3.11.43 \(\int \frac {4 e x+e (160 x-40 x^2) \log (4)+e (-16+4 x) \log (4-x)+e^{\frac {e^{\frac {x}{e}}}{4}} (4 e+e (80-20 x) \log (4)+e^{\frac {x}{e}} (20 x-5 x^2) \log (4)+e^{\frac {x}{e}} (-4+x) \log (4-x))}{e (-16+4 x)} \, dx\) [1043]

3.11.43.1 Optimal result

Integrand size = 105, antiderivative size = 30 \[ \int \frac {4 e x+e \left (160 x-40 x^2\right ) \log (4)+e (-16+4 x) \log (4-x)+e^{\frac {e^{\frac {x}{e}}}{4}} \left (4 e+e (80-20 x) \log (4)+e^{\frac {x}{e}} \left (20 x-5 x^2\right ) \log (4)+e^{\frac {x}{e}} (-4+x) \log (4-x)\right )}{e (-16+4 x)} \, dx=25+\left (e^{\frac {e^{\frac {x}{e}}}{4}}+x\right ) (-5 x \log (4)+\log (4-x)) \]

output

(ln(-x+4)-10*x*ln(2))*(x+exp(1/4*exp(x/exp(1))))+25
 

3.11.43.2 Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.67 \[ \int \frac {4 e x+e \left (160 x-40 x^2\right ) \log (4)+e (-16+4 x) \log (4-x)+e^{\frac {e^{\frac {x}{e}}}{4}} \left (4 e+e (80-20 x) \log (4)+e^{\frac {x}{e}} \left (20 x-5 x^2\right ) \log (4)+e^{\frac {x}{e}} (-4+x) \log (4-x)\right )}{e (-16+4 x)} \, dx=-5 x \left (e^{\frac {e^{\frac {x}{e}}}{4}}+x\right ) \log (4)+\left (-4+e^{\frac {e^{\frac {x}{e}}}{4}}+x\right ) \log (4-x)+4 \log (-4+x) \]

input

Integrate[(4*E*x + E*(160*x - 40*x^2)*Log[4] + E*(-16 + 4*x)*Log[4 - x] + 
E^(E^(x/E)/4)*(4*E + E*(80 - 20*x)*Log[4] + E^(x/E)*(20*x - 5*x^2)*Log[4] 
+ E^(x/E)*(-4 + x)*Log[4 - x]))/(E*(-16 + 4*x)),x]
 

output

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

3.11.43.3 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.

\(\displaystyle \int \frac {e^{\frac {e^{\frac {x}{e}}}{4}} \left (e^{\frac {x}{e}} \left (20 x-5 x^2\right ) \log (4)+e (80-20 x) \log (4)+e^{\frac {x}{e}} (x-4) \log (4-x)+4 e\right )+e \left (160 x-40 x^2\right ) \log (4)+4 e x+e (4 x-16) \log (4-x)}{e (4 x-16)} \, dx\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 7293

\(\displaystyle -\frac {\int \left (e^{\frac {x}{e}+\frac {e^{\frac {x}{e}}}{4}} (5 x \log (4)-\log (4-x))+\frac {4 e \left (-10 \log (4) x^2+\log (4-x) x+(1+40 \log (4)) x-5 e^{\frac {e^{\frac {x}{e}}}{4}} \log (4) x-4 \log (4-x)+e^{\frac {e^{\frac {x}{e}}}{4}} (1+20 \log (4))\right )}{4-x}\right )dx}{4 e}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {4 e \int \frac {e^{\frac {e^{\frac {x}{e}}}{4}}}{4-x}dx+4 \int \frac {e^{\frac {1}{4} \left (4+e^{\frac {x}{e}}\right )}}{x-4}dx+5 \log (4) \int e^{\frac {x}{e}+\frac {e^{\frac {x}{e}}}{4}} xdx+20 e^2 \log (4) \operatorname {ExpIntegralEi}\left (\frac {e^{\frac {x}{e}}}{4}\right )+20 e x^2 \log (4)-4 e^{\frac {e^{\frac {x}{e}}}{4}+1} \log (4-x)+4 e (4-x) \log (4-x)-16 e \log (4-x)}{4 e}\)

