3.97.35 \(\int \frac {e^{2 x} ((-32+32 x) \log ^2(6)-96 \log ^3(6))+e^x ((-144 x+144 x^2) \log (6)+(432-864 x) \log ^2(6)+1296 \log ^3(6))}{-x^3+9 x^2 \log (6)-27 x \log ^2(6)+27 \log ^3(6)} \, dx\)

Optimal. Leaf size=24 \[ 5-\left (18-\frac {4 e^x}{3-\frac {x}{\log (6)}}\right )^2 \]

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Rubi [A]  time = 0.74, antiderivative size = 35, normalized size of antiderivative = 1.46, number of steps used = 6, number of rules used = 4, integrand size = 86, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {6688, 12, 6742, 2197} \begin {gather*} -\frac {16 e^{2 x} \log ^2(6)}{(x-3 \log (6))^2}-\frac {144 e^x \log (6)}{x-3 \log (6)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(2*x)*((-32 + 32*x)*Log[6]^2 - 96*Log[6]^3) + E^x*((-144*x + 144*x^2)*Log[6] + (432 - 864*x)*Log[6]^2 +
1296*Log[6]^3))/(-x^3 + 9*x^2*Log[6] - 27*x*Log[6]^2 + 27*Log[6]^3),x]

[Out]

(-144*E^x*Log[6])/(x - 3*Log[6]) - (16*E^(2*x)*Log[6]^2)/(x - 3*Log[6])^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 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {16 e^x \log (6) (1-x+3 \log (6)) \left (9 x+\left (-27+2 e^x\right ) \log (6)\right )}{(x-3 \log (6))^3} \, dx\\ &=(16 \log (6)) \int \frac {e^x (1-x+3 \log (6)) \left (9 x+\left (-27+2 e^x\right ) \log (6)\right )}{(x-3 \log (6))^3} \, dx\\ &=(16 \log (6)) \int \left (-\frac {9 e^x (-1+x-3 \log (6))}{(x-3 \log (6))^2}-\frac {2 e^{2 x} (-1+x-3 \log (6)) \log (6)}{(x-3 \log (6))^3}\right ) \, dx\\ &=-\left ((144 \log (6)) \int \frac {e^x (-1+x-3 \log (6))}{(x-3 \log (6))^2} \, dx\right )-\left (32 \log ^2(6)\right ) \int \frac {e^{2 x} (-1+x-3 \log (6))}{(x-3 \log (6))^3} \, dx\\ &=-\frac {144 e^x \log (6)}{x-3 \log (6)}-\frac {16 e^{2 x} \log ^2(6)}{(x-3 \log (6))^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.28, size = 27, normalized size = 1.12 \begin {gather*} -\frac {16 e^x \log (6) \left (9 x+\left (-27+e^x\right ) \log (6)\right )}{(x-3 \log (6))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*x)*((-32 + 32*x)*Log[6]^2 - 96*Log[6]^3) + E^x*((-144*x + 144*x^2)*Log[6] + (432 - 864*x)*Log[
6]^2 + 1296*Log[6]^3))/(-x^3 + 9*x^2*Log[6] - 27*x*Log[6]^2 + 27*Log[6]^3),x]

[Out]

(-16*E^x*Log[6]*(9*x + (-27 + E^x)*Log[6]))/(x - 3*Log[6])^2

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fricas [A]  time = 0.89, size = 44, normalized size = 1.83 \begin {gather*} -\frac {16 \, {\left (e^{\left (2 \, x\right )} \log \relax (6)^{2} + 9 \, {\left (x \log \relax (6) - 3 \, \log \relax (6)^{2}\right )} e^{x}\right )}}{x^{2} - 6 \, x \log \relax (6) + 9 \, \log \relax (6)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-96*log(6)^3+(32*x-32)*log(6)^2)*exp(x)^2+(1296*log(6)^3+(-864*x+432)*log(6)^2+(144*x^2-144*x)*log
(6))*exp(x))/(27*log(6)^3-27*x*log(6)^2+9*x^2*log(6)-x^3),x, algorithm="fricas")

[Out]

-16*(e^(2*x)*log(6)^2 + 9*(x*log(6) - 3*log(6)^2)*e^x)/(x^2 - 6*x*log(6) + 9*log(6)^2)

