3.72.49 \(\int \frac {-e^4 x-e^4 \log (16)}{x^3 \log (2)+2 x^2 \log (2) \log (16) \log (3 x)+x \log (2) \log ^2(16) \log ^2(3 x)} \, dx\)

Optimal. Leaf size=19 \[ \frac {e^4}{\log (2) (x+\log (16) \log (3 x))} \]

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Rubi [A]  time = 0.19, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {6688, 12, 6686} \begin {gather*} \frac {e^4}{\log (2) (x+\log (16) \log (3 x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-(E^4*x) - E^4*Log[16])/(x^3*Log[2] + 2*x^2*Log[2]*Log[16]*Log[3*x] + x*Log[2]*Log[16]^2*Log[3*x]^2),x]

[Out]

E^4/(Log[2]*(x + Log[16]*Log[3*x]))

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 {e^4 (-x-\log (16))}{x \log (2) (x+\log (16) \log (3 x))^2} \, dx\\ &=\frac {e^4 \int \frac {-x-\log (16)}{x (x+\log (16) \log (3 x))^2} \, dx}{\log (2)}\\ &=\frac {e^4}{\log (2) (x+\log (16) \log (3 x))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 19, normalized size = 1.00 \begin {gather*} \frac {e^4}{\log (2) (x+\log (16) \log (3 x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-(E^4*x) - E^4*Log[16])/(x^3*Log[2] + 2*x^2*Log[2]*Log[16]*Log[3*x] + x*Log[2]*Log[16]^2*Log[3*x]^2
),x]

[Out]

E^4/(Log[2]*(x + Log[16]*Log[3*x]))

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fricas [A]  time = 0.72, size = 20, normalized size = 1.05 \begin {gather*} \frac {e^{4}}{4 \, \log \relax (2)^{2} \log \left (3 \, x\right ) + x \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*exp(2)^2*log(2)-x*exp(2)^2)/(16*x*log(2)^3*log(3*x)^2+8*x^2*log(2)^2*log(3*x)+x^3*log(2)),x, alg
orithm="fricas")

[Out]

e^4/(4*log(2)^2*log(3*x) + x*log(2))

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giac [A]  time = 0.32, size = 20, normalized size = 1.05 \begin {gather*} \frac {e^{4}}{4 \, \log \relax (2)^{2} \log \left (3 \, x\right ) + x \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*exp(2)^2*log(2)-x*exp(2)^2)/(16*x*log(2)^3*log(3*x)^2+8*x^2*log(2)^2*log(3*x)+x^3*log(2)),x, alg
orithm="giac")

[Out]

e^4/(4*log(2)^2*log(3*x) + x*log(2))

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maple [A]  time = 0.08, size = 20, normalized size = 1.05

method	result	size
risch	\(\frac {{\mathrm e}^{4}}{\ln \relax (2) \left (x +4 \ln \left (3 x \right ) \ln \relax (2)\right )}\)	\(20\)
norman	\(\frac {{\mathrm e}^{4}}{\ln \relax (2) \left (x +4 \ln \left (3 x \right ) \ln \relax (2)\right )}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*exp(2)^2*ln(2)-x*exp(2)^2)/(16*x*ln(2)^3*ln(3*x)^2+8*x^2*ln(2)^2*ln(3*x)+x^3*ln(2)),x,method=_RETURNVE
RBOSE)

[Out]

1/ln(2)/(x+4*ln(3*x)*ln(2))*exp(4)

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maxima [A]  time = 0.49, size = 26, normalized size = 1.37 \begin {gather*} \frac {e^{4}}{4 \, \log \relax (3) \log \relax (2)^{2} + 4 \, \log \relax (2)^{2} \log \relax (x) + x \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*exp(2)^2*log(2)-x*exp(2)^2)/(16*x*log(2)^3*log(3*x)^2+8*x^2*log(2)^2*log(3*x)+x^3*log(2)),x, alg
orithm="maxima")

[Out]

e^4/(4*log(3)*log(2)^2 + 4*log(2)^2*log(x) + x*log(2))

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mupad [B]  time = 4.36, size = 19, normalized size = 1.00 \begin {gather*} \frac {{\mathrm {e}}^4}{\ln \relax (2)\,\left (x+4\,\ln \left (3\,x\right )\,\ln \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*exp(4)*log(2) + x*exp(4))/(x^3*log(2) + 8*x^2*log(3*x)*log(2)^2 + 16*x*log(3*x)^2*log(2)^3),x)

[Out]

exp(4)/(log(2)*(x + 4*log(3*x)*log(2)))

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sympy [A]  time = 0.15, size = 19, normalized size = 1.00 \begin {gather*} \frac {e^{4}}{x \log {\relax (2 )} + 4 \log {\relax (2 )}^{2} \log {\left (3 x \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*exp(2)**2*ln(2)-x*exp(2)**2)/(16*x*ln(2)**3*ln(3*x)**2+8*x**2*ln(2)**2*ln(3*x)+x**3*ln(2)),x)

[Out]

exp(4)/(x*log(2) + 4*log(2)**2*log(3*x))

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