3.85.91 \(\int \frac {-10+e^4 (-2-x)-9 x-x^2}{5 x+e^4 x+x^2} \, dx\)

Optimal. Leaf size=21 \[ -23-x-\log \left (2 x^2 \left (5+e^4+x\right )^2\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 17, normalized size of antiderivative = 0.81, number of steps used = 4, number of rules used = 3, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {6, 1593, 1820} \begin {gather*} -x-2 \log (x)-2 \log \left (x+e^4+5\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-10 + E^4*(-2 - x) - 9*x - x^2)/(5*x + E^4*x + x^2),x]

[Out]

-x - 2*Log[x] - 2*Log[5 + E^4 + x]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1820

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a +
b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-10+e^4 (-2-x)-9 x-x^2}{\left (5+e^4\right ) x+x^2} \, dx\\ &=\int \frac {-10+e^4 (-2-x)-9 x-x^2}{x \left (5+e^4+x\right )} \, dx\\ &=\int \left (-1-\frac {2}{x}-\frac {2}{5+e^4+x}\right ) \, dx\\ &=-x-2 \log (x)-2 \log \left (5+e^4+x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 17, normalized size = 0.81 \begin {gather*} -x-2 \log (x)-2 \log \left (5+e^4+x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10 + E^4*(-2 - x) - 9*x - x^2)/(5*x + E^4*x + x^2),x]

[Out]

-x - 2*Log[x] - 2*Log[5 + E^4 + x]

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fricas [A]  time = 0.54, size = 18, normalized size = 0.86 \begin {gather*} -x - 2 \, \log \left (x^{2} + x e^{4} + 5 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x-2)*exp(4)-x^2-9*x-10)/(x*exp(4)+x^2+5*x),x, algorithm="fricas")

[Out]

-x - 2*log(x^2 + x*e^4 + 5*x)

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giac [A]  time = 0.12, size = 18, normalized size = 0.86 \begin {gather*} -x - 2 \, \log \left ({\left | x + e^{4} + 5 \right |}\right ) - 2 \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x-2)*exp(4)-x^2-9*x-10)/(x*exp(4)+x^2+5*x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.39, size = 17, normalized size = 0.81

method	result	size
default	\(-x -2 \ln \relax (x )-2 \ln \left ({\mathrm e}^{4}+x +5\right )\)	\(17\)
norman	\(-x -2 \ln \relax (x )-2 \ln \left ({\mathrm e}^{4}+x +5\right )\)	\(17\)
risch	\(-x -2 \ln \left (x^{2}+\left (5+{\mathrm e}^{4}\right ) x \right )\)	\(18\)
meijerg	\(\left (-{\mathrm e}^{4}-9\right ) \ln \left (1+\frac {x}{5+{\mathrm e}^{4}}\right )-\left (5+{\mathrm e}^{4}\right ) \left (\frac {x}{5+{\mathrm e}^{4}}-\ln \left (1+\frac {x}{5+{\mathrm e}^{4}}\right )\right )-\frac {2 \,{\mathrm e}^{4} \left (\ln \relax (x )-\ln \left (5+{\mathrm e}^{4}\right )-\ln \left (1+\frac {x}{5+{\mathrm e}^{4}}\right )\right )}{5+{\mathrm e}^{4}}-\frac {10 \left (\ln \relax (x )-\ln \left (5+{\mathrm e}^{4}\right )-\ln \left (1+\frac {x}{5+{\mathrm e}^{4}}\right )\right )}{5+{\mathrm e}^{4}}\)	\(112\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x-2)*exp(4)-x^2-9*x-10)/(x*exp(4)+x^2+5*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.41, size = 16, normalized size = 0.76 \begin {gather*} -x - 2 \, \log \left (x + e^{4} + 5\right ) - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x-2)*exp(4)-x^2-9*x-10)/(x*exp(4)+x^2+5*x),x, algorithm="maxima")

[Out]

-x - 2*log(x + e^4 + 5) - 2*log(x)

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mupad [B]  time = 0.13, size = 17, normalized size = 0.81 \begin {gather*} -x-2\,\ln \left (x^2+\left ({\mathrm {e}}^4+5\right )\,x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(9*x + exp(4)*(x + 2) + x^2 + 10)/(5*x + x*exp(4) + x^2),x)

[Out]

- x - 2*log(x*(exp(4) + 5) + x^2)

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sympy [A]  time = 0.23, size = 15, normalized size = 0.71 \begin {gather*} - x - 2 \log {\left (x^{2} + x \left (5 + e^{4}\right ) \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x-2)*exp(4)-x**2-9*x-10)/(x*exp(4)+x**2+5*x),x)

[Out]

-x - 2*log(x**2 + x*(5 + exp(4)))

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