3.88.31 \(\int \frac {27+2 x+e^{5+x} (2+2 x)}{4+27 x+2 e^{5+x} x+x^2} \, dx\)

Optimal. Leaf size=18 \[ \log \left (4+2 x+x \left (25+2 e^{5+x}+x\right )\right ) \]

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Rubi [A]  time = 0.05, antiderivative size = 17, normalized size of antiderivative = 0.94, number of steps used = 1, number of rules used = 1, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {6684} \begin {gather*} \log \left (x^2+2 e^{x+5} x+27 x+4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(27 + 2*x + E^(5 + x)*(2 + 2*x))/(4 + 27*x + 2*E^(5 + x)*x + x^2),x]

[Out]

Log[4 + 27*x + 2*E^(5 + x)*x + 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} &=\log \left (4+27 x+2 e^{5+x} x+x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 17, normalized size = 0.94 \begin {gather*} \log \left (4+27 x+2 e^{5+x} x+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(27 + 2*x + E^(5 + x)*(2 + 2*x))/(4 + 27*x + 2*E^(5 + x)*x + x^2),x]

[Out]

Log[4 + 27*x + 2*E^(5 + x)*x + x^2]

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fricas [A]  time = 0.57, size = 23, normalized size = 1.28 \begin {gather*} \log \relax (x) + \log \left (\frac {x^{2} + 2 \, x e^{\left (x + 5\right )} + 27 \, x + 4}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+2)*exp(5)*exp(x)+2*x+27)/(2*x*exp(5)*exp(x)+x^2+27*x+4),x, algorithm="fricas")

[Out]

log(x) + log((x^2 + 2*x*e^(x + 5) + 27*x + 4)/x)

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giac [A]  time = 0.16, size = 16, normalized size = 0.89 \begin {gather*} \log \left (x^{2} + 2 \, x e^{\left (x + 5\right )} + 27 \, x + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+2)*exp(5)*exp(x)+2*x+27)/(2*x*exp(5)*exp(x)+x^2+27*x+4),x, algorithm="giac")

[Out]

log(x^2 + 2*x*e^(x + 5) + 27*x + 4)

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maple [A]  time = 0.07, size = 17, normalized size = 0.94

method	result	size
norman	\(\ln \left (2 x \,{\mathrm e}^{5} {\mathrm e}^{x}+x^{2}+27 x +4\right )\)	\(17\)
risch	\(\ln \relax (x )+\ln \left ({\mathrm e}^{x}+\frac {\left (x^{2}+27 x +4\right ) {\mathrm e}^{-5}}{2 x}\right )\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x+2)*exp(5)*exp(x)+2*x+27)/(2*x*exp(5)*exp(x)+x^2+27*x+4),x,method=_RETURNVERBOSE)

[Out]

ln(2*x*exp(5)*exp(x)+x^2+27*x+4)

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maxima [A]  time = 0.38, size = 26, normalized size = 1.44 \begin {gather*} \log \relax (x) + \log \left (\frac {{\left (x^{2} + 2 \, x e^{\left (x + 5\right )} + 27 \, x + 4\right )} e^{\left (-5\right )}}{2 \, x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+2)*exp(5)*exp(x)+2*x+27)/(2*x*exp(5)*exp(x)+x^2+27*x+4),x, algorithm="maxima")

[Out]

log(x) + log(1/2*(x^2 + 2*x*e^(x + 5) + 27*x + 4)*e^(-5)/x)

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mupad [B]  time = 5.37, size = 16, normalized size = 0.89 \begin {gather*} \ln \left (27\,x+2\,x\,{\mathrm {e}}^{x+5}+x^2+4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + exp(5)*exp(x)*(2*x + 2) + 27)/(27*x + x^2 + 2*x*exp(5)*exp(x) + 4),x)

[Out]

log(27*x + 2*x*exp(x + 5) + x^2 + 4)

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sympy [A]  time = 0.22, size = 22, normalized size = 1.22 \begin {gather*} \log {\relax (x )} + \log {\left (e^{x} + \frac {x^{2} + 27 x + 4}{2 x e^{5}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+2)*exp(5)*exp(x)+2*x+27)/(2*x*exp(5)*exp(x)+x**2+27*x+4),x)

[Out]

log(x) + log(exp(x) + (x**2 + 27*x + 4)*exp(-5)/(2*x))

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