3.12.96 \(\int \frac {e^x (3 x-2 x^2-x^3+e^5 (x-x^2))+e^x (-3+e^5 (-1+x)+2 x+x^2) \log (3+e^5+x)+\frac {(e^x (-x-4 x^2-x^3+e^5 (-x-x^2))+e^x (-3+e^5 (-1+x)+2 x+x^2) \log (3+e^5+x)) (x^2-2 x \log (3+e^5+x)+\log ^2(3+e^5+x))}{e}}{-3 x^3-e^5 x^3-x^4+(3 x^2+e^5 x^2+x^3) \log (3+e^5+x)} \, dx\) [1196]

3.12.96.1 Optimal result
3.12.96.2 Mathematica [A] (verified)
3.12.96.3 Rubi [F]
3.12.96.4 Maple [B] (verified)
3.12.96.5 Fricas [A] (verification not implemented)
3.12.96.6 Sympy [A] (verification not implemented)
3.12.96.7 Maxima [A] (verification not implemented)
3.12.96.8 Giac [A] (verification not implemented)
3.12.96.9 Mupad [F(-1)]

3.12.96.1 Optimal result

Integrand size = 187, antiderivative size = 26 \[ \int \frac {e^x \left (3 x-2 x^2-x^3+e^5 \left (x-x^2\right )\right )+e^x \left (-3+e^5 (-1+x)+2 x+x^2\right ) \log \left (3+e^5+x\right )+\frac {\left (e^x \left (-x-4 x^2-x^3+e^5 \left (-x-x^2\right )\right )+e^x \left (-3+e^5 (-1+x)+2 x+x^2\right ) \log \left (3+e^5+x\right )\right ) \left (x^2-2 x \log \left (3+e^5+x\right )+\log ^2\left (3+e^5+x\right )\right )}{e}}{-3 x^3-e^5 x^3-x^4+\left (3 x^2+e^5 x^2+x^3\right ) \log \left (3+e^5+x\right )} \, dx=\frac {e^x \left (1+\frac {\left (x-\log \left (3+e^5+x\right )\right )^2}{e}\right )}{x} \]

output
exp(x)*(1+exp(ln((x-ln(exp(5)+3+x))^2)-1))/x
 
3.12.96.2 Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \frac {e^x \left (3 x-2 x^2-x^3+e^5 \left (x-x^2\right )\right )+e^x \left (-3+e^5 (-1+x)+2 x+x^2\right ) \log \left (3+e^5+x\right )+\frac {\left (e^x \left (-x-4 x^2-x^3+e^5 \left (-x-x^2\right )\right )+e^x \left (-3+e^5 (-1+x)+2 x+x^2\right ) \log \left (3+e^5+x\right )\right ) \left (x^2-2 x \log \left (3+e^5+x\right )+\log ^2\left (3+e^5+x\right )\right )}{e}}{-3 x^3-e^5 x^3-x^4+\left (3 x^2+e^5 x^2+x^3\right ) \log \left (3+e^5+x\right )} \, dx=\frac {e^{-1+x} \left (e+x^2-2 x \log \left (3+e^5+x\right )+\log ^2\left (3+e^5+x\right )\right )}{x} \]

input
Integrate[(E^x*(3*x - 2*x^2 - x^3 + E^5*(x - x^2)) + E^x*(-3 + E^5*(-1 + x 
) + 2*x + x^2)*Log[3 + E^5 + x] + ((E^x*(-x - 4*x^2 - x^3 + E^5*(-x - x^2) 
) + E^x*(-3 + E^5*(-1 + x) + 2*x + x^2)*Log[3 + E^5 + x])*(x^2 - 2*x*Log[3 
 + E^5 + x] + Log[3 + E^5 + x]^2))/E)/(-3*x^3 - E^5*x^3 - x^4 + (3*x^2 + E 
^5*x^2 + x^3)*Log[3 + E^5 + x]),x]
 
output
(E^(-1 + x)*(E + x^2 - 2*x*Log[3 + E^5 + x] + Log[3 + E^5 + x]^2))/x
 
3.12.96.3 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.

