3.25.6 \(\int \frac {9 x^2+3 x^3+(-3 x^2+3 x^3) \log (-1+x)+e^{e^x x} (-25 x+(-25+25 x+e^x (25 x-25 x^3)) \log (-1+x))}{-3 x^2+3 x^3} \, dx\) [2406]

3.25.6.1 Optimal result

Integrand size = 76, antiderivative size = 22 \[ \int \frac {9 x^2+3 x^3+\left (-3 x^2+3 x^3\right ) \log (-1+x)+e^{e^x x} \left (-25 x+\left (-25+25 x+e^x \left (25 x-25 x^3\right )\right ) \log (-1+x)\right )}{-3 x^2+3 x^3} \, dx=\left (3-\frac {25 e^{e^x x}}{3 x}+x\right ) \log (-1+x) \]

output

ln(-1+x)*(3+x-25/3*exp(exp(x)*x)/x)
 

3.25.6.2 Mathematica [A] (verified)

Time = 0.60 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68 \[ \int \frac {9 x^2+3 x^3+\left (-3 x^2+3 x^3\right ) \log (-1+x)+e^{e^x x} \left (-25 x+\left (-25+25 x+e^x \left (25 x-25 x^3\right )\right ) \log (-1+x)\right )}{-3 x^2+3 x^3} \, dx=\frac {1}{3} \left (12 \log (1-x)+\frac {\left (-25 e^{e^x x}+3 (-1+x) x\right ) \log (-1+x)}{x}\right ) \]

input

Integrate[(9*x^2 + 3*x^3 + (-3*x^2 + 3*x^3)*Log[-1 + x] + E^(E^x*x)*(-25*x 
 + (-25 + 25*x + E^x*(25*x - 25*x^3))*Log[-1 + x]))/(-3*x^2 + 3*x^3),x]
 

output

(12*Log[1 - x] + ((-25*E^(E^x*x) + 3*(-1 + x)*x)*Log[-1 + x])/x)/3
 

3.25.6.3 Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(74\) vs. \(2(22)=44\).

Time = 1.23 (sec) , antiderivative size = 74, normalized size of antiderivative = 3.36, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {2026, 7293, 2009}

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.

\(\displaystyle \int \frac {3 x^3+e^{e^x x} \left (\left (e^x \left (25 x-25 x^3\right )+25 x-25\right ) \log (x-1)-25 x\right )+9 x^2+\left (3 x^3-3 x^2\right ) \log (x-1)}{3 x^3-3 x^2} \, dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {3 x^3+e^{e^x x} \left (\left (e^x \left (25 x-25 x^3\right )+25 x-25\right ) \log (x-1)-25 x\right )+9 x^2+\left (3 x^3-3 x^2\right ) \log (x-1)}{x^2 (3 x-3)}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {x+x \log (x-1)-\log (x-1)+3}{x-1}-\frac {25 e^{e^x x} \left (e^x x^3 \log (x-1)+x-e^x x \log (x-1)-x \log (x-1)+\log (x-1)\right )}{3 (x-1) x^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {25 e^{e^x x} \left (e^x x \log (x-1)-e^x x^3 \log (x-1)\right )}{3 (1-x) x^2 \left (e^x x+e^x\right )}+4 \log (1-x)-(1-x) \log (x-1)\)

input

Int[(9*x^2 + 3*x^3 + (-3*x^2 + 3*x^3)*Log[-1 + x] + E^(E^x*x)*(-25*x + (-2 
5 + 25*x + E^x*(25*x - 25*x^3))*Log[-1 + x]))/(-3*x^2 + 3*x^3),x]
 

output

4*Log[1 - x] - (1 - x)*Log[-1 + x] - (25*E^(E^x*x)*(E^x*x*Log[-1 + x] - E^ 
x*x^3*Log[-1 + x]))/(3*(1 - x)*x^2*(E^x + E^x*x))
 

3.25.6.3.1 Defintions of rubi rules used

rule 2009

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

rule 2026

Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p 
*r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ 
erQ[p] &&  !MonomialQ[Px, x] && (ILtQ[p, 0] ||  !PolyQ[u, x])
 

rule 7293

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

3.25.6.4 Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27

method	result	size
risch	\(\ln \left (-1+x \right ) x +3 \ln \left (-1+x \right )-\frac {25 \ln \left (-1+x \right ) {\mathrm e}^{{\mathrm e}^{x} x}}{3 x}\)	\(28\)
parallelrisch	\(-\frac {-6 \ln \left (-1+x \right ) x^{2}-18 \ln \left (-1+x \right ) x +50 \ln \left (-1+x \right ) {\mathrm e}^{{\mathrm e}^{x} x}}{6 x}\)	\(34\)

input

int(((((-25*x^3+25*x)*exp(x)+25*x-25)*ln(-1+x)-25*x)*exp(exp(x)*x)+(3*x^3- 
3*x^2)*ln(-1+x)+3*x^3+9*x^2)/(3*x^3-3*x^2),x,method=_RETURNVERBOSE)
 

output

ln(-1+x)*x+3*ln(-1+x)-25/3*ln(-1+x)/x*exp(exp(x)*x)
 

