\(\int \frac {5 e^{3+2 x}-5 x+(5 x-10 x^2) \log (x)}{(e^{3+2 x} x-x^2) \log (x) \log ^2(\frac {e^{2 e^{2 e^3}-2 x} (e^{3+2 x}-x)}{\log (x)})} \, dx\) [8389]

Optimal result

Integrand size = 81, antiderivative size = 33 \[ \int \frac {5 e^{3+2 x}-5 x+\left (5 x-10 x^2\right ) \log (x)}{\left (e^{3+2 x} x-x^2\right ) \log (x) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{3+2 x}-x\right )}{\log (x)}\right )} \, dx=\frac {5}{\log \left (\frac {e^{2 e^{2 e^3}} \left (e^3-e^{-2 x} x\right )}{\log (x)}\right )} \]

[Out]

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

Rubi [F]

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

[In]

Int[(5*E^(3 + 2*x) - 5*x + (5*x - 10*x^2)*Log[x])/((E^(3 + 2*x)*x - x^2)*Log[x]*Log[(E^(2*E^(2*E^3) - 2*x)*(E^
(3 + 2*x) - x))/Log[x]]^2),x]

[Out]

5*Defer[Int][1/((E^(3 + 2*x) - x)*Log[(E^(2*E^(2*E^3) - 2*x)*(E^(3 + 2*x) - x))/Log[x]]^2), x] - 10*Defer[Int]
[x/((E^(3 + 2*x) - x)*Log[(E^(2*E^(2*E^3) - 2*x)*(E^(3 + 2*x) - x))/Log[x]]^2), x] + 5*Defer[Int][1/(x*Log[x]*
Log[(E^(2*E^(2*E^3) - 2*x)*(E^(3 + 2*x) - x))/Log[x]]^2), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {5 \left (e^{3+2 x}-x+x \log (x)-2 x^2 \log (x)\right )}{\left (e^{3+2 x} x-x^2\right ) \log (x) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{3+2 x}-x\right )}{\log (x)}\right )} \, dx \\ & = 5 \int \frac {e^{3+2 x}-x+x \log (x)-2 x^2 \log (x)}{\left (e^{3+2 x} x-x^2\right ) \log (x) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{3+2 x}-x\right )}{\log (x)}\right )} \, dx \\ & = 5 \int \left (\frac {-1+2 x}{\left (-e^{3+2 x}+x\right ) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{3+2 x}-x\right )}{\log (x)}\right )}+\frac {1}{x \log (x) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{3+2 x}-x\right )}{\log (x)}\right )}\right ) \, dx \\ & = 5 \int \frac {-1+2 x}{\left (-e^{3+2 x}+x\right ) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{3+2 x}-x\right )}{\log (x)}\right )} \, dx+5 \int \frac {1}{x \log (x) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{3+2 x}-x\right )}{\log (x)}\right )} \, dx \\ & = 5 \int \left (\frac {1}{\left (e^{3+2 x}-x\right ) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{3+2 x}-x\right )}{\log (x)}\right )}-\frac {2 x}{\left (e^{3+2 x}-x\right ) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{3+2 x}-x\right )}{\log (x)}\right )}\right ) \, dx+5 \int \frac {1}{x \log (x) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{3+2 x}-x\right )}{\log (x)}\right )} \, dx \\ & = 5 \int \frac {1}{\left (e^{3+2 x}-x\right ) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{3+2 x}-x\right )}{\log (x)}\right )} \, dx+5 \int \frac {1}{x \log (x) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{3+2 x}-x\right )}{\log (x)}\right )} \, dx-10 \int \frac {x}{\left (e^{3+2 x}-x\right ) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{3+2 x}-x\right )}{\log (x)}\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09 \[ \int \frac {5 e^{3+2 x}-5 x+\left (5 x-10 x^2\right ) \log (x)}{\left (e^{3+2 x} x-x^2\right ) \log (x) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{3+2 x}-x\right )}{\log (x)}\right )} \, dx=\frac {5}{\log \left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{3+2 x}-x\right )}{\log (x)}\right )} \]

[In]

Integrate[(5*E^(3 + 2*x) - 5*x + (5*x - 10*x^2)*Log[x])/((E^(3 + 2*x)*x - x^2)*Log[x]*Log[(E^(2*E^(2*E^3) - 2*
x)*(E^(3 + 2*x) - x))/Log[x]]^2),x]

[Out]

5/Log[(E^(2*E^(2*E^3) - 2*x)*(E^(3 + 2*x) - x))/Log[x]]

Maple [A] (verified)

Time = 12.35 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09

method	result	size
parallelrisch	\(\frac {5}{\ln \left (\frac {\left ({\mathrm e}^{3} {\mathrm e}^{2 x}-x \right ) {\mathrm e}^{2 \,{\mathrm e}^{2 \,{\mathrm e}^{3}}} {\mathrm e}^{-2 x}}{\ln \left (x \right )}\right )}\)	\(36\)
risch	\(\frac {10 i}{\pi \,\operatorname {csgn}\left (i {\mathrm e}^{-2 x}\right ) \operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{3+2 x}+x \right )}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{-2 x} \left (-{\mathrm e}^{3+2 x}+x \right )}{\ln \left (x \right )}\right )-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{-2 x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{-2 x} \left (-{\mathrm e}^{3+2 x}+x \right )}{\ln \left (x \right )}\right )^{2}+\pi \,\operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (i \left (-{\mathrm e}^{3+2 x}+x \right )\right ) \operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{3+2 x}+x \right )}{\ln \left (x \right )}\right )-\pi \,\operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right ) {\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{3+2 x}+x \right )}{\ln \left (x \right )}\right )}^{2}+\pi \,\operatorname {csgn}\left (i \left (-{\mathrm e}^{3+2 x}+x \right )\right ) {\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{3+2 x}+x \right )}{\ln \left (x \right )}\right )}^{2}-\pi {\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{3+2 x}+x \right )}{\ln \left (x \right )}\right )}^{3}+\pi \,\operatorname {csgn}\left (\frac {i \left (-{\mathrm e}^{3+2 x}+x \right )}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{-2 x} \left (-{\mathrm e}^{3+2 x}+x \right )}{\ln \left (x \right )}\right )^{2}-\pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{-2 x} \left (-{\mathrm e}^{3+2 x}+x \right )}{\ln \left (x \right )}\right )^{3}+2 i \ln \left ({\mathrm e}^{3+2 x}-x \right )-2 i \ln \left (\ln \left (x \right )\right )-2 i \ln \left ({\mathrm e}^{2 x}\right )+4 i {\mathrm e}^{2 \,{\mathrm e}^{3}}}\)	\(331\)

