Integrand size = 212, antiderivative size = 33 \[ \int \frac {12 x^2 \log (2)+3 e^{2/x} x^2 \log (2)+3 e^{\frac {2 \left (x^3+x^3 \log (2)\right )}{\log (2)}} x^2 \log (2)+e^{\frac {1}{x}} \left (5+12 x^2\right ) \log (2)+e^{\frac {x^3+x^3 \log (2)}{\log (2)}} \left (15 x^4-6 e^{\frac {1}{x}} x^2 \log (2)+\left (-12 x^2+15 x^4\right ) \log (2)\right )}{12 x^2 \log (2)+12 e^{\frac {1}{x}} x^2 \log (2)+3 e^{2/x} x^2 \log (2)+3 e^{\frac {2 \left (x^3+x^3 \log (2)\right )}{\log (2)}} x^2 \log (2)+e^{\frac {x^3+x^3 \log (2)}{\log (2)}} \left (-12 x^2 \log (2)-6 e^{\frac {1}{x}} x^2 \log (2)\right )} \, dx=\frac {5}{3 \left (2+e^{\frac {1}{x}}-e^{\frac {x^2 (x+x \log (2))}{\log (2)}}\right )}+x \] Output:
5/3/(exp(1/x)-exp(x^2*(x*ln(2)+x)/ln(2))+2)+x
\[ \int \frac {12 x^2 \log (2)+3 e^{2/x} x^2 \log (2)+3 e^{\frac {2 \left (x^3+x^3 \log (2)\right )}{\log (2)}} x^2 \log (2)+e^{\frac {1}{x}} \left (5+12 x^2\right ) \log (2)+e^{\frac {x^3+x^3 \log (2)}{\log (2)}} \left (15 x^4-6 e^{\frac {1}{x}} x^2 \log (2)+\left (-12 x^2+15 x^4\right ) \log (2)\right )}{12 x^2 \log (2)+12 e^{\frac {1}{x}} x^2 \log (2)+3 e^{2/x} x^2 \log (2)+3 e^{\frac {2 \left (x^3+x^3 \log (2)\right )}{\log (2)}} x^2 \log (2)+e^{\frac {x^3+x^3 \log (2)}{\log (2)}} \left (-12 x^2 \log (2)-6 e^{\frac {1}{x}} x^2 \log (2)\right )} \, dx=\int \frac {12 x^2 \log (2)+3 e^{2/x} x^2 \log (2)+3 e^{\frac {2 \left (x^3+x^3 \log (2)\right )}{\log (2)}} x^2 \log (2)+e^{\frac {1}{x}} \left (5+12 x^2\right ) \log (2)+e^{\frac {x^3+x^3 \log (2)}{\log (2)}} \left (15 x^4-6 e^{\frac {1}{x}} x^2 \log (2)+\left (-12 x^2+15 x^4\right ) \log (2)\right )}{12 x^2 \log (2)+12 e^{\frac {1}{x}} x^2 \log (2)+3 e^{2/x} x^2 \log (2)+3 e^{\frac {2 \left (x^3+x^3 \log (2)\right )}{\log (2)}} x^2 \log (2)+e^{\frac {x^3+x^3 \log (2)}{\log (2)}} \left (-12 x^2 \log (2)-6 e^{\frac {1}{x}} x^2 \log (2)\right )} \, dx \] Input:
Integrate[(12*x^2*Log[2] + 3*E^(2/x)*x^2*Log[2] + 3*E^((2*(x^3 + x^3*Log[2 ]))/Log[2])*x^2*Log[2] + E^x^(-1)*(5 + 12*x^2)*Log[2] + E^((x^3 + x^3*Log[ 2])/Log[2])*(15*x^4 - 6*E^x^(-1)*x^2*Log[2] + (-12*x^2 + 15*x^4)*Log[2]))/ (12*x^2*Log[2] + 12*E^x^(-1)*x^2*Log[2] + 3*E^(2/x)*x^2*Log[2] + 3*E^((2*( x^3 + x^3*Log[2]))/Log[2])*x^2*Log[2] + E^((x^3 + x^3*Log[2])/Log[2])*(-12 *x^2*Log[2] - 6*E^x^(-1)*x^2*Log[2])),x]
Output:
Integrate[(12*x^2*Log[2] + 3*E^(2/x)*x^2*Log[2] + 3*E^((2*(x^3 + x^3*Log[2 ]))/Log[2])*x^2*Log[2] + E^x^(-1)*(5 + 12*x^2)*Log[2] + E^((x^3 + x^3*Log[ 2])/Log[2])*(15*x^4 - 6*E^x^(-1)*x^2*Log[2] + (-12*x^2 + 15*x^4)*Log[2]))/ (12*x^2*Log[2] + 12*E^x^(-1)*x^2*Log[2] + 3*E^(2/x)*x^2*Log[2] + 3*E^((2*( x^3 + x^3*Log[2]))/Log[2])*x^2*Log[2] + E^((x^3 + x^3*Log[2])/Log[2])*(-12 *x^2*Log[2] - 6*E^x^(-1)*x^2*Log[2])), 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.
