3.12.30 \(\int \frac {-1-4 x+e^{e^{10 x^2}-2 e^{5 x^2} x \log (\frac {1}{x+4 x^2})+x^2 \log ^2(\frac {1}{x+4 x^2})} (e^{5 x^2} (2+16 x)+e^{10 x^2} (20 x+80 x^2)+(-2 x-16 x^2+e^{5 x^2} (-2-8 x-20 x^2-80 x^3)) \log (\frac {1}{x+4 x^2})+(2 x+8 x^2) \log ^2(\frac {1}{x+4 x^2}))}{1+4 x} \, dx\) [1130]

3.12.30.1 Optimal result

Integrand size = 156, antiderivative size = 31 \[ \int \frac {-1-4 x+e^{e^{10 x^2}-2 e^{5 x^2} x \log \left (\frac {1}{x+4 x^2}\right )+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (e^{5 x^2} (2+16 x)+e^{10 x^2} \left (20 x+80 x^2\right )+\left (-2 x-16 x^2+e^{5 x^2} \left (-2-8 x-20 x^2-80 x^3\right )\right ) \log \left (\frac {1}{x+4 x^2}\right )+\left (2 x+8 x^2\right ) \log ^2\left (\frac {1}{x+4 x^2}\right )\right )}{1+4 x} \, dx=-1+e^{\left (-e^{5 x^2}+x \log \left (\frac {1}{x+4 x^2}\right )\right )^2}-x \]

output

exp((x*ln(1/(4*x^2+x))-exp(5*x^2))^2)-1-x
 

3.12.30.2 Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.65 \[ \int \frac {-1-4 x+e^{e^{10 x^2}-2 e^{5 x^2} x \log \left (\frac {1}{x+4 x^2}\right )+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (e^{5 x^2} (2+16 x)+e^{10 x^2} \left (20 x+80 x^2\right )+\left (-2 x-16 x^2+e^{5 x^2} \left (-2-8 x-20 x^2-80 x^3\right )\right ) \log \left (\frac {1}{x+4 x^2}\right )+\left (2 x+8 x^2\right ) \log ^2\left (\frac {1}{x+4 x^2}\right )\right )}{1+4 x} \, dx=-x+e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (\frac {1}{x+4 x^2}\right )^{-2 e^{5 x^2} x} \]

input

Integrate[(-1 - 4*x + E^(E^(10*x^2) - 2*E^(5*x^2)*x*Log[(x + 4*x^2)^(-1)] 
+ x^2*Log[(x + 4*x^2)^(-1)]^2)*(E^(5*x^2)*(2 + 16*x) + E^(10*x^2)*(20*x + 
80*x^2) + (-2*x - 16*x^2 + E^(5*x^2)*(-2 - 8*x - 20*x^2 - 80*x^3))*Log[(x 
+ 4*x^2)^(-1)] + (2*x + 8*x^2)*Log[(x + 4*x^2)^(-1)]^2))/(1 + 4*x),x]
 

output

-x + E^(E^(10*x^2) + x^2*Log[(x + 4*x^2)^(-1)]^2)/((x + 4*x^2)^(-1))^(2*E^ 
(5*x^2)*x)
 

3.12.30.3 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.

\(\displaystyle \int \frac {\left (e^{5 x^2} (16 x+2)+e^{10 x^2} \left (80 x^2+20 x\right )+\left (8 x^2+2 x\right ) \log ^2\left (\frac {1}{4 x^2+x}\right )+\left (-16 x^2+e^{5 x^2} \left (-80 x^3-20 x^2-8 x-2\right )-2 x\right ) \log \left (\frac {1}{4 x^2+x}\right )\right ) \exp \left (e^{10 x^2}+x^2 \log ^2\left (\frac {1}{4 x^2+x}\right )-2 e^{5 x^2} x \log \left (\frac {1}{4 x^2+x}\right )\right )-4 x-1}{4 x+1} \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {2 \left (\frac {1}{x (4 x+1)}\right )^{-2 e^{5 x^2} x} e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{4 x^2+x}\right )} \left (40 e^{5 x^2} x^2+10 e^{5 x^2} x-4 x \log \left (\frac {1}{4 x^2+x}\right )-\log \left (\frac {1}{4 x^2+x}\right )+8 x+1\right ) \left (e^{5 x^2}-x \log \left (\frac {1}{4 x^2+x}\right )\right )}{4 x+1}-1\right )dx\)

