\(\int \frac {e^{-\frac {48+48 x}{-5+x}} (288 e^2+100 x^3+248 x^4+4 x^5+e (-100 x-536 x^2-4 x^3))}{25-10 x+x^2} \, dx\) [495]

Optimal result

Integrand size = 62, antiderivative size = 22 \[ \int \frac {e^{-\frac {48+48 x}{-5+x}} \left (288 e^2+100 x^3+248 x^4+4 x^5+e \left (-100 x-536 x^2-4 x^3\right )\right )}{25-10 x+x^2} \, dx=e^{-\frac {48 (1+x)}{-5+x}} \left (-e+x^2\right )^2 \] Output:

(-exp(1)+x^2)^2/exp(3*(1+x)/(1/16*x-5/16))
 

Mathematica [A] (verified)

Time = 3.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-\frac {48+48 x}{-5+x}} \left (288 e^2+100 x^3+248 x^4+4 x^5+e \left (-100 x-536 x^2-4 x^3\right )\right )}{25-10 x+x^2} \, dx=e^{-\frac {48 (1+x)}{-5+x}} \left (e-x^2\right )^2 \] Input:

Integrate[(288*E^2 + 100*x^3 + 248*x^4 + 4*x^5 + E*(-100*x - 536*x^2 - 4*x 
^3))/(E^((48 + 48*x)/(-5 + x))*(25 - 10*x + x^2)),x]
 

Output:

(E - x^2)^2/E^((48*(1 + 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.

\(\displaystyle \int \frac {e^{-\frac {48 x+48}{x-5}} \left (4 x^5+248 x^4+100 x^3+e \left (-4 x^3-536 x^2-100 x\right )+288 e^2\right )}{x^2-10 x+25} \, dx\)

\(\Big \downarrow \) 7277

\(\displaystyle 4 \int \frac {e^{\frac {48 (x+1)}{5-x}} \left (x^5+62 x^4+25 x^3-e \left (x^3+134 x^2+25 x\right )+72 e^2\right )}{(5-x)^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle 4 \int \left (e^{\frac {48 (x+1)}{5-x}} x^3+72 e^{\frac {48 (x+1)}{5-x}} x^2-(-720+e) e^{\frac {48 (x+1)}{5-x}} x-\frac {1440 (-25+e) e^{\frac {48 (x+1)}{5-x}}}{x-5}+\frac {72 (-25+e)^2 e^{\frac {48 (x+1)}{5-x}}}{(x-5)^2}-72 e^{\frac {48 (x+1)}{5-x}} (-75+2 e)\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 4 \left (\int e^{\frac {48 (x+1)}{5-x}} x^3dx+72 \int e^{\frac {48 (x+1)}{5-x}} x^2dx+72 (75-2 e) \int e^{\frac {48 (x+1)}{5-x}}dx+(720-e) \int e^{\frac {48 (x+1)}{5-x}} xdx-\frac {1440 (25-e) \operatorname {ExpIntegralEi}\left (\frac {288}{5-x}\right )}{e^{48}}+\frac {1}{4} (25-e)^2 e^{\frac {288}{5-x}-48}\right )\)

Input:

Int[(288*E^2 + 100*x^3 + 248*x^4 + 4*x^5 + E*(-100*x - 536*x^2 - 4*x^3))/( 
E^((48 + 48*x)/(-5 + x))*(25 - 10*x + x^2)),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18

