\(\int \frac {-15+12 e^{-5+x}-24 x}{-16+4 e^{-5+x}-5 x-4 x^2} \, dx\) [1346]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 32, antiderivative size = 24 \[ \int \frac {-15+12 e^{-5+x}-24 x}{-16+4 e^{-5+x}-5 x-4 x^2} \, dx=3 \log \left (\frac {2}{3} \left (-4+e^{-5+x}-\frac {5 x}{4}-x^2\right )\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {6816} \[ \int \frac {-15+12 e^{-5+x}-24 x}{-16+4 e^{-5+x}-5 x-4 x^2} \, dx=3 \log \left (4 e^5 x^2+5 e^5 x-4 e^x+16 e^5\right ) \]

[In]

Int[(-15 + 12*E^(-5 + x) - 24*x)/(-16 + 4*E^(-5 + x) - 5*x - 4*x^2),x]

[Out]

3*Log[16*E^5 - 4*E^x + 5*E^5*x + 4*E^5*x^2]

Rule 6816

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

Rubi steps \begin{align*} \text {integral}& = 3 \log \left (16 e^5-4 e^x+5 e^5 x+4 e^5 x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {-15+12 e^{-5+x}-24 x}{-16+4 e^{-5+x}-5 x-4 x^2} \, dx=3 \log \left (-16+4 e^{-5+x}-5 x-4 x^2\right ) \]

[In]

Integrate[(-15 + 12*E^(-5 + x) - 24*x)/(-16 + 4*E^(-5 + x) - 5*x - 4*x^2),x]

[Out]

3*Log[-16 + 4*E^(-5 + x) - 5*x - 4*x^2]

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75

method result size
parallelrisch \(3 \ln \left (x^{2}+\frac {5 x}{4}-{\mathrm e}^{-5+x}+4\right )\) \(18\)
derivativedivides \(3 \ln \left (4 \,{\mathrm e}^{-5+x}-4 x^{2}-5 x -16\right )\) \(20\)
default \(3 \ln \left (4 \,{\mathrm e}^{-5+x}-4 x^{2}-5 x -16\right )\) \(20\)
norman \(3 \ln \left (4 x^{2}+5 x -4 \,{\mathrm e}^{-5+x}+16\right )\) \(20\)
risch \(15+3 \ln \left (-x^{2}-\frac {5 x}{4}+{\mathrm e}^{-5+x}-4\right )\) \(20\)

[In]

int((12*exp(-5+x)-24*x-15)/(4*exp(-5+x)-4*x^2-5*x-16),x,method=_RETURNVERBOSE)

[Out]

3*ln(x^2+5/4*x-exp(-5+x)+4)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {-15+12 e^{-5+x}-24 x}{-16+4 e^{-5+x}-5 x-4 x^2} \, dx=3 \, \log \left (-4 \, x^{2} - 5 \, x + 4 \, e^{\left (x - 5\right )} - 16\right ) \]

[In]

integrate((12*exp(-5+x)-24*x-15)/(4*exp(-5+x)-4*x^2-5*x-16),x, algorithm="fricas")

[Out]

3*log(-4*x^2 - 5*x + 4*e^(x - 5) - 16)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {-15+12 e^{-5+x}-24 x}{-16+4 e^{-5+x}-5 x-4 x^2} \, dx=3 \log {\left (- x^{2} - \frac {5 x}{4} + e^{x - 5} - 4 \right )} \]

[In]

integrate((12*exp(-5+x)-24*x-15)/(4*exp(-5+x)-4*x**2-5*x-16),x)

[Out]

3*log(-x**2 - 5*x/4 + exp(x - 5) - 4)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {-15+12 e^{-5+x}-24 x}{-16+4 e^{-5+x}-5 x-4 x^2} \, dx=3 \, \log \left (4 \, x^{2} + 5 \, x - 4 \, e^{\left (x - 5\right )} + 16\right ) \]

[In]

integrate((12*exp(-5+x)-24*x-15)/(4*exp(-5+x)-4*x^2-5*x-16),x, algorithm="maxima")

[Out]

3*log(4*x^2 + 5*x - 4*e^(x - 5) + 16)

Giac [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {-15+12 e^{-5+x}-24 x}{-16+4 e^{-5+x}-5 x-4 x^2} \, dx=3 \, \log \left (-4 \, x^{2} e^{5} - 5 \, x e^{5} - 16 \, e^{5} + 4 \, e^{x}\right ) \]

[In]

integrate((12*exp(-5+x)-24*x-15)/(4*exp(-5+x)-4*x^2-5*x-16),x, algorithm="giac")

[Out]

3*log(-4*x^2*e^5 - 5*x*e^5 - 16*e^5 + 4*e^x)

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {-15+12 e^{-5+x}-24 x}{-16+4 e^{-5+x}-5 x-4 x^2} \, dx=3\,\ln \left (\frac {5\,x}{4}-{\mathrm {e}}^{x-5}+x^2+4\right ) \]

[In]

int((24*x - 12*exp(x - 5) + 15)/(5*x - 4*exp(x - 5) + 4*x^2 + 16),x)

[Out]

3*log((5*x)/4 - exp(x - 5) + x^2 + 4)