\(\displaystyle \int \frac {3 e^{2/x} x^2 \log (2)+12 x^2 \log (2)+e^{\frac {1}{x}} \left (12 x^2+5\right ) \log (2)+3 x^2 \log (2) e^{\frac {2 \left (x^3+x^3 \log (2)\right )}{\log (2)}}+e^{\frac {x^3+x^3 \log (2)}{\log (2)}} \left (15 x^4-6 e^{\frac {1}{x}} x^2 \log (2)+\left (15 x^4-12 x^2\right ) \log (2)\right )}{3 e^{2/x} x^2 \log (2)+12 e^{\frac {1}{x}} x^2 \log (2)+12 x^2 \log (2)+3 x^2 \log (2) e^{\frac {2 \left (x^3+x^3 \log (2)\right )}{\log (2)}}+e^{\frac {x^3+x^3 \log (2)}{\log (2)}} \left (-6 e^{\frac {1}{x}} x^2 \log (2)-12 x^2 \log (2)\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {3 e^{2/x} x^2 \log (2)+12 x^2 \log (2)+e^{\frac {1}{x}} \left (12 x^2+5\right ) \log (2)+3 x^2 \log (2) e^{\frac {2 \left (x^3+x^3 \log (2)\right )}{\log (2)}}+e^{\frac {x^3+x^3 \log (2)}{\log (2)}} \left (15 x^4-6 e^{\frac {1}{x}} x^2 \log (2)+\left (15 x^4-12 x^2\right ) \log (2)\right )}{3 x^2 \log (2) \left (-e^{x^3 \left (1+\frac {1}{\log (2)}\right )}+e^{\frac {1}{x}}+2\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {3 e^{2/x} \log (2) x^2+3 (2 e)^{\frac {2 x^3}{\log (2)}} \log (2) x^2+12 \log (2) x^2+3 (2 e)^{\frac {x^3}{\log (2)}} \left (5 x^4-2 e^{\frac {1}{x}} \log (2) x^2-\left (4 x^2-5 x^4\right ) \log (2)\right )+e^{\frac {1}{x}} \left (12 x^2+5\right ) \log (2)}{\left (2+e^{\frac {1}{x}}-e^{x^3 \left (1+\frac {1}{\log (2)}\right )}\right )^2 x^2}dx}{3 \log (2)}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {\int \left (\frac {\log (2) \left (3 e^{2/x} x^2+12 e^{\frac {1}{x}} x^2+12 x^2+5 e^{\frac {1}{x}}\right )}{\left (2+e^{\frac {1}{x}}-e^{x^3 \left (1+\frac {1}{\log (2)}\right )}\right )^2 x^2}+\frac {3 (2 e)^{\frac {x^3}{\log (2)}} \left (5 (1+\log (2)) x^2-2 e^{\frac {1}{x}} \log (2)-4 \log (2)\right )}{\left (2+e^{\frac {1}{x}}-e^{x^3 \left (1+\frac {1}{\log (2)}\right )}\right )^2}+\frac {3 (2 e)^{\frac {2 x^3}{\log (2)}} \log (2)}{\left (2+e^{\frac {1}{x}}-e^{x^3 \left (1+\frac {1}{\log (2)}\right )}\right )^2}\right )dx}{3 \log (2)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {12 \log (2) \int \frac {1}{\left (2+e^{\frac {1}{x}}-e^{x^3 \left (1+\frac {1}{\log (2)}\right )}\right )^2}dx+12 \log (2) \int \frac {e^{\frac {1}{x}}}{\left (2+e^{\frac {1}{x}}-e^{x^3 \left (1+\frac {1}{\log (2)}\right )}\right )^2}dx+3 \log (2) \int \frac {e^{2/x}}{\left (2+e^{\frac {1}{x}}-e^{x^3 \left (1+\frac {1}{\log (2)}\right )}\right )^2}dx-6 \log (2) \int \frac {e^{\left (1+\frac {1}{\log (2)}\right ) x^3+\frac {1}{x}}}{\left (2+e^{\frac {1}{x}}-e^{x^3 \left (1+\frac {1}{\log (2)}\right )}\right )^2}dx-12 \log (2) \int \frac {e^{x^3 \left (1+\frac {1}{\log (2)}\right )}}{\left (2+e^{\frac {1}{x}}-e^{x^3 \left (1+\frac {1}{\log (2)}\right )}\right )^2}dx+3 \log (2) \int \frac {(2 e)^{\frac {2 x^3}{\log (2)}}}{\left (2+e^{\frac {1}{x}}-e^{x^3 \left (1+\frac {1}{\log (2)}\right )}\right )^2}dx+5 \log (2) \int \frac {e^{\frac {1}{x}}}{\left (2+e^{\frac {1}{x}}-e^{x^3 \left (1+\frac {1}{\log (2)}\right )}\right )^2 x^2}dx+15 (1+\log (2)) \int \frac {(2 e)^{\frac {x^3}{\log (2)}} x^2}{\left (2+e^{\frac {1}{x}}-e^{x^3 \left (1+\frac {1}{\log (2)}\right )}\right )^2}dx}{3 \log (2)}\) |
Input:
Int[(12*x^2*Log[2] + 3*E^(2/x)*x^2*Log[2] + 3*E^((2*(x^3 + x^3*Log[2]))/Lo g[2])*x^2*Log[2] + E^x^(-1)*(5 + 12*x^2)*Log[2] + E^((x^3 + x^3*Log[2])/Lo g[2])*(15*x^4 - 6*E^x^(-1)*x^2*Log[2] + (-12*x^2 + 15*x^4)*Log[2]))/(12*x^ 2*Log[2] + 12*E^x^(-1)*x^2*Log[2] + 3*E^(2/x)*x^2*Log[2] + 3*E^((2*(x^3 + x^3*Log[2]))/Log[2])*x^2*Log[2] + E^((x^3 + x^3*Log[2])/Log[2])*(-12*x^2*L og[2] - 6*E^x^(-1)*x^2*Log[2])),x]
Output:
$Aborted
Time = 3.67 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85
method | result | size |
risch | \(x +\frac {5}{3 \left ({\mathrm e}^{\frac {1}{x}}-{\mathrm e}^{\frac {x^{3} \left (1+\ln \left (2\right )\right )}{\ln \left (2\right )}}+2\right )}\) | \(28\) |
parallelrisch | \(\frac {3 x \ln \left (2\right ) {\mathrm e}^{\frac {1}{x}}-3 \ln \left (2\right ) {\mathrm e}^{\frac {x^{3} \left (1+\ln \left (2\right )\right )}{\ln \left (2\right )}} x +6 x \ln \left (2\right )+5 \ln \left (2\right )}{3 \ln \left (2\right ) \left ({\mathrm e}^{\frac {1}{x}}-{\mathrm e}^{\frac {x^{3} \left (1+\ln \left (2\right )\right )}{\ln \left (2\right )}}+2\right )}\) | \(67\) |
Input:
int((3*x^2*ln(2)*exp((x^3*ln(2)+x^3)/ln(2))^2+(-6*x^2*ln(2)*exp(1/x)+(15*x ^4-12*x^2)*ln(2)+15*x^4)*exp((x^3*ln(2)+x^3)/ln(2))+3*x^2*ln(2)*exp(1/x)^2 +(12*x^2+5)*ln(2)*exp(1/x)+12*x^2*ln(2))/(3*x^2*ln(2)*exp((x^3*ln(2)+x^3)/ ln(2))^2+(-6*x^2*ln(2)*exp(1/x)-12*x^2*ln(2))*exp((x^3*ln(2)+x^3)/ln(2))+3 *x^2*ln(2)*exp(1/x)^2+12*x^2*ln(2)*exp(1/x)+12*x^2*ln(2)),x,method=_RETURN VERBOSE)
Output:
x+5/3/(exp(1/x)-exp(x^3*(1+ln(2))/ln(2))+2)
Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (29) = 58\).