\(\Big \downarrow \) 7299

\(\displaystyle \int \left (\frac {2 \left (\frac {1}{x (4 x+1)}\right )^{-2 e^{5 x^2} x} e^{e^{10 x^2}+x^2 \log ^2\left (\frac {1}{4 x^2+x}\right )} \left (40 e^{5 x^2} x^2+10 e^{5 x^2} x-4 x \log \left (\frac {1}{4 x^2+x}\right )-\log \left (\frac {1}{4 x^2+x}\right )+8 x+1\right ) \left (e^{5 x^2}-x \log \left (\frac {1}{4 x^2+x}\right )\right )}{4 x+1}-1\right )dx\)

input

Int[(-1 - 4*x + E^(E^(10*x^2) - 2*E^(5*x^2)*x*Log[(x + 4*x^2)^(-1)] + x^2* 
Log[(x + 4*x^2)^(-1)]^2)*(E^(5*x^2)*(2 + 16*x) + E^(10*x^2)*(20*x + 80*x^2 
) + (-2*x - 16*x^2 + E^(5*x^2)*(-2 - 8*x - 20*x^2 - 80*x^3))*Log[(x + 4*x^ 
2)^(-1)] + (2*x + 8*x^2)*Log[(x + 4*x^2)^(-1)]^2))/(1 + 4*x),x]
 

output

$Aborted
 

3.12.30.3.1 Defintions of rubi rules used

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

rule 7299

Int[u_, x_] :> CannotIntegrate[u, x]
 

3.12.30.4 Maple [A] (verified)

Time = 2.24 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.77

method	result	size
parallelrisch	\(\frac {1}{8}-x +{\mathrm e}^{x^{2} \ln \left (\frac {1}{\left (1+4 x \right ) x}\right )^{2}-2 x \,{\mathrm e}^{5 x^{2}} \ln \left (\frac {1}{\left (1+4 x \right ) x}\right )+{\mathrm e}^{10 x^{2}}}\)	\(55\)
risch	\(\text {Expression too large to display}\)	\(863\)

input

int((((8*x^2+2*x)*ln(1/(4*x^2+x))^2+((-80*x^3-20*x^2-8*x-2)*exp(5*x^2)-16* 
x^2-2*x)*ln(1/(4*x^2+x))+(80*x^2+20*x)*exp(5*x^2)^2+(16*x+2)*exp(5*x^2))*e 
xp(x^2*ln(1/(4*x^2+x))^2-2*x*exp(5*x^2)*ln(1/(4*x^2+x))+exp(5*x^2)^2)-4*x- 
1)/(1+4*x),x,method=_RETURNVERBOSE)
 

output

1/8-x+exp(x^2*ln(1/(1+4*x)/x)^2-2*x*exp(5*x^2)*ln(1/(1+4*x)/x)+exp(5*x^2)^ 
2)
 

3.12.30.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.52 \[ \int \frac {-1-4 x+e^{e^{10 x^2}-2 e^{5 x^2} x \log \left (\frac {1}{x+4 x^2}\right )+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (e^{5 x^2} (2+16 x)+e^{10 x^2} \left (20 x+80 x^2\right )+\left (-2 x-16 x^2+e^{5 x^2} \left (-2-8 x-20 x^2-80 x^3\right )\right ) \log \left (\frac {1}{x+4 x^2}\right )+\left (2 x+8 x^2\right ) \log ^2\left (\frac {1}{x+4 x^2}\right )\right )}{1+4 x} \, dx=-x + e^{\left (x^{2} \log \left (\frac {1}{4 \, x^{2} + x}\right )^{2} - 2 \, x e^{\left (5 \, x^{2}\right )} \log \left (\frac {1}{4 \, x^{2} + x}\right ) + e^{\left (10 \, x^{2}\right )}\right )} \]

input

integrate((((8*x^2+2*x)*log(1/(4*x^2+x))^2+((-80*x^3-20*x^2-8*x-2)*exp(5*x 
^2)-16*x^2-2*x)*log(1/(4*x^2+x))+(80*x^2+20*x)*exp(5*x^2)^2+(16*x+2)*exp(5 
*x^2))*exp(x^2*log(1/(4*x^2+x))^2-2*x*exp(5*x^2)*log(1/(4*x^2+x))+exp(5*x^ 
2)^2)-4*x-1)/(1+4*x),x, algorithm=\
 

output

-x + e^(x^2*log(1/(4*x^2 + x))^2 - 2*x*e^(5*x^2)*log(1/(4*x^2 + x)) + e^(1 
0*x^2))
 

