3.45.86 \(\int \frac {6+13 x+6 x^2+x^3+e (2+x)+(-2 x^2-x^3) \log (\frac {e^{e^4}}{x})}{2 x^2+x^3} \, dx\)

Optimal. Leaf size=32 \[ \frac {-3-e+x}{x}-(5+x) \log \left (\frac {e^{e^4}}{x}\right )-\log (2+x) \]

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Rubi [A]  time = 0.30, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1593, 6742, 1620, 2295} \begin {gather*} -\frac {3+e}{x}-x \log \left (\frac {e^{e^4}}{x}\right )+5 \log (x)-\log (x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(6 + 13*x + 6*x^2 + x^3 + E*(2 + x) + (-2*x^2 - x^3)*Log[E^E^4/x])/(2*x^2 + x^3),x]

[Out]

-((3 + E)/x) - x*Log[E^E^4/x] + 5*Log[x] - Log[2 + 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 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {6+13 x+6 x^2+x^3+e (2+x)+\left (-2 x^2-x^3\right ) \log \left (\frac {e^{e^4}}{x}\right )}{x^2 (2+x)} \, dx\\ &=\int \left (\frac {2 (3+e)+(13+e) x+6 x^2+x^3}{x^2 (2+x)}-\log \left (\frac {e^{e^4}}{x}\right )\right ) \, dx\\ &=\int \frac {2 (3+e)+(13+e) x+6 x^2+x^3}{x^2 (2+x)} \, dx-\int \log \left (\frac {e^{e^4}}{x}\right ) \, dx\\ &=-x-x \log \left (\frac {e^{e^4}}{x}\right )+\int \left (1+\frac {1}{-2-x}+\frac {3+e}{x^2}+\frac {5}{x}\right ) \, dx\\ &=-\frac {3+e}{x}-x \log \left (\frac {e^{e^4}}{x}\right )+5 \log (x)-\log (2+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 35, normalized size = 1.09 \begin {gather*} -\frac {3}{x}-\frac {e}{x}-e^4 x-x \log \left (\frac {1}{x}\right )+5 \log (x)-\log (2+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6 + 13*x + 6*x^2 + x^3 + E*(2 + x) + (-2*x^2 - x^3)*Log[E^E^4/x])/(2*x^2 + x^3),x]

[Out]

-3/x - E/x - E^4*x - x*Log[x^(-1)] + 5*Log[x] - Log[2 + x]

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fricas [A]  time = 0.57, size = 32, normalized size = 1.00 \begin {gather*} -\frac {x^{2} \log \left (\frac {e^{\left (e^{4}\right )}}{x}\right ) + x \log \left (x + 2\right ) - 5 \, x \log \relax (x) + e + 3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3-2*x^2)*log(exp(exp(4))/x)+(2+x)*exp(1)+x^3+6*x^2+13*x+6)/(x^3+2*x^2),x, algorithm="fricas")

[Out]

-(x^2*log(e^(e^4)/x) + x*log(x + 2) - 5*x*log(x) + e + 3)/x

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giac [B]  time = 0.15, size = 140, normalized size = 4.38 \begin {gather*} -\frac {1}{2} \, {\left (2 \, e^{\left (2 \, e^{4}\right )} \log \left (\frac {e^{\left (e^{4}\right )}}{x}\right ) + \frac {e^{\left (2 \, e^{4} + 1\right )} \log \left (\frac {2 \, e^{\left (e^{4} + 1\right )}}{x} + e^{\left (e^{4} + 1\right )}\right )}{x} + \frac {2 \, e^{\left (2 \, e^{4}\right )} \log \left (\frac {2 \, e^{\left (e^{4}\right )}}{x} + e^{\left (e^{4}\right )}\right )}{x} - \frac {e^{\left (2 \, e^{4} + 1\right )} \log \left (\frac {2 \, e^{\left (e^{4}\right )}}{x} + e^{\left (e^{4}\right )}\right )}{x} + \frac {8 \, e^{\left (2 \, e^{4}\right )} \log \left (\frac {e^{\left (e^{4}\right )}}{x}\right )}{x} + \frac {6 \, e^{\left (2 \, e^{4}\right )}}{x^{2}} + \frac {2 \, e^{\left (2 \, e^{4} + 1\right )}}{x^{2}}\right )} x e^{\left (-2 \, e^{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3-2*x^2)*log(exp(exp(4))/x)+(2+x)*exp(1)+x^3+6*x^2+13*x+6)/(x^3+2*x^2),x, algorithm="giac")

