\(\int \frac {e^{\frac {2 (4 x^2+x^3+\log (\frac {15}{40+3 e^x}))}{4 x^2+x^3}} (e^x (-24 x-6 x^2)+(-640+e^x (-48-18 x)-240 x) \log (\frac {15}{40+3 e^x}))}{640 x^3+320 x^4+40 x^5+e^x (48 x^3+24 x^4+3 x^5)} \, dx\) [1414]

Optimal result

Integrand size = 116, antiderivative size = 30 \[ \int \frac {e^{\frac {2 \left (4 x^2+x^3+\log \left (\frac {15}{40+3 e^x}\right )\right )}{4 x^2+x^3}} \left (e^x \left (-24 x-6 x^2\right )+\left (-640+e^x (-48-18 x)-240 x\right ) \log \left (\frac {15}{40+3 e^x}\right )\right )}{640 x^3+320 x^4+40 x^5+e^x \left (48 x^3+24 x^4+3 x^5\right )} \, dx=e^{\frac {2 \left (x+\frac {\log \left (\frac {5}{\frac {40}{3}+e^x}\right )}{x (4+x)}\right )}{x}} \] Output:

exp((x+ln(5/(exp(x)+40/3))/(4+x)/x)/x)^2
 

Mathematica [F]

\[ \int \frac {e^{\frac {2 \left (4 x^2+x^3+\log \left (\frac {15}{40+3 e^x}\right )\right )}{4 x^2+x^3}} \left (e^x \left (-24 x-6 x^2\right )+\left (-640+e^x (-48-18 x)-240 x\right ) \log \left (\frac {15}{40+3 e^x}\right )\right )}{640 x^3+320 x^4+40 x^5+e^x \left (48 x^3+24 x^4+3 x^5\right )} \, dx=\int \frac {e^{\frac {2 \left (4 x^2+x^3+\log \left (\frac {15}{40+3 e^x}\right )\right )}{4 x^2+x^3}} \left (e^x \left (-24 x-6 x^2\right )+\left (-640+e^x (-48-18 x)-240 x\right ) \log \left (\frac {15}{40+3 e^x}\right )\right )}{640 x^3+320 x^4+40 x^5+e^x \left (48 x^3+24 x^4+3 x^5\right )} \, dx \] Input:

Integrate[(E^((2*(4*x^2 + x^3 + Log[15/(40 + 3*E^x)]))/(4*x^2 + x^3))*(E^x 
*(-24*x - 6*x^2) + (-640 + E^x*(-48 - 18*x) - 240*x)*Log[15/(40 + 3*E^x)]) 
)/(640*x^3 + 320*x^4 + 40*x^5 + E^x*(48*x^3 + 24*x^4 + 3*x^5)),x]
 

Output:

Integrate[(E^((2*(4*x^2 + x^3 + Log[15/(40 + 3*E^x)]))/(4*x^2 + x^3))*(E^x 
*(-24*x - 6*x^2) + (-640 + E^x*(-48 - 18*x) - 240*x)*Log[15/(40 + 3*E^x)]) 
)/(640*x^3 + 320*x^4 + 40*x^5 + E^x*(48*x^3 + 24*x^4 + 3*x^5)), 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.