3.12.30.6 Sympy [A] (verification not implemented)

Time = 0.70 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \[ \int \frac {-1-4 x+e^{e^{10 x^2}-2 e^{5 x^2} x \log \left (\frac {1}{x+4 x^2}\right )+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (e^{5 x^2} (2+16 x)+e^{10 x^2} \left (20 x+80 x^2\right )+\left (-2 x-16 x^2+e^{5 x^2} \left (-2-8 x-20 x^2-80 x^3\right )\right ) \log \left (\frac {1}{x+4 x^2}\right )+\left (2 x+8 x^2\right ) \log ^2\left (\frac {1}{x+4 x^2}\right )\right )}{1+4 x} \, dx=- x + e^{x^{2} \log {\left (\frac {1}{4 x^{2} + x} \right )}^{2} - 2 x e^{5 x^{2}} \log {\left (\frac {1}{4 x^{2} + x} \right )} + e^{10 x^{2}}} \]

input

integrate((((8*x**2+2*x)*ln(1/(4*x**2+x))**2+((-80*x**3-20*x**2-8*x-2)*exp 
(5*x**2)-16*x**2-2*x)*ln(1/(4*x**2+x))+(80*x**2+20*x)*exp(5*x**2)**2+(16*x 
+2)*exp(5*x**2))*exp(x**2*ln(1/(4*x**2+x))**2-2*x*exp(5*x**2)*ln(1/(4*x**2 
+x))+exp(5*x**2)**2)-4*x-1)/(1+4*x),x)
 

output

-x + exp(x**2*log(1/(4*x**2 + x))**2 - 2*x*exp(5*x**2)*log(1/(4*x**2 + x)) 
 + exp(10*x**2))
 

3.12.30.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (29) = 58\).

Time = 0.34 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.29 \[ \int \frac {-1-4 x+e^{e^{10 x^2}-2 e^{5 x^2} x \log \left (\frac {1}{x+4 x^2}\right )+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (e^{5 x^2} (2+16 x)+e^{10 x^2} \left (20 x+80 x^2\right )+\left (-2 x-16 x^2+e^{5 x^2} \left (-2-8 x-20 x^2-80 x^3\right )\right ) \log \left (\frac {1}{x+4 x^2}\right )+\left (2 x+8 x^2\right ) \log ^2\left (\frac {1}{x+4 x^2}\right )\right )}{1+4 x} \, dx=-x + e^{\left (x^{2} \log \left (4 \, x + 1\right )^{2} + 2 \, x^{2} \log \left (4 \, x + 1\right ) \log \left (x\right ) + x^{2} \log \left (x\right )^{2} + 2 \, x e^{\left (5 \, x^{2}\right )} \log \left (4 \, x + 1\right ) + 2 \, x e^{\left (5 \, x^{2}\right )} \log \left (x\right ) + e^{\left (10 \, x^{2}\right )}\right )} \]

input

integrate((((8*x^2+2*x)*log(1/(4*x^2+x))^2+((-80*x^3-20*x^2-8*x-2)*exp(5*x 
^2)-16*x^2-2*x)*log(1/(4*x^2+x))+(80*x^2+20*x)*exp(5*x^2)^2+(16*x+2)*exp(5 
*x^2))*exp(x^2*log(1/(4*x^2+x))^2-2*x*exp(5*x^2)*log(1/(4*x^2+x))+exp(5*x^ 
2)^2)-4*x-1)/(1+4*x),x, algorithm=\
 

output

-x + e^(x^2*log(4*x + 1)^2 + 2*x^2*log(4*x + 1)*log(x) + x^2*log(x)^2 + 2* 
x*e^(5*x^2)*log(4*x + 1) + 2*x*e^(5*x^2)*log(x) + e^(10*x^2))
 

