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

Optimal. Leaf size=24 \[ \log (5)+\log \left (4 \left (e^5+x\right ) \log \left (x-e^{e^2} x^2\right )\right ) \]

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Rubi [F]  time = 0.52, 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 {-e^5-x+e^{e^2} \left (2 e^5 x+2 x^2\right )+\left (-x+e^{e^2} x^2\right ) \log \left (x-e^{e^2} x^2\right )}{\left (-e^5 x-x^2+e^{e^2} \left (e^5 x^2+x^3\right )\right ) \log \left (x-e^{e^2} x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {1}{e^5+x}+\frac {-1+2 e^{e^2} x}{x \left (-1+e^{e^2} x\right ) \log \left (x \left (1-e^{e^2} x\right )\right )}\right ) \, dx\\ &=\log \left (e^5+x\right )+\int \frac {-1+2 e^{e^2} x}{x \left (-1+e^{e^2} x\right ) \log \left (x \left (1-e^{e^2} x\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 21, normalized size = 0.88 \begin {gather*} \log \left (e^5+x\right )+\log \left (\log \left (x-e^{e^2} x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[E^5 + x] + Log[Log[x - E^E^2*x^2]]

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fricas [A]  time = 0.54, size = 18, normalized size = 0.75 \begin {gather*} \log \left (x + e^{5}\right ) + \log \left (\log \left (-x^{2} e^{\left (e^{2}\right )} + x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {2 \, {\left (x^{2} + x e^{5}\right )} e^{\left (e^{2}\right )} + {\left (x^{2} e^{\left (e^{2}\right )} - x\right )} \log \left (-x^{2} e^{\left (e^{2}\right )} + x\right ) - x - e^{5}}{{\left (x^{2} + x e^{5} - {\left (x^{3} + x^{2} e^{5}\right )} e^{\left (e^{2}\right )}\right )} \log \left (-x^{2} e^{\left (e^{2}\right )} + x\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.12, size = 19, normalized size = 0.79

method	result	size
default	\(\ln \left ({\mathrm e}^{5}+x \right )+\ln \left (\ln \left (-x^{2} {\mathrm e}^{{\mathrm e}^{2}}+x \right )\right )\)	\(19\)
norman	\(\ln \left ({\mathrm e}^{5}+x \right )+\ln \left (\ln \left (-x^{2} {\mathrm e}^{{\mathrm e}^{2}}+x \right )\right )\)	\(19\)
risch	\(\ln \left ({\mathrm e}^{5}+x \right )+\ln \left (\ln \left (-x^{2} {\mathrm e}^{{\mathrm e}^{2}}+x \right )\right )\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(exp(5)+x)+ln(ln(-x^2*exp(exp(2))+x))

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maxima [A]  time = 0.42, size = 19, normalized size = 0.79 \begin {gather*} \log \left (x + e^{5}\right ) + \log \left (\log \left (-x e^{\left (e^{2}\right )} + 1\right ) + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x + e^5) + log(log(-x*e^(e^2) + 1) + log(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {x+{\mathrm {e}}^5+\ln \left (x-x^2\,{\mathrm {e}}^{{\mathrm {e}}^2}\right )\,\left (x-x^2\,{\mathrm {e}}^{{\mathrm {e}}^2}\right )-{\mathrm {e}}^{{\mathrm {e}}^2}\,\left (2\,x^2+2\,{\mathrm {e}}^5\,x\right )}{\ln \left (x-x^2\,{\mathrm {e}}^{{\mathrm {e}}^2}\right )\,\left (x\,{\mathrm {e}}^5+x^2-{\mathrm {e}}^{{\mathrm {e}}^2}\,\left (x^3+{\mathrm {e}}^5\,x^2\right )\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.25, size = 19, normalized size = 0.79 \begin {gather*} \log {\left (x + e^{5} \right )} + \log {\left (\log {\left (- x^{2} e^{e^{2}} + x \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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