Integrand size = 215, antiderivative size = 28 \[ \int \frac {(1+2 x)^{\frac {3 x}{-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)}} \left (-18 x \log (x)+(-9-18 x) \log (x) \log (1+2 x)+e^{\log ^2(\log (x))} \left (-6 x-12 x^2\right ) \log ^2(1+2 x) \log (\log (x))\right )}{(9+18 x) \log (x)+\left (12 x+24 x^2\right ) \log (x) \log (1+2 x)+e^{2 \log ^2(\log (x))} \left (x^2+2 x^3\right ) \log (x) \log ^2(1+2 x)+\left (4 x^2+8 x^3\right ) \log (x) \log ^2(1+2 x)+e^{\log ^2(\log (x))} \left (\left (-6 x-12 x^2\right ) \log (x) \log (1+2 x)+\left (-4 x^2-8 x^3\right ) \log (x) \log ^2(1+2 x)\right )} \, dx=e^{\frac {3}{-2+e^{\log ^2(\log (x))}-\frac {3}{x \log (1+2 x)}}} \] Output:
exp(3/(exp(ln(ln(x))^2)-3/x/ln(1+2*x)-2))
Time = 0.16 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {(1+2 x)^{\frac {3 x}{-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)}} \left (-18 x \log (x)+(-9-18 x) \log (x) \log (1+2 x)+e^{\log ^2(\log (x))} \left (-6 x-12 x^2\right ) \log ^2(1+2 x) \log (\log (x))\right )}{(9+18 x) \log (x)+\left (12 x+24 x^2\right ) \log (x) \log (1+2 x)+e^{2 \log ^2(\log (x))} \left (x^2+2 x^3\right ) \log (x) \log ^2(1+2 x)+\left (4 x^2+8 x^3\right ) \log (x) \log ^2(1+2 x)+e^{\log ^2(\log (x))} \left (\left (-6 x-12 x^2\right ) \log (x) \log (1+2 x)+\left (-4 x^2-8 x^3\right ) \log (x) \log ^2(1+2 x)\right )} \, dx=(1+2 x)^{\frac {3 x}{-3+\left (-2+e^{\log ^2(\log (x))}\right ) x \log (1+2 x)}} \] Input:
Integrate[((1 + 2*x)^((3*x)/(-3 - 2*x*Log[1 + 2*x] + E^Log[Log[x]]^2*x*Log [1 + 2*x]))*(-18*x*Log[x] + (-9 - 18*x)*Log[x]*Log[1 + 2*x] + E^Log[Log[x] ]^2*(-6*x - 12*x^2)*Log[1 + 2*x]^2*Log[Log[x]]))/((9 + 18*x)*Log[x] + (12* x + 24*x^2)*Log[x]*Log[1 + 2*x] + E^(2*Log[Log[x]]^2)*(x^2 + 2*x^3)*Log[x] *Log[1 + 2*x]^2 + (4*x^2 + 8*x^3)*Log[x]*Log[1 + 2*x]^2 + E^Log[Log[x]]^2* ((-6*x - 12*x^2)*Log[x]*Log[1 + 2*x] + (-4*x^2 - 8*x^3)*Log[x]*Log[1 + 2*x ]^2)),x]
Output:
(1 + 2*x)^((3*x)/(-3 + (-2 + E^Log[Log[x]]^2)*x*Log[1 + 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.
