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

Optimal. Leaf size=33 \[ \frac {5}{\log \left (\frac {e^{2 e^{2 e^3}} \left (e^3-e^{-2 x} x\right )}{\log (x)}\right )} \]

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Rubi [F]  time = 2.24, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.29, size = 36, normalized size = 1.09 \begin {gather*} \frac {5}{\log \left (\frac {e^{2 e^{2 e^3}-2 x} \left (e^{3+2 x}-x\right )}{\log (x)}\right )} \end {gather*}

Antiderivative was successfully verified.

[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]]

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fricas [A]  time = 0.99, size = 38, normalized size = 1.15 \begin {gather*} \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 \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 0.21, size = 43, normalized size = 1.30 \begin {gather*} \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 \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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maple [C]  time = 0.24, size = 331, normalized size = 10.03

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

Verification of antiderivative is not currently implemented for this CAS.

[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]

10*I/(Pi*csgn(I*exp(-2*x))*csgn(I/ln(x)*(-exp(2*x+3)+x))*csgn(I*exp(-2*x)/ln(x)*(-exp(2*x+3)+x))-Pi*csgn(I*exp
(-2*x))*csgn(I*exp(-2*x)/ln(x)*(-exp(2*x+3)+x))^2+Pi*csgn(I*(-exp(2*x+3)+x))*csgn(I/ln(x)*(-exp(2*x+3)+x))^2+P
i*csgn(I*(-exp(2*x+3)+x))*csgn(I/ln(x)*(-exp(2*x+3)+x))*csgn(I/ln(x))-Pi*csgn(I/ln(x)*(-exp(2*x+3)+x))^3-Pi*cs
gn(I/ln(x)*(-exp(2*x+3)+x))^2*csgn(I/ln(x))+Pi*csgn(I/ln(x)*(-exp(2*x+3)+x))*csgn(I*exp(-2*x)/ln(x)*(-exp(2*x+
3)+x))^2-Pi*csgn(I*exp(-2*x)/ln(x)*(-exp(2*x+3)+x))^3+2*I*ln(exp(2*x+3)-x)-2*I*ln(ln(x))-2*I*ln(exp(2*x))+4*I*
exp(2*exp(3)))

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maxima [A]  time = 0.41, size = 31, normalized size = 0.94 \begin {gather*} -\frac {5}{2 \, x - 2 \, e^{\left (2 \, e^{3}\right )} - \log \left (-x + e^{\left (2 \, x + 3\right )}\right ) + \log \left (\log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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mupad [B]  time = 6.09, size = 33, normalized size = 1.00 \begin {gather*} \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 \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.75, size = 31, normalized size = 0.94 \begin {gather*} \frac {5}{\log {\left (\frac {\left (- x + e^{3} e^{2 x}\right ) e^{- 2 x} e^{2 e^{2 e^{3}}}}{\log {\relax (x )}} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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