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giac [A]  time = 0.17, size = 44, normalized size = 1.83 \begin {gather*} -\frac {16 \, {\left (9 \, x e^{x} \log \relax (6) + e^{\left (2 \, x\right )} \log \relax (6)^{2} - 27 \, e^{x} \log \relax (6)^{2}\right )}}{x^{2} - 6 \, x \log \relax (6) + 9 \, \log \relax (6)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-96*log(6)^3+(32*x-32)*log(6)^2)*exp(x)^2+(1296*log(6)^3+(-864*x+432)*log(6)^2+(144*x^2-144*x)*log
(6))*exp(x))/(27*log(6)^3-27*x*log(6)^2+9*x^2*log(6)-x^3),x, algorithm="giac")

[Out]

-16*(9*x*e^x*log(6) + e^(2*x)*log(6)^2 - 27*e^x*log(6)^2)/(x^2 - 6*x*log(6) + 9*log(6)^2)

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maple [A]  time = 0.10, size = 34, normalized size = 1.42

method	result	size
default	\(-\frac {16 \ln \relax (6)^{2} {\mathrm e}^{2 x}}{\left (-3 \ln \relax (6)+x \right )^{2}}-\frac {144 \ln \relax (6) {\mathrm e}^{x}}{-3 \ln \relax (6)+x}\)	\(34\)
norman	\(\frac {432 \ln \relax (6)^{2} {\mathrm e}^{x}-16 \ln \relax (6)^{2} {\mathrm e}^{2 x}-144 \,{\mathrm e}^{x} \ln \relax (6) x}{\left (3 \ln \relax (6)-x \right )^{2}}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-96*ln(6)^3+(32*x-32)*ln(6)^2)*exp(x)^2+(1296*ln(6)^3+(-864*x+432)*ln(6)^2+(144*x^2-144*x)*ln(6))*exp(x)
)/(27*ln(6)^3-27*x*ln(6)^2+9*x^2*ln(6)-x^3),x,method=_RETURNVERBOSE)

[Out]

-16*ln(6)^2*exp(x)^2/(-3*ln(6)+x)^2-144*ln(6)*exp(x)/(-3*ln(6)+x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {279936 \, E_{3}\left (-x + 3 \, \log \relax (6)\right ) \log \relax (6)^{3}}{{\left (x - 3 \, \log \relax (6)\right )}^{2}} + \frac {93312 \, E_{3}\left (-x + 3 \, \log \relax (6)\right ) \log \relax (6)^{2}}{{\left (x - 3 \, \log \relax (6)\right )}^{2}} + \frac {16 \, {\left ({\left (3 \, \log \relax (3)^{3} + 9 \, \log \relax (3)^{2} \log \relax (2) + 9 \, \log \relax (3) \log \relax (2)^{2} + 3 \, \log \relax (2)^{3} - {\left (\log \relax (3)^{2} + 2 \, \log \relax (3) \log \relax (2) + \log \relax (2)^{2}\right )} x\right )} e^{\left (2 \, x\right )} - 9 \, {\left (x^{2} {\left (\log \relax (3) + \log \relax (2)\right )} - 6 \, {\left (\log \relax (3)^{2} + 2 \, \log \relax (3) \log \relax (2) + \log \relax (2)^{2}\right )} x\right )} e^{x}\right )}}{x^{3} - 9 \, x^{2} {\left (\log \relax (3) + \log \relax (2)\right )} - 27 \, \log \relax (3)^{3} - 81 \, \log \relax (3)^{2} \log \relax (2) - 81 \, \log \relax (3) \log \relax (2)^{2} - 27 \, \log \relax (2)^{3} + 27 \, {\left (\log \relax (3)^{2} + 2 \, \log \relax (3) \log \relax (2) + \log \relax (2)^{2}\right )} x} + 16 \, \int \frac {27 \, {\left (6 \, \log \relax (3)^{3} + 18 \, \log \relax (3)^{2} \log \relax (2) + 18 \, \log \relax (3) \log \relax (2)^{2} + 6 \, \log \relax (2)^{3} + {\left (\log \relax (3)^{2} + 2 \, \log \relax (3) \log \relax (2) + \log \relax (2)^{2}\right )} x\right )} e^{x}}{x^{4} - 12 \, x^{3} {\left (\log \relax (3) + \log \relax (2)\right )} + 81 \, \log \relax (3)^{4} + 324 \, \log \relax (3)^{3} \log \relax (2) + 486 \, \log \relax (3)^{2} \log \relax (2)^{2} + 324 \, \log \relax (3) \log \relax (2)^{3} + 81 \, \log \relax (2)^{4} + 54 \, {\left (\log \relax (3)^{2} + 2 \, \log \relax (3) \log \relax (2) + \log \relax (2)^{2}\right )} x^{2} - 108 \, {\left (\log \relax (3)^{3} + 3 \, \log \relax (3)^{2} \log \relax (2) + 3 \, \log \relax (3) \log \relax (2)^{2} + \log \relax (2)^{3}\right )} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-96*log(6)^3+(32*x-32)*log(6)^2)*exp(x)^2+(1296*log(6)^3+(-864*x+432)*log(6)^2+(144*x^2-144*x)*log
(6))*exp(x))/(27*log(6)^3-27*x*log(6)^2+9*x^2*log(6)-x^3),x, algorithm="maxima")

