3.98.38 \(\int \frac {-5-22 x-4 x^2+2 e^{-2-x} (16+3 x)}{-5 x-x^2+2 e^{-2-x} (5+x)} \, dx\)

Optimal. Leaf size=28 \[ 1+2 x+\log \left (e^{2 x} \left (2 e^{-2-x}-x\right ) (5+x)\right ) \]

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Rubi [F]  time = 1.12, 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 {-5-22 x-4 x^2+2 e^{-2-x} (16+3 x)}{-5 x-x^2+2 e^{-2-x} (5+x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-5 - 22*x - 4*x^2 + 2*E^(-2 - x)*(16 + 3*x))/(-5*x - x^2 + 2*E^(-2 - x)*(5 + x)),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{2+x} \left (-5-22 x-4 x^2+2 e^{-2-x} (16+3 x)\right )}{(5+x) \left (2-e^{2+x} x\right )} \, dx\\ &=\int \frac {32+6 x-e^{2+x} \left (5+22 x+4 x^2\right )}{(5+x) \left (2-e^{2+x} x\right )} \, dx\\ &=\int \left (\frac {2 (1+x)}{x \left (-2+e^{2+x} x\right )}+\frac {5+22 x+4 x^2}{x (5+x)}\right ) \, dx\\ &=2 \int \frac {1+x}{x \left (-2+e^{2+x} x\right )} \, dx+\int \frac {5+22 x+4 x^2}{x (5+x)} \, dx\\ &=2 \int \left (\frac {1}{-2+e^{2+x} x}+\frac {1}{x \left (-2+e^{2+x} x\right )}\right ) \, dx+\int \left (4+\frac {1}{x}+\frac {1}{5+x}\right ) \, dx\\ &=4 x+\log (x)+\log (5+x)+2 \int \frac {1}{-2+e^{2+x} x} \, dx+2 \int \frac {1}{x \left (-2+e^{2+x} x\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 19, normalized size = 0.68 \begin {gather*} 3 x+\log (5+x)+\log \left (2-e^{2+x} x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5 - 22*x - 4*x^2 + 2*E^(-2 - x)*(16 + 3*x))/(-5*x - x^2 + 2*E^(-2 - x)*(5 + x)),x]

[Out]

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

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fricas [A]  time = 0.86, size = 21, normalized size = 0.75 \begin {gather*} 4 \, x + \log \left (x + 5\right ) + \log \left (-x + e^{\left (-x + \log \relax (2) - 2\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x+16)*exp(log(2)-x-2)-4*x^2-22*x-5)/((5+x)*exp(log(2)-x-2)-x^2-5*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.23, size = 21, normalized size = 0.75 \begin {gather*} 4 \, x + \log \left (-x e^{2} + 2 \, e^{\left (-x\right )}\right ) + \log \left (x + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x+16)*exp(log(2)-x-2)-4*x^2-22*x-5)/((5+x)*exp(log(2)-x-2)-x^2-5*x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.15, size = 22, normalized size = 0.79

method	result	size
norman	\(4 x +\ln \left (5+x \right )+\ln \left (x -{\mathrm e}^{\ln \relax (2)-x -2}\right )\)	\(22\)
risch	\(4 x +\ln \left (5+x \right )-\ln \relax (2)+2+\ln \left (2 \,{\mathrm e}^{-x -2}-x \right )\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*x+16)*exp(ln(2)-x-2)-4*x^2-22*x-5)/((5+x)*exp(ln(2)-x-2)-x^2-5*x),x,method=_RETURNVERBOSE)

[Out]

4*x+ln(5+x)+ln(x-exp(ln(2)-x-2))

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maxima [A]  time = 0.46, size = 25, normalized size = 0.89 \begin {gather*} 3 \, x + \log \left (x + 5\right ) + \log \relax (x) + \log \left (\frac {{\left (x e^{\left (x + 2\right )} - 2\right )} e^{\left (-2\right )}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x+16)*exp(log(2)-x-2)-4*x^2-22*x-5)/((5+x)*exp(log(2)-x-2)-x^2-5*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.96, size = 19, normalized size = 0.68 \begin {gather*} 4\,x+\ln \left (x-2\,{\mathrm {e}}^{-x-2}\right )+\ln \left (x+5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((22*x - exp(log(2) - x - 2)*(3*x + 16) + 4*x^2 + 5)/(5*x - exp(log(2) - x - 2)*(x + 5) + x^2),x)

[Out]

4*x + log(x - 2*exp(- x - 2)) + log(x + 5)

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sympy [A]  time = 0.17, size = 19, normalized size = 0.68 \begin {gather*} 4 x + \log {\left (- \frac {x}{2} + e^{- x - 2} \right )} + \log {\left (x + 5 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x+16)*exp(ln(2)-x-2)-4*x**2-22*x-5)/((5+x)*exp(ln(2)-x-2)-x**2-5*x),x)

[Out]

4*x + log(-x/2 + exp(-x - 2)) + log(x + 5)

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