method	result	size
risch	\(\left (x^{4}-2 x^{2} {\mathrm e}+{\mathrm e}^{2}\right ) {\mathrm e}^{-\frac {48 \left (1+x \right )}{-5+x}}\)	\(26\)
gosper	\(\left (x^{4}-2 x^{2} {\mathrm e}+{\mathrm e}^{2}\right ) {\mathrm e}^{-\frac {48 \left (1+x \right )}{-5+x}}\)	\(30\)
parallelrisch	\(\frac {\left (x^{5}+{\mathrm e}^{2} x -5 x^{4}-5 \,{\mathrm e}^{2}+10 x^{2} {\mathrm e}-2 x^{3} {\mathrm e}\right ) {\mathrm e}^{-\frac {48 \left (1+x \right )}{-5+x}}}{-5+x}\)	\(55\)
norman	\(\frac {\left (x^{5}+{\mathrm e}^{2} x -5 x^{4}-5 \,{\mathrm e}^{2}+10 x^{2} {\mathrm e}-2 x^{3} {\mathrm e}\right ) {\mathrm e}^{-\frac {48 x +48}{-5+x}}}{-5+x}\)	\(56\)
derivativedivides	\(43700 \,{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-5+x \right )-\frac {{\mathrm e}^{-48-\frac {288}{-5+x}} \left (\left (48+\frac {288}{-5+x}\right )^{3}-145 \left (48+\frac {288}{-5+x}\right )^{2}+223482+\frac {2018880}{-5+x}\right ) \left (-5+x \right )^{4}}{6}+58 \,{\mathrm e}^{-48-\frac {288}{-5+x}} \left (\left (48+\frac {288}{-5+x}\right )^{2}-2302-\frac {27936}{-5+x}\right ) \left (-5+x \right )^{3}-3030 \,{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-1+\frac {288}{-5+x}\right ) \left (-5+x \right )^{2}+625 \,{\mathrm e}^{-48-\frac {288}{-5+x}}+10368 \,{\mathrm e} \left (-\frac {{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-5+x \right )}{288}+{\mathrm e}^{-48} \operatorname {expIntegral}_{1}\left (\frac {288}{-5+x}\right )\right )-2304 \,{\mathrm e}^{2} \left (-\frac {{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-5+x \right )}{288}+{\mathrm e}^{-48} \operatorname {expIntegral}_{1}\left (\frac {288}{-5+x}\right )\right )+331776 \,{\mathrm e} \left (\frac {{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-1+\frac {288}{-5+x}\right ) \left (-5+x \right )^{2}}{165888}-\frac {{\mathrm e}^{-48} \operatorname {expIntegral}_{1}\left (\frac {288}{-5+x}\right )}{2}\right )+960 \,{\mathrm e} \left (-\frac {{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-5+x \right )}{6}+47 \,{\mathrm e}^{-48} \operatorname {expIntegral}_{1}\left (\frac {288}{-5+x}\right )\right )+50 \,{\mathrm e} \left (-{\mathrm e}^{-48-\frac {288}{-5+x}}-8 \,{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-5+x \right )+2208 \,{\mathrm e}^{-48} \operatorname {expIntegral}_{1}\left (\frac {288}{-5+x}\right )\right )+96 \,{\mathrm e}^{2} \left (-\frac {{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-5+x \right )}{6}+47 \,{\mathrm e}^{-48} \operatorname {expIntegral}_{1}\left (\frac {288}{-5+x}\right )\right )-{\mathrm e}^{2} \left (-{\mathrm e}^{-48-\frac {288}{-5+x}}-8 \,{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-5+x \right )+2208 \,{\mathrm e}^{-48} \operatorname {expIntegral}_{1}\left (\frac {288}{-5+x}\right )\right )\)	\(412\)
default	\(43700 \,{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-5+x \right )-\frac {{\mathrm e}^{-48-\frac {288}{-5+x}} \left (\left (48+\frac {288}{-5+x}\right )^{3}-145 \left (48+\frac {288}{-5+x}\right )^{2}+223482+\frac {2018880}{-5+x}\right ) \left (-5+x \right )^{4}}{6}+58 \,{\mathrm e}^{-48-\frac {288}{-5+x}} \left (\left (48+\frac {288}{-5+x}\right )^{2}-2302-\frac {27936}{-5+x}\right ) \left (-5+x \right )^{3}-3030 \,{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-1+\frac {288}{-5+x}\right ) \left (-5+x \right )^{2}+625 \,{\mathrm e}^{-48-\frac {288}{-5+x}}+10368 \,{\mathrm e} \left (-\frac {{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-5+x \right )}{288}+{\mathrm e}^{-48} \operatorname {expIntegral}_{1}\left (\frac {288}{-5+x}\right )\right )-2304 \,{\mathrm e}^{2} \left (-\frac {{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-5+x \right )}{288}+{\mathrm e}^{-48} \operatorname {expIntegral}_{1}\left (\frac {288}{-5+x}\right )\right )+331776 \,{\mathrm e} \left (\frac {{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-1+\frac {288}{-5+x}\right ) \left (-5+x \right )^{2}}{165888}-\frac {{\mathrm e}^{-48} \operatorname {expIntegral}_{1}\left (\frac {288}{-5+x}\right )}{2}\right )+960 \,{\mathrm e} \left (-\frac {{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-5+x \right )}{6}+47 \,{\mathrm e}^{-48} \operatorname {expIntegral}_{1}\left (\frac {288}{-5+x}\right )\right )+50 \,{\mathrm e} \left (-{\mathrm e}^{-48-\frac {288}{-5+x}}-8 \,{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-5+x \right )+2208 \,{\mathrm e}^{-48} \operatorname {expIntegral}_{1}\left (\frac {288}{-5+x}\right )\right )+96 \,{\mathrm e}^{2} \left (-\frac {{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-5+x \right )}{6}+47 \,{\mathrm e}^{-48} \operatorname {expIntegral}_{1}\left (\frac {288}{-5+x}\right )\right )-{\mathrm e}^{2} \left (-{\mathrm e}^{-48-\frac {288}{-5+x}}-8 \,{\mathrm e}^{-48-\frac {288}{-5+x}} \left (-5+x \right )+2208 \,{\mathrm e}^{-48} \operatorname {expIntegral}_{1}\left (\frac {288}{-5+x}\right )\right )\)	\(412\)

