3.70.7 \(\int \frac {e^{e^{e^3+x}} (2 x+e^{e^3+x} x^2)}{\log (5)} \, dx\)

Optimal. Leaf size=17 \[ \frac {e^{e^{e^3+x}} x^2}{\log (5)} \]

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Rubi [A]  time = 0.03, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {12, 2288} \begin {gather*} \frac {e^{e^{x+e^3}} x^2}{\log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^E^(E^3 + x)*(2*x + E^(E^3 + x)*x^2))/Log[5],x]

[Out]

(E^E^(E^3 + x)*x^2)/Log[5]

Rule 12

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

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 17, normalized size = 1.00 \begin {gather*} \frac {e^{e^{e^3+x}} x^2}{\log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^E^(E^3 + x)*(2*x + E^(E^3 + x)*x^2))/Log[5],x]

[Out]

(E^E^(E^3 + x)*x^2)/Log[5]

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fricas [A]  time = 0.63, size = 14, normalized size = 0.82 \begin {gather*} \frac {x^{2} e^{\left (e^{\left (x + e^{3}\right )}\right )}}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2*e^(e^(x + e^3))/log(5)

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giac [A]  time = 0.23, size = 14, normalized size = 0.82 \begin {gather*} \frac {x^{2} e^{\left (e^{\left (x + e^{3}\right )}\right )}}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2*e^(e^(x + e^3))/log(5)

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maple [A]  time = 0.05, size = 15, normalized size = 0.88

method	result	size
norman	\(\frac {x^{2} {\mathrm e}^{{\mathrm e}^{{\mathrm e}^{3}+x}}}{\ln \relax (5)}\)	\(15\)
risch	\(\frac {x^{2} {\mathrm e}^{{\mathrm e}^{{\mathrm e}^{3}+x}}}{\ln \relax (5)}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2*exp(exp(exp(3)+x))/ln(5)

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maxima [A]  time = 0.40, size = 14, normalized size = 0.82 \begin {gather*} \frac {x^{2} e^{\left (e^{\left (x + e^{3}\right )}\right )}}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2*e^(e^(x + e^3))/log(5)

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mupad [B]  time = 0.08, size = 15, normalized size = 0.88 \begin {gather*} \frac {x^2\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^3}\,{\mathrm {e}}^x}}{\ln \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(x + exp(3)))*(2*x + x^2*exp(x + exp(3))))/log(5),x)

[Out]

(x^2*exp(exp(exp(3))*exp(x)))/log(5)

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sympy [A]  time = 0.16, size = 14, normalized size = 0.82 \begin {gather*} \frac {x^{2} e^{e^{x + e^{3}}}}{\log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2*exp(exp(3)+x)+2*x)*exp(exp(exp(3)+x))/ln(5),x)

[Out]

x**2*exp(exp(x + exp(3)))/log(5)

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