3.46.99 \(\int \frac {(4+2 e^4+2 \log (x)) \log (e^8 x+2 e^4 x \log (x)+x \log ^2(x))}{3 e^4 x+e^4 x \log (2)+(3 x+x \log (2)) \log (x)+(e^4 x+x \log (x)) \log ^2(e^8 x+2 e^4 x \log (x)+x \log ^2(x))} \, dx\)

Optimal. Leaf size=18 \[ \log \left (3+\log (2)+\log ^2\left (x \left (e^4+\log (x)\right )^2\right )\right ) \]

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Rubi [A]  time = 0.57, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 6, integrand size = 94, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.064, Rules used = {6, 6688, 12, 6696, 203, 6684} \begin {gather*} \log \left (\log ^2\left (x \left (\log (x)+e^4\right )^2\right )+3+\log (2)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((4 + 2*E^4 + 2*Log[x])*Log[E^8*x + 2*E^4*x*Log[x] + x*Log[x]^2])/(3*E^4*x + E^4*x*Log[2] + (3*x + x*Log[2
])*Log[x] + (E^4*x + x*Log[x])*Log[E^8*x + 2*E^4*x*Log[x] + x*Log[x]^2]^2),x]

[Out]

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

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 6684

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

Rule 6688

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

Rule 6696

Int[(u_.)*((a_.) + (b_.)*(y_)^(n_))^(p_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Dist[q, Subst[In
t[(a + b*x^n)^p, x], x, y], x] /;  !FalseQ[q]] /; FreeQ[{a, b, n, p}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (4+2 e^4+2 \log (x)\right ) \log \left (e^8 x+2 e^4 x \log (x)+x \log ^2(x)\right )}{e^4 x (3+\log (2))+(3 x+x \log (2)) \log (x)+\left (e^4 x+x \log (x)\right ) \log ^2\left (e^8 x+2 e^4 x \log (x)+x \log ^2(x)\right )} \, dx\\ &=\int \frac {2 \left (2 \left (1+\frac {e^4}{2}\right )+\log (x)\right ) \log \left (x \left (e^4+\log (x)\right )^2\right )}{x \left (e^4+\log (x)\right ) \left (3 \left (1+\frac {\log (2)}{3}\right )+\log ^2\left (x \left (e^4+\log (x)\right )^2\right )\right )} \, dx\\ &=2 \int \frac {\left (2 \left (1+\frac {e^4}{2}\right )+\log (x)\right ) \log \left (x \left (e^4+\log (x)\right )^2\right )}{x \left (e^4+\log (x)\right ) \left (3 \left (1+\frac {\log (2)}{3}\right )+\log ^2\left (x \left (e^4+\log (x)\right )^2\right )\right )} \, dx\\ &=\log \left (3+\log (2)+\log ^2\left (x \left (e^4+\log (x)\right )^2\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 18, normalized size = 1.00 \begin {gather*} \log \left (3+\log (2)+\log ^2\left (x \left (e^4+\log (x)\right )^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((4 + 2*E^4 + 2*Log[x])*Log[E^8*x + 2*E^4*x*Log[x] + x*Log[x]^2])/(3*E^4*x + E^4*x*Log[2] + (3*x + x
*Log[2])*Log[x] + (E^4*x + x*Log[x])*Log[E^8*x + 2*E^4*x*Log[x] + x*Log[x]^2]^2),x]

[Out]

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

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fricas [A]  time = 0.59, size = 26, normalized size = 1.44 \begin {gather*} \log \left (\log \left (2 \, x e^{4} \log \relax (x) + x \log \relax (x)^{2} + x e^{8}\right )^{2} + \log \relax (2) + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(x)+2*exp(4)+4)*log(x*log(x)^2+2*x*exp(4)*log(x)+x*exp(4)^2)/((x*log(x)+x*exp(4))*log(x*log(x)
^2+2*x*exp(4)*log(x)+x*exp(4)^2)^2+(x*log(2)+3*x)*log(x)+x*exp(4)*log(2)+3*x*exp(4)),x, algorithm="fricas")

[Out]

log(log(2*x*e^4*log(x) + x*log(x)^2 + x*e^8)^2 + log(2) + 3)

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giac [B]  time = 0.68, size = 43, normalized size = 2.39 \begin {gather*} \log \left (\log \left (2 \, e^{4} \log \relax (x) + \log \relax (x)^{2} + e^{8}\right )^{2} + 2 \, \log \left (2 \, e^{4} \log \relax (x) + \log \relax (x)^{2} + e^{8}\right ) \log \relax (x) + \log \relax (x)^{2} + \log \relax (2) + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(x)+2*exp(4)+4)*log(x*log(x)^2+2*x*exp(4)*log(x)+x*exp(4)^2)/((x*log(x)+x*exp(4))*log(x*log(x)
^2+2*x*exp(4)*log(x)+x*exp(4)^2)^2+(x*log(2)+3*x)*log(x)+x*exp(4)*log(2)+3*x*exp(4)),x, algorithm="giac")

[Out]

log(log(2*e^4*log(x) + log(x)^2 + e^8)^2 + 2*log(2*e^4*log(x) + log(x)^2 + e^8)*log(x) + log(x)^2 + log(2) + 3
)

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maple [A]  time = 0.65, size = 29, normalized size = 1.61

method	result	size
norman	\(\ln \left (\ln \left (x \ln \relax (x )^{2}+2 x \,{\mathrm e}^{4} \ln \relax (x )+x \,{\mathrm e}^{8}\right )^{2}+\ln \relax (2)+3\right )\)	\(29\)
risch	\(\text {Expression too large to display}\)	\(1283\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(ln(x*ln(x)^2+2*x*exp(4)*ln(x)+x*exp(4)^2)^2+ln(2)+3)

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maxima [A]  time = 0.48, size = 30, normalized size = 1.67 \begin {gather*} \log \left (\frac {1}{4} \, \log \relax (x)^{2} + \log \relax (x) \log \left (e^{4} + \log \relax (x)\right ) + \log \left (e^{4} + \log \relax (x)\right )^{2} + \frac {1}{4} \, \log \relax (2) + \frac {3}{4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(x)+2*exp(4)+4)*log(x*log(x)^2+2*x*exp(4)*log(x)+x*exp(4)^2)/((x*log(x)+x*exp(4))*log(x*log(x)
^2+2*x*exp(4)*log(x)+x*exp(4)^2)^2+(x*log(2)+3*x)*log(x)+x*exp(4)*log(2)+3*x*exp(4)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 6.02, size = 26, normalized size = 1.44 \begin {gather*} \ln \left ({\ln \left (x\,{\ln \relax (x)}^2+2\,x\,{\mathrm {e}}^4\,\ln \relax (x)+x\,{\mathrm {e}}^8\right )}^2+\ln \relax (2)+3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(log(2) + log(x*log(x)^2 + x*exp(8) + 2*x*exp(4)*log(x))^2 + 3)

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sympy [A]  time = 0.62, size = 31, normalized size = 1.72 \begin {gather*} \log {\left (\log {\left (x \log {\relax (x )}^{2} + 2 x e^{4} \log {\relax (x )} + x e^{8} \right )}^{2} + \log {\relax (2 )} + 3 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(log(x*log(x)**2 + 2*x*exp(4)*log(x) + x*exp(8))**2 + log(2) + 3)

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