3.92.97 \(\int \frac {(-3-3 \log (x)) \log (\frac {3+16 x \log (-2+e^5) \log (x)}{4 x \log (-2+e^5) \log (x)})}{6 x \log (x)+32 x^2 \log (-2+e^5) \log ^2(x)} \, dx\) [9197]

Optimal. Leaf size=28 \[ \frac {1}{4} \log ^2\left (4+\frac {3}{4 x \log \left (-2+e^5\right ) \log (x)}\right ) \]

[Out]

1/4*ln(4+3/4/x/ln(x)/ln(exp(5)-2))^2

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Rubi [A]
time = 1.43, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 7, integrand size = 63, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6873, 12, 6874, 2339, 29, 6820, 6818} \begin {gather*} \frac {1}{4} \log ^2\left (\frac {3}{4 x \log \left (e^5-2\right ) \log (x)}+4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-3 - 3*Log[x])*Log[(3 + 16*x*Log[-2 + E^5]*Log[x])/(4*x*Log[-2 + E^5]*Log[x])])/(6*x*Log[x] + 32*x^2*Log
[-2 + E^5]*Log[x]^2),x]

[Out]

Log[4 + 3/(4*x*Log[-2 + E^5]*Log[x])]^2/4

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2339

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 6818

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 6820

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

Rule 6873

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

Rule 6874

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 (-1-\log (x)) \log \left (\frac {3+16 x \log \left (-2+e^5\right ) \log (x)}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{2 x \log (x) \left (3+16 x \log \left (-2+e^5\right ) \log (x)\right )} \, dx\\ &=\frac {3}{2} \int \frac {(-1-\log (x)) \log \left (\frac {3+16 x \log \left (-2+e^5\right ) \log (x)}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{x \log (x) \left (3+16 x \log \left (-2+e^5\right ) \log (x)\right )} \, dx\\ &=\frac {3}{2} \int \frac {(-1-\log (x)) \log \left (4+\frac {3}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{x \log (x) \left (3+16 x \log \left (-2+e^5\right ) \log (x)\right )} \, dx\\ &=\frac {1}{4} \log ^2\left (4+\frac {3}{4 x \log \left (-2+e^5\right ) \log (x)}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.04, size = 28, normalized size = 1.00 \begin {gather*} \frac {1}{4} \log ^2\left (4+\frac {3}{4 x \log \left (-2+e^5\right ) \log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-3 - 3*Log[x])*Log[(3 + 16*x*Log[-2 + E^5]*Log[x])/(4*x*Log[-2 + E^5]*Log[x])])/(6*x*Log[x] + 32*x
^2*Log[-2 + E^5]*Log[x]^2),x]

[Out]

Log[4 + 3/(4*x*Log[-2 + E^5]*Log[x])]^2/4

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(96\) vs. \(2(23)=46\).
time = 11.71, size = 97, normalized size = 3.46

method	result	size
default	\(\frac {\ln \left (x \right ) \ln \left (\ln \left ({\mathrm e}^{5}-2\right )\right )}{2}+\frac {\ln \left (\ln \left ({\mathrm e}^{5}-2\right )\right ) \ln \left (\ln \left (x \right )\right )}{2}-\frac {\ln \left (\ln \left ({\mathrm e}^{5}-2\right )\right ) \ln \left (16 x \ln \left (x \right ) \ln \left ({\mathrm e}^{5}-2\right )+3\right )}{2}+\ln \left (2\right ) \ln \left (x \right )+\ln \left (2\right ) \ln \left (\ln \left (x \right )\right )-\ln \left (2\right ) \ln \left (16 x \ln \left (x \right ) \ln \left ({\mathrm e}^{5}-2\right )+3\right )+\frac {\ln \left (\frac {16 x \ln \left (x \right ) \ln \left ({\mathrm e}^{5}-2\right )+3}{x \ln \left (x \right )}\right )^{2}}{4}\)	\(97\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3*ln(x)-3)*ln(1/4*(16*x*ln(x)*ln(exp(5)-2)+3)/x/ln(x)/ln(exp(5)-2))/(32*x^2*ln(x)^2*ln(exp(5)-2)+6*x*ln(
x)),x,method=_RETURNVERBOSE)

