3.99.35 \(\int \frac {14-4 x}{5+e^{-3+x}-2 x} \, dx\)

Optimal. Leaf size=21 \[ 2 \left (-2+e^3+x-\log \left (5+e^{-3+x}-2 x\right )\right ) \]

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Rubi [F]  time = 0.18, 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 {14-4 x}{5+e^{-3+x}-2 x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(14 - 4*x)/(5 + E^(-3 + x) - 2*x),x]

[Out]

-14*E^3*Defer[Int][(-5*E^3 - E^x + 2*E^3*x)^(-1), x] + 4*E^3*Defer[Int][x/(-5*E^3 - E^x + 2*E^3*x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {14 e^3}{-5 e^3-e^x+2 e^3 x}+\frac {4 e^3 x}{-5 e^3-e^x+2 e^3 x}\right ) \, dx\\ &=\left (4 e^3\right ) \int \frac {x}{-5 e^3-e^x+2 e^3 x} \, dx-\left (14 e^3\right ) \int \frac {1}{-5 e^3-e^x+2 e^3 x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 22, normalized size = 1.05 \begin {gather*} 2 x-2 \log \left (5 e^3+e^x-2 e^3 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(14 - 4*x)/(5 + E^(-3 + x) - 2*x),x]

[Out]

2*x - 2*Log[5*E^3 + E^x - 2*E^3*x]

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fricas [A]  time = 0.68, size = 16, normalized size = 0.76 \begin {gather*} 2 \, x - 2 \, \log \left (-2 \, x + e^{\left (x - 3\right )} + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x+14)/(exp(x-3)-2*x+5),x, algorithm="fricas")

[Out]

2*x - 2*log(-2*x + e^(x - 3) + 5)

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giac [A]  time = 0.20, size = 21, normalized size = 1.00 \begin {gather*} 2 \, x - 2 \, \log \left (2 \, x e^{3} - 5 \, e^{3} - e^{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x+14)/(exp(x-3)-2*x+5),x, algorithm="giac")

[Out]

2*x - 2*log(2*x*e^3 - 5*e^3 - e^x)

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maple [A]  time = 0.05, size = 18, normalized size = 0.86

method	result	size
risch	\(-6+2 x -2 \ln \left ({\mathrm e}^{x -3}-2 x +5\right )\)	\(18\)
norman	\(2 x -2 \ln \left (-{\mathrm e}^{x -3}+2 x -5\right )\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*x+14)/(exp(x-3)-2*x+5),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.48, size = 19, normalized size = 0.90 \begin {gather*} 2 \, x - 2 \, \log \left (-2 \, x e^{3} + 5 \, e^{3} + e^{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x+14)/(exp(x-3)-2*x+5),x, algorithm="maxima")

[Out]

2*x - 2*log(-2*x*e^3 + 5*e^3 + e^x)

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mupad [B]  time = 5.74, size = 18, normalized size = 0.86 \begin {gather*} 2\,x-2\,\ln \left (2\,x-{\mathrm {e}}^{x-3}-5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*x - 14)/(exp(x - 3) - 2*x + 5),x)

[Out]

2*x - 2*log(2*x - exp(x - 3) - 5)

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sympy [A]  time = 0.10, size = 15, normalized size = 0.71 \begin {gather*} 2 x - 2 \log {\left (- 2 x + e^{x - 3} + 5 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x+14)/(exp(x-3)-2*x+5),x)

[Out]

2*x - 2*log(-2*x + exp(x - 3) + 5)

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