| \(\displaystyle \int \frac {(2 x+1)^{\frac {3 x}{x e^{\log ^2(\log (x))} \log (2 x+1)-2 x \log (2 x+1)-3}} \left (\left (-12 x^2-6 x\right ) e^{\log ^2(\log (x))} \log (\log (x)) \log ^2(2 x+1)+(-18 x-9) \log (x) \log (2 x+1)-18 x \log (x)\right )}{\left (24 x^2+12 x\right ) \log (x) \log (2 x+1)+\left (2 x^3+x^2\right ) e^{2 \log ^2(\log (x))} \log (x) \log ^2(2 x+1)+\left (8 x^3+4 x^2\right ) \log (x) \log ^2(2 x+1)+e^{\log ^2(\log (x))} \left (\left (-12 x^2-6 x\right ) \log (x) \log (2 x+1)+\left (-8 x^3-4 x^2\right ) \log (x) \log ^2(2 x+1)\right )+(18 x+9) \log (x)} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {(2 x+1)^{\frac {3 x}{x \left (e^{\log ^2(\log (x))}-2\right ) \log (2 x+1)-3}-1} \left (-6 x (2 x+1) e^{\log ^2(\log (x))} \log (\log (x)) \log ^2(2 x+1)-9 \log (x) (2 x+(2 x+1) \log (2 x+1))\right )}{\log (x) \left (3-x \left (e^{\log ^2(\log (x))}-2\right ) \log (2 x+1)\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (-\frac {3 \left (8 x^2 \log (\log (x)) \log ^2(2 x+1)+4 x \log (\log (x)) \log ^2(2 x+1)+6 x \log (x) \log (2 x+1)+3 \log (x) \log (2 x+1)+12 x \log (\log (x)) \log (2 x+1)+6 \log (\log (x)) \log (2 x+1)+6 x \log (x)\right ) (2 x+1)^{\frac {3 x}{x \left (e^{\log ^2(\log (x))}-2\right ) \log (2 x+1)-3}-1}}{\log (x) \left (x e^{\log ^2(\log (x))} \log (2 x+1)-2 x \log (2 x+1)-3\right )^2}-\frac {6 \log (2 x+1) \log (\log (x)) (2 x+1)^{\frac {3 x}{x \left (e^{\log ^2(\log (x))}-2\right ) \log (2 x+1)-3}}}{\log (x) \left (x e^{\log ^2(\log (x))} \log (2 x+1)-2 x \log (2 x+1)-3\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -24 \int \frac {x^2 (2 x+1)^{\frac {3 x}{\left (-2+e^{\log ^2(\log (x))}\right ) x \log (2 x+1)-3}-1} \log ^2(2 x+1) \log (\log (x))}{\log (x) \left (e^{\log ^2(\log (x))} x \log (2 x+1)-2 x \log (2 x+1)-3\right )^2}dx-18 \int \frac {x (2 x+1)^{\frac {3 x}{\left (-2+e^{\log ^2(\log (x))}\right ) x \log (2 x+1)-3}-1}}{\left (e^{\log ^2(\log (x))} x \log (2 x+1)-2 x \log (2 x+1)-3\right )^2}dx-9 \int \frac {(2 x+1)^{\frac {3 x}{\left (-2+e^{\log ^2(\log (x))}\right ) x \log (2 x+1)-3}-1} \log (2 x+1)}{\left (e^{\log ^2(\log (x))} x \log (2 x+1)-2 x \log (2 x+1)-3\right )^2}dx-18 \int \frac {x (2 x+1)^{\frac {3 x}{\left (-2+e^{\log ^2(\log (x))}\right ) x \log (2 x+1)-3}-1} \log (2 x+1)}{\left (e^{\log ^2(\log (x))} x \log (2 x+1)-2 x \log (2 x+1)-3\right )^2}dx-18 \int \frac {(2 x+1)^{\frac {3 x}{\left (-2+e^{\log ^2(\log (x))}\right ) x \log (2 x+1)-3}-1} \log (2 x+1) \log (\log (x))}{\log (x) \left (e^{\log ^2(\log (x))} x \log (2 x+1)-2 x \log (2 x+1)-3\right )^2}dx-36 \int \frac {x (2 x+1)^{\frac {3 x}{\left (-2+e^{\log ^2(\log (x))}\right ) x \log (2 x+1)-3}-1} \log (2 x+1) \log (\log (x))}{\log (x) \left (e^{\log ^2(\log (x))} x \log (2 x+1)-2 x \log (2 x+1)-3\right )^2}dx-12 \int \frac {x (2 x+1)^{\frac {3 x}{\left (-2+e^{\log ^2(\log (x))}\right ) x \log (2 x+1)-3}-1} \log ^2(2 x+1) \log (\log (x))}{\log (x) \left (e^{\log ^2(\log (x))} x \log (2 x+1)-2 x \log (2 x+1)-3\right )^2}dx-6 \int \frac {(2 x+1)^{\frac {3 x}{\left (-2+e^{\log ^2(\log (x))}\right ) x \log (2 x+1)-3}} \log (2 x+1) \log (\log (x))}{\log (x) \left (e^{\log ^2(\log (x))} x \log (2 x+1)-2 x \log (2 x+1)-3\right )}dx\) |
Input:
Int[((1 + 2*x)^((3*x)/(-3 - 2*x*Log[1 + 2*x] + E^Log[Log[x]]^2*x*Log[1 + 2 *x]))*(-18*x*Log[x] + (-9 - 18*x)*Log[x]*Log[1 + 2*x] + E^Log[Log[x]]^2*(- 6*x - 12*x^2)*Log[1 + 2*x]^2*Log[Log[x]]))/((9 + 18*x)*Log[x] + (12*x + 24 *x^2)*Log[x]*Log[1 + 2*x] + E^(2*Log[Log[x]]^2)*(x^2 + 2*x^3)*Log[x]*Log[1 + 2*x]^2 + (4*x^2 + 8*x^3)*Log[x]*Log[1 + 2*x]^2 + E^Log[Log[x]]^2*((-6*x - 12*x^2)*Log[x]*Log[1 + 2*x] + (-4*x^2 - 8*x^3)*Log[x]*Log[1 + 2*x]^2)), x]
Output:
$Aborted
Time = 0.07 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32
\[\left (1+2 x \right )^{\frac {3 x}{x \ln \left (1+2 x \right ) {\mathrm e}^{\ln \left (\ln \left (x \right )\right )^{2}}-2 x \ln \left (1+2 x \right )-3}}\]
Input:
int(((-12*x^2-6*x)*ln(1+2*x)^2*ln(ln(x))*exp(ln(ln(x))^2)+(-18*x-9)*ln(x)* ln(1+2*x)-18*x*ln(x))*exp(3*x*ln(1+2*x)/(x*ln(1+2*x)*exp(ln(ln(x))^2)-2*x* ln(1+2*x)-3))/((2*x^3+x^2)*ln(x)*ln(1+2*x)^2*exp(ln(ln(x))^2)^2+((-8*x^3-4 *x^2)*ln(x)*ln(1+2*x)^2+(-12*x^2-6*x)*ln(x)*ln(1+2*x))*exp(ln(ln(x))^2)+(8 *x^3+4*x^2)*ln(x)*ln(1+2*x)^2+(24*x^2+12*x)*ln(x)*ln(1+2*x)+(18*x+9)*ln(x) ),x)
Output:
(1+2*x)^(3*x/(x*ln(1+2*x)*exp(ln(ln(x))^2)-2*x*ln(1+2*x)-3))
Time = 0.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {(1+2 x)^{\frac {3 x}{-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)}} \left (-18 x \log (x)+(-9-18 x) \log (x) \log (1+2 x)+e^{\log ^2(\log (x))} \left (-6 x-12 x^2\right ) \log ^2(1+2 x) \log (\log (x))\right )}{(9+18 x) \log (x)+\left (12 x+24 x^2\right ) \log (x) \log (1+2 x)+e^{2 \log ^2(\log (x))} \left (x^2+2 x^3\right ) \log (x) \log ^2(1+2 x)+\left (4 x^2+8 x^3\right ) \log (x) \log ^2(1+2 x)+e^{\log ^2(\log (x))} \left (\left (-6 x-12 x^2\right ) \log (x) \log (1+2 x)+\left (-4 x^2-8 x^3\right ) \log (x) \log ^2(1+2 x)\right )} \, dx={\left (2 \, x + 1\right )}^{\frac {3 \, x}{x e^{\left (\log \left (\log \left (x\right )\right )^{2}\right )} \log \left (2 \, x + 1\right ) - 2 \, x \log \left (2 \, x + 1\right ) - 3}} \] Input:
integrate(((-12*x^2-6*x)*log(1+2*x)^2*log(log(x))*exp(log(log(x))^2)+(-18* x-9)*log(x)*log(1+2*x)-18*x*log(x))*exp(3*x*log(1+2*x)/(x*log(1+2*x)*exp(l og(log(x))^2)-2*x*log(1+2*x)-3))/((2*x^3+x^2)*log(x)*log(1+2*x)^2*exp(log( log(x))^2)^2+((-8*x^3-4*x^2)*log(x)*log(1+2*x)^2+(-12*x^2-6*x)*log(x)*log( 1+2*x))*exp(log(log(x))^2)+(8*x^3+4*x^2)*log(x)*log(1+2*x)^2+(24*x^2+12*x) *log(x)*log(1+2*x)+(18*x+9)*log(x)),x, algorithm="fricas")
