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]

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

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Rubi [A]  time = 0.01, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, 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 12

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

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]

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*}

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Mathematica [A]  time = 0.01, 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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fricas [A]  time = 0.41, size = 11, normalized size = 0.48 \[ \frac {1}{4} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} \]

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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giac [A]  time = 1.22, size = 11, normalized size = 0.48 \[ \frac {1}{4} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} \]

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)

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maple [A]  time = 0.02, size = 28, normalized size = 1.22 \[ \frac {x \,{\mathrm e}^{2 x}}{2}-\frac {{\mathrm e}^{2 x}}{4}+\frac {\left (-x +\ln \left ({\mathrm e}^{x}\right )\right ) {\mathrm e}^{2 x}}{2} \]

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 = 0.44, size = 11, normalized size = 0.48 \[ \frac {1}{4} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} \]

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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mupad [B]  time = 0.04, size = 11, normalized size = 0.48 \[ \frac {{\mathrm {e}}^{2\,x}\,\left (2\,x-1\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(exp(2*x)*(2*x - 1))/4

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sympy [A]  time = 0.10, size = 10, normalized size = 0.43 \[ \frac {\left (2 x - 1\right ) e^{2 x}}{4} \]

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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