[Out]

1/2*ln(x)*ln(ln(exp(5)-2))+1/2*ln(ln(exp(5)-2))*ln(ln(x))-1/2*ln(ln(exp(5)-2))*ln(16*x*ln(x)*ln(exp(5)-2)+3)+l
n(2)*ln(x)+ln(2)*ln(ln(x))-ln(2)*ln(16*x*ln(x)*ln(exp(5)-2)+3)+1/4*ln((16*x*ln(x)*ln(exp(5)-2)+3)/x/ln(x))^2

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*log(x)-3)*log(1/4*(16*x*log(x)*log(exp(5)-2)+3)/x/log(x)/log(exp(5)-2))/(32*x^2*log(x)^2*log(exp
(5)-2)+6*x*log(x)),x, algorithm="maxima")

[Out]

-3/2*integrate((log(x) + 1)*log(1/4*(16*x*log(x)*log(e^5 - 2) + 3)/(x*log(x)*log(e^5 - 2)))/(16*x^2*log(x)^2*l
og(e^5 - 2) + 3*x*log(x)), x)

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Fricas [A]
time = 0.42, size = 33, normalized size = 1.18 \begin {gather*} \frac {1}{4} \, \log \left (\frac {16 \, x \log \left (x\right ) \log \left (e^{5} - 2\right ) + 3}{4 \, x \log \left (x\right ) \log \left (e^{5} - 2\right )}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*log(x)-3)*log(1/4*(16*x*log(x)*log(exp(5)-2)+3)/x/log(x)/log(exp(5)-2))/(32*x^2*log(x)^2*log(exp
(5)-2)+6*x*log(x)),x, algorithm="fricas")

[Out]

1/4*log(1/4*(16*x*log(x)*log(e^5 - 2) + 3)/(x*log(x)*log(e^5 - 2)))^2

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Sympy [A]
time = 0.13, size = 32, normalized size = 1.14 \begin {gather*} \frac {\log {\left (\frac {4 x \log {\left (x \right )} \log {\left (-2 + e^{5} \right )} + \frac {3}{4}}{x \log {\left (x \right )} \log {\left (-2 + e^{5} \right )}} \right )}^{2}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*ln(x)-3)*ln(1/4*(16*x*ln(x)*ln(exp(5)-2)+3)/x/ln(x)/ln(exp(5)-2))/(32*x**2*ln(x)**2*ln(exp(5)-2)
+6*x*ln(x)),x)

[Out]

log((4*x*log(x)*log(-2 + exp(5)) + 3/4)/(x*log(x)*log(-2 + exp(5))))**2/4

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*log(x)-3)*log(1/4*(16*x*log(x)*log(exp(5)-2)+3)/x/log(x)/log(exp(5)-2))/(32*x^2*log(x)^2*log(exp
(5)-2)+6*x*log(x)),x, algorithm="giac")

[Out]

integrate(-3/2*(log(x) + 1)*log(1/4*(16*x*log(x)*log(e^5 - 2) + 3)/(x*log(x)*log(e^5 - 2)))/(16*x^2*log(x)^2*l
og(e^5 - 2) + 3*x*log(x)), x)

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Mupad [B]
time = 7.54, size = 32, normalized size = 1.14 \begin {gather*} \frac {{\ln \left (\frac {4\,x\,\ln \left ({\mathrm {e}}^5-2\right )\,\ln \left (x\right )+\frac {3}{4}}{x\,\ln \left ({\mathrm {e}}^5-2\right )\,\ln \left (x\right )}\right )}^2}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log((4*x*log(exp(5) - 2)*log(x) + 3/4)/(x*log(exp(5) - 2)*log(x)))*(3*log(x) + 3))/(6*x*log(x) + 32*x^2*
log(exp(5) - 2)*log(x)^2),x)

[Out]

log((4*x*log(exp(5) - 2)*log(x) + 3/4)/(x*log(exp(5) - 2)*log(x)))^2/4

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