3.60.95 \(\int \frac {-128+e^x (-128-128 x)}{81+324 x+432 x^2+192 x^3+192 e^{3 x} x^3+(324+864 x+576 x^2) \log (3)+(432+576 x) \log ^2(3)+192 \log ^3(3)+e^{2 x} (432 x^2+576 x^3+576 x^2 \log (3))+e^x (324 x+864 x^2+576 x^3+(864 x+1152 x^2) \log (3)+576 x \log ^2(3))} \, dx\)

Optimal. Leaf size=18 \[ \frac {1}{3 \left (\frac {3}{4}+x+e^x x+\log (3)\right )^2} \]

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Rubi [A]  time = 0.27, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 129, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {6688, 12, 6686} \begin {gather*} \frac {16}{3 \left (4 \left (e^x+1\right ) x+3+\log (81)\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-128 + E^x*(-128 - 128*x))/(81 + 324*x + 432*x^2 + 192*x^3 + 192*E^(3*x)*x^3 + (324 + 864*x + 576*x^2)*Lo
g[3] + (432 + 576*x)*Log[3]^2 + 192*Log[3]^3 + E^(2*x)*(432*x^2 + 576*x^3 + 576*x^2*Log[3]) + E^x*(324*x + 864
*x^2 + 576*x^3 + (864*x + 1152*x^2)*Log[3] + 576*x*Log[3]^2)),x]

[Out]

16/(3*(3 + 4*(1 + E^x)*x + Log[81])^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 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {128 \left (-1-e^x (1+x)\right )}{3 \left (4 \left (1+e^x\right ) x+3 \left (1+\frac {\log (81)}{3}\right )\right )^3} \, dx\\ &=\frac {128}{3} \int \frac {-1-e^x (1+x)}{\left (4 \left (1+e^x\right ) x+3 \left (1+\frac {\log (81)}{3}\right )\right )^3} \, dx\\ &=\frac {16}{3 \left (3+4 \left (1+e^x\right ) x+\log (81)\right )^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 18, normalized size = 1.00 \begin {gather*} \frac {16}{3 \left (3+4 \left (1+e^x\right ) x+\log (81)\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-128 + E^x*(-128 - 128*x))/(81 + 324*x + 432*x^2 + 192*x^3 + 192*E^(3*x)*x^3 + (324 + 864*x + 576*x
^2)*Log[3] + (432 + 576*x)*Log[3]^2 + 192*Log[3]^3 + E^(2*x)*(432*x^2 + 576*x^3 + 576*x^2*Log[3]) + E^x*(324*x
+ 864*x^2 + 576*x^3 + (864*x + 1152*x^2)*Log[3] + 576*x*Log[3]^2)),x]

[Out]

16/(3*(3 + 4*(1 + E^x)*x + Log[81])^2)

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fricas [B]  time = 0.94, size = 56, normalized size = 3.11 \begin {gather*} \frac {16}{3 \, {\left (16 \, x^{2} e^{\left (2 \, x\right )} + 16 \, x^{2} + 8 \, {\left (4 \, x^{2} + 4 \, x \log \relax (3) + 3 \, x\right )} e^{x} + 8 \, {\left (4 \, x + 3\right )} \log \relax (3) + 16 \, \log \relax (3)^{2} + 24 \, x + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-128*x-128)*exp(x)-128)/(192*x^3*exp(x)^3+(576*x^2*log(3)+576*x^3+432*x^2)*exp(x)^2+(576*x*log(3)^
2+(1152*x^2+864*x)*log(3)+576*x^3+864*x^2+324*x)*exp(x)+192*log(3)^3+(576*x+432)*log(3)^2+(576*x^2+864*x+324)*
log(3)+192*x^3+432*x^2+324*x+81),x, algorithm="fricas")

[Out]

16/3/(16*x^2*e^(2*x) + 16*x^2 + 8*(4*x^2 + 4*x*log(3) + 3*x)*e^x + 8*(4*x + 3)*log(3) + 16*log(3)^2 + 24*x + 9
)

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giac [B]  time = 0.24, size = 57, normalized size = 3.17 \begin {gather*} \frac {16}{3 \, {\left (16 \, x^{2} e^{\left (2 \, x\right )} + 32 \, x^{2} e^{x} + 32 \, x e^{x} \log \relax (3) + 16 \, x^{2} + 24 \, x e^{x} + 32 \, x \log \relax (3) + 16 \, \log \relax (3)^{2} + 24 \, x + 24 \, \log \relax (3) + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-128*x-128)*exp(x)-128)/(192*x^3*exp(x)^3+(576*x^2*log(3)+576*x^3+432*x^2)*exp(x)^2+(576*x*log(3)^
2+(1152*x^2+864*x)*log(3)+576*x^3+864*x^2+324*x)*exp(x)+192*log(3)^3+(576*x+432)*log(3)^2+(576*x^2+864*x+324)*
log(3)+192*x^3+432*x^2+324*x+81),x, algorithm="giac")

[Out]

16/3/(16*x^2*e^(2*x) + 32*x^2*e^x + 32*x*e^x*log(3) + 16*x^2 + 24*x*e^x + 32*x*log(3) + 16*log(3)^2 + 24*x + 2
4*log(3) + 9)

