3.68.90 \(\int \frac {8+e^3-4 x}{12+e^3 (-5+x)+8 x-2 x^2} \, dx\)

Optimal. Leaf size=18 \[ \log \left (2-(-5+x) \left (2-e^3+2 x\right )\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 21, normalized size of antiderivative = 1.17, number of steps used = 1, number of rules used = 1, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {1587} \begin {gather*} \log \left (-2 x^2+8 x-e^3 (5-x)+12\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(8 + E^3 - 4*x)/(12 + E^3*(-5 + x) + 8*x - 2*x^2),x]

[Out]

Log[12 - E^3*(5 - x) + 8*x - 2*x^2]

Rule 1587

Int[(Pp_)/(Qq_), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[(Coeff[Pp, x, p]*Log[RemoveConte
nt[Qq, x]])/(q*Coeff[Qq, x, q]), x] /; EqQ[p, q - 1] && EqQ[Pp, Simplify[(Coeff[Pp, x, p]*D[Qq, x])/(q*Coeff[Q
q, x, q])]]] /; PolyQ[Pp, x] && PolyQ[Qq, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log \left (12-e^3 (5-x)+8 x-2 x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 1.00 \begin {gather*} \log \left (12+e^3 (-5+x)+8 x-2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8 + E^3 - 4*x)/(12 + E^3*(-5 + x) + 8*x - 2*x^2),x]

[Out]

Log[12 + E^3*(-5 + x) + 8*x - 2*x^2]

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fricas [A]  time = 0.46, size = 18, normalized size = 1.00 \begin {gather*} \log \left (2 \, x^{2} - {\left (x - 5\right )} e^{3} - 8 \, x - 12\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(3)-4*x+8)/((x-5)*exp(3)-2*x^2+8*x+12),x, algorithm="fricas")

[Out]

log(2*x^2 - (x - 5)*e^3 - 8*x - 12)

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giac [A]  time = 0.19, size = 21, normalized size = 1.17 \begin {gather*} \log \left ({\left | 2 \, x^{2} - x e^{3} - 8 \, x + 5 \, e^{3} - 12 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(3)-4*x+8)/((x-5)*exp(3)-2*x^2+8*x+12),x, algorithm="giac")

[Out]

log(abs(2*x^2 - x*e^3 - 8*x + 5*e^3 - 12))

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maple [A]  time = 0.21, size = 18, normalized size = 1.00

method	result	size
derivativedivides	\(\ln \left (\left (x -5\right ) {\mathrm e}^{3}-2 x^{2}+8 x +12\right )\)	\(18\)
default	\(\ln \left (x \,{\mathrm e}^{3}-2 x^{2}-5 \,{\mathrm e}^{3}+8 x +12\right )\)	\(20\)
norman	\(\ln \left (x \,{\mathrm e}^{3}-2 x^{2}-5 \,{\mathrm e}^{3}+8 x +12\right )\)	\(20\)
risch	\(\ln \left (2 x^{2}+x \left (-8-{\mathrm e}^{3}\right )+5 \,{\mathrm e}^{3}-12\right )\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(3)-4*x+8)/((x-5)*exp(3)-2*x^2+8*x+12),x,method=_RETURNVERBOSE)

[Out]

ln((x-5)*exp(3)-2*x^2+8*x+12)

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maxima [A]  time = 0.36, size = 18, normalized size = 1.00 \begin {gather*} \log \left (2 \, x^{2} - {\left (x - 5\right )} e^{3} - 8 \, x - 12\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(3)-4*x+8)/((x-5)*exp(3)-2*x^2+8*x+12),x, algorithm="maxima")

[Out]

log(2*x^2 - (x - 5)*e^3 - 8*x - 12)

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mupad [B]  time = 0.15, size = 17, normalized size = 0.94 \begin {gather*} \ln \left (\frac {5\,{\mathrm {e}}^3}{2}-\frac {x\,\left ({\mathrm {e}}^3+8\right )}{2}+x^2-6\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(3) - 4*x + 8)/(8*x + exp(3)*(x - 5) - 2*x^2 + 12),x)

[Out]

log((5*exp(3))/2 - (x*(exp(3) + 8))/2 + x^2 - 6)

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sympy [A]  time = 0.21, size = 20, normalized size = 1.11 \begin {gather*} \log {\left (2 x^{2} + x \left (- e^{3} - 8\right ) - 12 + 5 e^{3} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(3)-4*x+8)/((x-5)*exp(3)-2*x**2+8*x+12),x)

[Out]

log(2*x**2 + x*(-exp(3) - 8) - 12 + 5*exp(3))

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