[In]

int(((-10*x^2+5*x)*ln(x)+5*exp(3)*exp(2*x)-5*x)/(x*exp(3)*exp(2*x)-x^2)/ln(x)/ln((exp(3)*exp(2*x)-x)*exp(exp(e
xp(3))^2)^2/exp(2*x)/ln(x))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.15 \[ \int \frac {5 e^{3+2 x}-5 x+\left (5 x-10 x^2\right ) \log (x)}{\left (e^{3+2 x} x-x^2\right ) \log (x) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{3+2 x}-x\right )}{\log (x)}\right )} \, dx=\frac {5}{\log \left (-\frac {x e^{\left (-2 \, x + 2 \, e^{\left (2 \, e^{3}\right )}\right )} - e^{\left (2 \, e^{\left (2 \, e^{3}\right )} + 3\right )}}{\log \left (x\right )}\right )} \]

[In]

integrate(((-10*x^2+5*x)*log(x)+5*exp(3)*exp(2*x)-5*x)/(x*exp(3)*exp(2*x)-x^2)/log(x)/log((exp(3)*exp(2*x)-x)*
exp(exp(exp(3))^2)^2/exp(2*x)/log(x))^2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {5 e^{3+2 x}-5 x+\left (5 x-10 x^2\right ) \log (x)}{\left (e^{3+2 x} x-x^2\right ) \log (x) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{3+2 x}-x\right )}{\log (x)}\right )} \, dx=\frac {5}{\log {\left (\frac {\left (- x + e^{3} e^{2 x}\right ) e^{- 2 x} e^{2 e^{2 e^{3}}}}{\log {\left (x \right )}} \right )}} \]

[In]

integrate(((-10*x**2+5*x)*ln(x)+5*exp(3)*exp(2*x)-5*x)/(x*exp(3)*exp(2*x)-x**2)/ln(x)/ln((exp(3)*exp(2*x)-x)*e
xp(exp(exp(3))**2)**2/exp(2*x)/ln(x))**2,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {5 e^{3+2 x}-5 x+\left (5 x-10 x^2\right ) \log (x)}{\left (e^{3+2 x} x-x^2\right ) \log (x) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{3+2 x}-x\right )}{\log (x)}\right )} \, dx=-\frac {5}{2 \, x - 2 \, e^{\left (2 \, e^{3}\right )} - \log \left (-x + e^{\left (2 \, x + 3\right )}\right ) + \log \left (\log \left (x\right )\right )} \]

[In]

integrate(((-10*x^2+5*x)*log(x)+5*exp(3)*exp(2*x)-5*x)/(x*exp(3)*exp(2*x)-x^2)/log(x)/log((exp(3)*exp(2*x)-x)*
exp(exp(exp(3))^2)^2/exp(2*x)/log(x))^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.30 \[ \int \frac {5 e^{3+2 x}-5 x+\left (5 x-10 x^2\right ) \log (x)}{\left (e^{3+2 x} x-x^2\right ) \log (x) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{3+2 x}-x\right )}{\log (x)}\right )} \, dx=\frac {5}{\log \left (-{\left (x e^{\left (2 \, e^{\left (2 \, e^{3}\right )}\right )} - e^{\left (2 \, x + 2 \, e^{\left (2 \, e^{3}\right )} + 3\right )}\right )} e^{\left (-2 \, x\right )}\right ) - \log \left (\log \left (x\right )\right )} \]

[In]

integrate(((-10*x^2+5*x)*log(x)+5*exp(3)*exp(2*x)-5*x)/(x*exp(3)*exp(2*x)-x^2)/log(x)/log((exp(3)*exp(2*x)-x)*
exp(exp(exp(3))^2)^2/exp(2*x)/log(x))^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.69 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int \frac {5 e^{3+2 x}-5 x+\left (5 x-10 x^2\right ) \log (x)}{\left (e^{3+2 x} x-x^2\right ) \log (x) \log ^2\left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{3+2 x}-x\right )}{\log (x)}\right )} \, dx=\frac {5}{\ln \left (-\frac {{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^{2\,{\mathrm {e}}^{2\,{\mathrm {e}}^3}}\,\left (x-{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^3\right )}{\ln \left (x\right )}\right )} \]

[In]

int(-(5*exp(2*x)*exp(3) - 5*x + log(x)*(5*x - 10*x^2))/(log(-(exp(-2*x)*exp(2*exp(2*exp(3)))*(x - exp(2*x)*exp
(3)))/log(x))^2*log(x)*(x^2 - x*exp(2*x)*exp(3))),x)

[Out]

5/log(-(exp(-2*x)*exp(2*exp(2*exp(3)))*(x - exp(2*x)*exp(3)))/log(x))