3.12.30.8 Giac [F]

\[ \int \frac {-1-4 x+e^{e^{10 x^2}-2 e^{5 x^2} x \log \left (\frac {1}{x+4 x^2}\right )+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (e^{5 x^2} (2+16 x)+e^{10 x^2} \left (20 x+80 x^2\right )+\left (-2 x-16 x^2+e^{5 x^2} \left (-2-8 x-20 x^2-80 x^3\right )\right ) \log \left (\frac {1}{x+4 x^2}\right )+\left (2 x+8 x^2\right ) \log ^2\left (\frac {1}{x+4 x^2}\right )\right )}{1+4 x} \, dx=\int { \frac {2 \, {\left ({\left (4 \, x^{2} + x\right )} \log \left (\frac {1}{4 \, x^{2} + x}\right )^{2} + 10 \, {\left (4 \, x^{2} + x\right )} e^{\left (10 \, x^{2}\right )} + {\left (8 \, x + 1\right )} e^{\left (5 \, x^{2}\right )} - {\left (8 \, x^{2} + {\left (40 \, x^{3} + 10 \, x^{2} + 4 \, x + 1\right )} e^{\left (5 \, x^{2}\right )} + x\right )} \log \left (\frac {1}{4 \, x^{2} + x}\right )\right )} e^{\left (x^{2} \log \left (\frac {1}{4 \, x^{2} + x}\right )^{2} - 2 \, x e^{\left (5 \, x^{2}\right )} \log \left (\frac {1}{4 \, x^{2} + x}\right ) + e^{\left (10 \, x^{2}\right )}\right )} - 4 \, x - 1}{4 \, x + 1} \,d x } \]

input

integrate((((8*x^2+2*x)*log(1/(4*x^2+x))^2+((-80*x^3-20*x^2-8*x-2)*exp(5*x 
^2)-16*x^2-2*x)*log(1/(4*x^2+x))+(80*x^2+20*x)*exp(5*x^2)^2+(16*x+2)*exp(5 
*x^2))*exp(x^2*log(1/(4*x^2+x))^2-2*x*exp(5*x^2)*log(1/(4*x^2+x))+exp(5*x^ 
2)^2)-4*x-1)/(1+4*x),x, algorithm=\
 

output

integrate((2*((4*x^2 + x)*log(1/(4*x^2 + x))^2 + 10*(4*x^2 + x)*e^(10*x^2) 
 + (8*x + 1)*e^(5*x^2) - (8*x^2 + (40*x^3 + 10*x^2 + 4*x + 1)*e^(5*x^2) + 
x)*log(1/(4*x^2 + x)))*e^(x^2*log(1/(4*x^2 + x))^2 - 2*x*e^(5*x^2)*log(1/( 
4*x^2 + x)) + e^(10*x^2)) - 4*x - 1)/(4*x + 1), x)
 

3.12.30.9 Mupad [B] (verification not implemented)

Time = 12.92 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.61 \[ \int \frac {-1-4 x+e^{e^{10 x^2}-2 e^{5 x^2} x \log \left (\frac {1}{x+4 x^2}\right )+x^2 \log ^2\left (\frac {1}{x+4 x^2}\right )} \left (e^{5 x^2} (2+16 x)+e^{10 x^2} \left (20 x+80 x^2\right )+\left (-2 x-16 x^2+e^{5 x^2} \left (-2-8 x-20 x^2-80 x^3\right )\right ) \log \left (\frac {1}{x+4 x^2}\right )+\left (2 x+8 x^2\right ) \log ^2\left (\frac {1}{x+4 x^2}\right )\right )}{1+4 x} \, dx=\frac {{\mathrm {e}}^{{\mathrm {e}}^{10\,x^2}}\,{\mathrm {e}}^{x^2\,{\ln \left (\frac {1}{4\,x^2+x}\right )}^2}}{{\left (\frac {1}{4\,x^2+x}\right )}^{2\,x\,{\mathrm {e}}^{5\,x^2}}}-x \]

input

int(-(4*x - exp(exp(10*x^2) + x^2*log(1/(x + 4*x^2))^2 - 2*x*exp(5*x^2)*lo 
g(1/(x + 4*x^2)))*(log(1/(x + 4*x^2))^2*(2*x + 8*x^2) + exp(5*x^2)*(16*x + 
 2) - log(1/(x + 4*x^2))*(2*x + exp(5*x^2)*(8*x + 20*x^2 + 80*x^3 + 2) + 1 
6*x^2) + exp(10*x^2)*(20*x + 80*x^2)) + 1)/(4*x + 1),x)
 

output

(exp(exp(10*x^2))*exp(x^2*log(1/(x + 4*x^2))^2))/(1/(x + 4*x^2))^(2*x*exp( 
5*x^2)) - x