3.48.90 \(\int \frac {4+2 x^3+x^2 \log (9 e^{2 x})-4 \log (2 x)}{81 x^2+x^3 \log (9 e^{2 x})+4 x \log (2 x)} \, dx\)

Optimal. Leaf size=25 \[ \log \left (1+x \log \left (9 e^{2 x}\right )+4 \left (20+\frac {\log (2 x)}{x}\right )\right ) \]

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Rubi [A]  time = 0.22, antiderivative size = 28, normalized size of antiderivative = 1.12, number of steps used = 3, number of rules used = 2, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {6742, 6684} \begin {gather*} \log \left (x^2 \log \left (9 e^{2 x}\right )+81 x+4 \log (2 x)\right )-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4 + 2*x^3 + x^2*Log[9*E^(2*x)] - 4*Log[2*x])/(81*x^2 + x^3*Log[9*E^(2*x)] + 4*x*Log[2*x]),x]

[Out]

-Log[x] + Log[81*x + x^2*Log[9*E^(2*x)] + 4*Log[2*x]]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

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 \left (-\frac {1}{x}+\frac {4+81 x+2 x^3+2 x^2 \log \left (9 e^{2 x}\right )}{x \left (81 x+x^2 \log \left (9 e^{2 x}\right )+4 \log (2 x)\right )}\right ) \, dx\\ &=-\log (x)+\int \frac {4+81 x+2 x^3+2 x^2 \log \left (9 e^{2 x}\right )}{x \left (81 x+x^2 \log \left (9 e^{2 x}\right )+4 \log (2 x)\right )} \, dx\\ &=-\log (x)+\log \left (81 x+x^2 \log \left (9 e^{2 x}\right )+4 \log (2 x)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.37, size = 28, normalized size = 1.12 \begin {gather*} -\log (x)+\log \left (81 x+x^2 \log \left (9 e^{2 x}\right )+4 \log (2 x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 + 2*x^3 + x^2*Log[9*E^(2*x)] - 4*Log[2*x])/(81*x^2 + x^3*Log[9*E^(2*x)] + 4*x*Log[2*x]),x]

[Out]

-Log[x] + Log[81*x + x^2*Log[9*E^(2*x)] + 4*Log[2*x]]

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fricas [A]  time = 0.83, size = 30, normalized size = 1.20 \begin {gather*} \log \left (2 \, x^{3} + 2 \, x^{2} \log \relax (3) + 81 \, x + 4 \, \log \left (2 \, x\right )\right ) - \log \left (2 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(9*exp(x)^2)-4*log(2*x)+2*x^3+4)/(x^3*log(9*exp(x)^2)+4*x*log(2*x)+81*x^2),x, algorithm="fri
cas")

[Out]

log(2*x^3 + 2*x^2*log(3) + 81*x + 4*log(2*x)) - log(2*x)

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giac [A]  time = 0.16, size = 30, normalized size = 1.20 \begin {gather*} \log \left (2 \, x^{3} + 2 \, x^{2} \log \relax (3) + 81 \, x + 4 \, \log \relax (2) + 4 \, \log \relax (x)\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(9*exp(x)^2)-4*log(2*x)+2*x^3+4)/(x^3*log(9*exp(x)^2)+4*x*log(2*x)+81*x^2),x, algorithm="gia
c")

[Out]

log(2*x^3 + 2*x^2*log(3) + 81*x + 4*log(2) + 4*log(x)) - log(x)

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maple [B]  time = 0.12, size = 53, normalized size = 2.12

method	result	size
default	\(-\ln \left (2 x \right )+\ln \left (2 x^{3}+2 x^{2} \left (\ln \left ({\mathrm e}^{x}\right )-x \right )+x^{2} \left (\ln \left (9 \,{\mathrm e}^{2 x}\right )-2 \ln \left ({\mathrm e}^{x}\right )\right )+4 \ln \left (2 x \right )+81 x \right )\)	\(53\)
risch	\(\ln \relax (x )+\ln \left (\ln \left ({\mathrm e}^{x}\right )-\frac {i \left (x^{2} \pi \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )-2 x^{2} \pi \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{2}+x^{2} \pi \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}+4 i \ln \relax (3) x^{2}+162 i x +8 i \ln \left (2 x \right )\right )}{4 x^{2}}\right )\)	\(93\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*ln(9*exp(x)^2)-4*ln(2*x)+2*x^3+4)/(x^3*ln(9*exp(x)^2)+4*x*ln(2*x)+81*x^2),x,method=_RETURNVERBOSE)

[Out]

-ln(2*x)+ln(2*x^3+2*x^2*(ln(exp(x))-x)+x^2*(ln(9*exp(x)^2)-2*ln(exp(x)))+4*ln(2*x)+81*x)

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maxima [A]  time = 0.49, size = 26, normalized size = 1.04 \begin {gather*} \log \left (\frac {1}{2} \, x^{3} + \frac {1}{2} \, x^{2} \log \relax (3) + \frac {81}{4} \, x + \log \relax (2) + \log \relax (x)\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(9*exp(x)^2)-4*log(2*x)+2*x^3+4)/(x^3*log(9*exp(x)^2)+4*x*log(2*x)+81*x^2),x, algorithm="max
ima")

[Out]

log(1/2*x^3 + 1/2*x^2*log(3) + 81/4*x + log(2) + log(x)) - log(x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {x^2\,\ln \left (9\,{\mathrm {e}}^{2\,x}\right )-4\,\ln \left (2\,x\right )+2\,x^3+4}{4\,x\,\ln \left (2\,x\right )+x^3\,\ln \left (9\,{\mathrm {e}}^{2\,x}\right )+81\,x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*log(9*exp(2*x)) - 4*log(2*x) + 2*x^3 + 4)/(4*x*log(2*x) + x^3*log(9*exp(2*x)) + 81*x^2),x)

[Out]

int((x^2*log(9*exp(2*x)) - 4*log(2*x) + 2*x^3 + 4)/(4*x*log(2*x) + x^3*log(9*exp(2*x)) + 81*x^2), x)

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sympy [A]  time = 0.35, size = 27, normalized size = 1.08 \begin {gather*} - \log {\relax (x )} + \log {\left (\frac {x^{3}}{2} + \frac {x^{2} \log {\relax (3 )}}{2} + \frac {81 x}{4} + \log {\left (2 x \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2*ln(9*exp(x)**2)-4*ln(2*x)+2*x**3+4)/(x**3*ln(9*exp(x)**2)+4*x*ln(2*x)+81*x**2),x)

[Out]

-log(x) + log(x**3/2 + x**2*log(3)/2 + 81*x/4 + log(2*x))

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