3.53.44 \(\int \frac {e^{e^4} (1+x+2 x^2)}{x} \, dx\)

Optimal. Leaf size=16 \[ e^{e^4} \left (3+x+x^2+\log (625)+\log (x)\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 25, normalized size of antiderivative = 1.56, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 14} \begin {gather*} e^{e^4} x^2+e^{e^4} x+e^{e^4} \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^E^4*(1 + x + 2*x^2))/x,x]

[Out]

E^E^4*x + E^E^4*x^2 + E^E^4*Log[x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 13, normalized size = 0.81 \begin {gather*} e^{e^4} \left (x+x^2+\log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^E^4*(1 + x + 2*x^2))/x,x]

[Out]

E^E^4*(x + x^2 + Log[x])

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fricas [A]  time = 0.63, size = 16, normalized size = 1.00 \begin {gather*} {\left (x^{2} + x\right )} e^{\left (e^{4}\right )} + e^{\left (e^{4}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+x+1)*exp(exp(4))/x,x, algorithm="fricas")

[Out]

(x^2 + x)*e^(e^4) + e^(e^4)*log(x)

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giac [A]  time = 0.14, size = 12, normalized size = 0.75 \begin {gather*} {\left (x^{2} + x + \log \left ({\left | x \right |}\right )\right )} e^{\left (e^{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+x+1)*exp(exp(4))/x,x, algorithm="giac")

[Out]

(x^2 + x + log(abs(x)))*e^(e^4)

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maple [A]  time = 0.04, size = 12, normalized size = 0.75

method	result	size
default	\({\mathrm e}^{{\mathrm e}^{4}} \left (\ln \relax (x )+x^{2}+x \right )\)	\(12\)
norman	\(x \,{\mathrm e}^{{\mathrm e}^{4}}+x^{2} {\mathrm e}^{{\mathrm e}^{4}}+{\mathrm e}^{{\mathrm e}^{4}} \ln \relax (x )\)	\(20\)
risch	\(x \,{\mathrm e}^{{\mathrm e}^{4}}+x^{2} {\mathrm e}^{{\mathrm e}^{4}}+{\mathrm e}^{{\mathrm e}^{4}} \ln \relax (x )\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2+x+1)*exp(exp(4))/x,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.35, size = 11, normalized size = 0.69 \begin {gather*} {\left (x^{2} + x + \log \relax (x)\right )} e^{\left (e^{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+x+1)*exp(exp(4))/x,x, algorithm="maxima")

[Out]

(x^2 + x + log(x))*e^(e^4)

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mupad [B]  time = 3.41, size = 11, normalized size = 0.69 \begin {gather*} {\mathrm {e}}^{{\mathrm {e}}^4}\,\left (x+\ln \relax (x)+x^2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(4))*(x + 2*x^2 + 1))/x,x)

[Out]

exp(exp(4))*(x + log(x) + x^2)

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sympy [A]  time = 0.10, size = 22, normalized size = 1.38 \begin {gather*} x^{2} e^{e^{4}} + x e^{e^{4}} + e^{e^{4}} \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**2+x+1)*exp(exp(4))/x,x)

[Out]

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

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