\(\displaystyle \int \frac {\left (e^x \left (-6 x^2-24 x\right )+\left (e^x (-18 x-48)-240 x-640\right ) \log \left (\frac {15}{3 e^x+40}\right )\right ) \exp \left (\frac {2 \left (x^3+4 x^2+\log \left (\frac {15}{3 e^x+40}\right )\right )}{x^3+4 x^2}\right )}{40 x^5+320 x^4+640 x^3+e^x \left (3 x^5+24 x^4+48 x^3\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {\left (e^x \left (-6 x^2-24 x\right )+\left (e^x (-18 x-48)-240 x-640\right ) \log \left (\frac {15}{3 e^x+40}\right )\right ) \exp \left (\frac {2 \left (x^3+4 x^2+\log \left (\frac {15}{3 e^x+40}\right )\right )}{x^2 (x+4)}\right )}{\left (3 e^x+40\right ) x^3 (x+4)^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {80 \exp \left (\frac {2 \left (x^3+4 x^2+\log \left (\frac {15}{3 e^x+40}\right )\right )}{x^2 (x+4)}\right )}{\left (3 e^x+40\right ) x^2 (x+4)}-\frac {2 \left (x^2+4 x+3 x \log \left (\frac {15}{3 e^x+40}\right )+8 \log \left (\frac {15}{3 e^x+40}\right )\right ) \exp \left (\frac {2 \left (x^3+4 x^2+\log \left (\frac {15}{3 e^x+40}\right )\right )}{x^2 (x+4)}\right )}{x^3 (x+4)^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {1}{2} \int \frac {\exp \left (\frac {2 \left (x^3+4 x^2+\log \left (\frac {15}{40+3 e^x}\right )\right )}{x^2 (x+4)}\right )}{x^2}dx+20 \int \frac {\exp \left (\frac {2 \left (x^3+4 x^2+\log \left (\frac {15}{40+3 e^x}\right )\right )}{x^2 (x+4)}\right )}{\left (40+3 e^x\right ) x^2}dx+\frac {1}{8} \int \frac {\exp \left (\frac {2 \left (x^3+4 x^2+\log \left (\frac {15}{40+3 e^x}\right )\right )}{x^2 (x+4)}\right )}{x}dx-5 \int \frac {\exp \left (\frac {2 \left (x^3+4 x^2+\log \left (\frac {15}{40+3 e^x}\right )\right )}{x^2 (x+4)}\right )}{\left (40+3 e^x\right ) x}dx-\frac {1}{8} \int \frac {\exp \left (\frac {2 \left (x^3+4 x^2+\log \left (\frac {15}{40+3 e^x}\right )\right )}{x^2 (x+4)}\right )}{x+4}dx+5 \int \frac {\exp \left (\frac {2 \left (x^3+4 x^2+\log \left (\frac {15}{40+3 e^x}\right )\right )}{x^2 (x+4)}\right )}{\left (40+3 e^x\right ) (x+4)}dx-\int \frac {\exp \left (\frac {2 \left (x^3+4 x^2+\log \left (\frac {15}{40+3 e^x}\right )\right )}{x^2 (x+4)}\right ) \log \left (\frac {15}{40+3 e^x}\right )}{x^3}dx+\frac {1}{8} \int \frac {\exp \left (\frac {2 \left (x^3+4 x^2+\log \left (\frac {15}{40+3 e^x}\right )\right )}{x^2 (x+4)}\right ) \log \left (\frac {15}{40+3 e^x}\right )}{x^2}dx-\frac {1}{8} \int \frac {\exp \left (\frac {2 \left (x^3+4 x^2+\log \left (\frac {15}{40+3 e^x}\right )\right )}{x^2 (x+4)}\right ) \log \left (\frac {15}{40+3 e^x}\right )}{(x+4)^2}dx\)

Input:

Int[(E^((2*(4*x^2 + x^3 + Log[15/(40 + 3*E^x)]))/(4*x^2 + x^3))*(E^x*(-24* 
x - 6*x^2) + (-640 + E^x*(-48 - 18*x) - 240*x)*Log[15/(40 + 3*E^x)]))/(640 
*x^3 + 320*x^4 + 40*x^5 + E^x*(48*x^3 + 24*x^4 + 3*x^5)),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 33.61 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13

method	result	size
risch	\(5^{\frac {2}{x^{2} \left (4+x \right )}} \left ({\mathrm e}^{x}+\frac {40}{3}\right )^{-\frac {2}{x^{2} \left (4+x \right )}} {\mathrm e}^{2}\)	\(34\)

Input:

int((((-18*x-48)*exp(x)-240*x-640)*ln(15/(3*exp(x)+40))+(-6*x^2-24*x)*exp( 
x))*exp((ln(15/(3*exp(x)+40))+x^3+4*x^2)/(x^3+4*x^2))^2/((3*x^5+24*x^4+48* 
x^3)*exp(x)+40*x^5+320*x^4+640*x^3),x,method=_RETURNVERBOSE)
 

Output:

(5^(1/x^2/(4+x)))^2*((exp(x)+40/3)^(-1/x^2/(4+x)))^2*exp(2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {e^{\frac {2 \left (4 x^2+x^3+\log \left (\frac {15}{40+3 e^x}\right )\right )}{4 x^2+x^3}} \left (e^x \left (-24 x-6 x^2\right )+\left (-640+e^x (-48-18 x)-240 x\right ) \log \left (\frac {15}{40+3 e^x}\right )\right )}{640 x^3+320 x^4+40 x^5+e^x \left (48 x^3+24 x^4+3 x^5\right )} \, dx=e^{\left (\frac {2 \, {\left (x^{3} + 4 \, x^{2} + \log \left (\frac {15}{3 \, e^{x} + 40}\right )\right )}}{x^{3} + 4 \, x^{2}}\right )} \] Input:

integrate((((-18*x-48)*exp(x)-240*x-640)*log(15/(3*exp(x)+40))+(-6*x^2-24* 
x)*exp(x))*exp((log(15/(3*exp(x)+40))+x^3+4*x^2)/(x^3+4*x^2))^2/((3*x^5+24 
*x^4+48*x^3)*exp(x)+40*x^5+320*x^4+640*x^3),x, algorithm="fricas")
 

Output:

e^(2*(x^3 + 4*x^2 + log(15/(3*e^x + 40)))/(x^3 + 4*x^2))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{\frac {2 \left (4 x^2+x^3+\log \left (\frac {15}{40+3 e^x}\right )\right )}{4 x^2+x^3}} \left (e^x \left (-24 x-6 x^2\right )+\left (-640+e^x (-48-18 x)-240 x\right ) \log \left (\frac {15}{40+3 e^x}\right )\right )}{640 x^3+320 x^4+40 x^5+e^x \left (48 x^3+24 x^4+3 x^5\right )} \, dx=\text {Timed out} \] Input:

integrate((((-18*x-48)*exp(x)-240*x-640)*ln(15/(3*exp(x)+40))+(-6*x**2-24* 
x)*exp(x))*exp((ln(15/(3*exp(x)+40))+x**3+4*x**2)/(x**3+4*x**2))**2/((3*x* 
*5+24*x**4+48*x**3)*exp(x)+40*x**5+320*x**4+640*x**3),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 87 vs. \(2 (28) = 56\).

Time = 0.36 (sec) , antiderivative size = 87, normalized size of antiderivative = 2.90 \[ \int \frac {e^{\frac {2 \left (4 x^2+x^3+\log \left (\frac {15}{40+3 e^x}\right )\right )}{4 x^2+x^3}} \left (e^x \left (-24 x-6 x^2\right )+\left (-640+e^x (-48-18 x)-240 x\right ) \log \left (\frac {15}{40+3 e^x}\right )\right )}{640 x^3+320 x^4+40 x^5+e^x \left (48 x^3+24 x^4+3 x^5\right )} \, dx=e^{\left (\frac {\log \left (5\right )}{8 \, {\left (x + 4\right )}} - \frac {\log \left (5\right )}{8 \, x} + \frac {\log \left (3\right )}{8 \, {\left (x + 4\right )}} - \frac {\log \left (3\right )}{8 \, x} - \frac {\log \left (3 \, e^{x} + 40\right )}{8 \, {\left (x + 4\right )}} + \frac {\log \left (3 \, e^{x} + 40\right )}{8 \, x} + \frac {\log \left (5\right )}{2 \, x^{2}} + \frac {\log \left (3\right )}{2 \, x^{2}} - \frac {\log \left (3 \, e^{x} + 40\right )}{2 \, x^{2}} + 2\right )} \] Input:

integrate((((-18*x-48)*exp(x)-240*x-640)*log(15/(3*exp(x)+40))+(-6*x^2-24* 
x)*exp(x))*exp((log(15/(3*exp(x)+40))+x^3+4*x^2)/(x^3+4*x^2))^2/((3*x^5+24 
*x^4+48*x^3)*exp(x)+40*x^5+320*x^4+640*x^3),x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (28) = 56\).

