Integrand size = 68, antiderivative size = 26 \[ \int \frac {-108+144 x-36 x^2+\left (54-144 x+54 x^2\right ) \log \left (x^2\right )+e^{-5+e^x} \left (-18+18 x+\left (9-18 x+e^x \left (9 x-9 x^2\right )\right ) \log \left (x^2\right )\right )}{\log ^2\left (x^2\right )} \, dx=\frac {3 (3-3 x) \left (6+e^{-5+e^x}-2 x\right ) x}{\log \left (x^2\right )} \] Output:
3*(exp(exp(x)-5)+6-2*x)*x/ln(x^2)*(-3*x+3)
Time = 0.37 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {-108+144 x-36 x^2+\left (54-144 x+54 x^2\right ) \log \left (x^2\right )+e^{-5+e^x} \left (-18+18 x+\left (9-18 x+e^x \left (9 x-9 x^2\right )\right ) \log \left (x^2\right )\right )}{\log ^2\left (x^2\right )} \, dx=\frac {9 \left (-e^{e^x}+2 e^5 (-3+x)\right ) (-1+x) x}{e^5 \log \left (x^2\right )} \] Input:
Integrate[(-108 + 144*x - 36*x^2 + (54 - 144*x + 54*x^2)*Log[x^2] + E^(-5 + E^x)*(-18 + 18*x + (9 - 18*x + E^x*(9*x - 9*x^2))*Log[x^2]))/Log[x^2]^2, x]
Output:
(9*(-E^E^x + 2*E^5*(-3 + x))*(-1 + x)*x)/(E^5*Log[x^2])
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.
\(\displaystyle \int \frac {-36 x^2+\left (54 x^2-144 x+54\right ) \log \left (x^2\right )+e^{e^x-5} \left (\left (e^x \left (9 x-9 x^2\right )-18 x+9\right ) \log \left (x^2\right )+18 x-18\right )+144 x-108}{\log ^2\left (x^2\right )} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {9 \left (-4 e^5 x^2+6 e^5 x^2 \log \left (x^2\right )-2 e^{e^x} x \log \left (x^2\right )-16 e^5 x \log \left (x^2\right )+e^{e^x} \log \left (x^2\right )+6 e^5 \log \left (x^2\right )+2 e^{e^x} x+16 e^5 x-2 e^{e^x}-12 e^5\right )}{e^5 \log ^2\left (x^2\right )}-\frac {9 e^{x+e^x-5} (x-1) x}{\log \left (x^2\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {18 \int \frac {e^{e^x}}{\log ^2\left (x^2\right )}dx}{e^5}+\frac {18 \int \frac {e^{e^x} x}{\log ^2\left (x^2\right )}dx}{e^5}+\frac {9 \int \frac {e^{e^x}}{\log \left (x^2\right )}dx}{e^5}-\frac {18 \int \frac {e^{e^x} x}{\log \left (x^2\right )}dx}{e^5}+9 \int \frac {e^{x+e^x-5} x}{\log \left (x^2\right )}dx-9 \int \frac {e^{x+e^x-5} x^2}{\log \left (x^2\right )}dx-\frac {72 x^2}{\log \left (x^2\right )}+\frac {54 x}{\log \left (x^2\right )}+\frac {18 x^3}{\log \left (x^2\right )}\) |
Input:
Int[(-108 + 144*x - 36*x^2 + (54 - 144*x + 54*x^2)*Log[x^2] + E^(-5 + E^x) *(-18 + 18*x + (9 - 18*x + E^x*(9*x - 9*x^2))*Log[x^2]))/Log[x^2]^2,x]
Output:
$Aborted
Time = 1.91 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58
method | result | size |
parallelrisch | \(\frac {36 x^{3}-18 \,{\mathrm e}^{{\mathrm e}^{x}-5} x^{2}-144 x^{2}+18 \,{\mathrm e}^{{\mathrm e}^{x}-5} x +108 x}{2 \ln \left (x^{2}\right )}\) | \(41\) |
Input:
int(((((-9*x^2+9*x)*exp(x)-18*x+9)*ln(x^2)+18*x-18)*exp(exp(x)-5)+(54*x^2- 144*x+54)*ln(x^2)-36*x^2+144*x-108)/ln(x^2)^2,x,method=_RETURNVERBOSE)
Output:
1/2*(36*x^3-18*exp(exp(x)-5)*x^2-144*x^2+18*exp(exp(x)-5)*x+108*x)/ln(x^2)
Time = 0.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {-108+144 x-36 x^2+\left (54-144 x+54 x^2\right ) \log \left (x^2\right )+e^{-5+e^x} \left (-18+18 x+\left (9-18 x+e^x \left (9 x-9 x^2\right )\right ) \log \left (x^2\right )\right )}{\log ^2\left (x^2\right )} \, dx=\frac {9 \, {\left (2 \, x^{3} - 8 \, x^{2} - {\left (x^{2} - x\right )} e^{\left (e^{x} - 5\right )} + 6 \, x\right )}}{\log \left (x^{2}\right )} \] Input:
integrate(((((-9*x^2+9*x)*exp(x)-18*x+9)*log(x^2)+18*x-18)*exp(exp(x)-5)+( 54*x^2-144*x+54)*log(x^2)-36*x^2+144*x-108)/log(x^2)^2,x, algorithm="frica s")
Output:
9*(2*x^3 - 8*x^2 - (x^2 - x)*e^(e^x - 5) + 6*x)/log(x^2)
