3.173 \(\int e^{2 x} \log (e^x) \, dx\)

Optimal. Leaf size=23 \[ \frac{1}{2} e^{2 x} \log \left (e^x\right )-\frac{e^{2 x}}{4} \]

[Out]

-E^(2*x)/4 + (E^(2*x)*Log[E^x])/2

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Rubi [A]  time = 0.012565, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {2194, 2554, 12} \[ \frac{1}{2} e^{2 x} \log \left (e^x\right )-\frac{e^{2 x}}{4} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*x)*Log[E^x],x]

[Out]

-E^(2*x)/4 + (E^(2*x)*Log[E^x])/2

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 12

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

Rubi steps

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

Mathematica [A]  time = 0.004753, size = 17, normalized size = 0.74 \[ \frac{1}{4} e^{2 x} \left (2 \log \left (e^x\right )-1\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*x)*Log[E^x],x]

[Out]

(E^(2*x)*(-1 + 2*Log[E^x]))/4

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Maple [A]  time = 0.001, size = 28, normalized size = 1.2 \begin{align*}{\frac{{{\rm e}^{2\,x}}x}{2}}-{\frac{{{\rm e}^{2\,x}}}{4}}+{\frac{{{\rm e}^{2\,x}} \left ( \ln \left ({{\rm e}^{x}} \right ) -x \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*x)*ln(exp(x)),x)

[Out]

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

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Maxima [A]  time = 0.951424, size = 15, normalized size = 0.65 \begin{align*} \frac{1}{4} \,{\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*x)*log(exp(x)),x, algorithm="maxima")

[Out]

1/4*(2*x - 1)*e^(2*x)

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Fricas [A]  time = 1.91165, size = 31, normalized size = 1.35 \begin{align*} \frac{1}{4} \,{\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*x)*log(exp(x)),x, algorithm="fricas")

[Out]

1/4*(2*x - 1)*e^(2*x)

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Sympy [A]  time = 0.088962, size = 10, normalized size = 0.43 \begin{align*} \frac{\left (2 x - 1\right ) e^{2 x}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*x)*ln(exp(x)),x)

[Out]

(2*x - 1)*exp(2*x)/4

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Giac [A]  time = 1.07369, size = 15, normalized size = 0.65 \begin{align*} \frac{1}{4} \,{\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*x)*log(exp(x)),x, algorithm="giac")

[Out]

1/4*(2*x - 1)*e^(2*x)