Time = 0.12 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.79 \[ \int \frac {12 x^2 \log (2)+3 e^{2/x} x^2 \log (2)+3 e^{\frac {2 \left (x^3+x^3 \log (2)\right )}{\log (2)}} x^2 \log (2)+e^{\frac {1}{x}} \left (5+12 x^2\right ) \log (2)+e^{\frac {x^3+x^3 \log (2)}{\log (2)}} \left (15 x^4-6 e^{\frac {1}{x}} x^2 \log (2)+\left (-12 x^2+15 x^4\right ) \log (2)\right )}{12 x^2 \log (2)+12 e^{\frac {1}{x}} x^2 \log (2)+3 e^{2/x} x^2 \log (2)+3 e^{\frac {2 \left (x^3+x^3 \log (2)\right )}{\log (2)}} x^2 \log (2)+e^{\frac {x^3+x^3 \log (2)}{\log (2)}} \left (-12 x^2 \log (2)-6 e^{\frac {1}{x}} x^2 \log (2)\right )} \, dx=\frac {3 \, x e^{\left (\frac {x^{3} \log \left (2\right ) + x^{3}}{\log \left (2\right )}\right )} - 3 \, x e^{\frac {1}{x}} - 6 \, x - 5}{3 \, {\left (e^{\left (\frac {x^{3} \log \left (2\right ) + x^{3}}{\log \left (2\right )}\right )} - e^{\frac {1}{x}} - 2\right )}} \] Input:
integrate((3*x^2*log(2)*exp((x^3*log(2)+x^3)/log(2))^2+(-6*x^2*log(2)*exp( 1/x)+(15*x^4-12*x^2)*log(2)+15*x^4)*exp((x^3*log(2)+x^3)/log(2))+3*x^2*log (2)*exp(1/x)^2+(12*x^2+5)*log(2)*exp(1/x)+12*x^2*log(2))/(3*x^2*log(2)*exp ((x^3*log(2)+x^3)/log(2))^2+(-6*x^2*log(2)*exp(1/x)-12*x^2*log(2))*exp((x^ 3*log(2)+x^3)/log(2))+3*x^2*log(2)*exp(1/x)^2+12*x^2*log(2)*exp(1/x)+12*x^ 2*log(2)),x, algorithm="fricas")
Output:
1/3*(3*x*e^((x^3*log(2) + x^3)/log(2)) - 3*x*e^(1/x) - 6*x - 5)/(e^((x^3*l og(2) + x^3)/log(2)) - e^(1/x) - 2)
Time = 0.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int \frac {12 x^2 \log (2)+3 e^{2/x} x^2 \log (2)+3 e^{\frac {2 \left (x^3+x^3 \log (2)\right )}{\log (2)}} x^2 \log (2)+e^{\frac {1}{x}} \left (5+12 x^2\right ) \log (2)+e^{\frac {x^3+x^3 \log (2)}{\log (2)}} \left (15 x^4-6 e^{\frac {1}{x}} x^2 \log (2)+\left (-12 x^2+15 x^4\right ) \log (2)\right )}{12 x^2 \log (2)+12 e^{\frac {1}{x}} x^2 \log (2)+3 e^{2/x} x^2 \log (2)+3 e^{\frac {2 \left (x^3+x^3 \log (2)\right )}{\log (2)}} x^2 \log (2)+e^{\frac {x^3+x^3 \log (2)}{\log (2)}} \left (-12 x^2 \log (2)-6 e^{\frac {1}{x}} x^2 \log (2)\right )} \, dx=x - \frac {5}{- 3 e^{\frac {1}{x}} + 3 e^{\frac {x^{3} \log {\left (2 \right )} + x^{3}}{\log {\left (2 \right )}}} - 6} \] Input:
integrate((3*x**2*ln(2)*exp((x**3*ln(2)+x**3)/ln(2))**2+(-6*x**2*ln(2)*exp (1/x)+(15*x**4-12*x**2)*ln(2)+15*x**4)*exp((x**3*ln(2)+x**3)/ln(2))+3*x**2 *ln(2)*exp(1/x)**2+(12*x**2+5)*ln(2)*exp(1/x)+12*x**2*ln(2))/(3*x**2*ln(2) *exp((x**3*ln(2)+x**3)/ln(2))**2+(-6*x**2*ln(2)*exp(1/x)-12*x**2*ln(2))*ex p((x**3*ln(2)+x**3)/ln(2))+3*x**2*ln(2)*exp(1/x)**2+12*x**2*ln(2)*exp(1/x) +12*x**2*ln(2)),x)
Output:
x - 5/(-3*exp(1/x) + 3*exp((x**3*log(2) + x**3)/log(2)) - 6)
Time = 0.16 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.61 \[ \int \frac {12 x^2 \log (2)+3 e^{2/x} x^2 \log (2)+3 e^{\frac {2 \left (x^3+x^3 \log (2)\right )}{\log (2)}} x^2 \log (2)+e^{\frac {1}{x}} \left (5+12 x^2\right ) \log (2)+e^{\frac {x^3+x^3 \log (2)}{\log (2)}} \left (15 x^4-6 e^{\frac {1}{x}} x^2 \log (2)+\left (-12 x^2+15 x^4\right ) \log (2)\right )}{12 x^2 \log (2)+12 e^{\frac {1}{x}} x^2 \log (2)+3 e^{2/x} x^2 \log (2)+3 e^{\frac {2 \left (x^3+x^3 \log (2)\right )}{\log (2)}} x^2 \log (2)+e^{\frac {x^3+x^3 \log (2)}{\log (2)}} \left (-12 x^2 \log (2)-6 e^{\frac {1}{x}} x^2 \log (2)\right )} \, dx=\frac {3 \, x e^{\left (x^{3} + \frac {x^{3}}{\log \left (2\right )}\right )} - 3 \, x e^{\frac {1}{x}} - 6 \, x - 5}{3 \, {\left (e^{\left (x^{3} + \frac {x^{3}}{\log \left (2\right )}\right )} - e^{\frac {1}{x}} - 2\right )}} \] Input:
integrate((3*x^2*log(2)*exp((x^3*log(2)+x^3)/log(2))^2+(-6*x^2*log(2)*exp( 1/x)+(15*x^4-12*x^2)*log(2)+15*x^4)*exp((x^3*log(2)+x^3)/log(2))+3*x^2*log (2)*exp(1/x)^2+(12*x^2+5)*log(2)*exp(1/x)+12*x^2*log(2))/(3*x^2*log(2)*exp ((x^3*log(2)+x^3)/log(2))^2+(-6*x^2*log(2)*exp(1/x)-12*x^2*log(2))*exp((x^ 3*log(2)+x^3)/log(2))+3*x^2*log(2)*exp(1/x)^2+12*x^2*log(2)*exp(1/x)+12*x^ 2*log(2)),x, algorithm="maxima")
Output:
1/3*(3*x*e^(x^3 + x^3/log(2)) - 3*x*e^(1/x) - 6*x - 5)/(e^(x^3 + x^3/log(2 )) - e^(1/x) - 2)
Leaf count of result is larger than twice the leaf count of optimal. 10848 vs. \(2 (29) = 58\).