[Out]

-1/2*(2*e^(2*e^4)*log(e^(e^4)/x) + e^(2*e^4 + 1)*log(2*e^(e^4 + 1)/x + e^(e^4 + 1))/x + 2*e^(2*e^4)*log(2*e^(e
^4)/x + e^(e^4))/x - e^(2*e^4 + 1)*log(2*e^(e^4)/x + e^(e^4))/x + 8*e^(2*e^4)*log(e^(e^4)/x)/x + 6*e^(2*e^4)/x
^2 + 2*e^(2*e^4 + 1)/x^2)*x*e^(-2*e^4)

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maple [A]  time = 0.16, size = 35, normalized size = 1.09

method	result	size
risch	\(-x \ln \left (\frac {{\mathrm e}^{{\mathrm e}^{4}}}{x}\right )+\frac {5 x \ln \relax (x )-x \ln \left (2+x \right )-{\mathrm e}-3}{x}\)	\(35\)
norman	\(\frac {-5 x \ln \left (\frac {{\mathrm e}^{{\mathrm e}^{4}}}{x}\right )-x^{2} \ln \left (\frac {{\mathrm e}^{{\mathrm e}^{4}}}{x}\right )-3-{\mathrm e}}{x}-\ln \left (2+x \right )\)	\(42\)
derivativedivides	\(-{\mathrm e}^{{\mathrm e}^{4}} \left (\frac {3 \,{\mathrm e}^{-{\mathrm e}^{4}}}{x}+\frac {{\mathrm e}^{-{\mathrm e}^{4}} {\mathrm e}}{x}+4 \left ({\mathrm e}^{-2 \,{\mathrm e}^{4}}\right )^{2} {\mathrm e}^{3 \,{\mathrm e}^{4}} \ln \left (\frac {{\mathrm e}^{{\mathrm e}^{4}}}{x}\right )-{\mathrm e}^{-3 \,{\mathrm e}^{4}} {\mathrm e}^{4 \,{\mathrm e}^{4}} {\mathrm e}^{-2 \,{\mathrm e}^{4}} x +5 \,{\mathrm e}^{-{\mathrm e}^{4}} \ln \left ({\mathrm e}^{{\mathrm e}^{4}}+\frac {2 \,{\mathrm e}^{{\mathrm e}^{4}}}{x}\right )-6 \,{\mathrm e}^{-3 \,{\mathrm e}^{4}} \ln \left ({\mathrm e}^{{\mathrm e}^{4}}+\frac {2 \,{\mathrm e}^{{\mathrm e}^{4}}}{x}\right ) {\mathrm e}^{2 \,{\mathrm e}^{4}}+2 \,{\mathrm e}^{-4 \,{\mathrm e}^{4}} \ln \left ({\mathrm e}^{{\mathrm e}^{4}}+\frac {2 \,{\mathrm e}^{{\mathrm e}^{4}}}{x}\right ) {\mathrm e}^{3 \,{\mathrm e}^{4}}+x \,{\mathrm e}^{-{\mathrm e}^{4}} \ln \left (\frac {{\mathrm e}^{{\mathrm e}^{4}}}{x}\right )+x \,{\mathrm e}^{-{\mathrm e}^{4}}\right )\)	\(163\)
default	\(-{\mathrm e}^{{\mathrm e}^{4}} \left (\frac {3 \,{\mathrm e}^{-{\mathrm e}^{4}}}{x}+\frac {{\mathrm e}^{-{\mathrm e}^{4}} {\mathrm e}}{x}+4 \left ({\mathrm e}^{-2 \,{\mathrm e}^{4}}\right )^{2} {\mathrm e}^{3 \,{\mathrm e}^{4}} \ln \left (\frac {{\mathrm e}^{{\mathrm e}^{4}}}{x}\right )-{\mathrm e}^{-3 \,{\mathrm e}^{4}} {\mathrm e}^{4 \,{\mathrm e}^{4}} {\mathrm e}^{-2 \,{\mathrm e}^{4}} x +5 \,{\mathrm e}^{-{\mathrm e}^{4}} \ln \left ({\mathrm e}^{{\mathrm e}^{4}}+\frac {2 \,{\mathrm e}^{{\mathrm e}^{4}}}{x}\right )-6 \,{\mathrm e}^{-3 \,{\mathrm e}^{4}} \ln \left ({\mathrm e}^{{\mathrm e}^{4}}+\frac {2 \,{\mathrm e}^{{\mathrm e}^{4}}}{x}\right ) {\mathrm e}^{2 \,{\mathrm e}^{4}}+2 \,{\mathrm e}^{-4 \,{\mathrm e}^{4}} \ln \left ({\mathrm e}^{{\mathrm e}^{4}}+\frac {2 \,{\mathrm e}^{{\mathrm e}^{4}}}{x}\right ) {\mathrm e}^{3 \,{\mathrm e}^{4}}+x \,{\mathrm e}^{-{\mathrm e}^{4}} \ln \left (\frac {{\mathrm e}^{{\mathrm e}^{4}}}{x}\right )+x \,{\mathrm e}^{-{\mathrm e}^{4}}\right )\)	\(163\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^3-2*x^2)*ln(exp(exp(4))/x)+(2+x)*exp(1)+x^3+6*x^2+13*x+6)/(x^3+2*x^2),x,method=_RETURNVERBOSE)