\(\displaystyle \int \frac {e^x \left (x^2+2 x+e^5 (x-1)-3\right ) \log \left (x+e^5+3\right )+e^x \left (-x^3-2 x^2+e^5 \left (x-x^2\right )+3 x\right )+\frac {\left (e^x \left (x^2+2 x+e^5 (x-1)-3\right ) \log \left (x+e^5+3\right )+e^x \left (-x^3-4 x^2+e^5 \left (-x^2-x\right )-x\right )\right ) \left (x^2+\log ^2\left (x+e^5+3\right )-2 x \log \left (x+e^5+3\right )\right )}{e}}{-x^4-e^5 x^3-3 x^3+\left (x^3+e^5 x^2+3 x^2\right ) \log \left (x+e^5+3\right )} \, dx\)

\(\Big \downarrow \) 6

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

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {e^{x-1} \left (e^5 (x+1) x^2+\left (x^2+4 x+1\right ) x^2+e \left (x^2+2 x-3\right )-2 \left (x^2+\left (3+e^5\right ) x-1\right ) x \log \left (x+e^5+3\right )+e^6 (x-1)+(x-1) \left (x+e^5+3\right ) \log ^2\left (x+e^5+3\right )\right )}{x^2 \left (x+e^5+3\right )}dx\)

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 \int \frac {\operatorname {ExpIntegralEi}(x)}{x+e^5+3}dx}{e \left (3+e^5\right )}+\frac {2 e^{-4-e^5} \int \frac {\operatorname {ExpIntegralEi}\left (x+e^5+3\right )}{x+e^5+3}dx}{3+e^5}-\int \frac {e^{x-1} \log ^2\left (x+e^5+3\right )}{x^2}dx+\int \frac {e^{x-1} \log ^2\left (x+e^5+3\right )}{x}dx+\frac {2 \operatorname {ExpIntegralEi}(x) \log \left (x+e^5+3\right )}{e \left (3+e^5\right )}-\frac {2 e^{-4-e^5} \operatorname {ExpIntegralEi}\left (x+e^5+3\right ) \log \left (x+e^5+3\right )}{3+e^5}+e^{x-1} x+\frac {e^x}{x}-2 e^{x-1} \log \left (x+e^5+3\right )\)

input
Int[(E^x*(3*x - 2*x^2 - x^3 + E^5*(x - x^2)) + E^x*(-3 + E^5*(-1 + x) + 2* 
x + x^2)*Log[3 + E^5 + x] + ((E^x*(-x - 4*x^2 - x^3 + E^5*(-x - x^2)) + E^ 
x*(-3 + E^5*(-1 + x) + 2*x + x^2)*Log[3 + E^5 + x])*(x^2 - 2*x*Log[3 + E^5 
 + x] + Log[3 + E^5 + x]^2))/E)/(-3*x^3 - E^5*x^3 - x^4 + (3*x^2 + E^5*x^2 
 + x^3)*Log[3 + E^5 + x]),x]
 
output
$Aborted
 

3.12.96.3.1 Defintions of rubi rules used

rule 6
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v 
+ (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] &&  !FreeQ[Fx, x]
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7239
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl 
erIntegrandQ[v, u, x]]
 

rule 7293
Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 
3.12.96.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(153\) vs. \(2(24)=48\).

Time = 48.36 (sec) , antiderivative size = 154, normalized size of antiderivative = 5.92

method result size
parallelrisch \(-\frac {-2 \,{\mathrm e}^{x} x^{2}+4 \ln \left ({\mathrm e}^{5}+3+x \right ) {\mathrm e}^{x} x +4 \ln \left ({\mathrm e}^{5}+3+x \right ) x \,{\mathrm e}^{x} {\mathrm e}^{\ln \left (\ln \left ({\mathrm e}^{5}+3+x \right )^{2}-2 x \ln \left ({\mathrm e}^{5}+3+x \right )+x^{2}\right )-1}-2 \,{\mathrm e}^{\ln \left (\ln \left ({\mathrm e}^{5}+3+x \right )^{2}-2 x \ln \left ({\mathrm e}^{5}+3+x \right )+x^{2}\right )-1} {\mathrm e}^{x} x^{2}-2 \,{\mathrm e}^{\ln \left (\ln \left ({\mathrm e}^{5}+3+x \right )^{2}-2 x \ln \left ({\mathrm e}^{5}+3+x \right )+x^{2}\right )-1} {\mathrm e}^{x} \ln \left ({\mathrm e}^{5}+3+x \right )^{2}-2 \,{\mathrm e}^{x} \ln \left ({\mathrm e}^{5}+3+x \right )^{2}}{2 x {\left (x -\ln \left ({\mathrm e}^{5}+3+x \right )\right )}^{2}}\) \(154\)

