3.2.23 \(\int \frac {(1+2 x)^{\frac {3 x}{-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)}} (-18 x \log (x)+(-9-18 x) \log (x) \log (1+2 x)+e^{\log ^2(\log (x))} (-6 x-12 x^2) \log ^2(1+2 x) \log (\log (x)))}{(9+18 x) \log (x)+(12 x+24 x^2) \log (x) \log (1+2 x)+e^{2 \log ^2(\log (x))} (x^2+2 x^3) \log (x) \log ^2(1+2 x)+(4 x^2+8 x^3) \log (x) \log ^2(1+2 x)+e^{\log ^2(\log (x))} ((-6 x-12 x^2) \log (x) \log (1+2 x)+(-4 x^2-8 x^3) \log (x) \log ^2(1+2 x))} \, dx\) [123]

3.2.23.1 Optimal result
3.2.23.2 Mathematica [A] (verified)
3.2.23.3 Rubi [F]
3.2.23.4 Maple [A] (verified)
3.2.23.5 Fricas [A] (verification not implemented)
3.2.23.6 Sympy [F(-1)]
3.2.23.7 Maxima [B] (verification not implemented)
3.2.23.8 Giac [F]
3.2.23.9 Mupad [B] (verification not implemented)

3.2.23.1 Optimal result

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))
 
3.2.23.2 Mathematica [A] (verified)

Time = 0.27 (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]))
 
3.2.23.3 Rubi [F]

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
 

3.2.23.3.1 Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7239
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl 
erIntegrandQ[v, u, x]]
 

rule 7293
Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 
3.2.23.4 Maple [A] (verified)

Time = 1.15 (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))
 
3.2.23.5 Fricas [A] (verification not implemented)

Time = 0.25 (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=\
 
output
(2*x + 1)^(3*x/(x*e^(log(log(x))^2)*log(2*x + 1) - 2*x*log(2*x + 1) - 3))
 
3.2.23.6 Sympy [F(-1)]

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
 
3.2.23.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (28) = 56\).

Time = 0.54 (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=\
 
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))
 
3.2.23.8 Giac [F]

\[ \int \frac {(1+2 x)^{\frac {3 x}{-3-2 x \log (1+2 x)+e^{\log ^2(\log (x))} x \log (1+2 x)}} \left (-18 x \log (x)+(-9-18 x) \log (x) \log (1+2 x)+e^{\log ^2(\log (x))} \left (-6 x-12 x^2\right ) \log ^2(1+2 x) \log (\log (x))\right )}{(9+18 x) \log (x)+\left (12 x+24 x^2\right ) \log (x) \log (1+2 x)+e^{2 \log ^2(\log (x))} \left (x^2+2 x^3\right ) \log (x) \log ^2(1+2 x)+\left (4 x^2+8 x^3\right ) \log (x) \log ^2(1+2 x)+e^{\log ^2(\log (x))} \left (\left (-6 x-12 x^2\right ) \log (x) \log (1+2 x)+\left (-4 x^2-8 x^3\right ) \log (x) \log ^2(1+2 x)\right )} \, dx=\int { -\frac {3 \, {\left (2 \, {\left (2 \, x^{2} + x\right )} e^{\left (\log \left (\log \left (x\right )\right )^{2}\right )} \log \left (2 \, x + 1\right )^{2} \log \left (\log \left (x\right )\right ) + 3 \, {\left (2 \, x + 1\right )} \log \left (2 \, x + 1\right ) \log \left (x\right ) + 6 \, x \log \left (x\right )\right )} {\left (2 \, x + 1\right )}^{\frac {3 \, x}{x e^{\left (\log \left (\log \left (x\right )\right )^{2}\right )} \log \left (2 \, x + 1\right ) - 2 \, x \log \left (2 \, x + 1\right ) - 3}}}{{\left (2 \, x^{3} + x^{2}\right )} e^{\left (2 \, \log \left (\log \left (x\right )\right )^{2}\right )} \log \left (2 \, x + 1\right )^{2} \log \left (x\right ) + 4 \, {\left (2 \, x^{3} + x^{2}\right )} \log \left (2 \, x + 1\right )^{2} \log \left (x\right ) + 12 \, {\left (2 \, x^{2} + x\right )} \log \left (2 \, x + 1\right ) \log \left (x\right ) - 2 \, {\left (2 \, {\left (2 \, x^{3} + x^{2}\right )} \log \left (2 \, x + 1\right )^{2} \log \left (x\right ) + 3 \, {\left (2 \, x^{2} + x\right )} \log \left (2 \, x + 1\right ) \log \left (x\right )\right )} e^{\left (\log \left (\log \left (x\right )\right )^{2}\right )} + 9 \, {\left (2 \, x + 1\right )} \log \left (x\right )} \,d x } \]

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=\
 
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)
 
3.2.23.9 Mupad [B] (verification not implemented)

Time = 12.11 (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))