\(\int \frac {(e^{1+x} (4+x))^{(3 x+3 x^2) \log (8)+x \log (8) \log (x)} ((15 x+18 x^2+3 x^3) \log (8)+(5 x+x^2) \log (8) \log (x)+((16+28 x+6 x^2) \log (8)+(4+x) \log (8) \log (x)) \log (e^{1+x} (4+x)))}{4+x} \, dx\) [2703]

Optimal result

Integrand size = 98, antiderivative size = 24 \[ \int \frac {\left (e^{1+x} (4+x)\right )^{\left (3 x+3 x^2\right ) \log (8)+x \log (8) \log (x)} \left (\left (15 x+18 x^2+3 x^3\right ) \log (8)+\left (5 x+x^2\right ) \log (8) \log (x)+\left (\left (16+28 x+6 x^2\right ) \log (8)+(4+x) \log (8) \log (x)\right ) \log \left (e^{1+x} (4+x)\right )\right )}{4+x} \, dx=e^{x \log (8) (3 (1+x)+\log (x)) \log \left (e^{1+x} (4+x)\right )} \] Output:

exp(3*ln((4+x)*exp(1+x))*x*(3*x+3+ln(x))*ln(2))
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {\left (e^{1+x} (4+x)\right )^{\left (3 x+3 x^2\right ) \log (8)+x \log (8) \log (x)} \left (\left (15 x+18 x^2+3 x^3\right ) \log (8)+\left (5 x+x^2\right ) \log (8) \log (x)+\left (\left (16+28 x+6 x^2\right ) \log (8)+(4+x) \log (8) \log (x)\right ) \log \left (e^{1+x} (4+x)\right )\right )}{4+x} \, dx=\left (e^{1+x} (4+x)\right )^{x \log (8) (3+3 x+\log (x))} \] Input:

Integrate[((E^(1 + x)*(4 + x))^((3*x + 3*x^2)*Log[8] + x*Log[8]*Log[x])*(( 
15*x + 18*x^2 + 3*x^3)*Log[8] + (5*x + x^2)*Log[8]*Log[x] + ((16 + 28*x + 
6*x^2)*Log[8] + (4 + x)*Log[8]*Log[x])*Log[E^(1 + x)*(4 + x)]))/(4 + x),x]
 

Output:

(E^(1 + x)*(4 + x))^(x*Log[8]*(3 + 3*x + Log[x]))
 

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.

\(\displaystyle \int \frac {\left (e^{x+1} (x+4)\right )^{\left (3 x^2+3 x\right ) \log (8)+x \log (8) \log (x)} \left (\left (x^2+5 x\right ) \log (8) \log (x)+\left (\left (6 x^2+28 x+16\right ) \log (8)+(x+4) \log (8) \log (x)\right ) \log \left (e^{x+1} (x+4)\right )+\left (3 x^3+18 x^2+15 x\right ) \log (8)\right )}{x+4} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {\left (e^{x+1} (x+4)\right )^{x \log (8) (3 x+\log (x)+3)} \left (\left (x^2+5 x\right ) \log (8) \log (x)+\left (\left (6 x^2+28 x+16\right ) \log (8)+(x+4) \log (8) \log (x)\right ) \log \left (e^{x+1} (x+4)\right )+\left (3 x^3+18 x^2+15 x\right ) \log (8)\right )}{x+4}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {x (x+5) \log (8) (3 x+\log (x)+3) \left (e^{x+1} (x+4)\right )^{x \log (8) (3 x+\log (x)+3)}}{x+4}+\log (8) (6 x+\log (x)+4) \log \left (e^{x+1} (x+4)\right ) \left (e^{x+1} (x+4)\right )^{x \log (8) (3 x+\log (x)+3)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 3 \log (8) \int x^2 \left (e^{x+1} (x+4)\right )^{x \log (8) (3 x+\log (x)+3)}dx-9 \log (8) \int \left (e^{x+1} (x+4)\right )^{x \log (8) (3 x+\log (x)+3)}dx+6 \log (8) \int x \left (e^{x+1} (x+4)\right )^{x \log (8) (3 x+\log (x)+3)}dx+36 \log (8) \int \frac {\left (e^{x+1} (x+4)\right )^{x \log (8) (3 x+\log (x)+3)}}{x+4}dx+\log (8) \int \left (e^{x+1} (x+4)\right )^{x \log (8) (3 x+\log (x)+3)} \log (x)dx+\log (8) \int x \left (e^{x+1} (x+4)\right )^{x \log (8) (3 x+\log (x)+3)} \log (x)dx-4 \log (8) \int \frac {\left (e^{x+1} (x+4)\right )^{x \log (8) (3 x+\log (x)+3)} \log (x)}{x+4}dx+4 \log (8) \int \left (e^{x+1} (x+4)\right )^{x \log (8) (3 x+\log (x)+3)} \log \left (e^{x+1} (x+4)\right )dx+6 \log (8) \int x \left (e^{x+1} (x+4)\right )^{x \log (8) (3 x+\log (x)+3)} \log \left (e^{x+1} (x+4)\right )dx+\log (8) \int \left (e^{x+1} (x+4)\right )^{x \log (8) (3 x+\log (x)+3)} \log (x) \log \left (e^{x+1} (x+4)\right )dx\)