input
int(((((-1+x)*exp(5)+x^2+2*x-3)*exp(x)*ln(exp(5)+3+x)+((-x^2-x)*exp(5)-x^3 
-4*x^2-x)*exp(x))*exp(ln(ln(exp(5)+3+x)^2-2*x*ln(exp(5)+3+x)+x^2)-1)+((-1+ 
x)*exp(5)+x^2+2*x-3)*exp(x)*ln(exp(5)+3+x)+((-x^2+x)*exp(5)-x^3-2*x^2+3*x) 
*exp(x))/((x^2*exp(5)+x^3+3*x^2)*ln(exp(5)+3+x)-x^3*exp(5)-x^4-3*x^3),x,me 
thod=_RETURNVERBOSE)
 
output
-1/2*(-2*exp(x)*x^2+4*ln(exp(5)+3+x)*exp(x)*x+4*ln(exp(5)+3+x)*x*exp(x)*ex 
p(ln(ln(exp(5)+3+x)^2-2*x*ln(exp(5)+3+x)+x^2)-1)-2*exp(ln(ln(exp(5)+3+x)^2 
-2*x*ln(exp(5)+3+x)+x^2)-1)*exp(x)*x^2-2*exp(ln(ln(exp(5)+3+x)^2-2*x*ln(ex 
p(5)+3+x)+x^2)-1)*exp(x)*ln(exp(5)+3+x)^2-2*exp(x)*ln(exp(5)+3+x)^2)/x/(x- 
ln(exp(5)+3+x))^2
 
3.12.96.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int \frac {e^x \left (3 x-2 x^2-x^3+e^5 \left (x-x^2\right )\right )+e^x \left (-3+e^5 (-1+x)+2 x+x^2\right ) \log \left (3+e^5+x\right )+\frac {\left (e^x \left (-x-4 x^2-x^3+e^5 \left (-x-x^2\right )\right )+e^x \left (-3+e^5 (-1+x)+2 x+x^2\right ) \log \left (3+e^5+x\right )\right ) \left (x^2-2 x \log \left (3+e^5+x\right )+\log ^2\left (3+e^5+x\right )\right )}{e}}{-3 x^3-e^5 x^3-x^4+\left (3 x^2+e^5 x^2+x^3\right ) \log \left (3+e^5+x\right )} \, dx=-\frac {{\left (2 \, x e^{x} \log \left (x + e^{5} + 3\right ) - e^{x} \log \left (x + e^{5} + 3\right )^{2} - {\left (x^{2} + e\right )} e^{x}\right )} e^{\left (-1\right )}}{x} \]

input
integrate(((((-1+x)*exp(5)+x^2+2*x-3)*exp(x)*log(exp(5)+3+x)+((-x^2-x)*exp 
(5)-x^3-4*x^2-x)*exp(x))*exp(log(log(exp(5)+3+x)^2-2*x*log(exp(5)+3+x)+x^2 
)-1)+((-1+x)*exp(5)+x^2+2*x-3)*exp(x)*log(exp(5)+3+x)+((-x^2+x)*exp(5)-x^3 
-2*x^2+3*x)*exp(x))/((x^2*exp(5)+x^3+3*x^2)*log(exp(5)+3+x)-x^3*exp(5)-x^4 
-3*x^3),x, algorithm=\
 
output
-(2*x*e^x*log(x + e^5 + 3) - e^x*log(x + e^5 + 3)^2 - (x^2 + e)*e^x)*e^(-1 
)/x
 
3.12.96.6 Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {e^x \left (3 x-2 x^2-x^3+e^5 \left (x-x^2\right )\right )+e^x \left (-3+e^5 (-1+x)+2 x+x^2\right ) \log \left (3+e^5+x\right )+\frac {\left (e^x \left (-x-4 x^2-x^3+e^5 \left (-x-x^2\right )\right )+e^x \left (-3+e^5 (-1+x)+2 x+x^2\right ) \log \left (3+e^5+x\right )\right ) \left (x^2-2 x \log \left (3+e^5+x\right )+\log ^2\left (3+e^5+x\right )\right )}{e}}{-3 x^3-e^5 x^3-x^4+\left (3 x^2+e^5 x^2+x^3\right ) \log \left (3+e^5+x\right )} \, dx=\frac {\left (x^{2} - 2 x \log {\left (x + 3 + e^{5} \right )} + \log {\left (x + 3 + e^{5} \right )}^{2} + e\right ) e^{x}}{e x} \]

