3.29.62 \(\int \frac {2^{-72+8 x} (50 x+200 x^2 \log (2))}{6561} \, dx\)

Optimal. Leaf size=14 \[ \frac {25\ 2^{-72+8 x} x^2}{6561} \]

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Rubi [A]  time = 0.10, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {12, 1593, 2196, 2176, 2194} \begin {gather*} \frac {25\ 2^{8 x-72} x^2}{6561} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2^(-72 + 8*x)*(50*x + 200*x^2*Log[2]))/6561,x]

[Out]

(25*2^(-72 + 8*x)*x^2)/6561

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} &=\frac {\int 2^{-72+8 x} \left (50 x+200 x^2 \log (2)\right ) \, dx}{6561}\\ &=\frac {\int 2^{-72+8 x} x (50+200 x \log (2)) \, dx}{6561}\\ &=\frac {\int \left (25\ 2^{-71+8 x} x+25\ 2^{-69+8 x} x^2 \log (2)\right ) \, dx}{6561}\\ &=\frac {25 \int 2^{-71+8 x} x \, dx}{6561}+\frac {(25 \log (2)) \int 2^{-69+8 x} x^2 \, dx}{6561}\\ &=\frac {25\ 2^{-72+8 x} x^2}{6561}+\frac {25\ 2^{-74+8 x} x}{6561 \log (2)}-\frac {25 \int 2^{-69+8 x} x \, dx}{26244}-\frac {25 \int 2^{-71+8 x} \, dx}{52488 \log (2)}\\ &=\frac {25\ 2^{-72+8 x} x^2}{6561}-\frac {25\ 2^{-77+8 x}}{6561 \log ^2(2)}+\frac {25 \int 2^{-69+8 x} \, dx}{209952 \log (2)}\\ &=\frac {25\ 2^{-72+8 x} x^2}{6561}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 20, normalized size = 1.43 \begin {gather*} \frac {25\ 2^{-74+8 x} x^2 \log (16)}{6561 \log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2^(-72 + 8*x)*(50*x + 200*x^2*Log[2]))/6561,x]

[Out]

(25*2^(-74 + 8*x)*x^2*Log[16])/(6561*Log[2])

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fricas [A]  time = 0.66, size = 12, normalized size = 0.86 \begin {gather*} \frac {25}{6561} \cdot 2^{8 \, x - 72} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6561*(200*x^2*log(2)+50*x)*exp((x-9)*log(2))^8,x, algorithm="fricas")

[Out]

25/6561*2^(8*x - 72)*x^2

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giac [A]  time = 0.24, size = 16, normalized size = 1.14 \begin {gather*} \frac {25}{6561} \, x^{2} e^{\left (8 \, x \log \relax (2) - 72 \, \log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6561*(200*x^2*log(2)+50*x)*exp((x-9)*log(2))^8,x, algorithm="giac")

[Out]

25/6561*x^2*e^(8*x*log(2) - 72*log(2))

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maple [A]  time = 0.12, size = 13, normalized size = 0.93

method	result	size
risch	\(\frac {25 \,2^{8 x} x^{2}}{30983446494107742247059456}\)	\(13\)
gosper	\(\frac {25 \,{\mathrm e}^{8 \left (x -9\right ) \ln \relax (2)} x^{2}}{6561}\)	\(15\)
norman	\(\frac {25 \,{\mathrm e}^{8 \left (x -9\right ) \ln \relax (2)} x^{2}}{6561}\)	\(15\)
derivativedivides	\(\frac {\frac {25 \,{\mathrm e}^{8 x \ln \relax (2)} \left (x \ln \relax (2)-9 \ln \relax (2)\right )}{1721302583005985680392192}+\frac {25 \,{\mathrm e}^{8 x \ln \relax (2)} \ln \relax (2)}{382511685112441262309376}+\frac {25 \,{\mathrm e}^{8 x \ln \relax (2)} \left (x \ln \relax (2)-9 \ln \relax (2)\right )^{2}}{30983446494107742247059456 \ln \relax (2)}}{\ln \relax (2)}\)	\(76\)
default	\(\frac {\frac {25 \,{\mathrm e}^{8 x \ln \relax (2)} \left (x \ln \relax (2)-9 \ln \relax (2)\right )}{1721302583005985680392192}+\frac {25 \,{\mathrm e}^{8 x \ln \relax (2)} \ln \relax (2)}{382511685112441262309376}+\frac {25 \,{\mathrm e}^{8 x \ln \relax (2)} \left (x \ln \relax (2)-9 \ln \relax (2)\right )^{2}}{30983446494107742247059456 \ln \relax (2)}}{\ln \relax (2)}\)	\(76\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/6561*(200*x^2*ln(2)+50*x)*exp((x-9)*ln(2))^8,x,method=_RETURNVERBOSE)

[Out]

25/6561*(2^(x-9))^8*x^2

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maxima [B]  time = 0.51, size = 46, normalized size = 3.29 \begin {gather*} \frac {25 \, {\left (32 \, x^{2} \log \relax (2)^{2} - 8 \, x \log \relax (2) + 1\right )} 2^{8 \, x}}{991470287811447751905902592 \, \log \relax (2)^{2}} + \frac {25 \, {\left (8 \, x \log \relax (2) - 1\right )} 2^{8 \, x}}{991470287811447751905902592 \, \log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6561*(200*x^2*log(2)+50*x)*exp((x-9)*log(2))^8,x, algorithm="maxima")

[Out]

25/991470287811447751905902592*(32*x^2*log(2)^2 - 8*x*log(2) + 1)*2^(8*x)/log(2)^2 + 25/9914702878114477519059
02592*(8*x*log(2) - 1)*2^(8*x)/log(2)^2

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mupad [B]  time = 1.69, size = 10, normalized size = 0.71 \begin {gather*} \frac {25\,2^{8\,x}\,x^2}{30983446494107742247059456} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(8*log(2)*(x - 9))*(50*x + 200*x^2*log(2)))/6561,x)

[Out]

(25*2^(8*x)*x^2)/30983446494107742247059456

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sympy [A]  time = 0.11, size = 15, normalized size = 1.07 \begin {gather*} \frac {25 x^{2} e^{8 \left (x - 9\right ) \log {\relax (2 )}}}{6561} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6561*(200*x**2*ln(2)+50*x)*exp((x-9)*ln(2))**8,x)

[Out]

25*x**2*exp(8*(x - 9)*log(2))/6561

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