[Out]

279936*exp_integral_e(3, -x + 3*log(6))*log(6)^3/(x - 3*log(6))^2 + 93312*exp_integral_e(3, -x + 3*log(6))*log
(6)^2/(x - 3*log(6))^2 + 16*((3*log(3)^3 + 9*log(3)^2*log(2) + 9*log(3)*log(2)^2 + 3*log(2)^3 - (log(3)^2 + 2*
log(3)*log(2) + log(2)^2)*x)*e^(2*x) - 9*(x^2*(log(3) + log(2)) - 6*(log(3)^2 + 2*log(3)*log(2) + log(2)^2)*x)
*e^x)/(x^3 - 9*x^2*(log(3) + log(2)) - 27*log(3)^3 - 81*log(3)^2*log(2) - 81*log(3)*log(2)^2 - 27*log(2)^3 + 2
7*(log(3)^2 + 2*log(3)*log(2) + log(2)^2)*x) + 16*integrate(27*(6*log(3)^3 + 18*log(3)^2*log(2) + 18*log(3)*lo
g(2)^2 + 6*log(2)^3 + (log(3)^2 + 2*log(3)*log(2) + log(2)^2)*x)*e^x/(x^4 - 12*x^3*(log(3) + log(2)) + 81*log(
3)^4 + 324*log(3)^3*log(2) + 486*log(3)^2*log(2)^2 + 324*log(3)*log(2)^3 + 81*log(2)^4 + 54*(log(3)^2 + 2*log(
3)*log(2) + log(2)^2)*x^2 - 108*(log(3)^3 + 3*log(3)^2*log(2) + 3*log(3)*log(2)^2 + log(2)^3)*x), x)

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mupad [B]  time = 0.45, size = 27, normalized size = 1.12 \begin {gather*} -\frac {16\,{\mathrm {e}}^x\,\ln \relax (6)\,\left (9\,x-27\,\ln \relax (6)+{\mathrm {e}}^x\,\ln \relax (6)\right )}{{\left (x-3\,\ln \relax (6)\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(log(6)*(144*x - 144*x^2) + log(6)^2*(864*x - 432) - 1296*log(6)^3) - exp(2*x)*(log(6)^2*(32*x - 3
2) - 96*log(6)^3))/(27*x*log(6)^2 - 9*x^2*log(6) - 27*log(6)^3 + x^3),x)

[Out]

-(16*exp(x)*log(6)*(9*x - 27*log(6) + exp(x)*log(6)))/(x - 3*log(6))^2

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sympy [B]  time = 0.20, size = 73, normalized size = 3.04 \begin {gather*} \frac {\left (- 16 x \log {\relax (6 )}^{2} + 48 \log {\relax (6 )}^{3}\right ) e^{2 x} + \left (- 144 x^{2} \log {\relax (6 )} + 864 x \log {\relax (6 )}^{2} - 1296 \log {\relax (6 )}^{3}\right ) e^{x}}{x^{3} - 9 x^{2} \log {\relax (6 )} + 27 x \log {\relax (6 )}^{2} - 27 \log {\relax (6 )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-96*ln(6)**3+(32*x-32)*ln(6)**2)*exp(x)**2+(1296*ln(6)**3+(-864*x+432)*ln(6)**2+(144*x**2-144*x)*l
n(6))*exp(x))/(27*ln(6)**3-27*x*ln(6)**2+9*x**2*ln(6)-x**3),x)

[Out]

((-16*x*log(6)**2 + 48*log(6)**3)*exp(2*x) + (-144*x**2*log(6) + 864*x*log(6)**2 - 1296*log(6)**3)*exp(x))/(x*
*3 - 9*x**2*log(6) + 27*x*log(6)**2 - 27*log(6)**3)

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