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

Optimal. Leaf size=24 \[ 3 \log \left (\frac {2}{3} \left (-4+e^{-5+x}-\frac {5 x}{4}-x^2\right )\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 28, normalized size of antiderivative = 1.17, number of steps used = 1, number of rules used = 1, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {6684} \begin {gather*} 3 \log \left (4 e^5 x^2+5 e^5 x-4 e^x+16 e^5\right ) \end {gather*}

Antiderivative was successfully verified.

[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 6684

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

Rubi steps

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

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Mathematica [A]  time = 0.19, size = 20, normalized size = 0.83 \begin {gather*} 3 \log \left (-16+4 e^{-5+x}-5 x-4 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[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]

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fricas [A]  time = 1.01, size = 19, normalized size = 0.79 \begin {gather*} 3 \, \log \left (-4 \, x^{2} - 5 \, x + 4 \, e^{\left (x - 5\right )} - 16\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.34, size = 24, normalized size = 1.00 \begin {gather*} 3 \, \log \left (-4 \, x^{2} e^{5} - 5 \, x e^{5} - 16 \, e^{5} + 4 \, e^{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*exp(x-5)-24*x-15)/(4*exp(x-5)-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)

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maple [A]  time = 0.03, size = 20, normalized size = 0.83

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*ln(4*exp(x-5)-4*x^2-5*x-16)

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maxima [A]  time = 0.35, size = 19, normalized size = 0.79 \begin {gather*} 3 \, \log \left (4 \, x^{2} + 5 \, x - 4 \, e^{\left (x - 5\right )} + 16\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*log(4*x^2 + 5*x - 4*e^(x - 5) + 16)

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mupad [B]  time = 0.98, size = 17, normalized size = 0.71 \begin {gather*} 3\,\ln \left (\frac {5\,x}{4}-{\mathrm {e}}^{x-5}+x^2+4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.14, size = 17, normalized size = 0.71 \begin {gather*} 3 \log {\left (- x^{2} - \frac {5 x}{4} + e^{x - 5} - 4 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*log(-x**2 - 5*x/4 + exp(x - 5) - 4)

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