\(\int \frac {e^{\frac {3 x+e^{200+40 x+2 x^2} x-\log (-2+e^2+e^5)}{x}} (e^{200+40 x+2 x^2} (40 x^2+4 x^3)+\log (-2+e^2+e^5))}{x^2} \, dx\) [8742]

Optimal result

Integrand size = 73, antiderivative size = 29 \[ \int \frac {e^{\frac {3 x+e^{200+40 x+2 x^2} x-\log \left (-2+e^2+e^5\right )}{x}} \left (e^{200+40 x+2 x^2} \left (40 x^2+4 x^3\right )+\log \left (-2+e^2+e^5\right )\right )}{x^2} \, dx=1+e^{3+e^{2 (10+x)^2}-\frac {\log \left (-2+e^2+e^5\right )}{x}} \]

[Out]

1+exp(exp((x+10)^2)^2-ln(exp(5)+exp(2)-2)/x+3)

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {6838} \[ \int \frac {e^{\frac {3 x+e^{200+40 x+2 x^2} x-\log \left (-2+e^2+e^5\right )}{x}} \left (e^{200+40 x+2 x^2} \left (40 x^2+4 x^3\right )+\log \left (-2+e^2+e^5\right )\right )}{x^2} \, dx=e^{\frac {e^{2 x^2+40 x+200} x+3 x}{x}} \left (-2+e^2+e^5\right )^{-1/x} \]

[In]

Int[(E^((3*x + E^(200 + 40*x + 2*x^2)*x - Log[-2 + E^2 + E^5])/x)*(E^(200 + 40*x + 2*x^2)*(40*x^2 + 4*x^3) + L
og[-2 + E^2 + E^5]))/x^2,x]

[Out]

E^((3*x + E^(200 + 40*x + 2*x^2)*x)/x)/(-2 + E^2 + E^5)^x^(-1)

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = e^{\frac {3 x+e^{200+40 x+2 x^2} x}{x}} \left (-2+e^2+e^5\right )^{-1/x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {e^{\frac {3 x+e^{200+40 x+2 x^2} x-\log \left (-2+e^2+e^5\right )}{x}} \left (e^{200+40 x+2 x^2} \left (40 x^2+4 x^3\right )+\log \left (-2+e^2+e^5\right )\right )}{x^2} \, dx=e^{3+e^{2 (10+x)^2}} \left (-2+e^2+e^5\right )^{-1/x} \]

[In]

Integrate[(E^((3*x + E^(200 + 40*x + 2*x^2)*x - Log[-2 + E^2 + E^5])/x)*(E^(200 + 40*x + 2*x^2)*(40*x^2 + 4*x^
3) + Log[-2 + E^2 + E^5]))/x^2,x]

[Out]

E^(3 + E^(2*(10 + x)^2))/(-2 + E^2 + E^5)^x^(-1)

Maple [A] (verified)

Time = 1.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10

method	result	size
norman	\({\mathrm e}^{\frac {-\ln \left ({\mathrm e}^{5}+{\mathrm e}^{2}-2\right )+x \,{\mathrm e}^{2 x^{2}+40 x +200}+3 x}{x}}\)	\(32\)
parallelrisch	\({\mathrm e}^{-\frac {-x \,{\mathrm e}^{2 x^{2}+40 x +200}+\ln \left ({\mathrm e}^{5}+{\mathrm e}^{2}-2\right )-3 x}{x}}\)	\(32\)
risch	\(\left ({\mathrm e}-1\right )^{-\frac {1}{x}} \left ({\mathrm e}^{4}+{\mathrm e}^{3}+{\mathrm e}^{2}+2 \,{\mathrm e}+2\right )^{-\frac {1}{x}} {\mathrm e}^{{\mathrm e}^{2 \left (x +10\right )^{2}}+3}\)	\(41\)

[In]

int((ln(exp(5)+exp(2)-2)+(4*x^3+40*x^2)*exp(x^2+20*x+100)^2)*exp((-ln(exp(5)+exp(2)-2)+x*exp(x^2+20*x+100)^2+3
*x)/x)/x^2,x,method=_RETURNVERBOSE)

[Out]

exp((-ln(exp(5)+exp(2)-2)+x*exp(x^2+20*x+100)^2+3*x)/x)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {e^{\frac {3 x+e^{200+40 x+2 x^2} x-\log \left (-2+e^2+e^5\right )}{x}} \left (e^{200+40 x+2 x^2} \left (40 x^2+4 x^3\right )+\log \left (-2+e^2+e^5\right )\right )}{x^2} \, dx=e^{\left (\frac {x e^{\left (2 \, x^{2} + 40 \, x + 200\right )} + 3 \, x - \log \left (e^{5} + e^{2} - 2\right )}{x}\right )} \]