Time = 0.13 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int \frac {-108+144 x-36 x^2+\left (54-144 x+54 x^2\right ) \log \left (x^2\right )+e^{-5+e^x} \left (-18+18 x+\left (9-18 x+e^x \left (9 x-9 x^2\right )\right ) \log \left (x^2\right )\right )}{\log ^2\left (x^2\right )} \, dx=\frac {\left (- 9 x^{2} + 9 x\right ) e^{e^{x} - 5}}{\log {\left (x^{2} \right )}} + \frac {18 x^{3} - 72 x^{2} + 54 x}{\log {\left (x^{2} \right )}} \] Input:
integrate(((((-9*x**2+9*x)*exp(x)-18*x+9)*ln(x**2)+18*x-18)*exp(exp(x)-5)+ (54*x**2-144*x+54)*ln(x**2)-36*x**2+144*x-108)/ln(x**2)**2,x)
Output:
(-9*x**2 + 9*x)*exp(exp(x) - 5)/log(x**2) + (18*x**3 - 72*x**2 + 54*x)/log (x**2)
Time = 0.06 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.54 \[ \int \frac {-108+144 x-36 x^2+\left (54-144 x+54 x^2\right ) \log \left (x^2\right )+e^{-5+e^x} \left (-18+18 x+\left (9-18 x+e^x \left (9 x-9 x^2\right )\right ) \log \left (x^2\right )\right )}{\log ^2\left (x^2\right )} \, dx=\frac {9 \, {\left (2 \, x^{3} e^{5} - 8 \, x^{2} e^{5} + 6 \, x e^{5} - {\left (x^{2} - x\right )} e^{\left (e^{x}\right )}\right )} e^{\left (-5\right )}}{2 \, \log \left (x\right )} \] Input:
integrate(((((-9*x^2+9*x)*exp(x)-18*x+9)*log(x^2)+18*x-18)*exp(exp(x)-5)+( 54*x^2-144*x+54)*log(x^2)-36*x^2+144*x-108)/log(x^2)^2,x, algorithm="maxim a")
Output:
9/2*(2*x^3*e^5 - 8*x^2*e^5 + 6*x*e^5 - (x^2 - x)*e^(e^x))*e^(-5)/log(x)
Time = 0.14 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.65 \[ \int \frac {-108+144 x-36 x^2+\left (54-144 x+54 x^2\right ) \log \left (x^2\right )+e^{-5+e^x} \left (-18+18 x+\left (9-18 x+e^x \left (9 x-9 x^2\right )\right ) \log \left (x^2\right )\right )}{\log ^2\left (x^2\right )} \, dx=\frac {9 \, {\left (2 \, x^{3} e^{5} - 8 \, x^{2} e^{5} - x^{2} e^{\left (e^{x}\right )} + 6 \, x e^{5} + x e^{\left (e^{x}\right )}\right )} e^{\left (-5\right )}}{\log \left (x^{2}\right )} \] Input:
integrate(((((-9*x^2+9*x)*exp(x)-18*x+9)*log(x^2)+18*x-18)*exp(exp(x)-5)+( 54*x^2-144*x+54)*log(x^2)-36*x^2+144*x-108)/log(x^2)^2,x, algorithm="giac" )
Output:
9*(2*x^3*e^5 - 8*x^2*e^5 - x^2*e^(e^x) + 6*x*e^5 + x*e^(e^x))*e^(-5)/log(x ^2)
Time = 2.55 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.73 \[ \int \frac {-108+144 x-36 x^2+\left (54-144 x+54 x^2\right ) \log \left (x^2\right )+e^{-5+e^x} \left (-18+18 x+\left (9-18 x+e^x \left (9 x-9 x^2\right )\right ) \log \left (x^2\right )\right )}{\log ^2\left (x^2\right )} \, dx=27\,x+\frac {18\,x\,\left (x^2-4\,x+3\right )-9\,x\,\ln \left (x^2\right )\,\left (3\,x^2-8\,x+3\right )}{\ln \left (x^2\right )}-72\,x^2+27\,x^3+\frac {{\mathrm {e}}^{{\mathrm {e}}^x-5}\,\left (9\,x-9\,x^2\right )}{\ln \left (x^2\right )} \] Input:
int((144*x + log(x^2)*(54*x^2 - 144*x + 54) + exp(exp(x) - 5)*(18*x + log( x^2)*(exp(x)*(9*x - 9*x^2) - 18*x + 9) - 18) - 36*x^2 - 108)/log(x^2)^2,x)
Output:
27*x + (18*x*(x^2 - 4*x + 3) - 9*x*log(x^2)*(3*x^2 - 8*x + 3))/log(x^2) - 72*x^2 + 27*x^3 + (exp(exp(x) - 5)*(9*x - 9*x^2))/log(x^2)
Time = 0.17 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.73 \[ \int \frac {-108+144 x-36 x^2+\left (54-144 x+54 x^2\right ) \log \left (x^2\right )+e^{-5+e^x} \left (-18+18 x+\left (9-18 x+e^x \left (9 x-9 x^2\right )\right ) \log \left (x^2\right )\right )}{\log ^2\left (x^2\right )} \, dx=\frac {9 x \left (-e^{e^{x}} x +e^{e^{x}}+2 e^{5} x^{2}-8 e^{5} x +6 e^{5}\right )}{\mathrm {log}\left (x^{2}\right ) e^{5}} \] Input:
int(((((-9*x^2+9*x)*exp(x)-18*x+9)*log(x^2)+18*x-18)*exp(exp(x)-5)+(54*x^2 -144*x+54)*log(x^2)-36*x^2+144*x-108)/log(x^2)^2,x)
Output:
(9*x*( - e**(e**x)*x + e**(e**x) + 2*e**5*x**2 - 8*e**5*x + 6*e**5))/(log( x**2)*e**5)