Time = 0.24 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.93 \[ \int \frac {e^{\frac {2 \left (4 x^2+x^3+\log \left (\frac {15}{40+3 e^x}\right )\right )}{4 x^2+x^3}} \left (e^x \left (-24 x-6 x^2\right )+\left (-640+e^x (-48-18 x)-240 x\right ) \log \left (\frac {15}{40+3 e^x}\right )\right )}{640 x^3+320 x^4+40 x^5+e^x \left (48 x^3+24 x^4+3 x^5\right )} \, dx=e^{\left (\frac {2 \, x^{3}}{x^{3} + 4 \, x^{2}} + \frac {8 \, x^{2}}{x^{3} + 4 \, x^{2}} + \frac {2 \, \log \left (\frac {15}{3 \, e^{x} + 40}\right )}{x^{3} + 4 \, x^{2}}\right )} \] Input:

integrate((((-18*x-48)*exp(x)-240*x-640)*log(15/(3*exp(x)+40))+(-6*x^2-24* 
x)*exp(x))*exp((log(15/(3*exp(x)+40))+x^3+4*x^2)/(x^3+4*x^2))^2/((3*x^5+24 
*x^4+48*x^3)*exp(x)+40*x^5+320*x^4+640*x^3),x, algorithm="giac")
 

Output:

e^(2*x^3/(x^3 + 4*x^2) + 8*x^2/(x^3 + 4*x^2) + 2*log(15/(3*e^x + 40))/(x^3 
 + 4*x^2))
 

Mupad [B] (verification not implemented)

Time = 3.23 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.90 \[ \int \frac {e^{\frac {2 \left (4 x^2+x^3+\log \left (\frac {15}{40+3 e^x}\right )\right )}{4 x^2+x^3}} \left (e^x \left (-24 x-6 x^2\right )+\left (-640+e^x (-48-18 x)-240 x\right ) \log \left (\frac {15}{40+3 e^x}\right )\right )}{640 x^3+320 x^4+40 x^5+e^x \left (48 x^3+24 x^4+3 x^5\right )} \, dx={\mathrm {e}}^{\frac {2\,x^3}{x^3+4\,x^2}}\,{\mathrm {e}}^{\frac {8\,x^2}{x^3+4\,x^2}}\,{\left (\frac {225}{{\left (3\,{\mathrm {e}}^x+40\right )}^2}\right )}^{\frac {1}{x^3+4\,x^2}} \] Input:

int(-(exp((2*(log(15/(3*exp(x) + 40)) + 4*x^2 + x^3))/(4*x^2 + x^3))*(exp( 
x)*(24*x + 6*x^2) + log(15/(3*exp(x) + 40))*(240*x + exp(x)*(18*x + 48) + 
640)))/(exp(x)*(48*x^3 + 24*x^4 + 3*x^5) + 640*x^3 + 320*x^4 + 40*x^5),x)
 

Output:

exp((2*x^3)/(4*x^2 + x^3))*exp((8*x^2)/(4*x^2 + x^3))*(225/(3*exp(x) + 40) 
^2)^(1/(4*x^2 + x^3))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {e^{\frac {2 \left (4 x^2+x^3+\log \left (\frac {15}{40+3 e^x}\right )\right )}{4 x^2+x^3}} \left (e^x \left (-24 x-6 x^2\right )+\left (-640+e^x (-48-18 x)-240 x\right ) \log \left (\frac {15}{40+3 e^x}\right )\right )}{640 x^3+320 x^4+40 x^5+e^x \left (48 x^3+24 x^4+3 x^5\right )} \, dx=e^{\frac {2 \,\mathrm {log}\left (\frac {15}{3 e^{x}+40}\right )}{x^{3}+4 x^{2}}} e^{2} \] Input:

int((((-18*x-48)*exp(x)-240*x-640)*log(15/(3*exp(x)+40))+(-6*x^2-24*x)*exp 
(x))*exp((log(15/(3*exp(x)+40))+x^3+4*x^2)/(x^3+4*x^2))^2/((3*x^5+24*x^4+4 
8*x^3)*exp(x)+40*x^5+320*x^4+640*x^3),x)
 

Output:

e**((2*log(15/(3*e**x + 40)))/(x**3 + 4*x**2))*e**2