input

Int[(4*E*x + E*(160*x - 40*x^2)*Log[4] + E*(-16 + 4*x)*Log[4 - x] + E^(E^( 
x/E)/4)*(4*E + E*(80 - 20*x)*Log[4] + E^(x/E)*(20*x - 5*x^2)*Log[4] + E^(x 
/E)*(-4 + x)*Log[4 - x]))/(E*(-16 + 4*x)),x]
 

output

$Aborted
 

3.11.43.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

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

rule 7293

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

3.11.43.4 Maple [A] (verified)

Time = 1.24 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.50

method	result	size
risch	\(-10 x^{2} \ln \left (2\right )+x \ln \left (-x +4\right )+\left (-10 x \,{\mathrm e} \ln \left (2\right )+{\mathrm e} \ln \left (-x +4\right )\right ) {\mathrm e}^{-1+\frac {{\mathrm e}^{{\mathrm e}^{-1} x}}{4}}\)	\(45\)
parallelrisch	\({\mathrm e}^{-1} \left (-10 \,{\mathrm e} \ln \left (2\right ) x^{2}-10 \,{\mathrm e}^{\frac {{\mathrm e}^{{\mathrm e}^{-1} x}}{4}} \ln \left (2\right ) {\mathrm e} x +{\mathrm e} \ln \left (-x +4\right ) x +{\mathrm e} \ln \left (-x +4\right ) {\mathrm e}^{\frac {{\mathrm e}^{{\mathrm e}^{-1} x}}{4}}+160 \,{\mathrm e} \ln \left (2\right )\right )\)	\(68\)

input

int((((x-4)*exp(x/exp(1))*ln(-x+4)+2*(-5*x^2+20*x)*ln(2)*exp(x/exp(1))+2*( 
-20*x+80)*exp(1)*ln(2)+4*exp(1))*exp(1/4*exp(x/exp(1)))+(4*x-16)*exp(1)*ln 
(-x+4)+2*(-40*x^2+160*x)*exp(1)*ln(2)+4*x*exp(1))/(4*x-16)/exp(1),x,method 
=_RETURNVERBOSE)
 

output

-10*x^2*ln(2)+x*ln(-x+4)+(-10*x*exp(1)*ln(2)+exp(1)*ln(-x+4))*exp(-1+1/4*e 
xp(exp(-1)*x))
 

3.11.43.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.33 \[ \int \frac {4 e x+e \left (160 x-40 x^2\right ) \log (4)+e (-16+4 x) \log (4-x)+e^{\frac {e^{\frac {x}{e}}}{4}} \left (4 e+e (80-20 x) \log (4)+e^{\frac {x}{e}} \left (20 x-5 x^2\right ) \log (4)+e^{\frac {x}{e}} (-4+x) \log (4-x)\right )}{e (-16+4 x)} \, dx=-10 \, x^{2} \log \left (2\right ) - {\left (10 \, x \log \left (2\right ) - \log \left (-x + 4\right )\right )} e^{\left (\frac {1}{4} \, e^{\left (x e^{\left (-1\right )}\right )}\right )} + x \log \left (-x + 4\right ) \]

input

integrate((((x-4)*exp(x/exp(1))*log(-x+4)+2*(-5*x^2+20*x)*log(2)*exp(x/exp 
(1))+2*(-20*x+80)*exp(1)*log(2)+4*exp(1))*exp(1/4*exp(x/exp(1)))+(4*x-16)* 
exp(1)*log(-x+4)+2*(-40*x^2+160*x)*exp(1)*log(2)+4*x*exp(1))/(4*x-16)/exp( 
1),x, algorithm=\
 

