}\right )}{-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3} \, dx=\int \frac {-4 x^3+e^{\frac {2 x+4 \log (3)}{x^2}} \left (30 x-6 x^2+x^3+(120-24 x) \log (3)\right )+\left (-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3\right ) \log \left (-4+e^{\frac {2 x+4 \log (3)}{x^2}}\right )}{-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3} \, dx \]
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]
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 |
\(\displaystyle \frac {1}{3} \left (-6 (5-\log (81)) \text {Subst}\left (\int \frac {1}{-2+9^{x^2} e^x}dx,x,\frac {1}{x}\right )+6 (5-\log (81)) \text {Subst}\left (\int \frac {1}{2+9^{x^2} e^x}dx,x,\frac {1}{x}\right )-6 \log (81) \text {Subst}\left (\int \frac {e^{\log (81) x^2+2 x}}{-4+81^{x^2} e^{2 x}}dx,x,\frac {1}{x}\right )-6 \int \frac {1}{\left (-2+9^{\frac {1}{x^2}} e^{\frac {1}{x}}\right ) x}dx+6 \int \frac {1}{\left (2+9^{\frac {1}{x^2}} e^{\frac {1}{x}}\right ) x}dx+6 \int \frac {81^{\frac {1}{x^2}} e^{2/x}}{\left (-4+81^{\frac {1}{x^2}} e^{2/x}\right ) x}dx+30 \log (81) \int \frac {1}{\left (-2+9^{\frac {1}{x^2}} e^{\frac {1}{x}}\right ) x^3}dx-30 \log (81) \int \frac {1}{\left (2+9^{\frac {1}{x^2}} e^{\frac {1}{x}}\right ) x^3}dx+3 x \log \left (81^{\frac {1}{x^2}} e^{2/x}-4\right )-\frac {60 \log (3)}{x^2}+x-6 \log (x)-\frac {6 (5-\log (81))}{x}\right )\) |
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]
3.4.65.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 6.85 (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 | \(\frac {x^{3} \ln \left ({\mathrm e}^{\frac {4 \ln \left (3\right )+2 x}{x^{2}}}-4\right )-5 x^{2} \ln \left ({\mathrm e}^{\frac {4 \ln \left (3\right )+2 x}{x^{2}}}-4\right )+\frac {x^{3}}{3}}{x^{2}}\) | \(52\) |
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)
Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {-4 x^3+e^{\frac {2 x+4 \log (3)}{x^2}} \left (30 x-6 x^2+x^3+(120-24 x) \log (3)\right )+\left (-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3\right ) \log \left (-4+e^{\frac {2 x+4 \log (3)}{x^2}}\right )}{-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3} \, dx={\left (x - 5\right )} \log \left (e^{\left (\frac {2 \, {\left (x + 2 \, \log \left (3\right )\right )}}{x^{2}}\right )} - 4\right ) + \frac {1}{3} \, x \]
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=\
Time = 0.20 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \[ \int \frac {-4 x^3+e^{\frac {2 x+4 \log (3)}{x^2}} \left (30 x-6 x^2+x^3+(120-24 x) \log (3)\right )+\left (-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3\right ) \log \left (-4+e^{\frac {2 x+4 \log (3)}{x^2}}\right )}{-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3} \, dx=x \log {\left (e^{\frac {2 x + 4 \log {\left (3 \right )}}{x^{2}}} - 4 \right )} + \frac {x}{3} - 5 \log {\left (e^{\frac {2 x + 4 \log {\left (3 \right )}}{x^{2}}} - 4 \right )} \]
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)
Leaf count of result is larger than twice the leaf count of optimal. 99 vs. \(2 (27) = 54\).
Time = 0.32 (sec) , antiderivative size = 99, normalized size of antiderivative = 3.81 \[ \int \frac {-4 x^3+e^{\frac {2 x+4 \log (3)}{x^2}} \left (30 x-6 x^2+x^3+(120-24 x) \log (3)\right )+\left (-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3\right ) \log \left (-4+e^{\frac {2 x+4 \log (3)}{x^2}}\right )}{-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3} \, dx=\frac {3 \, x^{2} \log \left (e^{\left (\frac {1}{x} + \frac {2 \, \log \left (3\right )}{x^{2}}\right )} + 2\right ) + 3 \, x^{2} \log \left (e^{\left (\frac {1}{x} + \frac {2 \, \log \left (3\right )}{x^{2}}\right )} - 2\right ) + x^{2} - 30}{3 \, x} - 5 \, \log \left ({\left (e^{\left (\frac {1}{x} + \frac {2 \, \log \left (3\right )}{x^{2}}\right )} + 2\right )} e^{\left (-\frac {1}{x}\right )}\right ) - 5 \, \log \left ({\left (e^{\left (\frac {1}{x} + \frac {2 \, \log \left (3\right )}{x^{2}}\right )} - 2\right )} e^{\left (-\frac {1}{x}\right )}\right ) \]
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=\
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 = 0.37 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \[ \int \frac {-4 x^3+e^{\frac {2 x+4 \log (3)}{x^2}} \left (30 x-6 x^2+x^3+(120-24 x) \log (3)\right )+\left (-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3\right ) \log \left (-4+e^{\frac {2 x+4 \log (3)}{x^2}}\right )}{-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3} \, dx=x \log \left (e^{\left (\frac {2 \, {\left (x + 2 \, \log \left (3\right )\right )}}{x^{2}}\right )} - 4\right ) + \frac {1}{3} \, x - 5 \, \log \left (e^{\left (\frac {2 \, {\left (x + 2 \, \log \left (3\right )\right )}}{x^{2}}\right )} - 4\right ) \]
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=\
Time = 9.22 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {-4 x^3+e^{\frac {2 x+4 \log (3)}{x^2}} \left (30 x-6 x^2+x^3+(120-24 x) \log (3)\right )+\left (-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3\right ) \log \left (-4+e^{\frac {2 x+4 \log (3)}{x^2}}\right )}{-12 x^3+3 e^{\frac {2 x+4 \log (3)}{x^2}} x^3} \, dx=\frac {x}{3}-5\,\ln \left (3^{\frac {4}{x^2}}\,{\mathrm {e}}^{2/x}-4\right )+x\,\ln \left (3^{\frac {4}{x^2}}\,{\mathrm {e}}^{2/x}-4\right ) \]