3.56.7 \(\int \frac {-2 e^{8+2 x}+3 \log (5)}{3 \log (5)} \, dx\) [5507]

Optimal. Leaf size=18 \[ -6+x-\frac {e^{2 (4+x)}}{3 \log (5)} \]

[Out]

x-1/3*exp(2*x+8)/ln(5)-6

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Rubi [A]
time = 0.01, antiderivative size = 17, normalized size of antiderivative = 0.94, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {12, 2225} \begin {gather*} x-\frac {e^{2 x+8}}{3 \log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*E^(8 + 2*x) + 3*Log[5])/(3*Log[5]),x]

[Out]

x - E^(8 + 2*x)/(3*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 2225

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} &=\frac {\int \left (-2 e^{8+2 x}+3 \log (5)\right ) \, dx}{3 \log (5)}\\ &=x-\frac {2 \int e^{8+2 x} \, dx}{3 \log (5)}\\ &=x-\frac {e^{8+2 x}}{3 \log (5)}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.03, size = 15, normalized size = 0.83 \begin {gather*} x-\frac {e^{8+2 x}}{\log (125)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*E^(8 + 2*x) + 3*Log[5])/(3*Log[5]),x]

[Out]

x - E^(8 + 2*x)/Log[125]

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Maple [A]
time = 0.33, size = 21, normalized size = 1.17

method	result	size
norman	\(x -\frac {{\mathrm e}^{2 x +8}}{3 \ln \left (5\right )}\)	\(15\)
risch	\(x -\frac {{\mathrm e}^{2 x +8}}{3 \ln \left (5\right )}\)	\(15\)
default	\(\frac {-{\mathrm e}^{2 x +8}+3 x \ln \left (5\right )}{3 \ln \left (5\right )}\)	\(21\)
derivativedivides	\(\frac {-2 \,{\mathrm e}^{2 x +8}+3 \ln \left (5\right ) \ln \left ({\mathrm e}^{2 x +8}\right )}{6 \ln \left (5\right )}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*(-2*exp(2*x+8)+3*ln(5))/ln(5),x,method=_RETURNVERBOSE)

[Out]

1/3/ln(5)*(-exp(2*x+8)+3*x*ln(5))

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Maxima [A]
time = 0.27, size = 20, normalized size = 1.11 \begin {gather*} \frac {3 \, x \log \left (5\right ) - e^{\left (2 \, x + 8\right )}}{3 \, \log \left (5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-2*exp(2*x+8)+3*log(5))/log(5),x, algorithm="maxima")

[Out]

1/3*(3*x*log(5) - e^(2*x + 8))/log(5)

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Fricas [A]
time = 0.36, size = 20, normalized size = 1.11 \begin {gather*} \frac {3 \, x \log \left (5\right ) - e^{\left (2 \, x + 8\right )}}{3 \, \log \left (5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-2*exp(2*x+8)+3*log(5))/log(5),x, algorithm="fricas")

[Out]

1/3*(3*x*log(5) - e^(2*x + 8))/log(5)

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Sympy [A]
time = 0.05, size = 12, normalized size = 0.67 \begin {gather*} x - \frac {e^{2 x + 8}}{3 \log {\left (5 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-2*exp(2*x+8)+3*ln(5))/ln(5),x)

[Out]

x - exp(2*x + 8)/(3*log(5))

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Giac [A]
time = 0.41, size = 20, normalized size = 1.11 \begin {gather*} \frac {3 \, x \log \left (5\right ) - e^{\left (2 \, x + 8\right )}}{3 \, \log \left (5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-2*exp(2*x+8)+3*log(5))/log(5),x, algorithm="giac")

[Out]

1/3*(3*x*log(5) - e^(2*x + 8))/log(5)

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Mupad [B]
time = 0.08, size = 14, normalized size = 0.78 \begin {gather*} x-\frac {{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^8}{3\,\ln \left (5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(5) - (2*exp(2*x + 8))/3)/log(5),x)

[Out]

x - (exp(2*x)*exp(8))/(3*log(5))

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