output

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

3.11.43.6 Sympy [A] (verification not implemented)

Time = 17.47 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {4 e x+e \left (160 x-40 x^2\right ) \log (4)+e (-16+4 x) \log (4-x)+e^{\frac {e^{\frac {x}{e}}}{4}} \left (4 e+e (80-20 x) \log (4)+e^{\frac {x}{e}} \left (20 x-5 x^2\right ) \log (4)+e^{\frac {x}{e}} (-4+x) \log (4-x)\right )}{e (-16+4 x)} \, dx=- 10 x^{2} \log {\left (2 \right )} + x \log {\left (4 - x \right )} + \left (- 10 x \log {\left (2 \right )} + \log {\left (4 - x \right )}\right ) e^{\frac {e^{\frac {x}{e}}}{4}} \]

input

integrate((((x-4)*exp(x/exp(1))*ln(-x+4)+2*(-5*x**2+20*x)*ln(2)*exp(x/exp( 
1))+2*(-20*x+80)*exp(1)*ln(2)+4*exp(1))*exp(1/4*exp(x/exp(1)))+(4*x-16)*ex 
p(1)*ln(-x+4)+2*(-40*x**2+160*x)*exp(1)*ln(2)+4*x*exp(1))/(4*x-16)/exp(1), 
x)
 

output

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

3.11.43.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 155 vs. \(2 (28) = 56\).

Time = 0.34 (sec) , antiderivative size = 155, normalized size of antiderivative = 5.17 \[ \int \frac {4 e x+e \left (160 x-40 x^2\right ) \log (4)+e (-16+4 x) \log (4-x)+e^{\frac {e^{\frac {x}{e}}}{4}} \left (4 e+e (80-20 x) \log (4)+e^{\frac {x}{e}} \left (20 x-5 x^2\right ) \log (4)+e^{\frac {x}{e}} (-4+x) \log (4-x)\right )}{e (-16+4 x)} \, dx=-{\left (10 \, {\left (x^{2} + 8 \, x + 32 \, \log \left (x - 4\right )\right )} e \log \left (2\right ) - 80 \, {\left (x + 4 \, \log \left (x - 4\right )\right )} e \log \left (2\right ) - {\left (x + 4 \, \log \left (x - 4\right )\right )} e \log \left (-x + 4\right ) + 4 \, e \log \left (x - 4\right ) \log \left (-x + 4\right ) + {\left (2 \, \log \left (x - 4\right )^{2} + x + 4 \, \log \left (x - 4\right )\right )} e - 2 \, {\left (2 \, \log \left (x - 4\right ) \log \left (-x + 4\right ) - \log \left (-x + 4\right )^{2}\right )} e - {\left (x + 4 \, \log \left (x - 4\right )\right )} e + {\left (10 \, x e \log \left (2\right ) - e \log \left (-x + 4\right )\right )} e^{\left (\frac {1}{4} \, e^{\left (x e^{\left (-1\right )}\right )}\right )}\right )} e^{\left (-1\right )} \]

input

integrate((((x-4)*exp(x/exp(1))*log(-x+4)+2*(-5*x^2+20*x)*log(2)*exp(x/exp 
(1))+2*(-20*x+80)*exp(1)*log(2)+4*exp(1))*exp(1/4*exp(x/exp(1)))+(4*x-16)* 
exp(1)*log(-x+4)+2*(-40*x^2+160*x)*exp(1)*log(2)+4*x*exp(1))/(4*x-16)/exp( 
1),x, algorithm=\
 

output

-(10*(x^2 + 8*x + 32*log(x - 4))*e*log(2) - 80*(x + 4*log(x - 4))*e*log(2) 
 - (x + 4*log(x - 4))*e*log(-x + 4) + 4*e*log(x - 4)*log(-x + 4) + (2*log( 
x - 4)^2 + x + 4*log(x - 4))*e - 2*(2*log(x - 4)*log(-x + 4) - log(-x + 4) 
^2)*e - (x + 4*log(x - 4))*e + (10*x*e*log(2) - e*log(-x + 4))*e^(1/4*e^(x 
*e^(-1))))*e^(-1)
 

