3.49.2 \(\int e^{-2 x} (54 x-54 x^2) \log (5) \, dx\)

Optimal. Leaf size=14 \[ 15+27 e^{-2 x} x^2 \log (5) \]

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Rubi [A]  time = 0.06, antiderivative size = 12, normalized size of antiderivative = 0.86, number of steps used = 9, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {12, 1593, 2196, 2176, 2194} \begin {gather*} 27 e^{-2 x} x^2 \log (5) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((54*x - 54*x^2)*Log[5])/E^(2*x),x]

[Out]

(27*x^2*Log[5])/E^(2*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 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

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 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log (5) \int e^{-2 x} \left (54 x-54 x^2\right ) \, dx\\ &=\log (5) \int e^{-2 x} (54-54 x) x \, dx\\ &=\log (5) \int \left (54 e^{-2 x} x-54 e^{-2 x} x^2\right ) \, dx\\ &=(54 \log (5)) \int e^{-2 x} x \, dx-(54 \log (5)) \int e^{-2 x} x^2 \, dx\\ &=-27 e^{-2 x} x \log (5)+27 e^{-2 x} x^2 \log (5)+(27 \log (5)) \int e^{-2 x} \, dx-(54 \log (5)) \int e^{-2 x} x \, dx\\ &=-\frac {27}{2} e^{-2 x} \log (5)+27 e^{-2 x} x^2 \log (5)-(27 \log (5)) \int e^{-2 x} \, dx\\ &=27 e^{-2 x} x^2 \log (5)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 12, normalized size = 0.86 \begin {gather*} 27 e^{-2 x} x^2 \log (5) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((54*x - 54*x^2)*Log[5])/E^(2*x),x]

[Out]

(27*x^2*Log[5])/E^(2*x)

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fricas [A]  time = 0.81, size = 11, normalized size = 0.79 \begin {gather*} 27 \, x^{2} e^{\left (-2 \, x\right )} \log \relax (5) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-54*x^2+54*x)*log(5)/exp(x)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 11, normalized size = 0.79 \begin {gather*} 27 \, x^{2} e^{\left (-2 \, x\right )} \log \relax (5) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-54*x^2+54*x)*log(5)/exp(x)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 12, normalized size = 0.86

method	result	size
gosper	\(27 \ln \relax (5) x^{2} {\mathrm e}^{-2 x}\)	\(12\)
default	\(27 \ln \relax (5) x^{2} {\mathrm e}^{-2 x}\)	\(12\)
norman	\(27 \ln \relax (5) x^{2} {\mathrm e}^{-2 x}\)	\(12\)
risch	\(27 \ln \relax (5) x^{2} {\mathrm e}^{-2 x}\)	\(12\)
meijerg	\(-\frac {27 \ln \relax (5) \left (2-\frac {\left (12 x^{2}+12 x +6\right ) {\mathrm e}^{-2 x}}{3}\right )}{4}+\frac {27 \ln \relax (5) \left (1-\frac {\left (4 x +2\right ) {\mathrm e}^{-2 x}}{2}\right )}{2}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-54*x^2+54*x)*ln(5)/exp(x)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.35, size = 31, normalized size = 2.21 \begin {gather*} \frac {27}{2} \, {\left ({\left (2 \, x^{2} + 2 \, x + 1\right )} e^{\left (-2 \, x\right )} - {\left (2 \, x + 1\right )} e^{\left (-2 \, x\right )}\right )} \log \relax (5) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-54*x^2+54*x)*log(5)/exp(x)^2,x, algorithm="maxima")

[Out]

27/2*((2*x^2 + 2*x + 1)*e^(-2*x) - (2*x + 1)*e^(-2*x))*log(5)

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mupad [B]  time = 0.05, size = 11, normalized size = 0.79 \begin {gather*} 27\,x^2\,{\mathrm {e}}^{-2\,x}\,\ln \relax (5) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-2*x)*log(5)*(54*x - 54*x^2),x)

[Out]

27*x^2*exp(-2*x)*log(5)

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sympy [A]  time = 0.10, size = 12, normalized size = 0.86 \begin {gather*} 27 x^{2} e^{- 2 x} \log {\relax (5 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-54*x**2+54*x)*ln(5)/exp(x)**2,x)

[Out]

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

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