Output:
(2*x + 1)^(3*x/(x*e^(log(log(x))^2)*log(2*x + 1) - 2*x*log(2*x + 1) - 3))
Timed out. \[ \int \frac {(1+2 x)^{\frac {3 x}{-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)}} \left (-18 x \log (x)+(-9-18 x) \log (x) \log (1+2 x)+e^{\log ^2(\log (x))} \left (-6 x-12 x^2\right ) \log ^2(1+2 x) \log (\log (x))\right )}{(9+18 x) \log (x)+\left (12 x+24 x^2\right ) \log (x) \log (1+2 x)+e^{2 \log ^2(\log (x))} \left (x^2+2 x^3\right ) \log (x) \log ^2(1+2 x)+\left (4 x^2+8 x^3\right ) \log (x) \log ^2(1+2 x)+e^{\log ^2(\log (x))} \left (\left (-6 x-12 x^2\right ) \log (x) \log (1+2 x)+\left (-4 x^2-8 x^3\right ) \log (x) \log ^2(1+2 x)\right )} \, dx=\text {Timed out} \] Input:
integrate(((-12*x**2-6*x)*ln(1+2*x)**2*ln(ln(x))*exp(ln(ln(x))**2)+(-18*x- 9)*ln(x)*ln(1+2*x)-18*x*ln(x))*exp(3*x*ln(1+2*x)/(x*ln(1+2*x)*exp(ln(ln(x) )**2)-2*x*ln(1+2*x)-3))/((2*x**3+x**2)*ln(x)*ln(1+2*x)**2*exp(ln(ln(x))**2 )**2+((-8*x**3-4*x**2)*ln(x)*ln(1+2*x)**2+(-12*x**2-6*x)*ln(x)*ln(1+2*x))* exp(ln(ln(x))**2)+(8*x**3+4*x**2)*ln(x)*ln(1+2*x)**2+(24*x**2+12*x)*ln(x)* ln(1+2*x)+(18*x+9)*ln(x)),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (28) = 56\).
Time = 0.40 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.29 \[ \int \frac {(1+2 x)^{\frac {3 x}{-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)}} \left (-18 x \log (x)+(-9-18 x) \log (x) \log (1+2 x)+e^{\log ^2(\log (x))} \left (-6 x-12 x^2\right ) \log ^2(1+2 x) \log (\log (x))\right )}{(9+18 x) \log (x)+\left (12 x+24 x^2\right ) \log (x) \log (1+2 x)+e^{2 \log ^2(\log (x))} \left (x^2+2 x^3\right ) \log (x) \log ^2(1+2 x)+\left (4 x^2+8 x^3\right ) \log (x) \log ^2(1+2 x)+e^{\log ^2(\log (x))} \left (\left (-6 x-12 x^2\right ) \log (x) \log (1+2 x)+\left (-4 x^2-8 x^3\right ) \log (x) \log ^2(1+2 x)\right )} \, dx=e^{\left (\frac {9}{x e^{\left (2 \, \log \left (\log \left (x\right )\right )^{2}\right )} \log \left (2 \, x + 1\right ) - {\left (4 \, x \log \left (2 \, x + 1\right ) + 3\right )} e^{\left (\log \left (\log \left (x\right )\right )^{2}\right )} + 4 \, x \log \left (2 \, x + 1\right ) + 6} + \frac {3}{e^{\left (\log \left (\log \left (x\right )\right )^{2}\right )} - 2}\right )} \] Input:
integrate(((-12*x^2-6*x)*log(1+2*x)^2*log(log(x))*exp(log(log(x))^2)+(-18* x-9)*log(x)*log(1+2*x)-18*x*log(x))*exp(3*x*log(1+2*x)/(x*log(1+2*x)*exp(l og(log(x))^2)-2*x*log(1+2*x)-3))/((2*x^3+x^2)*log(x)*log(1+2*x)^2*exp(log( log(x))^2)^2+((-8*x^3-4*x^2)*log(x)*log(1+2*x)^2+(-12*x^2-6*x)*log(x)*log( 1+2*x))*exp(log(log(x))^2)+(8*x^3+4*x^2)*log(x)*log(1+2*x)^2+(24*x^2+12*x) *log(x)*log(1+2*x)+(18*x+9)*log(x)),x, algorithm="maxima")
Output:
e^(9/(x*e^(2*log(log(x))^2)*log(2*x + 1) - (4*x*log(2*x + 1) + 3)*e^(log(l og(x))^2) + 4*x*log(2*x + 1) + 6) + 3/(e^(log(log(x))^2) - 2))
\[ \int \frac {(1+2 x)^{\frac {3 x}{-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)}} \left (-18 x \log (x)+(-9-18 x) \log (x) \log (1+2 x)+e^{\log ^2(\log (x))} \left (-6 x-12 x^2\right ) \log ^2(1+2 x) \log (\log (x))\right )}{(9+18 x) \log (x)+\left (12 x+24 x^2\right ) \log (x) \log (1+2 x)+e^{2 \log ^2(\log (x))} \left (x^2+2 x^3\right ) \log (x) \log ^2(1+2 x)+\left (4 x^2+8 x^3\right ) \log (x) \log ^2(1+2 x)+e^{\log ^2(\log (x))} \left (\left (-6 x-12 x^2\right ) \log (x) \log (1+2 x)+\left (-4 x^2-8 x^3\right ) \log (x) \log ^2(1+2 x)\right )} \, dx=\int { -\frac {3 \, {\left (2 \, {\left (2 \, x^{2} + x\right )} e^{\left (\log \left (\log \left (x\right )\right )^{2}\right )} \log \left (2 \, x + 1\right )^{2} \log \left (\log \left (x\right )\right ) + 3 \, {\left (2 \, x + 1\right )} \log \left (2 \, x + 1\right ) \log \left (x\right ) + 6 \, x \log \left (x\right )\right )} {\left (2 \, x + 1\right )}^{\frac {3 \, x}{x e^{\left (\log \left (\log \left (x\right )\right )^{2}\right )} \log \left (2 \, x + 1\right ) - 2 \, x \log \left (2 \, x + 1\right ) - 3}}}{{\left (2 \, x^{3} + x^{2}\right )} e^{\left (2 \, \log \left (\log \left (x\right )\right )^{2}\right )} \log \left (2 \, x + 1\right )^{2} \log \left (x\right ) + 4 \, {\left (2 \, x^{3} + x^{2}\right )} \log \left (2 \, x + 1\right )^{2} \log \left (x\right ) + 12 \, {\left (2 \, x^{2} + x\right )} \log \left (2 \, x + 1\right ) \log \left (x\right ) - 2 \, {\left (2 \, {\left (2 \, x^{3} + x^{2}\right )} \log \left (2 \, x + 1\right )^{2} \log \left (x\right ) + 3 \, {\left (2 \, x^{2} + x\right )} \log \left (2 \, x + 1\right ) \log \left (x\right )\right )} e^{\left (\log \left (\log \left (x\right )\right )^{2}\right )} + 9 \, {\left (2 \, x + 1\right )} \log \left (x\right )} \,d x } \] Input:
integrate(((-12*x^2-6*x)*log(1+2*x)^2*log(log(x))*exp(log(log(x))^2)+(-18* x-9)*log(x)*log(1+2*x)-18*x*log(x))*exp(3*x*log(1+2*x)/(x*log(1+2*x)*exp(l og(log(x))^2)-2*x*log(1+2*x)-3))/((2*x^3+x^2)*log(x)*log(1+2*x)^2*exp(log( log(x))^2)^2+((-8*x^3-4*x^2)*log(x)*log(1+2*x)^2+(-12*x^2-6*x)*log(x)*log( 1+2*x))*exp(log(log(x))^2)+(8*x^3+4*x^2)*log(x)*log(1+2*x)^2+(24*x^2+12*x) *log(x)*log(1+2*x)+(18*x+9)*log(x)),x, algorithm="giac")
Output:
integrate(-3*(2*(2*x^2 + x)*e^(log(log(x))^2)*log(2*x + 1)^2*log(log(x)) + 3*(2*x + 1)*log(2*x + 1)*log(x) + 6*x*log(x))*(2*x + 1)^(3*x/(x*e^(log(lo g(x))^2)*log(2*x + 1) - 2*x*log(2*x + 1) - 3))/((2*x^3 + x^2)*e^(2*log(log (x))^2)*log(2*x + 1)^2*log(x) + 4*(2*x^3 + x^2)*log(2*x + 1)^2*log(x) + 12 *(2*x^2 + x)*log(2*x + 1)*log(x) - 2*(2*(2*x^3 + x^2)*log(2*x + 1)^2*log(x ) + 3*(2*x^2 + x)*log(2*x + 1)*log(x))*e^(log(log(x))^2) + 9*(2*x + 1)*log (x)), x)
