3.11.80 \(\int \frac {(324-162 e^4) \log (4) \log (x)+(162-81 e^4) \log (4) \log ^2(x)}{164025+(4050 x-810 x \log (4)) \log ^2(x)+(25 x^2-10 x^2 \log (4)+x^2 \log ^2(4)) \log ^4(x)} \, dx\)

Optimal. Leaf size=24 \[ \frac {-2+e^4}{-5+\frac {x \log (4)}{x+\frac {81}{\log ^2(x)}}} \]

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Rubi [A]  time = 0.27, antiderivative size = 35, normalized size of antiderivative = 1.46, number of steps used = 3, number of rules used = 4, integrand size = 72, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {6688, 12, 6742, 6686} \begin {gather*} -\frac {81 \left (2-e^4\right ) \log (4)}{(5-\log (4)) \left (x (5-\log (4)) \log ^2(x)+405\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((324 - 162*E^4)*Log[4]*Log[x] + (162 - 81*E^4)*Log[4]*Log[x]^2)/(164025 + (4050*x - 810*x*Log[4])*Log[x]^
2 + (25*x^2 - 10*x^2*Log[4] + x^2*Log[4]^2)*Log[x]^4),x]

[Out]

(-81*(2 - E^4)*Log[4])/((5 - Log[4])*(405 + x*(5 - Log[4])*Log[x]^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]]

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 {81 \left (2-e^4\right ) \log (4) \log (x) (2+\log (x))}{\left (405-x (-5+\log (4)) \log ^2(x)\right )^2} \, dx\\ &=\left (81 \left (2-e^4\right ) \log (4)\right ) \int \frac {\log (x) (2+\log (x))}{\left (405-x (-5+\log (4)) \log ^2(x)\right )^2} \, dx\\ &=-\frac {81 \left (2-e^4\right ) \log (4)}{(5-\log (4)) \left (405+x (5-\log (4)) \log ^2(x)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 29, normalized size = 1.21 \begin {gather*} \frac {81 \left (-2+e^4\right ) \log (4)}{(-5+\log (4)) \left (-405+x (-5+\log (4)) \log ^2(x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((324 - 162*E^4)*Log[4]*Log[x] + (162 - 81*E^4)*Log[4]*Log[x]^2)/(164025 + (4050*x - 810*x*Log[4])*L
og[x]^2 + (25*x^2 - 10*x^2*Log[4] + x^2*Log[4]^2)*Log[x]^4),x]

[Out]

(81*(-2 + E^4)*Log[4])/((-5 + Log[4])*(-405 + x*(-5 + Log[4])*Log[x]^2))

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fricas [A]  time = 1.11, size = 37, normalized size = 1.54 \begin {gather*} \frac {162 \, {\left (e^{4} - 2\right )} \log \relax (2)}{{\left (4 \, x \log \relax (2)^{2} - 20 \, x \log \relax (2) + 25 \, x\right )} \log \relax (x)^{2} - 810 \, \log \relax (2) + 2025} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-81*exp(4)+162)*log(2)*log(x)^2+2*(-162*exp(4)+324)*log(2)*log(x))/((4*x^2*log(2)^2-20*x^2*log(2
)+25*x^2)*log(x)^4+(-1620*x*log(2)+4050*x)*log(x)^2+164025),x, algorithm="fricas")

[Out]

162*(e^4 - 2)*log(2)/((4*x*log(2)^2 - 20*x*log(2) + 25*x)*log(x)^2 - 810*log(2) + 2025)

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giac [A]  time = 0.84, size = 47, normalized size = 1.96 \begin {gather*} \frac {162 \, {\left (e^{4} \log \relax (2) - 2 \, \log \relax (2)\right )}}{4 \, x \log \relax (2)^{2} \log \relax (x)^{2} - 20 \, x \log \relax (2) \log \relax (x)^{2} + 25 \, x \log \relax (x)^{2} - 810 \, \log \relax (2) + 2025} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-81*exp(4)+162)*log(2)*log(x)^2+2*(-162*exp(4)+324)*log(2)*log(x))/((4*x^2*log(2)^2-20*x^2*log(2
)+25*x^2)*log(x)^4+(-1620*x*log(2)+4050*x)*log(x)^2+164025),x, algorithm="giac")

[Out]

162*(e^4*log(2) - 2*log(2))/(4*x*log(2)^2*log(x)^2 - 20*x*log(2)*log(x)^2 + 25*x*log(x)^2 - 810*log(2) + 2025)

