3.32.9 \(\int \frac {-1-3 x+2 x^2}{16 e^{3+x}-x-2 x^2} \, dx\)

Optimal. Leaf size=20 \[ \log \left (8-\frac {1}{2} e^{-3-x} x (1+2 x)\right ) \]

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Rubi [F]  time = 0.26, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-1-3 x+2 x^2}{16 e^{3+x}-x-2 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-1 - 3*x + 2*x^2)/(16*E^(3 + x) - x - 2*x^2),x]

[Out]

-Defer[Int][(16*E^(3 + x) - x - 2*x^2)^(-1), x] + 3*Defer[Int][x/(-16*E^(3 + x) + x + 2*x^2), x] - 2*Defer[Int
][x^2/(-16*E^(3 + x) + x + 2*x^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {1}{16 e^{3+x}-x-2 x^2}+\frac {3 x}{-16 e^{3+x}+x+2 x^2}-\frac {2 x^2}{-16 e^{3+x}+x+2 x^2}\right ) \, dx\\ &=-\left (2 \int \frac {x^2}{-16 e^{3+x}+x+2 x^2} \, dx\right )+3 \int \frac {x}{-16 e^{3+x}+x+2 x^2} \, dx-\int \frac {1}{16 e^{3+x}-x-2 x^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 21, normalized size = 1.05 \begin {gather*} -x+\log \left (16 e^{3+x}-x-2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 - 3*x + 2*x^2)/(16*E^(3 + x) - x - 2*x^2),x]

[Out]

-x + Log[16*E^(3 + x) - x - 2*x^2]

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fricas [A]  time = 0.50, size = 20, normalized size = 1.00 \begin {gather*} -x + \log \left (-2 \, x^{2} - x + 16 \, e^{\left (x + 3\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2-3*x-1)/(16*exp(x)*exp(3)-2*x^2-x),x, algorithm="fricas")

[Out]

-x + log(-2*x^2 - x + 16*e^(x + 3))

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giac [A]  time = 0.19, size = 18, normalized size = 0.90 \begin {gather*} -x + \log \left (2 \, x^{2} + x - 16 \, e^{\left (x + 3\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2-3*x-1)/(16*exp(x)*exp(3)-2*x^2-x),x, algorithm="giac")

[Out]

-x + log(2*x^2 + x - 16*e^(x + 3))

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maple [A]  time = 0.06, size = 19, normalized size = 0.95

method	result	size
risch	\(-x +\ln \left ({\mathrm e}^{x}-\frac {\left (2 x +1\right ) x \,{\mathrm e}^{-3}}{16}\right )\)	\(19\)
norman	\(-x +\ln \left (16 \,{\mathrm e}^{x} {\mathrm e}^{3}-2 x^{2}-x \right )\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2-3*x-1)/(16*exp(x)*exp(3)-2*x^2-x),x,method=_RETURNVERBOSE)

[Out]

-x+ln(exp(x)-1/16*(2*x+1)*x*exp(-3))

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maxima [A]  time = 0.50, size = 22, normalized size = 1.10 \begin {gather*} -x + \log \left (-\frac {1}{16} \, {\left (2 \, x^{2} + x - 16 \, e^{\left (x + 3\right )}\right )} e^{\left (-3\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2-3*x-1)/(16*exp(x)*exp(3)-2*x^2-x),x, algorithm="maxima")

[Out]

-x + log(-1/16*(2*x^2 + x - 16*e^(x + 3))*e^(-3))

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mupad [B]  time = 0.09, size = 18, normalized size = 0.90 \begin {gather*} \ln \left (x-16\,{\mathrm {e}}^{x+3}+2\,x^2\right )-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x - 2*x^2 + 1)/(x - 16*exp(3)*exp(x) + 2*x^2),x)

[Out]

log(x - 16*exp(x + 3) + 2*x^2) - x

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sympy [A]  time = 0.15, size = 19, normalized size = 0.95 \begin {gather*} - x + \log {\left (\frac {- 2 x^{2} - x}{16 e^{3}} + e^{x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**2-3*x-1)/(16*exp(x)*exp(3)-2*x**2-x),x)

[Out]

-x + log((-2*x**2 - x)*exp(-3)/16 + exp(x))

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