\(\int \frac {e^{-\frac {e^x}{\log (-9+x)}} (e^x x+e^x (9 x-x^2) \log (-9+x)+(9-x) \log ^2(-9+x))}{(-45 x^2+5 x^3+e (-9 x^2+x^3)) \log ^2(-9+x)} \, dx\) [6530]

Optimal result

Integrand size = 79, antiderivative size = 26 \[ \int \frac {e^{-\frac {e^x}{\log (-9+x)}} \left (e^x x+e^x \left (9 x-x^2\right ) \log (-9+x)+(9-x) \log ^2(-9+x)\right )}{\left (-45 x^2+5 x^3+e \left (-9 x^2+x^3\right )\right ) \log ^2(-9+x)} \, dx=\frac {e^{-\frac {e^x}{\log (-9+x)}}+5 x}{(5+e) x} \]

[Out]

(5*x+exp(-exp(x)/ln(x-9)))/x/(exp(1)+5)

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(99\) vs. \(2(26)=52\).

Time = 0.21 (sec) , antiderivative size = 99, normalized size of antiderivative = 3.81, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.013, Rules used = {2326} \[ \int \frac {e^{-\frac {e^x}{\log (-9+x)}} \left (e^x x+e^x \left (9 x-x^2\right ) \log (-9+x)+(9-x) \log ^2(-9+x)\right )}{\left (-45 x^2+5 x^3+e \left (-9 x^2+x^3\right )\right ) \log ^2(-9+x)} \, dx=\frac {e^{-\frac {e^x}{\log (x-9)}} \left (e^x \left (9 x-x^2\right ) \log (x-9)+e^x x\right )}{\left (-5 x^3+45 x^2+e \left (9 x^2-x^3\right )\right ) \left (\frac {e^x}{(9-x) \log ^2(x-9)}+\frac {e^x}{\log (x-9)}\right ) \log ^2(x-9)} \]

[In]

Int[(E^x*x + E^x*(9*x - x^2)*Log[-9 + x] + (9 - x)*Log[-9 + x]^2)/(E^(E^x/Log[-9 + x])*(-45*x^2 + 5*x^3 + E*(-
9*x^2 + x^3))*Log[-9 + x]^2),x]

[Out]

(E^x*x + E^x*(9*x - x^2)*Log[-9 + x])/(E^(E^x/Log[-9 + x])*(45*x^2 - 5*x^3 + E*(9*x^2 - x^3))*(E^x/((9 - x)*Lo
g[-9 + x]^2) + E^x/Log[-9 + x])*Log[-9 + x]^2)

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \frac {e^{-\frac {e^x}{\log (-9+x)}} \left (e^x x+e^x \left (9 x-x^2\right ) \log (-9+x)\right )}{\left (45 x^2-5 x^3+e \left (9 x^2-x^3\right )\right ) \left (\frac {e^x}{(9-x) \log ^2(-9+x)}+\frac {e^x}{\log (-9+x)}\right ) \log ^2(-9+x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {e^{-\frac {e^x}{\log (-9+x)}} \left (e^x x+e^x \left (9 x-x^2\right ) \log (-9+x)+(9-x) \log ^2(-9+x)\right )}{\left (-45 x^2+5 x^3+e \left (-9 x^2+x^3\right )\right ) \log ^2(-9+x)} \, dx=\frac {e^{-\frac {e^x}{\log (-9+x)}}}{(5+e) x} \]

[In]

Integrate[(E^x*x + E^x*(9*x - x^2)*Log[-9 + x] + (9 - x)*Log[-9 + x]^2)/(E^(E^x/Log[-9 + x])*(-45*x^2 + 5*x^3
+ E*(-9*x^2 + x^3))*Log[-9 + x]^2),x]

[Out]

1/(E^(E^x/Log[-9 + x])*(5 + E)*x)

Maple [A] (verified)

Time = 9.74 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85

method	result	size
risch	\(\frac {{\mathrm e}^{-\frac {{\mathrm e}^{x}}{\ln \left (x -9\right )}}}{x \left ({\mathrm e}+5\right )}\)	\(22\)
parallelrisch	\(\frac {{\mathrm e}^{-\frac {{\mathrm e}^{x}}{\ln \left (x -9\right )}}}{x \left ({\mathrm e}+5\right )}\)	\(22\)

[In]

int(((9-x)*ln(x-9)^2+(-x^2+9*x)*exp(x)*ln(x-9)+exp(x)*x)*exp(-exp(x)/ln(x-9))/((x^3-9*x^2)*exp(1)+5*x^3-45*x^2
)/ln(x-9)^2,x,method=_RETURNVERBOSE)