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maple [A]  time = 0.15, size = 38, normalized size = 1.58

method	result	size
norman	\(\frac {\left (\frac {2 \,{\mathrm e}^{4} \ln \relax (2)}{5}-\frac {4 \ln \relax (2)}{5}\right ) x \ln \relax (x )^{2}}{2 \ln \relax (2) \ln \relax (x )^{2} x -5 x \ln \relax (x )^{2}-405}\)	\(38\)
risch	\(\frac {162 \ln \relax (2) {\mathrm e}^{4}}{\left (2 \ln \relax (2)-5\right ) \left (2 \ln \relax (2) \ln \relax (x )^{2} x -5 x \ln \relax (x )^{2}-405\right )}-\frac {324 \ln \relax (2)}{\left (2 \ln \relax (2)-5\right ) \left (2 \ln \relax (2) \ln \relax (x )^{2} x -5 x \ln \relax (x )^{2}-405\right )}\)	\(68\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*(-81*exp(4)+162)*ln(2)*ln(x)^2+2*(-162*exp(4)+324)*ln(2)*ln(x))/((4*x^2*ln(2)^2-20*x^2*ln(2)+25*x^2)*ln
(x)^4+(-1620*x*ln(2)+4050*x)*ln(x)^2+164025),x,method=_RETURNVERBOSE)

[Out]

(2/5*exp(4)*ln(2)-4/5*ln(2))*x*ln(x)^2/(2*ln(2)*ln(x)^2*x-5*x*ln(x)^2-405)

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maxima [A]  time = 0.48, size = 38, normalized size = 1.58 \begin {gather*} \frac {162 \, {\left (e^{4} \log \relax (2) - 2 \, \log \relax (2)\right )}}{{\left (4 \, \log \relax (2)^{2} - 20 \, \log \relax (2) + 25\right )} x \log \relax (x)^{2} - 810 \, \log \relax (2) + 2025} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-81*exp(4)+162)*log(2)*log(x)^2+2*(-162*exp(4)+324)*log(2)*log(x))/((4*x^2*log(2)^2-20*x^2*log(2
)+25*x^2)*log(x)^4+(-1620*x*log(2)+4050*x)*log(x)^2+164025),x, algorithm="maxima")

[Out]

162*(e^4*log(2) - 2*log(2))/((4*log(2)^2 - 20*log(2) + 25)*x*log(x)^2 - 810*log(2) + 2025)

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mupad [B]  time = 3.11, size = 45, normalized size = 1.88 \begin {gather*} \frac {x^2\,{\ln \relax (x)}^2\,\left (\frac {4\,\ln \relax (2)}{5}-\frac {2\,{\mathrm {e}}^4\,\ln \relax (2)}{5}\right )}{405\,x+5\,x^2\,{\ln \relax (x)}^2-2\,x^2\,\ln \relax (2)\,{\ln \relax (x)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*log(2)*log(x)^2*(81*exp(4) - 162) + 2*log(2)*log(x)*(162*exp(4) - 324))/(log(x)^2*(4050*x - 1620*x*log
(2)) + log(x)^4*(4*x^2*log(2)^2 - 20*x^2*log(2) + 25*x^2) + 164025),x)

[Out]

(x^2*log(x)^2*((4*log(2))/5 - (2*exp(4)*log(2))/5))/(405*x + 5*x^2*log(x)^2 - 2*x^2*log(2)*log(x)^2)

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sympy [B]  time = 0.24, size = 42, normalized size = 1.75 \begin {gather*} \frac {- 324 \log {\relax (2 )} + 162 e^{4} \log {\relax (2 )}}{\left (- 20 x \log {\relax (2 )} + 4 x \log {\relax (2 )}^{2} + 25 x\right ) \log {\relax (x )}^{2} - 810 \log {\relax (2 )} + 2025} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-81*exp(4)+162)*ln(2)*ln(x)**2+2*(-162*exp(4)+324)*ln(2)*ln(x))/((4*x**2*ln(2)**2-20*x**2*ln(2)+
25*x**2)*ln(x)**4+(-1620*x*ln(2)+4050*x)*ln(x)**2+164025),x)

[Out]

(-324*log(2) + 162*exp(4)*log(2))/((-20*x*log(2) + 4*x*log(2)**2 + 25*x)*log(x)**2 - 810*log(2) + 2025)

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