3.73.99 \(\int \frac {2-e^9+x-x^2+(2 x-e^9 x) \log (x)}{-2 x+x^2+(-2 x+e^9 x) \log (x)} \, dx\)

Optimal. Leaf size=28 \[ 5-x+\log \left (\frac {1}{\log (2) \left (2-x+\left (2-e^9\right ) \log (x)\right )}\right ) \]

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Rubi [A]  time = 0.26, antiderivative size = 22, normalized size of antiderivative = 0.79, number of steps used = 4, number of rules used = 3, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {6741, 6742, 6684} \begin {gather*} -x-\log \left (-x+\left (2-e^9\right ) \log (x)+2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-x - Log[2 - x + (2 - E^9)*Log[x]]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-2 \left (1-\frac {e^9}{2}\right )-x+x^2-\left (2 x-e^9 x\right ) \log (x)}{2 x-x^2-\left (-2 x+e^9 x\right ) \log (x)} \, dx\\ &=\int \left (-1+\frac {-2+e^9+x}{x \left (2-x+2 \left (1-\frac {e^9}{2}\right ) \log (x)\right )}\right ) \, dx\\ &=-x+\int \frac {-2+e^9+x}{x \left (2-x+2 \left (1-\frac {e^9}{2}\right ) \log (x)\right )} \, dx\\ &=-x-\log \left (2-x+\left (2-e^9\right ) \log (x)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.23, size = 23, normalized size = 0.82 \begin {gather*} -x-\log \left (2-x+2 \log (x)-e^9 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-x - Log[2 - x + 2*Log[x] - E^9*Log[x]]

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fricas [A]  time = 1.10, size = 17, normalized size = 0.61 \begin {gather*} -x - \log \left ({\left (e^{9} - 2\right )} \log \relax (x) + x - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x - log((e^9 - 2)*log(x) + x - 2)

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giac [A]  time = 0.17, size = 19, normalized size = 0.68 \begin {gather*} -x - \log \left (e^{9} \log \relax (x) + x - 2 \, \log \relax (x) - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x - log(e^9*log(x) + x - 2*log(x) - 2)

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maple [A]  time = 0.03, size = 20, normalized size = 0.71

method	result	size
norman	\(-x -\ln \left (\ln \relax (x ) {\mathrm e}^{9}-2 \ln \relax (x )+x -2\right )\)	\(20\)
risch	\(-x -\ln \left (\ln \relax (x )+\frac {x -2}{{\mathrm e}^{9}-2}\right )\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x-ln(ln(x)*exp(9)-2*ln(x)+x-2)

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maxima [A]  time = 0.38, size = 24, normalized size = 0.86 \begin {gather*} -x - \log \left (\frac {{\left (e^{9} - 2\right )} \log \relax (x) + x - 2}{e^{9} - 2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x - log(((e^9 - 2)*log(x) + x - 2)/(e^9 - 2))

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mupad [B]  time = 4.52, size = 19, normalized size = 0.68 \begin {gather*} -x-\ln \left (x-2\,\ln \relax (x)+{\mathrm {e}}^9\,\ln \relax (x)-2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x - exp(9) + log(x)*(2*x - x*exp(9)) - x^2 + 2)/(2*x + log(x)*(2*x - x*exp(9)) - x^2),x)

[Out]

- x - log(x - 2*log(x) + exp(9)*log(x) - 2)

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sympy [A]  time = 0.22, size = 15, normalized size = 0.54 \begin {gather*} - x - \log {\left (\frac {x - 2}{-2 + e^{9}} + \log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(9)+2*x)*ln(x)-exp(9)-x**2+x+2)/((x*exp(9)-2*x)*ln(x)+x**2-2*x),x)

[Out]

-x - log((x - 2)/(-2 + exp(9)) + log(x))

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