Input:

int((288*exp(1)^2+(-4*x^3-536*x^2-100*x)*exp(1)+4*x^5+248*x^4+100*x^3)/(x^ 
2-10*x+25)/exp((48*x+48)/(-5+x)),x,method=_RETURNVERBOSE)
 

Output:

(x^4-2*x^2*exp(1)+exp(2))*exp(-48*(1+x)/(-5+x))
                                                                                    
                                                                                    
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {e^{-\frac {48+48 x}{-5+x}} \left (288 e^2+100 x^3+248 x^4+4 x^5+e \left (-100 x-536 x^2-4 x^3\right )\right )}{25-10 x+x^2} \, dx={\left (x^{4} - 2 \, x^{2} e + e^{2}\right )} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5}\right )} \] Input:

integrate((288*exp(1)^2+(-4*x^3-536*x^2-100*x)*exp(1)+4*x^5+248*x^4+100*x^ 
3)/(x^2-10*x+25)/exp((48*x+48)/(-5+x)),x, algorithm="fricas")
 

Output:

(x^4 - 2*x^2*e + e^2)*e^(-48*(x + 1)/(x - 5))
 

Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {e^{-\frac {48+48 x}{-5+x}} \left (288 e^2+100 x^3+248 x^4+4 x^5+e \left (-100 x-536 x^2-4 x^3\right )\right )}{25-10 x+x^2} \, dx=\left (x^{4} - 2 e x^{2} + e^{2}\right ) e^{- \frac {48 x + 48}{x - 5}} \] Input:

integrate((288*exp(1)**2+(-4*x**3-536*x**2-100*x)*exp(1)+4*x**5+248*x**4+1 
00*x**3)/(x**2-10*x+25)/exp((48*x+48)/(-5+x)),x)
 

Output:

(x**4 - 2*E*x**2 + exp(2))*exp(-(48*x + 48)/(x - 5))
 

Maxima [F]

\[ \int \frac {e^{-\frac {48+48 x}{-5+x}} \left (288 e^2+100 x^3+248 x^4+4 x^5+e \left (-100 x-536 x^2-4 x^3\right )\right )}{25-10 x+x^2} \, dx=\int { \frac {4 \, {\left (x^{5} + 62 \, x^{4} + 25 \, x^{3} - {\left (x^{3} + 134 \, x^{2} + 25 \, x\right )} e + 72 \, e^{2}\right )} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5}\right )}}{x^{2} - 10 \, x + 25} \,d x } \] Input:

integrate((288*exp(1)^2+(-4*x^3-536*x^2-100*x)*exp(1)+4*x^5+248*x^4+100*x^ 
3)/(x^2-10*x+25)/exp((48*x+48)/(-5+x)),x, algorithm="maxima")
 

Output:

(x^4 - 2*x^2*e)*e^(-288/(x - 5) - 48) + 288*integrate(e^(-288/(x - 5))/(x^ 
2*e^46 - 10*x*e^46 + 25*e^46), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 375 vs. \(2 (22) = 44\).

Time = 0.23 (sec) , antiderivative size = 375, normalized size of antiderivative = 17.05 \[ \int \frac {e^{-\frac {48+48 x}{-5+x}} \left (288 e^2+100 x^3+248 x^4+4 x^5+e \left (-100 x-536 x^2-4 x^3\right )\right )}{25-10 x+x^2} \, dx=\frac {\frac {{\left (x + 1\right )}^{4} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5} + 2\right )}}{{\left (x - 5\right )}^{4}} - \frac {4 \, {\left (x + 1\right )}^{3} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5} + 2\right )}}{{\left (x - 5\right )}^{3}} + \frac {6 \, {\left (x + 1\right )}^{2} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5} + 2\right )}}{{\left (x - 5\right )}^{2}} - \frac {4 \, {\left (x + 1\right )} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5} + 2\right )}}{x - 5} - \frac {50 \, {\left (x + 1\right )}^{4} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5} + 1\right )}}{{\left (x - 5\right )}^{4}} + \frac {80 \, {\left (x + 1\right )}^{3} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5} + 1\right )}}{{\left (x - 5\right )}^{3}} - \frac {12 \, {\left (x + 1\right )}^{2} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5} + 1\right )}}{{\left (x - 5\right )}^{2}} - \frac {16 \, {\left (x + 1\right )} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5} + 1\right )}}{x - 5} + \frac {625 \, {\left (x + 1\right )}^{4} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5}\right )}}{{\left (x - 5\right )}^{4}} + \frac {500 \, {\left (x + 1\right )}^{3} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5}\right )}}{{\left (x - 5\right )}^{3}} + \frac {150 \, {\left (x + 1\right )}^{2} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5}\right )}}{{\left (x - 5\right )}^{2}} + \frac {20 \, {\left (x + 1\right )} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5}\right )}}{x - 5} + e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5} + 2\right )} - 2 \, e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5} + 1\right )} + e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5}\right )}}{\frac {{\left (x + 1\right )}^{4}}{{\left (x - 5\right )}^{4}} - \frac {4 \, {\left (x + 1\right )}^{3}}{{\left (x - 5\right )}^{3}} + \frac {6 \, {\left (x + 1\right )}^{2}}{{\left (x - 5\right )}^{2}} - \frac {4 \, {\left (x + 1\right )}}{x - 5} + 1} \] Input:

integrate((288*exp(1)^2+(-4*x^3-536*x^2-100*x)*exp(1)+4*x^5+248*x^4+100*x^ 
3)/(x^2-10*x+25)/exp((48*x+48)/(-5+x)),x, algorithm="giac")
 

Output:

((x + 1)^4*e^(-48*(x + 1)/(x - 5) + 2)/(x - 5)^4 - 4*(x + 1)^3*e^(-48*(x + 
 1)/(x - 5) + 2)/(x - 5)^3 + 6*(x + 1)^2*e^(-48*(x + 1)/(x - 5) + 2)/(x - 
5)^2 - 4*(x + 1)*e^(-48*(x + 1)/(x - 5) + 2)/(x - 5) - 50*(x + 1)^4*e^(-48 
*(x + 1)/(x - 5) + 1)/(x - 5)^4 + 80*(x + 1)^3*e^(-48*(x + 1)/(x - 5) + 1) 
/(x - 5)^3 - 12*(x + 1)^2*e^(-48*(x + 1)/(x - 5) + 1)/(x - 5)^2 - 16*(x + 
1)*e^(-48*(x + 1)/(x - 5) + 1)/(x - 5) + 625*(x + 1)^4*e^(-48*(x + 1)/(x - 
 5))/(x - 5)^4 + 500*(x + 1)^3*e^(-48*(x + 1)/(x - 5))/(x - 5)^3 + 150*(x 
+ 1)^2*e^(-48*(x + 1)/(x - 5))/(x - 5)^2 + 20*(x + 1)*e^(-48*(x + 1)/(x - 
5))/(x - 5) + e^(-48*(x + 1)/(x - 5) + 2) - 2*e^(-48*(x + 1)/(x - 5) + 1) 
+ e^(-48*(x + 1)/(x - 5)))/((x + 1)^4/(x - 5)^4 - 4*(x + 1)^3/(x - 5)^3 + 
6*(x + 1)^2/(x - 5)^2 - 4*(x + 1)/(x - 5) + 1)
 

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {e^{-\frac {48+48 x}{-5+x}} \left (288 e^2+100 x^3+248 x^4+4 x^5+e \left (-100 x-536 x^2-4 x^3\right )\right )}{25-10 x+x^2} \, dx={\mathrm {e}}^{-\frac {48\,x}{x-5}-\frac {48}{x-5}}\,{\left (\mathrm {e}-x^2\right )}^2 \] Input:

int((exp(-(48*x + 48)/(x - 5))*(288*exp(2) - exp(1)*(100*x + 536*x^2 + 4*x 
^3) + 100*x^3 + 248*x^4 + 4*x^5))/(x^2 - 10*x + 25),x)
 

Output:

exp(- (48*x)/(x - 5) - 48/(x - 5))*(exp(1) - x^2)^2
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {e^{-\frac {48+48 x}{-5+x}} \left (288 e^2+100 x^3+248 x^4+4 x^5+e \left (-100 x-536 x^2-4 x^3\right )\right )}{25-10 x+x^2} \, dx=\frac {x^{4}-2 e \,x^{2}+e^{2}}{e^{\frac {288}{-5+x}} e^{48}} \] Input:

int((288*exp(1)^2+(-4*x^3-536*x^2-100*x)*exp(1)+4*x^5+248*x^4+100*x^3)/(x^ 
2-10*x+25)/exp((48*x+48)/(-5+x)),x)
 

Output:

(e**2 - 2*e*x**2 + x**4)/(e**(288/(x - 5))*e**48)