input
integrate(((((-1+x)*exp(5)+x**2+2*x-3)*exp(x)*ln(exp(5)+3+x)+((-x**2-x)*ex 
p(5)-x**3-4*x**2-x)*exp(x))*exp(ln(ln(exp(5)+3+x)**2-2*x*ln(exp(5)+3+x)+x* 
*2)-1)+((-1+x)*exp(5)+x**2+2*x-3)*exp(x)*ln(exp(5)+3+x)+((-x**2+x)*exp(5)- 
x**3-2*x**2+3*x)*exp(x))/((x**2*exp(5)+x**3+3*x**2)*ln(exp(5)+3+x)-x**3*ex 
p(5)-x**4-3*x**3),x)
 
output
(x**2 - 2*x*log(x + 3 + exp(5)) + log(x + 3 + exp(5))**2 + E)*exp(-1)*exp( 
x)/x
 
3.12.96.7 Maxima [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int \frac {e^x \left (3 x-2 x^2-x^3+e^5 \left (x-x^2\right )\right )+e^x \left (-3+e^5 (-1+x)+2 x+x^2\right ) \log \left (3+e^5+x\right )+\frac {\left (e^x \left (-x-4 x^2-x^3+e^5 \left (-x-x^2\right )\right )+e^x \left (-3+e^5 (-1+x)+2 x+x^2\right ) \log \left (3+e^5+x\right )\right ) \left (x^2-2 x \log \left (3+e^5+x\right )+\log ^2\left (3+e^5+x\right )\right )}{e}}{-3 x^3-e^5 x^3-x^4+\left (3 x^2+e^5 x^2+x^3\right ) \log \left (3+e^5+x\right )} \, dx=-\frac {{\left (2 \, x e^{x} \log \left (x + e^{5} + 3\right ) - e^{x} \log \left (x + e^{5} + 3\right )^{2} - {\left (x^{2} + e\right )} e^{x}\right )} e^{\left (-1\right )}}{x} \]

input
integrate(((((-1+x)*exp(5)+x^2+2*x-3)*exp(x)*log(exp(5)+3+x)+((-x^2-x)*exp 
(5)-x^3-4*x^2-x)*exp(x))*exp(log(log(exp(5)+3+x)^2-2*x*log(exp(5)+3+x)+x^2 
)-1)+((-1+x)*exp(5)+x^2+2*x-3)*exp(x)*log(exp(5)+3+x)+((-x^2+x)*exp(5)-x^3 
-2*x^2+3*x)*exp(x))/((x^2*exp(5)+x^3+3*x^2)*log(exp(5)+3+x)-x^3*exp(5)-x^4 
-3*x^3),x, algorithm=\
 
output
-(2*x*e^x*log(x + e^5 + 3) - e^x*log(x + e^5 + 3)^2 - (x^2 + e)*e^x)*e^(-1 
)/x
 
3.12.96.8 Giac [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \[ \int \frac {e^x \left (3 x-2 x^2-x^3+e^5 \left (x-x^2\right )\right )+e^x \left (-3+e^5 (-1+x)+2 x+x^2\right ) \log \left (3+e^5+x\right )+\frac {\left (e^x \left (-x-4 x^2-x^3+e^5 \left (-x-x^2\right )\right )+e^x \left (-3+e^5 (-1+x)+2 x+x^2\right ) \log \left (3+e^5+x\right )\right ) \left (x^2-2 x \log \left (3+e^5+x\right )+\log ^2\left (3+e^5+x\right )\right )}{e}}{-3 x^3-e^5 x^3-x^4+\left (3 x^2+e^5 x^2+x^3\right ) \log \left (3+e^5+x\right )} \, dx=\frac {{\left (x^{2} e^{x} - 2 \, x e^{x} \log \left (x + e^{5} + 3\right ) + e^{x} \log \left (x + e^{5} + 3\right )^{2} + e^{\left (x + 1\right )}\right )} e^{\left (-1\right )}}{x} \]