3.11.43.8 Giac [F]

\[ \int \frac {4 e x+e \left (160 x-40 x^2\right ) \log (4)+e (-16+4 x) \log (4-x)+e^{\frac {e^{\frac {x}{e}}}{4}} \left (4 e+e (80-20 x) \log (4)+e^{\frac {x}{e}} \left (20 x-5 x^2\right ) \log (4)+e^{\frac {x}{e}} (-4+x) \log (4-x)\right )}{e (-16+4 x)} \, dx=\int { -\frac {{\left (80 \, {\left (x^{2} - 4 \, x\right )} e \log \left (2\right ) - 4 \, {\left (x - 4\right )} e \log \left (-x + 4\right ) - 4 \, x e + {\left (40 \, {\left (x - 4\right )} e \log \left (2\right ) + 10 \, {\left (x^{2} - 4 \, x\right )} e^{\left (x e^{\left (-1\right )}\right )} \log \left (2\right ) - {\left (x - 4\right )} e^{\left (x e^{\left (-1\right )}\right )} \log \left (-x + 4\right ) - 4 \, e\right )} e^{\left (\frac {1}{4} \, e^{\left (x e^{\left (-1\right )}\right )}\right )}\right )} e^{\left (-1\right )}}{4 \, {\left (x - 4\right )}} \,d x } \]

input

integrate((((x-4)*exp(x/exp(1))*log(-x+4)+2*(-5*x^2+20*x)*log(2)*exp(x/exp 
(1))+2*(-20*x+80)*exp(1)*log(2)+4*exp(1))*exp(1/4*exp(x/exp(1)))+(4*x-16)* 
exp(1)*log(-x+4)+2*(-40*x^2+160*x)*exp(1)*log(2)+4*x*exp(1))/(4*x-16)/exp( 
1),x, algorithm=\
 

output

integrate(-1/4*(80*(x^2 - 4*x)*e*log(2) - 4*(x - 4)*e*log(-x + 4) - 4*x*e 
+ (40*(x - 4)*e*log(2) + 10*(x^2 - 4*x)*e^(x*e^(-1))*log(2) - (x - 4)*e^(x 
*e^(-1))*log(-x + 4) - 4*e)*e^(1/4*e^(x*e^(-1))))*e^(-1)/(x - 4), x)
 

3.11.43.9 Mupad [B] (verification not implemented)

Time = 0.67 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {4 e x+e \left (160 x-40 x^2\right ) \log (4)+e (-16+4 x) \log (4-x)+e^{\frac {e^{\frac {x}{e}}}{4}} \left (4 e+e (80-20 x) \log (4)+e^{\frac {x}{e}} \left (20 x-5 x^2\right ) \log (4)+e^{\frac {x}{e}} (-4+x) \log (4-x)\right )}{e (-16+4 x)} \, dx=\left (x+{\mathrm {e}}^{\frac {{\mathrm {e}}^{x\,{\mathrm {e}}^{-1}}}{4}}\right )\,\left (\ln \left (4-x\right )-10\,x\,\ln \left (2\right )\right ) \]

input

int((exp(-1)*(4*x*exp(1) + exp(exp(x*exp(-1))/4)*(4*exp(1) - 2*exp(1)*log( 
2)*(20*x - 80) + exp(x*exp(-1))*log(4 - x)*(x - 4) + 2*exp(x*exp(-1))*log( 
2)*(20*x - 5*x^2)) + exp(1)*log(4 - x)*(4*x - 16) + 2*exp(1)*log(2)*(160*x 
 - 40*x^2)))/(4*x - 16),x)
 

output

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