3.100.25 \(\int \frac {6+e^3-x-2 \log (6+e^3-x)}{6 x+e^3 x-x^2+(6+e^3-x) \log ^2(6+e^3-x)} \, dx\)

Optimal. Leaf size=18 \[ \log \left (x+\log ^2\left (e^3+2 (3-x)+x\right )\right ) \]

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Rubi [A]  time = 0.16, antiderivative size = 14, normalized size of antiderivative = 0.78, number of steps used = 3, number of rules used = 3, integrand size = 56, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {6, 6741, 6684} \begin {gather*} \log \left (x+\log ^2\left (-x+e^3+6\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(6 + E^3 - x - 2*Log[6 + E^3 - x])/(6*x + E^3*x - x^2 + (6 + E^3 - x)*Log[6 + E^3 - x]^2),x]

[Out]

Log[x + Log[6 + E^3 - x]^2]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {6+e^3-x-2 \log \left (6+e^3-x\right )}{\left (6+e^3\right ) x-x^2+\left (6+e^3-x\right ) \log ^2\left (6+e^3-x\right )} \, dx\\ &=\int \frac {6 \left (1+\frac {e^3}{6}\right )-x-2 \log \left (6+e^3-x\right )}{\left (6+e^3-x\right ) \left (x+\log ^2\left (6+e^3-x\right )\right )} \, dx\\ &=\log \left (x+\log ^2\left (6+e^3-x\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.37, size = 14, normalized size = 0.78 \begin {gather*} \log \left (x+\log ^2\left (6+e^3-x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6 + E^3 - x - 2*Log[6 + E^3 - x])/(6*x + E^3*x - x^2 + (6 + E^3 - x)*Log[6 + E^3 - x]^2),x]

[Out]

Log[x + Log[6 + E^3 - x]^2]

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fricas [A]  time = 1.05, size = 13, normalized size = 0.72 \begin {gather*} \log \left (\log \left (-x + e^{3} + 6\right )^{2} + x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(exp(3)-x+6)+exp(3)-x+6)/((exp(3)-x+6)*log(exp(3)-x+6)^2+x*exp(3)-x^2+6*x),x, algorithm="fric
as")

[Out]

log(log(-x + e^3 + 6)^2 + x)

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giac [A]  time = 0.26, size = 13, normalized size = 0.72 \begin {gather*} \log \left (\log \left (-x + e^{3} + 6\right )^{2} + x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(exp(3)-x+6)+exp(3)-x+6)/((exp(3)-x+6)*log(exp(3)-x+6)^2+x*exp(3)-x^2+6*x),x, algorithm="giac
")

[Out]

log(log(-x + e^3 + 6)^2 + x)

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maple [A]  time = 0.08, size = 14, normalized size = 0.78

method	result	size
norman	\(\ln \left (\ln \left ({\mathrm e}^{3}-x +6\right )^{2}+x \right )\)	\(14\)
risch	\(\ln \left (\ln \left ({\mathrm e}^{3}-x +6\right )^{2}+x \right )\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*ln(exp(3)-x+6)+exp(3)-x+6)/((exp(3)-x+6)*ln(exp(3)-x+6)^2+x*exp(3)-x^2+6*x),x,method=_RETURNVERBOSE)

[Out]

ln(ln(exp(3)-x+6)^2+x)

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maxima [A]  time = 0.38, size = 13, normalized size = 0.72 \begin {gather*} \log \left (\log \left (-x + e^{3} + 6\right )^{2} + x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(exp(3)-x+6)+exp(3)-x+6)/((exp(3)-x+6)*log(exp(3)-x+6)^2+x*exp(3)-x^2+6*x),x, algorithm="maxi
ma")

[Out]

log(log(-x + e^3 + 6)^2 + x)

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mupad [B]  time = 7.18, size = 13, normalized size = 0.72 \begin {gather*} \ln \left ({\ln \left ({\mathrm {e}}^3-x+6\right )}^2+x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x + 2*log(exp(3) - x + 6) - exp(3) - 6)/(6*x + x*exp(3) + log(exp(3) - x + 6)^2*(exp(3) - x + 6) - x^2),
x)

[Out]

log(x + log(exp(3) - x + 6)^2)

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sympy [A]  time = 0.30, size = 12, normalized size = 0.67 \begin {gather*} \log {\left (x + \log {\left (- x + 6 + e^{3} \right )}^{2} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*ln(exp(3)-x+6)+exp(3)-x+6)/((exp(3)-x+6)*ln(exp(3)-x+6)**2+x*exp(3)-x**2+6*x),x)

[Out]

log(x + log(-x + 6 + exp(3))**2)

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