[Out]

-x*ln(exp(exp(4))/x)+(5*x*ln(x)-x*ln(2+x)-exp(1)-3)/x

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maxima [A]  time = 0.40, size = 56, normalized size = 1.75 \begin {gather*} -x e^{4} - \frac {1}{2} \, {\left (\frac {2}{x} - \log \left (x + 2\right ) + \log \relax (x)\right )} e - \frac {1}{2} \, {\left (\log \left (x + 2\right ) - \log \relax (x)\right )} e + x \log \relax (x) - \frac {3}{x} - \log \left (x + 2\right ) + 5 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3-2*x^2)*log(exp(exp(4))/x)+(2+x)*exp(1)+x^3+6*x^2+13*x+6)/(x^3+2*x^2),x, algorithm="maxima")

[Out]

-x*e^4 - 1/2*(2/x - log(x + 2) + log(x))*e - 1/2*(log(x + 2) - log(x))*e + x*log(x) - 3/x - log(x + 2) + 5*log
(x)

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mupad [B]  time = 3.09, size = 32, normalized size = 1.00 \begin {gather*} -\ln \left (x+2\right )-5\,\ln \left (\frac {1}{x}\right )-x\,\left (\ln \left (\frac {1}{x}\right )+{\mathrm {e}}^4\right )-\frac {\mathrm {e}+3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((13*x - log(exp(exp(4))/x)*(2*x^2 + x^3) + exp(1)*(x + 2) + 6*x^2 + x^3 + 6)/(2*x^2 + x^3),x)

[Out]

- log(x + 2) - 5*log(1/x) - x*(log(1/x) + exp(4)) - (exp(1) + 3)/x

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sympy [A]  time = 0.27, size = 27, normalized size = 0.84 \begin {gather*} - x \log {\left (\frac {e^{e^{4}}}{x} \right )} + 5 \log {\relax (x )} - \log {\left (x + 2 \right )} + \frac {-3 - e}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**3-2*x**2)*ln(exp(exp(4))/x)+(2+x)*exp(1)+x**3+6*x**2+13*x+6)/(x**3+2*x**2),x)

[Out]

-x*log(exp(exp(4))/x) + 5*log(x) - log(x + 2) + (-3 - E)/x

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