3.25.6.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36 \[ \int \frac {9 x^2+3 x^3+\left (-3 x^2+3 x^3\right ) \log (-1+x)+e^{e^x x} \left (-25 x+\left (-25+25 x+e^x \left (25 x-25 x^3\right )\right ) \log (-1+x)\right )}{-3 x^2+3 x^3} \, dx=\frac {3 \, {\left (x^{2} + 3 \, x\right )} \log \left (x - 1\right ) - 25 \, e^{\left (x e^{x}\right )} \log \left (x - 1\right )}{3 \, x} \]

input

integrate(((((-25*x^3+25*x)*exp(x)+25*x-25)*log(-1+x)-25*x)*exp(exp(x)*x)+ 
(3*x^3-3*x^2)*log(-1+x)+3*x^3+9*x^2)/(3*x^3-3*x^2),x, algorithm=\
 

output

1/3*(3*(x^2 + 3*x)*log(x - 1) - 25*e^(x*e^x)*log(x - 1))/x
 

3.25.6.6 Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {9 x^2+3 x^3+\left (-3 x^2+3 x^3\right ) \log (-1+x)+e^{e^x x} \left (-25 x+\left (-25+25 x+e^x \left (25 x-25 x^3\right )\right ) \log (-1+x)\right )}{-3 x^2+3 x^3} \, dx=x \log {\left (x - 1 \right )} + 3 \log {\left (x - 1 \right )} - \frac {25 e^{x e^{x}} \log {\left (x - 1 \right )}}{3 x} \]

input

integrate(((((-25*x**3+25*x)*exp(x)+25*x-25)*ln(-1+x)-25*x)*exp(exp(x)*x)+ 
(3*x**3-3*x**2)*ln(-1+x)+3*x**3+9*x**2)/(3*x**3-3*x**2),x)
 

output

x*log(x - 1) + 3*log(x - 1) - 25*exp(x*exp(x))*log(x - 1)/(3*x)
 

3.25.6.7 Maxima [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.82 \[ \int \frac {9 x^2+3 x^3+\left (-3 x^2+3 x^3\right ) \log (-1+x)+e^{e^x x} \left (-25 x+\left (-25+25 x+e^x \left (25 x-25 x^3\right )\right ) \log (-1+x)\right )}{-3 x^2+3 x^3} \, dx={\left (x + \log \left (x - 1\right )\right )} \log \left (x - 1\right ) - \log \left (x - 1\right )^{2} - \frac {25 \, e^{\left (x e^{x}\right )} \log \left (x - 1\right )}{3 \, x} + 3 \, \log \left (x - 1\right ) \]

input

integrate(((((-25*x^3+25*x)*exp(x)+25*x-25)*log(-1+x)-25*x)*exp(exp(x)*x)+ 
(3*x^3-3*x^2)*log(-1+x)+3*x^3+9*x^2)/(3*x^3-3*x^2),x, algorithm=\
 

output

(x + log(x - 1))*log(x - 1) - log(x - 1)^2 - 25/3*e^(x*e^x)*log(x - 1)/x + 
 3*log(x - 1)
 

3.25.6.8 Giac [F]

\[ \int \frac {9 x^2+3 x^3+\left (-3 x^2+3 x^3\right ) \log (-1+x)+e^{e^x x} \left (-25 x+\left (-25+25 x+e^x \left (25 x-25 x^3\right )\right ) \log (-1+x)\right )}{-3 x^2+3 x^3} \, dx=\int { \frac {3 \, x^{3} + 9 \, x^{2} - 25 \, {\left ({\left ({\left (x^{3} - x\right )} e^{x} - x + 1\right )} \log \left (x - 1\right ) + x\right )} e^{\left (x e^{x}\right )} + 3 \, {\left (x^{3} - x^{2}\right )} \log \left (x - 1\right )}{3 \, {\left (x^{3} - x^{2}\right )}} \,d x } \]

input

integrate(((((-25*x^3+25*x)*exp(x)+25*x-25)*log(-1+x)-25*x)*exp(exp(x)*x)+ 
(3*x^3-3*x^2)*log(-1+x)+3*x^3+9*x^2)/(3*x^3-3*x^2),x, algorithm=\
 

output

integrate(1/3*(3*x^3 + 9*x^2 - 25*(((x^3 - x)*e^x - x + 1)*log(x - 1) + x) 
*e^(x*e^x) + 3*(x^3 - x^2)*log(x - 1))/(x^3 - x^2), x)
 

3.25.6.9 Mupad [F(-1)]

Timed out. \[ \int \frac {9 x^2+3 x^3+\left (-3 x^2+3 x^3\right ) \log (-1+x)+e^{e^x x} \left (-25 x+\left (-25+25 x+e^x \left (25 x-25 x^3\right )\right ) \log (-1+x)\right )}{-3 x^2+3 x^3} \, dx=\int \frac {{\mathrm {e}}^{x\,{\mathrm {e}}^x}\,\left (25\,x-\ln \left (x-1\right )\,\left (25\,x+{\mathrm {e}}^x\,\left (25\,x-25\,x^3\right )-25\right )\right )+\ln \left (x-1\right )\,\left (3\,x^2-3\,x^3\right )-9\,x^2-3\,x^3}{3\,x^2-3\,x^3} \,d x \]

input

int((exp(x*exp(x))*(25*x - log(x - 1)*(25*x + exp(x)*(25*x - 25*x^3) - 25) 
) + log(x - 1)*(3*x^2 - 3*x^3) - 9*x^2 - 3*x^3)/(3*x^2 - 3*x^3),x)
 

output

int((exp(x*exp(x))*(25*x - log(x - 1)*(25*x + exp(x)*(25*x - 25*x^3) - 25) 
) + log(x - 1)*(3*x^2 - 3*x^3) - 9*x^2 - 3*x^3)/(3*x^2 - 3*x^3), x)