3.71.79 \(\int -\frac {3 e^{\frac {-9 x-\log (2)+3 \log (5)}{-3 x+\log (5)}} \log (2)}{9 x^2-6 x \log (5)+\log ^2(5)} \, dx\)

Optimal. Leaf size=16 \[ e^{3-\frac {\log (2)}{-3 x+\log (5)}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 23, normalized size of antiderivative = 1.44, number of steps used = 4, number of rules used = 4, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {12, 27, 2230, 2209} \begin {gather*} \frac {3 e^3 \log (2) 2^{\frac {1}{3 x-\log (5)}}}{\log (8)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3*E^((-9*x - Log[2] + 3*Log[5])/(-3*x + Log[5]))*Log[2])/(9*x^2 - 6*x*Log[5] + Log[5]^2),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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]

Rule 2230

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left ((3 \log (2)) \int \frac {e^{\frac {-9 x-\log (2)+3 \log (5)}{-3 x+\log (5)}}}{9 x^2-6 x \log (5)+\log ^2(5)} \, dx\right )\\ &=-\left ((3 \log (2)) \int \frac {e^{\frac {-9 x-\log (2)+3 \log (5)}{-3 x+\log (5)}}}{(3 x-\log (5))^2} \, dx\right )\\ &=-\left ((3 \log (2)) \int \frac {\exp \left (3+\frac {-9 \log (5)+3 (-\log (2)+3 \log (5))}{3 (-3 x+\log (5))}\right )}{(3 x-\log (5))^2} \, dx\right )\\ &=\frac {3\ 2^{\frac {1}{3 x-\log (5)}} e^3 \log (2)}{\log (8)}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 23, normalized size = 1.44 \begin {gather*} \frac {3\ 2^{\frac {1}{3 x-\log (5)}} e^3 \log (2)}{\log (8)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*E^((-9*x - Log[2] + 3*Log[5])/(-3*x + Log[5]))*Log[2])/(9*x^2 - 6*x*Log[5] + Log[5]^2),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.66, size = 22, normalized size = 1.38 \begin {gather*} e^{\left (\frac {9 \, x - 3 \, \log \relax (5) + \log \relax (2)}{3 \, x - \log \relax (5)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-3*log(2)*exp((3*log(5)-log(2)-9*x)/(log(5)-3*x))/(log(5)^2-6*x*log(5)+9*x^2),x, algorithm="fricas")

[Out]

e^((9*x - 3*log(5) + log(2))/(3*x - log(5)))

________________________________________________________________________________________

giac [B]  time = 0.19, size = 42, normalized size = 2.62 \begin {gather*} e^{\left (\frac {9 \, x}{3 \, x - \log \relax (5)} - \frac {3 \, \log \relax (5)}{3 \, x - \log \relax (5)} + \frac {\log \relax (2)}{3 \, x - \log \relax (5)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-3*log(2)*exp((3*log(5)-log(2)-9*x)/(log(5)-3*x))/(log(5)^2-6*x*log(5)+9*x^2),x, algorithm="giac")

[Out]

e^(9*x/(3*x - log(5)) - 3*log(5)/(3*x - log(5)) + log(2)/(3*x - log(5)))

________________________________________________________________________________________

maple [A]  time = 0.16, size = 17, normalized size = 1.06

method	result	size
derivativedivides	\({\mathrm e}^{3+\frac {\ln \relax (2)}{-\ln \relax (5)+3 x}}\)	\(17\)
default	\({\mathrm e}^{3+\frac {\ln \relax (2)}{-\ln \relax (5)+3 x}}\)	\(17\)
risch	\({\mathrm e}^{-\frac {-3 \ln \relax (5)+\ln \relax (2)+9 x}{\ln \relax (5)-3 x}}\)	\(22\)
gosper	\({\mathrm e}^{\frac {3 \ln \relax (5)-\ln \relax (2)-9 x}{\ln \relax (5)-3 x}}\)	\(23\)
norman	\(\frac {\ln \relax (5) {\mathrm e}^{\frac {3 \ln \relax (5)-\ln \relax (2)-9 x}{\ln \relax (5)-3 x}}-3 x \,{\mathrm e}^{\frac {3 \ln \relax (5)-\ln \relax (2)-9 x}{\ln \relax (5)-3 x}}}{\ln \relax (5)-3 x}\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-3*ln(2)*exp((3*ln(5)-ln(2)-9*x)/(ln(5)-3*x))/(ln(5)^2-6*x*ln(5)+9*x^2),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.46, size = 16, normalized size = 1.00 \begin {gather*} e^{\left (\frac {\log \relax (2)}{3 \, x - \log \relax (5)} + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-3*log(2)*exp((3*log(5)-log(2)-9*x)/(log(5)-3*x))/(log(5)^2-6*x*log(5)+9*x^2),x, algorithm="maxima")

[Out]

e^(log(2)/(3*x - log(5)) + 3)

________________________________________________________________________________________

mupad [B]  time = 4.74, size = 27, normalized size = 1.69 \begin {gather*} {\left (\frac {2}{125}\right )}^{\frac {1}{3\,x-\ln \relax (5)}}\,{\mathrm {e}}^{\frac {9\,x}{3\,x-\ln \relax (5)}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*exp((9*x + log(2) - 3*log(5))/(3*x - log(5)))*log(2))/(log(5)^2 - 6*x*log(5) + 9*x^2),x)

[Out]

(2/125)^(1/(3*x - log(5)))*exp((9*x)/(3*x - log(5)))

________________________________________________________________________________________

sympy [A]  time = 0.22, size = 19, normalized size = 1.19 \begin {gather*} e^{\frac {- 9 x - \log {\relax (2 )} + 3 \log {\relax (5 )}}{- 3 x + \log {\relax (5 )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-3*ln(2)*exp((3*ln(5)-ln(2)-9*x)/(ln(5)-3*x))/(ln(5)**2-6*x*ln(5)+9*x**2),x)

[Out]

exp((-9*x - log(2) + 3*log(5))/(-3*x + log(5)))

________________________________________________________________________________________