input
integrate(((((-1+x)*exp(5)+x^2+2*x-3)*exp(x)*log(exp(5)+3+x)+((-x^2-x)*exp 
(5)-x^3-4*x^2-x)*exp(x))*exp(log(log(exp(5)+3+x)^2-2*x*log(exp(5)+3+x)+x^2 
)-1)+((-1+x)*exp(5)+x^2+2*x-3)*exp(x)*log(exp(5)+3+x)+((-x^2+x)*exp(5)-x^3 
-2*x^2+3*x)*exp(x))/((x^2*exp(5)+x^3+3*x^2)*log(exp(5)+3+x)-x^3*exp(5)-x^4 
-3*x^3),x, algorithm=\
 
output
(x^2*e^x - 2*x*e^x*log(x + e^5 + 3) + e^x*log(x + e^5 + 3)^2 + e^(x + 1))* 
e^(-1)/x
 
3.12.96.9 Mupad [F(-1)]

Timed out. \[ \int \frac {e^x \left (3 x-2 x^2-x^3+e^5 \left (x-x^2\right )\right )+e^x \left (-3+e^5 (-1+x)+2 x+x^2\right ) \log \left (3+e^5+x\right )+\frac {\left (e^x \left (-x-4 x^2-x^3+e^5 \left (-x-x^2\right )\right )+e^x \left (-3+e^5 (-1+x)+2 x+x^2\right ) \log \left (3+e^5+x\right )\right ) \left (x^2-2 x \log \left (3+e^5+x\right )+\log ^2\left (3+e^5+x\right )\right )}{e}}{-3 x^3-e^5 x^3-x^4+\left (3 x^2+e^5 x^2+x^3\right ) \log \left (3+e^5+x\right )} \, dx=-\int \frac {{\mathrm {e}}^x\,\left (3\,x+{\mathrm {e}}^5\,\left (x-x^2\right )-2\,x^2-x^3\right )-{\mathrm {e}}^{\ln \left (x^2-2\,x\,\ln \left (x+{\mathrm {e}}^5+3\right )+{\ln \left (x+{\mathrm {e}}^5+3\right )}^2\right )-1}\,\left ({\mathrm {e}}^x\,\left (x+{\mathrm {e}}^5\,\left (x^2+x\right )+4\,x^2+x^3\right )-{\mathrm {e}}^x\,\ln \left (x+{\mathrm {e}}^5+3\right )\,\left (2\,x+{\mathrm {e}}^5\,\left (x-1\right )+x^2-3\right )\right )+{\mathrm {e}}^x\,\ln \left (x+{\mathrm {e}}^5+3\right )\,\left (2\,x+{\mathrm {e}}^5\,\left (x-1\right )+x^2-3\right )}{x^3\,{\mathrm {e}}^5-\ln \left (x+{\mathrm {e}}^5+3\right )\,\left (x^2\,{\mathrm {e}}^5+3\,x^2+x^3\right )+3\,x^3+x^4} \,d x \]

input
int(-(exp(x)*(3*x + exp(5)*(x - x^2) - 2*x^2 - x^3) - exp(log(log(x + exp( 
5) + 3)^2 - 2*x*log(x + exp(5) + 3) + x^2) - 1)*(exp(x)*(x + exp(5)*(x + x 
^2) + 4*x^2 + x^3) - exp(x)*log(x + exp(5) + 3)*(2*x + exp(5)*(x - 1) + x^ 
2 - 3)) + exp(x)*log(x + exp(5) + 3)*(2*x + exp(5)*(x - 1) + x^2 - 3))/(x^ 
3*exp(5) - log(x + exp(5) + 3)*(x^2*exp(5) + 3*x^2 + x^3) + 3*x^3 + x^4),x 
)
 
output
-int((exp(x)*(3*x + exp(5)*(x - x^2) - 2*x^2 - x^3) - exp(log(log(x + exp( 
5) + 3)^2 - 2*x*log(x + exp(5) + 3) + x^2) - 1)*(exp(x)*(x + exp(5)*(x + x 
^2) + 4*x^2 + x^3) - exp(x)*log(x + exp(5) + 3)*(2*x + exp(5)*(x - 1) + x^ 
2 - 3)) + exp(x)*log(x + exp(5) + 3)*(2*x + exp(5)*(x - 1) + x^2 - 3))/(x^ 
3*exp(5) - log(x + exp(5) + 3)*(x^2*exp(5) + 3*x^2 + x^3) + 3*x^3 + x^4), 
x)