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

Optimal. Leaf size=19 \[ \frac {\log (5)}{1+\frac {3 e^{e^2} x}{\log (2)}} \]

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Rubi [A]  time = 0.01, antiderivative size = 18, normalized size of antiderivative = 0.95, number of steps used = 3, number of rules used = 3, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {12, 27, 32} \begin {gather*} \frac {\log (2) \log (5)}{3 e^{e^2} x+\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Log[2]*Log[5])/(3*E^E^2*x + Log[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 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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 18, normalized size = 0.95 \begin {gather*} \frac {\log (2) \log (5)}{3 e^{e^2} x+\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Log[2]*Log[5])/(3*E^E^2*x + Log[2])

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fricas [A]  time = 0.59, size = 16, normalized size = 0.84 \begin {gather*} \frac {\log \relax (5) \log \relax (2)}{3 \, x e^{\left (e^{2}\right )} + \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(5)*log(2)/(3*x*e^(e^2) + log(2))

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: -exp(exp(2))*ln(5)*ln(2)*3*1/6/sqrt(-exp
(2*exp(2))+exp(exp(2))^2)/ln(2)*ln(sqrt((18*sageVARx*exp(2*exp(2))+6*exp(exp(2))*ln(2))^2+(-6*sqrt(exp(2*exp(2
))-exp(exp(2))^2)*ln(

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maple [A]  time = 0.38, size = 17, normalized size = 0.89

method	result	size
gosper	\(\frac {\ln \relax (5) \ln \relax (2)}{3 x \,{\mathrm e}^{{\mathrm e}^{2}}+\ln \relax (2)}\)	\(17\)
risch	\(\frac {\ln \relax (5) \ln \relax (2)}{3 x \,{\mathrm e}^{{\mathrm e}^{2}}+\ln \relax (2)}\)	\(17\)
norman	\(-\frac {3 \ln \relax (5) {\mathrm e}^{{\mathrm e}^{2}} x}{3 x \,{\mathrm e}^{{\mathrm e}^{2}}+\ln \relax (2)}\)	\(20\)
meijerg	\(-\frac {3 \ln \relax (5) {\mathrm e}^{{\mathrm e}^{2}} x}{\ln \relax (2) \left (1+\frac {3 x \,{\mathrm e}^{{\mathrm e}^{2}}}{\ln \relax (2)}\right )}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.49, size = 25, normalized size = 1.32 \begin {gather*} \frac {e^{\left (e^{2}\right )} \log \relax (5) \log \relax (2)}{3 \, x e^{\left (2 \, e^{2}\right )} + e^{\left (e^{2}\right )} \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.16, size = 16, normalized size = 0.84 \begin {gather*} \frac {\ln \relax (2)\,\ln \relax (5)}{\ln \relax (2)+3\,x\,{\mathrm {e}}^{{\mathrm {e}}^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 0.14, size = 32, normalized size = 1.68 \begin {gather*} \frac {3 e^{e^{2}} \log {\relax (2 )} \log {\relax (5 )}}{9 x e^{2 e^{2}} + 3 e^{e^{2}} \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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