Time = 3.32 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {(1+2 x)^{\frac {3 x}{-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)}} \left (-18 x \log (x)+(-9-18 x) \log (x) \log (1+2 x)+e^{\log ^2(\log (x))} \left (-6 x-12 x^2\right ) \log ^2(1+2 x) \log (\log (x))\right )}{(9+18 x) \log (x)+\left (12 x+24 x^2\right ) \log (x) \log (1+2 x)+e^{2 \log ^2(\log (x))} \left (x^2+2 x^3\right ) \log (x) \log ^2(1+2 x)+\left (4 x^2+8 x^3\right ) \log (x) \log ^2(1+2 x)+e^{\log ^2(\log (x))} \left (\left (-6 x-12 x^2\right ) \log (x) \log (1+2 x)+\left (-4 x^2-8 x^3\right ) \log (x) \log ^2(1+2 x)\right )} \, dx={\mathrm {e}}^{-\frac {3\,x\,\ln \left (2\,x+1\right )}{2\,x\,\ln \left (2\,x+1\right )-x\,{\mathrm {e}}^{{\ln \left (\ln \left (x\right )\right )}^2}\,\ln \left (2\,x+1\right )+3}} \] Input:
int(-(exp(-(3*x*log(2*x + 1))/(2*x*log(2*x + 1) - x*exp(log(log(x))^2)*log (2*x + 1) + 3))*(18*x*log(x) + log(2*x + 1)*log(x)*(18*x + 9) + log(log(x) )*exp(log(log(x))^2)*log(2*x + 1)^2*(6*x + 12*x^2)))/(log(x)*(18*x + 9) - exp(log(log(x))^2)*(log(2*x + 1)^2*log(x)*(4*x^2 + 8*x^3) + log(2*x + 1)*l og(x)*(6*x + 12*x^2)) + log(2*x + 1)^2*log(x)*(4*x^2 + 8*x^3) + log(2*x + 1)*log(x)*(12*x + 24*x^2) + exp(2*log(log(x))^2)*log(2*x + 1)^2*log(x)*(x^ 2 + 2*x^3)),x)
Output:
exp(-(3*x*log(2*x + 1))/(2*x*log(2*x + 1) - x*exp(log(log(x))^2)*log(2*x + 1) + 3))
Time = 1.22 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {(1+2 x)^{\frac {3 x}{-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)}} \left (-18 x \log (x)+(-9-18 x) \log (x) \log (1+2 x)+e^{\log ^2(\log (x))} \left (-6 x-12 x^2\right ) \log ^2(1+2 x) \log (\log (x))\right )}{(9+18 x) \log (x)+\left (12 x+24 x^2\right ) \log (x) \log (1+2 x)+e^{2 \log ^2(\log (x))} \left (x^2+2 x^3\right ) \log (x) \log ^2(1+2 x)+\left (4 x^2+8 x^3\right ) \log (x) \log ^2(1+2 x)+e^{\log ^2(\log (x))} \left (\left (-6 x-12 x^2\right ) \log (x) \log (1+2 x)+\left (-4 x^2-8 x^3\right ) \log (x) \log ^2(1+2 x)\right )} \, dx=e^{\frac {3 \,\mathrm {log}\left (2 x +1\right ) x}{e^{\mathrm {log}\left (\mathrm {log}\left (x \right )\right )^{2}} \mathrm {log}\left (2 x +1\right ) x -2 \,\mathrm {log}\left (2 x +1\right ) x -3}} \] Input:
int(((-12*x^2-6*x)*log(1+2*x)^2*log(log(x))*exp(log(log(x))^2)+(-18*x-9)*l og(x)*log(1+2*x)-18*x*log(x))*exp(3*x*log(1+2*x)/(x*log(1+2*x)*exp(log(log (x))^2)-2*x*log(1+2*x)-3))/((2*x^3+x^2)*log(x)*log(1+2*x)^2*exp(log(log(x) )^2)^2+((-8*x^3-4*x^2)*log(x)*log(1+2*x)^2+(-12*x^2-6*x)*log(x)*log(1+2*x) )*exp(log(log(x))^2)+(8*x^3+4*x^2)*log(x)*log(1+2*x)^2+(24*x^2+12*x)*log(x )*log(1+2*x)+(18*x+9)*log(x)),x)
Output:
e**((3*log(2*x + 1)*x)/(e**(log(log(x))**2)*log(2*x + 1)*x - 2*log(2*x + 1 )*x - 3))