[In]

integrate((log(exp(5)+exp(2)-2)+(4*x^3+40*x^2)*exp(x^2+20*x+100)^2)*exp((-log(exp(5)+exp(2)-2)+x*exp(x^2+20*x+
100)^2+3*x)/x)/x^2,x, algorithm="fricas")

[Out]

e^((x*e^(2*x^2 + 40*x + 200) + 3*x - log(e^5 + e^2 - 2))/x)

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {3 x+e^{200+40 x+2 x^2} x-\log \left (-2+e^2+e^5\right )}{x}} \left (e^{200+40 x+2 x^2} \left (40 x^2+4 x^3\right )+\log \left (-2+e^2+e^5\right )\right )}{x^2} \, dx=e^{\frac {x e^{2 x^{2} + 40 x + 200} + 3 x - \log {\left (-2 + e^{2} + e^{5} \right )}}{x}} \]

[In]

integrate((ln(exp(5)+exp(2)-2)+(4*x**3+40*x**2)*exp(x**2+20*x+100)**2)*exp((-ln(exp(5)+exp(2)-2)+x*exp(x**2+20
*x+100)**2+3*x)/x)/x**2,x)

[Out]

exp((x*exp(2*x**2 + 40*x + 200) + 3*x - log(-2 + exp(2) + exp(5)))/x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \[ \int \frac {e^{\frac {3 x+e^{200+40 x+2 x^2} x-\log \left (-2+e^2+e^5\right )}{x}} \left (e^{200+40 x+2 x^2} \left (40 x^2+4 x^3\right )+\log \left (-2+e^2+e^5\right )\right )}{x^2} \, dx=e^{\left (-\frac {\log \left (e^{4} + e^{3} + e^{2} + 2 \, e + 2\right )}{x} - \frac {\log \left (e - 1\right )}{x} + e^{\left (2 \, x^{2} + 40 \, x + 200\right )} + 3\right )} \]

[In]

integrate((log(exp(5)+exp(2)-2)+(4*x^3+40*x^2)*exp(x^2+20*x+100)^2)*exp((-log(exp(5)+exp(2)-2)+x*exp(x^2+20*x+
100)^2+3*x)/x)/x^2,x, algorithm="maxima")

[Out]

e^(-log(e^4 + e^3 + e^2 + 2*e + 2)/x - log(e - 1)/x + e^(2*x^2 + 40*x + 200) + 3)

Giac [F]

\[ \int \frac {e^{\frac {3 x+e^{200+40 x+2 x^2} x-\log \left (-2+e^2+e^5\right )}{x}} \left (e^{200+40 x+2 x^2} \left (40 x^2+4 x^3\right )+\log \left (-2+e^2+e^5\right )\right )}{x^2} \, dx=\int { \frac {{\left (4 \, {\left (x^{3} + 10 \, x^{2}\right )} e^{\left (2 \, x^{2} + 40 \, x + 200\right )} + \log \left (e^{5} + e^{2} - 2\right )\right )} e^{\left (\frac {x e^{\left (2 \, x^{2} + 40 \, x + 200\right )} + 3 \, x - \log \left (e^{5} + e^{2} - 2\right )}{x}\right )}}{x^{2}} \,d x } \]

[In]

integrate((log(exp(5)+exp(2)-2)+(4*x^3+40*x^2)*exp(x^2+20*x+100)^2)*exp((-log(exp(5)+exp(2)-2)+x*exp(x^2+20*x+
100)^2+3*x)/x)/x^2,x, algorithm="giac")

[Out]

integrate((4*(x^3 + 10*x^2)*e^(2*x^2 + 40*x + 200) + log(e^5 + e^2 - 2))*e^((x*e^(2*x^2 + 40*x + 200) + 3*x -
log(e^5 + e^2 - 2))/x)/x^2, x)

Mupad [B] (verification not implemented)

Time = 14.22 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {3 x+e^{200+40 x+2 x^2} x-\log \left (-2+e^2+e^5\right )}{x}} \left (e^{200+40 x+2 x^2} \left (40 x^2+4 x^3\right )+\log \left (-2+e^2+e^5\right )\right )}{x^2} \, dx=\frac {{\mathrm {e}}^{{\mathrm {e}}^{40\,x}\,{\mathrm {e}}^{200}\,{\mathrm {e}}^{2\,x^2}}\,{\mathrm {e}}^3}{{\left ({\mathrm {e}}^2+{\mathrm {e}}^5-2\right )}^{1/x}} \]

[In]

int((exp((3*x - log(exp(2) + exp(5) - 2) + x*exp(40*x + 2*x^2 + 200))/x)*(log(exp(2) + exp(5) - 2) + exp(40*x
+ 2*x^2 + 200)*(40*x^2 + 4*x^3)))/x^2,x)

[Out]

(exp(exp(40*x)*exp(200)*exp(2*x^2))*exp(3))/(exp(2) + exp(5) - 2)^(1/x)