3.59.35 \(\int \frac {3\ 2^{-1+\frac {3 x^2}{2 e^5}} x \log (4)}{e^5} \, dx\)

Optimal. Leaf size=15 \[ 2^{\frac {3 x^2}{2 e^5}}+\log (6) \]

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Rubi [A]  time = 0.02, antiderivative size = 21, normalized size of antiderivative = 1.40, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {12, 2209} \begin {gather*} \frac {2^{\frac {3 x^2}{2 e^5}-1} \log (4)}{\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3*2^(-1 + (3*x^2)/(2*E^5))*x*Log[4])/E^5,x]

[Out]

(2^(-1 + (3*x^2)/(2*E^5))*Log[4])/Log[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 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {(3 \log (4)) \int 2^{-1+\frac {3 x^2}{2 e^5}} x \, dx}{e^5}\\ &=\frac {2^{-1+\frac {3 x^2}{2 e^5}} \log (4)}{\log (2)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 21, normalized size = 1.40 \begin {gather*} \frac {2^{-1+\frac {3 x^2}{2 e^5}} \log (64)}{\log (8)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3*2^(-1 + (3*x^2)/(2*E^5))*x*Log[4])/E^5,x]

[Out]

(2^(-1 + (3*x^2)/(2*E^5))*Log[64])/Log[8]

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fricas [A]  time = 0.72, size = 9, normalized size = 0.60 \begin {gather*} 2^{\frac {3}{2} \, x^{2} e^{\left (-5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(3*x*log(2)*exp(3/2*x^2*log(2)/exp(5))/exp(5),x, algorithm="fricas")

[Out]

2^(3/2*x^2*e^(-5))

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giac [A]  time = 0.12, size = 9, normalized size = 0.60 \begin {gather*} 2^{\frac {3}{2} \, x^{2} e^{\left (-5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(3*x*log(2)*exp(3/2*x^2*log(2)/exp(5))/exp(5),x, algorithm="giac")

[Out]

2^(3/2*x^2*e^(-5))

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maple [A]  time = 0.05, size = 10, normalized size = 0.67

method	result	size
risch	\(2^{\frac {3 \,{\mathrm e}^{-5} x^{2}}{2}}\)	\(10\)
gosper	\({\mathrm e}^{\frac {3 x^{2} {\mathrm e}^{-5} \ln \relax (2)}{2}}\)	\(13\)
derivativedivides	\({\mathrm e}^{\frac {3 x^{2} {\mathrm e}^{-5} \ln \relax (2)}{2}}\)	\(13\)
default	\({\mathrm e}^{\frac {3 x^{2} {\mathrm e}^{-5} \ln \relax (2)}{2}}\)	\(13\)
norman	\({\mathrm e}^{\frac {3 x^{2} {\mathrm e}^{-5} \ln \relax (2)}{2}}\)	\(13\)
meijerg	\(-1+{\mathrm e}^{\frac {3 x^{2} {\mathrm e}^{-5} \ln \relax (2)}{2}}\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(3*x*ln(2)*exp(3/2*x^2*ln(2)/exp(5))/exp(5),x,method=_RETURNVERBOSE)

[Out]

2^(3/2*exp(-5)*x^2)

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maxima [A]  time = 0.35, size = 9, normalized size = 0.60 \begin {gather*} 2^{\frac {3}{2} \, x^{2} e^{\left (-5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(3*x*log(2)*exp(3/2*x^2*log(2)/exp(5))/exp(5),x, algorithm="maxima")

[Out]

2^(3/2*x^2*e^(-5))

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mupad [B]  time = 0.04, size = 9, normalized size = 0.60 \begin {gather*} 2^{\frac {3\,x^2\,{\mathrm {e}}^{-5}}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2^((3*x^2*exp(-5))/2)

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sympy [A]  time = 0.10, size = 14, normalized size = 0.93 \begin {gather*} e^{\frac {3 x^{2} \log {\relax (2 )}}{2 e^{5}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(3*x*ln(2)*exp(3/2*x**2*ln(2)/exp(5))/exp(5),x)

[Out]

exp(3*x**2*exp(-5)*log(2)/2)

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