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maple [A]  time = 0.34, size = 19, normalized size = 1.06

method	result	size
norman	\(\frac {16}{3 \left (4 \,{\mathrm e}^{x} x +4 \ln \relax (3)+4 x +3\right )^{2}}\)	\(19\)
risch	\(\frac {16}{3 \left (4 \,{\mathrm e}^{x} x +4 \ln \relax (3)+4 x +3\right )^{2}}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-128*x-128)*exp(x)-128)/(192*x^3*exp(x)^3+(576*x^2*ln(3)+576*x^3+432*x^2)*exp(x)^2+(576*x*ln(3)^2+(1152*
x^2+864*x)*ln(3)+576*x^3+864*x^2+324*x)*exp(x)+192*ln(3)^3+(576*x+432)*ln(3)^2+(576*x^2+864*x+324)*ln(3)+192*x
^3+432*x^2+324*x+81),x,method=_RETURNVERBOSE)

[Out]

16/3/(4*exp(x)*x+4*ln(3)+4*x+3)^2

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maxima [B]  time = 0.51, size = 57, normalized size = 3.17 \begin {gather*} \frac {16}{3 \, {\left (16 \, x^{2} e^{\left (2 \, x\right )} + 16 \, x^{2} + 8 \, x {\left (4 \, \log \relax (3) + 3\right )} + 8 \, {\left (4 \, x^{2} + x {\left (4 \, \log \relax (3) + 3\right )}\right )} e^{x} + 16 \, \log \relax (3)^{2} + 24 \, \log \relax (3) + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-128*x-128)*exp(x)-128)/(192*x^3*exp(x)^3+(576*x^2*log(3)+576*x^3+432*x^2)*exp(x)^2+(576*x*log(3)^
2+(1152*x^2+864*x)*log(3)+576*x^3+864*x^2+324*x)*exp(x)+192*log(3)^3+(576*x+432)*log(3)^2+(576*x^2+864*x+324)*
log(3)+192*x^3+432*x^2+324*x+81),x, algorithm="maxima")

[Out]

16/3/(16*x^2*e^(2*x) + 16*x^2 + 8*x*(4*log(3) + 3) + 8*(4*x^2 + x*(4*log(3) + 3))*e^x + 16*log(3)^2 + 24*log(3
) + 9)

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mupad [B]  time = 5.29, size = 123, normalized size = 6.83 \begin {gather*} -\frac {\frac {128\,x}{4\,\ln \relax (3)+3}+\frac {256\,x^2}{{\left (4\,\ln \relax (3)+3\right )}^2}+\frac {512\,x^2\,{\mathrm {e}}^x}{{\left (4\,\ln \relax (3)+3\right )}^2}+\frac {256\,x^2\,{\mathrm {e}}^{2\,x}}{{\left (\ln \left (81\right )+3\right )}^2}+\frac {128\,x\,{\mathrm {e}}^x}{\ln \left (81\right )+3}}{72\,x+72\,\ln \relax (3)+96\,x^2\,{\mathrm {e}}^x+96\,x\,\ln \relax (3)+48\,x^2\,{\mathrm {e}}^{2\,x}+72\,x\,{\mathrm {e}}^x+48\,{\ln \relax (3)}^2+48\,x^2+96\,x\,{\mathrm {e}}^x\,\ln \relax (3)+27} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(128*x + 128) + 128)/(324*x + exp(2*x)*(576*x^2*log(3) + 432*x^2 + 576*x^3) + log(3)*(864*x + 576
*x^2 + 324) + log(3)^2*(576*x + 432) + 192*x^3*exp(3*x) + exp(x)*(324*x + log(3)*(864*x + 1152*x^2) + 576*x*lo
g(3)^2 + 864*x^2 + 576*x^3) + 192*log(3)^3 + 432*x^2 + 192*x^3 + 81),x)

[Out]

-((128*x)/(4*log(3) + 3) + (256*x^2)/(4*log(3) + 3)^2 + (512*x^2*exp(x))/(4*log(3) + 3)^2 + (256*x^2*exp(2*x))
/(log(81) + 3)^2 + (128*x*exp(x))/(log(81) + 3))/(72*x + 72*log(3) + 96*x^2*exp(x) + 96*x*log(3) + 48*x^2*exp(
2*x) + 72*x*exp(x) + 48*log(3)^2 + 48*x^2 + 96*x*exp(x)*log(3) + 27)

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sympy [B]  time = 0.30, size = 58, normalized size = 3.22 \begin {gather*} \frac {16}{48 x^{2} e^{2 x} + 48 x^{2} + 72 x + 96 x \log {\relax (3 )} + \left (96 x^{2} + 72 x + 96 x \log {\relax (3 )}\right ) e^{x} + 27 + 48 \log {\relax (3 )}^{2} + 72 \log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-128*x-128)*exp(x)-128)/(192*x**3*exp(x)**3+(576*x**2*ln(3)+576*x**3+432*x**2)*exp(x)**2+(576*x*ln
(3)**2+(1152*x**2+864*x)*ln(3)+576*x**3+864*x**2+324*x)*exp(x)+192*ln(3)**3+(576*x+432)*ln(3)**2+(576*x**2+864
*x+324)*ln(3)+192*x**3+432*x**2+324*x+81),x)

[Out]

16/(48*x**2*exp(2*x) + 48*x**2 + 72*x + 96*x*log(3) + (96*x**2 + 72*x + 96*x*log(3))*exp(x) + 27 + 48*log(3)**
2 + 72*log(3))

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