3.9.11 \(\int e^{-32 x} (-64-16 \log (\frac {16}{9})) \, dx\)

Optimal. Leaf size=15 \[ \frac {1}{2} e^{-32 x} \left (4+\log \left (\frac {16}{9}\right )\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 2194} \begin {gather*} \frac {1}{2} e^{-32 x} \left (4+\log \left (\frac {16}{9}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-64 - 16*Log[16/9])/E^(32*x),x]

[Out]

(4 + Log[16/9])/(2*E^(32*x))

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} &=-\left (\left (16 \left (4+\log \left (\frac {16}{9}\right )\right )\right ) \int e^{-32 x} \, dx\right )\\ &=\frac {1}{2} e^{-32 x} \left (4+\log \left (\frac {16}{9}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 15, normalized size = 1.00 \begin {gather*} \frac {1}{2} e^{-32 x} \left (4+\log \left (\frac {16}{9}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-64 - 16*Log[16/9])/E^(32*x),x]

[Out]

(4 + Log[16/9])/(2*E^(32*x))

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fricas [A]  time = 1.03, size = 10, normalized size = 0.67 \begin {gather*} \frac {1}{2} \, {\left (\log \left (\frac {16}{9}\right ) + 4\right )} e^{\left (-32 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*log(16/9)-64)/exp(32*x),x, algorithm="fricas")

[Out]

1/2*(log(16/9) + 4)*e^(-32*x)

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giac [A]  time = 0.40, size = 10, normalized size = 0.67 \begin {gather*} \frac {1}{2} \, {\left (\log \left (\frac {16}{9}\right ) + 4\right )} e^{\left (-32 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*log(16/9)-64)/exp(32*x),x, algorithm="giac")

[Out]

1/2*(log(16/9) + 4)*e^(-32*x)

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maple [A]  time = 0.03, size = 13, normalized size = 0.87

method	result	size
gosper	\(\frac {\left (\ln \left (\frac {16}{9}\right )+4\right ) {\mathrm e}^{-32 x}}{2}\)	\(13\)
derivativedivides	\(-\left (-\frac {\ln \left (\frac {16}{9}\right )}{2}-2\right ) {\mathrm e}^{-32 x}\)	\(15\)
default	\(-\frac {\left (-16 \ln \left (\frac {16}{9}\right )-64\right ) {\mathrm e}^{-32 x}}{32}\)	\(15\)
norman	\(\left (2 \ln \relax (2)-\ln \relax (3)+2\right ) {\mathrm e}^{-32 x}\)	\(18\)
meijerg	\(-\frac {\ln \left (\frac {16}{9}\right ) \left (1-{\mathrm e}^{-32 x}\right )}{2}-2+2 \,{\mathrm e}^{-32 x}\)	\(21\)
risch	\(2 \,{\mathrm e}^{-32 x} \ln \relax (2)-{\mathrm e}^{-32 x} \ln \relax (3)+2 \,{\mathrm e}^{-32 x}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-16*ln(16/9)-64)/exp(32*x),x,method=_RETURNVERBOSE)

[Out]

1/2*(ln(16/9)+4)/exp(32*x)

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maxima [A]  time = 0.37, size = 10, normalized size = 0.67 \begin {gather*} \frac {1}{2} \, {\left (\log \left (\frac {16}{9}\right ) + 4\right )} e^{\left (-32 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*log(16/9)-64)/exp(32*x),x, algorithm="maxima")

[Out]

1/2*(log(16/9) + 4)*e^(-32*x)

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mupad [B]  time = 0.56, size = 9, normalized size = 0.60 \begin {gather*} {\mathrm {e}}^{-32\,x}\,\left (\ln \left (\frac {4}{3}\right )+2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(-32*x)*(16*log(16/9) + 64),x)

[Out]

exp(-32*x)*(log(4/3) + 2)

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sympy [A]  time = 0.10, size = 14, normalized size = 0.93 \begin {gather*} \left (- \log {\relax (3 )} + 2 \log {\relax (2 )} + 2\right ) e^{- 32 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*ln(16/9)-64)/exp(32*x),x)

[Out]

(-log(3) + 2*log(2) + 2)*exp(-32*x)

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