Input:

Int[((E^(1 + x)*(4 + x))^((3*x + 3*x^2)*Log[8] + x*Log[8]*Log[x])*((15*x + 
 18*x^2 + 3*x^3)*Log[8] + (5*x + x^2)*Log[8]*Log[x] + ((16 + 28*x + 6*x^2) 
*Log[8] + (4 + x)*Log[8]*Log[x])*Log[E^(1 + x)*(4 + x)]))/(4 + x),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 17.43 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96

method	result	size
parallelrisch	\({\mathrm e}^{3 \ln \left (\left (4+x \right ) {\mathrm e}^{1+x}\right ) x \left (3 x +3+\ln \left (x \right )\right ) \ln \left (2\right )}\)	\(23\)
risch	\(2^{-\frac {3 x \left (3 x +3+\ln \left (x \right )\right ) \left (i \operatorname {csgn}\left (i {\mathrm e}^{1+x} \left (4+x \right )\right ) \pi -i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{1+x}\right )-i \pi \,\operatorname {csgn}\left (i \left (4+x \right )\right )+i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{1+x} \left (4+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{1+x}\right ) \operatorname {csgn}\left (i \left (4+x \right )\right )-2 \ln \left ({\mathrm e}^{1+x}\right )-2 \ln \left (4+x \right )\right )}{2}}\)	\(95\)

Input:

int(((3*(4+x)*ln(2)*ln(x)+3*(6*x^2+28*x+16)*ln(2))*ln((4+x)*exp(1+x))+3*(x 
^2+5*x)*ln(2)*ln(x)+3*(3*x^3+18*x^2+15*x)*ln(2))*exp((3*x*ln(2)*ln(x)+3*(3 
*x^2+3*x)*ln(2))*ln((4+x)*exp(1+x)))/(4+x),x,method=_RETURNVERBOSE)
 

Output:

exp(3*ln((4+x)*exp(1+x))*x*(3*x+3+ln(x))*ln(2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {\left (e^{1+x} (4+x)\right )^{\left (3 x+3 x^2\right ) \log (8)+x \log (8) \log (x)} \left (\left (15 x+18 x^2+3 x^3\right ) \log (8)+\left (5 x+x^2\right ) \log (8) \log (x)+\left (\left (16+28 x+6 x^2\right ) \log (8)+(4+x) \log (8) \log (x)\right ) \log \left (e^{1+x} (4+x)\right )\right )}{4+x} \, dx=\left ({\left (x + 4\right )} e^{\left (x + 1\right )}\right )^{3 \, x \log \left (2\right ) \log \left (x\right ) + 9 \, {\left (x^{2} + x\right )} \log \left (2\right )} \] Input:

integrate(((3*(4+x)*log(2)*log(x)+3*(6*x^2+28*x+16)*log(2))*log((4+x)*exp( 
1+x))+3*(x^2+5*x)*log(2)*log(x)+3*(3*x^3+18*x^2+15*x)*log(2))*exp((3*x*log 
(2)*log(x)+3*(3*x^2+3*x)*log(2))*log((4+x)*exp(1+x)))/(4+x),x, algorithm=" 
fricas")
 

Output:

((x + 4)*e^(x + 1))^(3*x*log(2)*log(x) + 9*(x^2 + x)*log(2))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (e^{1+x} (4+x)\right )^{\left (3 x+3 x^2\right ) \log (8)+x \log (8) \log (x)} \left (\left (15 x+18 x^2+3 x^3\right ) \log (8)+\left (5 x+x^2\right ) \log (8) \log (x)+\left (\left (16+28 x+6 x^2\right ) \log (8)+(4+x) \log (8) \log (x)\right ) \log \left (e^{1+x} (4+x)\right )\right )}{4+x} \, dx=\text {Timed out} \] Input:

integrate(((3*(4+x)*ln(2)*ln(x)+3*(6*x**2+28*x+16)*ln(2))*ln((4+x)*exp(1+x 
))+3*(x**2+5*x)*ln(2)*ln(x)+3*(3*x**3+18*x**2+15*x)*ln(2))*exp((3*x*ln(2)* 
ln(x)+3*(3*x**2+3*x)*ln(2))*ln((4+x)*exp(1+x)))/(4+x),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (22) = 44\).

Time = 0.25 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.83 \[ \int \frac {\left (e^{1+x} (4+x)\right )^{\left (3 x+3 x^2\right ) \log (8)+x \log (8) \log (x)} \left (\left (15 x+18 x^2+3 x^3\right ) \log (8)+\left (5 x+x^2\right ) \log (8) \log (x)+\left (\left (16+28 x+6 x^2\right ) \log (8)+(4+x) \log (8) \log (x)\right ) \log \left (e^{1+x} (4+x)\right )\right )}{4+x} \, dx=e^{\left (9 \, x^{3} \log \left (2\right ) + 9 \, x^{2} \log \left (2\right ) \log \left (x + 4\right ) + 3 \, x^{2} \log \left (2\right ) \log \left (x\right ) + 3 \, x \log \left (2\right ) \log \left (x + 4\right ) \log \left (x\right ) + 18 \, x^{2} \log \left (2\right ) + 9 \, x \log \left (2\right ) \log \left (x + 4\right ) + 3 \, x \log \left (2\right ) \log \left (x\right ) + 9 \, x \log \left (2\right )\right )} \] Input:

integrate(((3*(4+x)*log(2)*log(x)+3*(6*x^2+28*x+16)*log(2))*log((4+x)*exp( 
1+x))+3*(x^2+5*x)*log(2)*log(x)+3*(3*x^3+18*x^2+15*x)*log(2))*exp((3*x*log 
(2)*log(x)+3*(3*x^2+3*x)*log(2))*log((4+x)*exp(1+x)))/(4+x),x, algorithm=" 
maxima")
 

Output:

e^(9*x^3*log(2) + 9*x^2*log(2)*log(x + 4) + 3*x^2*log(2)*log(x) + 3*x*log( 
2)*log(x + 4)*log(x) + 18*x^2*log(2) + 9*x*log(2)*log(x + 4) + 3*x*log(2)* 
log(x) + 9*x*log(2))
 

Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {\left (e^{1+x} (4+x)\right )^{\left (3 x+3 x^2\right ) \log (8)+x \log (8) \log (x)} \left (\left (15 x+18 x^2+3 x^3\right ) \log (8)+\left (5 x+x^2\right ) \log (8) \log (x)+\left (\left (16+28 x+6 x^2\right ) \log (8)+(4+x) \log (8) \log (x)\right ) \log \left (e^{1+x} (4+x)\right )\right )}{4+x} \, dx={\left (x e^{\left (x + 1\right )} + 4 \, e^{\left (x + 1\right )}\right )}^{9 \, x^{2} \log \left (2\right ) + 3 \, x \log \left (2\right ) \log \left (x\right ) + 9 \, x \log \left (2\right )} \] Input:

integrate(((3*(4+x)*log(2)*log(x)+3*(6*x^2+28*x+16)*log(2))*log((4+x)*exp( 
1+x))+3*(x^2+5*x)*log(2)*log(x)+3*(3*x^3+18*x^2+15*x)*log(2))*exp((3*x*log 
(2)*log(x)+3*(3*x^2+3*x)*log(2))*log((4+x)*exp(1+x)))/(4+x),x, algorithm=" 
giac")
 

Output:

(x*e^(x + 1) + 4*e^(x + 1))^(9*x^2*log(2) + 3*x*log(2)*log(x) + 9*x*log(2) 
)
 

Mupad [B] (verification not implemented)

Time = 2.85 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.04 \[ \int \frac {\left (e^{1+x} (4+x)\right )^{\left (3 x+3 x^2\right ) \log (8)+x \log (8) \log (x)} \left (\left (15 x+18 x^2+3 x^3\right ) \log (8)+\left (5 x+x^2\right ) \log (8) \log (x)+\left (\left (16+28 x+6 x^2\right ) \log (8)+(4+x) \log (8) \log (x)\right ) \log \left (e^{1+x} (4+x)\right )\right )}{4+x} \, dx=x^{3\,x\,\ln \left (2\right )\,\ln \left (4\,\mathrm {e}\,{\mathrm {e}}^x+x\,\mathrm {e}\,{\mathrm {e}}^x\right )}\,{\left (4\,{\mathrm {e}}^{x+1}+x\,{\mathrm {e}}^{x+1}\right )}^{9\,\ln \left (2\right )\,x^2+9\,\ln \left (2\right )\,x} \] Input:

int((exp(log(exp(x + 1)*(x + 4))*(3*log(2)*(3*x + 3*x^2) + 3*x*log(2)*log( 
x)))*(log(exp(x + 1)*(x + 4))*(3*log(2)*(28*x + 6*x^2 + 16) + 3*log(2)*log 
(x)*(x + 4)) + 3*log(2)*(15*x + 18*x^2 + 3*x^3) + 3*log(2)*log(x)*(5*x + x 
^2)))/(x + 4),x)
 

Output:

x^(3*x*log(2)*log(4*exp(1)*exp(x) + x*exp(1)*exp(x)))*(4*exp(x + 1) + x*ex 
p(x + 1))^(9*x*log(2) + 9*x^2*log(2))
 

Reduce [F]

\[ \int \frac {\left (e^{1+x} (4+x)\right )^{\left (3 x+3 x^2\right ) \log (8)+x \log (8) \log (x)} \left (\left (15 x+18 x^2+3 x^3\right ) \log (8)+\left (5 x+x^2\right ) \log (8) \log (x)+\left (\left (16+28 x+6 x^2\right ) \log (8)+(4+x) \log (8) \log (x)\right ) \log \left (e^{1+x} (4+x)\right )\right )}{4+x} \, dx=\int \frac {\left (\left (3 \left (x +4\right ) \mathrm {log}\left (2\right ) \mathrm {log}\left (x \right )+3 \left (6 x^{2}+28 x +16\right ) \mathrm {log}\left (2\right )\right ) \mathrm {log}\left (\left (x +4\right ) {\mathrm e}^{x +1}\right )+3 \left (x^{2}+5 x \right ) \mathrm {log}\left (2\right ) \mathrm {log}\left (x \right )+3 \left (3 x^{3}+18 x^{2}+15 x \right ) \mathrm {log}\left (2\right )\right ) {\mathrm e}^{\left (3 x \,\mathrm {log}\left (2\right ) \mathrm {log}\left (x \right )+3 \left (3 x^{2}+3 x \right ) \mathrm {log}\left (2\right )\right ) \mathrm {log}\left (\left (x +4\right ) {\mathrm e}^{x +1}\right )}}{x +4}d x \] Input:

int(((3*(4+x)*log(2)*log(x)+3*(6*x^2+28*x+16)*log(2))*log((4+x)*exp(1+x))+ 
3*(x^2+5*x)*log(2)*log(x)+3*(3*x^3+18*x^2+15*x)*log(2))*exp((3*x*log(2)*lo 
g(x)+3*(3*x^2+3*x)*log(2))*log((4+x)*exp(1+x)))/(4+x),x)
 

Output:

int(((3*(4+x)*log(2)*log(x)+3*(6*x^2+28*x+16)*log(2))*log((4+x)*exp(1+x))+ 
3*(x^2+5*x)*log(2)*log(x)+3*(3*x^3+18*x^2+15*x)*log(2))*exp((3*x*log(2)*lo 
g(x)+3*(3*x^2+3*x)*log(2))*log((4+x)*exp(1+x)))/(4+x),x)