Time = 0.30 (sec) , antiderivative size = 10848, normalized size of antiderivative = 328.73 \[ \int \frac {12 x^2 \log (2)+3 e^{2/x} x^2 \log (2)+3 e^{\frac {2 \left (x^3+x^3 \log (2)\right )}{\log (2)}} x^2 \log (2)+e^{\frac {1}{x}} \left (5+12 x^2\right ) \log (2)+e^{\frac {x^3+x^3 \log (2)}{\log (2)}} \left (15 x^4-6 e^{\frac {1}{x}} x^2 \log (2)+\left (-12 x^2+15 x^4\right ) \log (2)\right )}{12 x^2 \log (2)+12 e^{\frac {1}{x}} x^2 \log (2)+3 e^{2/x} x^2 \log (2)+3 e^{\frac {2 \left (x^3+x^3 \log (2)\right )}{\log (2)}} x^2 \log (2)+e^{\frac {x^3+x^3 \log (2)}{\log (2)}} \left (-12 x^2 \log (2)-6 e^{\frac {1}{x}} x^2 \log (2)\right )} \, dx=\text {Too large to display} \] Input:
integrate((3*x^2*log(2)*exp((x^3*log(2)+x^3)/log(2))^2+(-6*x^2*log(2)*exp( 1/x)+(15*x^4-12*x^2)*log(2)+15*x^4)*exp((x^3*log(2)+x^3)/log(2))+3*x^2*log (2)*exp(1/x)^2+(12*x^2+5)*log(2)*exp(1/x)+12*x^2*log(2))/(3*x^2*log(2)*exp ((x^3*log(2)+x^3)/log(2))^2+(-6*x^2*log(2)*exp(1/x)-12*x^2*log(2))*exp((x^ 3*log(2)+x^3)/log(2))+3*x^2*log(2)*exp(1/x)^2+12*x^2*log(2)*exp(1/x)+12*x^ 2*log(2)),x, algorithm="giac")
Output:
1/3*(108*x^9*e^(4*(x^3*log(2) + x^3)/log(2))*log(2)^2 - 432*x^9*e^(3*(x^3* log(2) + x^3)/log(2))*log(2)^2 + 432*x^9*e^(2*(x^3*log(2) + x^3)/log(2))*l og(2)^2 + 54*x^9*e^(4*(x^3*log(2) + x^3)/log(2) + 1/x)*log(2)^2 - 54*x^9*e ^(3*(x^3*log(2) + x^3)/log(2) + 2/x)*log(2)^2 + 27*x^9*e^(3*(x^3*log(2) + x^3)/log(2) + 1/x + (x^4*log(2) + x^4 + log(2))/(x*log(2)))*log(2)^2 - 324 *x^9*e^(3*(x^3*log(2) + x^3)/log(2) + 1/x)*log(2)^2 + 54*x^9*e^(3*(x^3*log (2) + x^3)/log(2) + (x^4*log(2) + x^4 + log(2))/(x*log(2)))*log(2)^2 - 27* x^9*e^(2*(x^3*log(2) + x^3)/log(2) + 2/x + (x^4*log(2) + x^4 + log(2))/(x* log(2)))*log(2)^2 + 108*x^9*e^(2*(x^3*log(2) + x^3)/log(2) + 2/x)*log(2)^2 - 216*x^9*e^(2*(x^3*log(2) + x^3)/log(2) + 1/x + (x^4*log(2) + x^4 + log( 2))/(x*log(2)))*log(2)^2 + 432*x^9*e^(2*(x^3*log(2) + x^3)/log(2) + 1/x)*l og(2)^2 - 324*x^9*e^(2*(x^3*log(2) + x^3)/log(2) + (x^4*log(2) + x^4 + log (2))/(x*log(2)))*log(2)^2 + 108*x^9*e^((x^3*log(2) + x^3)/log(2) + 2/x + ( x^4*log(2) + x^4 + log(2))/(x*log(2)))*log(2)^2 - 27*x^9*e^((x^3*log(2) + x^3)/log(2) + 1/x + 2*(x^4*log(2) + x^4 + log(2))/(x*log(2)))*log(2)^2 + 4 32*x^9*e^((x^3*log(2) + x^3)/log(2) + 1/x + (x^4*log(2) + x^4 + log(2))/(x *log(2)))*log(2)^2 - 54*x^9*e^((x^3*log(2) + x^3)/log(2) + 2*(x^4*log(2) + x^4 + log(2))/(x*log(2)))*log(2)^2 + 432*x^9*e^((x^3*log(2) + x^3)/log(2) + (x^4*log(2) + x^4 + log(2))/(x*log(2)))*log(2)^2 + 27*x^9*e^(2/x + 2*(x ^4*log(2) + x^4 + log(2))/(x*log(2)))*log(2)^2 + 108*x^9*e^(1/x + 2*(x^...
Timed out. \[ \int \frac {12 x^2 \log (2)+3 e^{2/x} x^2 \log (2)+3 e^{\frac {2 \left (x^3+x^3 \log (2)\right )}{\log (2)}} x^2 \log (2)+e^{\frac {1}{x}} \left (5+12 x^2\right ) \log (2)+e^{\frac {x^3+x^3 \log (2)}{\log (2)}} \left (15 x^4-6 e^{\frac {1}{x}} x^2 \log (2)+\left (-12 x^2+15 x^4\right ) \log (2)\right )}{12 x^2 \log (2)+12 e^{\frac {1}{x}} x^2 \log (2)+3 e^{2/x} x^2 \log (2)+3 e^{\frac {2 \left (x^3+x^3 \log (2)\right )}{\log (2)}} x^2 \log (2)+e^{\frac {x^3+x^3 \log (2)}{\log (2)}} \left (-12 x^2 \log (2)-6 e^{\frac {1}{x}} x^2 \log (2)\right )} \, dx=\int \frac {12\,x^2\,\ln \left (2\right )-{\mathrm {e}}^{\frac {x^3\,\ln \left (2\right )+x^3}{\ln \left (2\right )}}\,\left (\ln \left (2\right )\,\left (12\,x^2-15\,x^4\right )-15\,x^4+6\,x^2\,{\mathrm {e}}^{1/x}\,\ln \left (2\right )\right )+3\,x^2\,{\mathrm {e}}^{\frac {2\,\left (x^3\,\ln \left (2\right )+x^3\right )}{\ln \left (2\right )}}\,\ln \left (2\right )+3\,x^2\,{\mathrm {e}}^{2/x}\,\ln \left (2\right )+{\mathrm {e}}^{1/x}\,\ln \left (2\right )\,\left (12\,x^2+5\right )}{12\,x^2\,\ln \left (2\right )-{\mathrm {e}}^{\frac {x^3\,\ln \left (2\right )+x^3}{\ln \left (2\right )}}\,\left (12\,x^2\,\ln \left (2\right )+6\,x^2\,{\mathrm {e}}^{1/x}\,\ln \left (2\right )\right )+3\,x^2\,{\mathrm {e}}^{\frac {2\,\left (x^3\,\ln \left (2\right )+x^3\right )}{\ln \left (2\right )}}\,\ln \left (2\right )+3\,x^2\,{\mathrm {e}}^{2/x}\,\ln \left (2\right )+12\,x^2\,{\mathrm {e}}^{1/x}\,\ln \left (2\right )} \,d x \] Input:
int((12*x^2*log(2) - exp((x^3*log(2) + x^3)/log(2))*(log(2)*(12*x^2 - 15*x ^4) - 15*x^4 + 6*x^2*exp(1/x)*log(2)) + 3*x^2*exp((2*(x^3*log(2) + x^3))/l og(2))*log(2) + 3*x^2*exp(2/x)*log(2) + exp(1/x)*log(2)*(12*x^2 + 5))/(12* x^2*log(2) - exp((x^3*log(2) + x^3)/log(2))*(12*x^2*log(2) + 6*x^2*exp(1/x )*log(2)) + 3*x^2*exp((2*(x^3*log(2) + x^3))/log(2))*log(2) + 3*x^2*exp(2/ x)*log(2) + 12*x^2*exp(1/x)*log(2)),x)
Output:
int((12*x^2*log(2) - exp((x^3*log(2) + x^3)/log(2))*(log(2)*(12*x^2 - 15*x ^4) - 15*x^4 + 6*x^2*exp(1/x)*log(2)) + 3*x^2*exp((2*(x^3*log(2) + x^3))/l og(2))*log(2) + 3*x^2*exp(2/x)*log(2) + exp(1/x)*log(2)*(12*x^2 + 5))/(12* x^2*log(2) - exp((x^3*log(2) + x^3)/log(2))*(12*x^2*log(2) + 6*x^2*exp(1/x )*log(2)) + 3*x^2*exp((2*(x^3*log(2) + x^3))/log(2))*log(2) + 3*x^2*exp(2/ x)*log(2) + 12*x^2*exp(1/x)*log(2)), x)
Time = 0.16 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.94 \[ \int \frac {12 x^2 \log (2)+3 e^{2/x} x^2 \log (2)+3 e^{\frac {2 \left (x^3+x^3 \log (2)\right )}{\log (2)}} x^2 \log (2)+e^{\frac {1}{x}} \left (5+12 x^2\right ) \log (2)+e^{\frac {x^3+x^3 \log (2)}{\log (2)}} \left (15 x^4-6 e^{\frac {1}{x}} x^2 \log (2)+\left (-12 x^2+15 x^4\right ) \log (2)\right )}{12 x^2 \log (2)+12 e^{\frac {1}{x}} x^2 \log (2)+3 e^{2/x} x^2 \log (2)+3 e^{\frac {2 \left (x^3+x^3 \log (2)\right )}{\log (2)}} x^2 \log (2)+e^{\frac {x^3+x^3 \log (2)}{\log (2)}} \left (-12 x^2 \log (2)-6 e^{\frac {1}{x}} x^2 \log (2)\right )} \, dx=\frac {3 e^{\frac {\mathrm {log}\left (2\right ) x^{3}+x^{3}}{\mathrm {log}\left (2\right )}} x -3 e^{\frac {1}{x}} x -6 x -5}{3 e^{\frac {\mathrm {log}\left (2\right ) x^{3}+x^{3}}{\mathrm {log}\left (2\right )}}-3 e^{\frac {1}{x}}-6} \] Input:
int((3*x^2*log(2)*exp((x^3*log(2)+x^3)/log(2))^2+(-6*x^2*log(2)*exp(1/x)+( 15*x^4-12*x^2)*log(2)+15*x^4)*exp((x^3*log(2)+x^3)/log(2))+3*x^2*log(2)*ex p(1/x)^2+(12*x^2+5)*log(2)*exp(1/x)+12*x^2*log(2))/(3*x^2*log(2)*exp((x^3* log(2)+x^3)/log(2))^2+(-6*x^2*log(2)*exp(1/x)-12*x^2*log(2))*exp((x^3*log( 2)+x^3)/log(2))+3*x^2*log(2)*exp(1/x)^2+12*x^2*log(2)*exp(1/x)+12*x^2*log( 2)),x)
Output:
(3*e**((log(2)*x**3 + x**3)/log(2))*x - 3*e**(1/x)*x - 6*x - 5)/(3*(e**((l og(2)*x**3 + x**3)/log(2)) - e**(1/x) - 2))