[Out]

1/x/(exp(1)+5)*exp(-exp(x)/ln(x-9))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {e^{-\frac {e^x}{\log (-9+x)}} \left (e^x x+e^x \left (9 x-x^2\right ) \log (-9+x)+(9-x) \log ^2(-9+x)\right )}{\left (-45 x^2+5 x^3+e \left (-9 x^2+x^3\right )\right ) \log ^2(-9+x)} \, dx=\frac {e^{\left (-\frac {e^{x}}{\log \left (x - 9\right )}\right )}}{x e + 5 \, x} \]

[In]

integrate(((9-x)*log(x-9)^2+(-x^2+9*x)*exp(x)*log(x-9)+exp(x)*x)*exp(-exp(x)/log(x-9))/((x^3-9*x^2)*exp(1)+5*x
^3-45*x^2)/log(x-9)^2,x, algorithm="fricas")

[Out]

e^(-e^x/log(x - 9))/(x*e + 5*x)

Sympy [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {e^{-\frac {e^x}{\log (-9+x)}} \left (e^x x+e^x \left (9 x-x^2\right ) \log (-9+x)+(9-x) \log ^2(-9+x)\right )}{\left (-45 x^2+5 x^3+e \left (-9 x^2+x^3\right )\right ) \log ^2(-9+x)} \, dx=\frac {e^{- \frac {e^{x}}{\log {\left (x - 9 \right )}}}}{e x + 5 x} \]

[In]

integrate(((9-x)*ln(x-9)**2+(-x**2+9*x)*exp(x)*ln(x-9)+exp(x)*x)*exp(-exp(x)/ln(x-9))/((x**3-9*x**2)*exp(1)+5*
x**3-45*x**2)/ln(x-9)**2,x)

[Out]

exp(-exp(x)/log(x - 9))/(E*x + 5*x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {e^{-\frac {e^x}{\log (-9+x)}} \left (e^x x+e^x \left (9 x-x^2\right ) \log (-9+x)+(9-x) \log ^2(-9+x)\right )}{\left (-45 x^2+5 x^3+e \left (-9 x^2+x^3\right )\right ) \log ^2(-9+x)} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(((9-x)*log(x-9)^2+(-x^2+9*x)*exp(x)*log(x-9)+exp(x)*x)*exp(-exp(x)/log(x-9))/((x^3-9*x^2)*exp(1)+5*x
^3-45*x^2)/log(x-9)^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {e^{-\frac {e^x}{\log (-9+x)}} \left (e^x x+e^x \left (9 x-x^2\right ) \log (-9+x)+(9-x) \log ^2(-9+x)\right )}{\left (-45 x^2+5 x^3+e \left (-9 x^2+x^3\right )\right ) \log ^2(-9+x)} \, dx=\frac {e^{\left (\frac {x \log \left (x - 9\right ) - e^{x}}{\log \left (x - 9\right )}\right )}}{x e^{\left (x + 1\right )} + 5 \, x e^{x}} \]

[In]

integrate(((9-x)*log(x-9)^2+(-x^2+9*x)*exp(x)*log(x-9)+exp(x)*x)*exp(-exp(x)/log(x-9))/((x^3-9*x^2)*exp(1)+5*x
^3-45*x^2)/log(x-9)^2,x, algorithm="giac")

[Out]

e^((x*log(x - 9) - e^x)/log(x - 9))/(x*e^(x + 1) + 5*x*e^x)

Mupad [B] (verification not implemented)

Time = 12.29 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {e^{-\frac {e^x}{\log (-9+x)}} \left (e^x x+e^x \left (9 x-x^2\right ) \log (-9+x)+(9-x) \log ^2(-9+x)\right )}{\left (-45 x^2+5 x^3+e \left (-9 x^2+x^3\right )\right ) \log ^2(-9+x)} \, dx=\frac {{\mathrm {e}}^{-\frac {{\mathrm {e}}^x}{\ln \left (x-9\right )}}}{x\,\left (\mathrm {e}+5\right )} \]

[In]

int(-(exp(-exp(x)/log(x - 9))*(x*exp(x) - log(x - 9)^2*(x - 9) + log(x - 9)*exp(x)*(9*x - x^2)))/(log(x - 9)^2
*(exp(1)*(9*x^2 - x^3) + 45*x^2 - 5*x^3)),x)

[Out]

exp(-exp(x)/log(x - 9))/(x*(exp(1) + 5))