3.72.35 \(\int 2 e^{2 x} \log (25) \, dx\)

Optimal. Leaf size=10 \[ -2+e^{2 x} \log (25) \]

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Rubi [A]  time = 0.00, antiderivative size = 8, normalized size of antiderivative = 0.80, number of steps used = 2, number of rules used = 2, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {12, 2194} \begin {gather*} e^{2 x} \log (25) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[2*E^(2*x)*Log[25],x]

[Out]

E^(2*x)*Log[25]

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=(2 \log (25)) \int e^{2 x} \, dx\\ &=e^{2 x} \log (25)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 11, normalized size = 1.10 \begin {gather*} \frac {1}{2} e^{2 x} \log (625) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[2*E^(2*x)*Log[25],x]

[Out]

(E^(2*x)*Log[625])/2

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fricas [A]  time = 0.91, size = 8, normalized size = 0.80 \begin {gather*} 2 \, e^{\left (2 \, x\right )} \log \relax (5) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(4*log(5)*exp(x)^2,x, algorithm="fricas")

[Out]

2*e^(2*x)*log(5)

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giac [A]  time = 0.12, size = 8, normalized size = 0.80 \begin {gather*} 2 \, e^{\left (2 \, x\right )} \log \relax (5) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(4*log(5)*exp(x)^2,x, algorithm="giac")

[Out]

2*e^(2*x)*log(5)

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maple [A]  time = 0.02, size = 9, normalized size = 0.90

method	result	size
gosper	\(2 \ln \relax (5) {\mathrm e}^{2 x}\)	\(9\)
derivativedivides	\(2 \ln \relax (5) {\mathrm e}^{2 x}\)	\(9\)
default	\(2 \ln \relax (5) {\mathrm e}^{2 x}\)	\(9\)
norman	\(2 \ln \relax (5) {\mathrm e}^{2 x}\)	\(9\)
risch	\(2 \ln \relax (5) {\mathrm e}^{2 x}\)	\(9\)
meijerg	\(-2 \ln \relax (5) \left (1-{\mathrm e}^{2 x}\right )\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(4*ln(5)*exp(x)^2,x,method=_RETURNVERBOSE)

[Out]

2*ln(5)*exp(x)^2

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maxima [A]  time = 0.36, size = 8, normalized size = 0.80 \begin {gather*} 2 \, e^{\left (2 \, x\right )} \log \relax (5) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(4*log(5)*exp(x)^2,x, algorithm="maxima")

[Out]

2*e^(2*x)*log(5)

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mupad [B]  time = 4.05, size = 8, normalized size = 0.80 \begin {gather*} 2\,{\mathrm {e}}^{2\,x}\,\ln \relax (5) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(4*exp(2*x)*log(5),x)

[Out]

2*exp(2*x)*log(5)

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sympy [A]  time = 0.06, size = 8, normalized size = 0.80 \begin {gather*} 2 e^{2 x} \log {\relax (5 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(4*ln(5)*exp(x)**2,x)

[Out]

2*exp(2*x)*log(5)

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