\(\int \frac {e^{\frac {3}{\log (1+\log (3 x))}} (12+3 x)+(x+12 x^3+3 x^4+(x+12 x^3+3 x^4) \log (3 x)) \log ^2(1+\log (3 x))}{(4 x+x^2+(4 x+x^2) \log (3 x)) \log ^2(1+\log (3 x))} \, dx\) [2329]

Optimal result

Integrand size = 90, antiderivative size = 23 \[ \int \frac {e^{\frac {3}{\log (1+\log (3 x))}} (12+3 x)+\left (x+12 x^3+3 x^4+\left (x+12 x^3+3 x^4\right ) \log (3 x)\right ) \log ^2(1+\log (3 x))}{\left (4 x+x^2+\left (4 x+x^2\right ) \log (3 x)\right ) \log ^2(1+\log (3 x))} \, dx=-e^{\frac {3}{\log (1+\log (3 x))}}+x^3+\log (4+x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {6820, 1864, 6838} \[ \int \frac {e^{\frac {3}{\log (1+\log (3 x))}} (12+3 x)+\left (x+12 x^3+3 x^4+\left (x+12 x^3+3 x^4\right ) \log (3 x)\right ) \log ^2(1+\log (3 x))}{\left (4 x+x^2+\left (4 x+x^2\right ) \log (3 x)\right ) \log ^2(1+\log (3 x))} \, dx=x^3-e^{\frac {3}{\log (\log (3 x)+1)}}+\log (x+4) \]

[In]

Int[(E^(3/Log[1 + Log[3*x]])*(12 + 3*x) + (x + 12*x^3 + 3*x^4 + (x + 12*x^3 + 3*x^4)*Log[3*x])*Log[1 + Log[3*x
]]^2)/((4*x + x^2 + (4*x + x^2)*Log[3*x])*Log[1 + Log[3*x]]^2),x]

[Out]

-E^(3/Log[1 + Log[3*x]]) + x^3 + Log[4 + x]

Rule 1864

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rule 6820

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

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1+12 x^2+3 x^3}{4+x}+\frac {3 e^{\frac {3}{\log (1+\log (3 x))}}}{x (1+\log (3 x)) \log ^2(1+\log (3 x))}\right ) \, dx \\ & = 3 \int \frac {e^{\frac {3}{\log (1+\log (3 x))}}}{x (1+\log (3 x)) \log ^2(1+\log (3 x))} \, dx+\int \frac {1+12 x^2+3 x^3}{4+x} \, dx \\ & = 3 \text {Subst}\left (\int \frac {e^{\frac {3}{\log (1+x)}}}{(1+x) \log ^2(1+x)} \, dx,x,\log (3 x)\right )+\int \left (3 x^2+\frac {1}{4+x}\right ) \, dx \\ & = -e^{\frac {3}{\log (1+\log (3 x))}}+x^3+\log (4+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {e^{\frac {3}{\log (1+\log (3 x))}} (12+3 x)+\left (x+12 x^3+3 x^4+\left (x+12 x^3+3 x^4\right ) \log (3 x)\right ) \log ^2(1+\log (3 x))}{\left (4 x+x^2+\left (4 x+x^2\right ) \log (3 x)\right ) \log ^2(1+\log (3 x))} \, dx=64-e^{\frac {3}{\log (1+\log (3 x))}}+x^3+\log (4+x) \]

[In]

Integrate[(E^(3/Log[1 + Log[3*x]])*(12 + 3*x) + (x + 12*x^3 + 3*x^4 + (x + 12*x^3 + 3*x^4)*Log[3*x])*Log[1 + L
og[3*x]]^2)/((4*x + x^2 + (4*x + x^2)*Log[3*x])*Log[1 + Log[3*x]]^2),x]

[Out]

64 - E^(3/Log[1 + Log[3*x]]) + x^3 + Log[4 + x]

Maple [A] (verified)

Time = 10.60 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00

method	result	size
risch	\(x^{3}+\ln \left (4+x \right )-{\mathrm e}^{\frac {3}{\ln \left (\ln \left (3 x \right )+1\right )}}\)	\(23\)
parallelrisch	\(x^{3}+\ln \left (4+x \right )-{\mathrm e}^{\frac {3}{\ln \left (\ln \left (3 x \right )+1\right )}}\)	\(23\)

[In]

int(((3*x+12)*exp(3/ln(ln(3*x)+1))+((3*x^4+12*x^3+x)*ln(3*x)+3*x^4+12*x^3+x)*ln(ln(3*x)+1)^2)/((x^2+4*x)*ln(3*
x)+x^2+4*x)/ln(ln(3*x)+1)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {3}{\log (1+\log (3 x))}} (12+3 x)+\left (x+12 x^3+3 x^4+\left (x+12 x^3+3 x^4\right ) \log (3 x)\right ) \log ^2(1+\log (3 x))}{\left (4 x+x^2+\left (4 x+x^2\right ) \log (3 x)\right ) \log ^2(1+\log (3 x))} \, dx=x^{3} - e^{\left (\frac {3}{\log \left (\log \left (3 \, x\right ) + 1\right )}\right )} + \log \left (x + 4\right ) \]

[In]

integrate(((3*x+12)*exp(3/log(log(3*x)+1))+((3*x^4+12*x^3+x)*log(3*x)+3*x^4+12*x^3+x)*log(log(3*x)+1)^2)/((x^2
+4*x)*log(3*x)+x^2+4*x)/log(log(3*x)+1)^2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {e^{\frac {3}{\log (1+\log (3 x))}} (12+3 x)+\left (x+12 x^3+3 x^4+\left (x+12 x^3+3 x^4\right ) \log (3 x)\right ) \log ^2(1+\log (3 x))}{\left (4 x+x^2+\left (4 x+x^2\right ) \log (3 x)\right ) \log ^2(1+\log (3 x))} \, dx=x^{3} - e^{\frac {3}{\log {\left (\log {\left (3 x \right )} + 1 \right )}}} + \log {\left (x + 4 \right )} \]

[In]

integrate(((3*x+12)*exp(3/ln(ln(3*x)+1))+((3*x**4+12*x**3+x)*ln(3*x)+3*x**4+12*x**3+x)*ln(ln(3*x)+1)**2)/((x**
2+4*x)*ln(3*x)+x**2+4*x)/ln(ln(3*x)+1)**2,x)

[Out]

x**3 - exp(3/log(log(3*x) + 1)) + log(x + 4)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (22) = 44\).

Time = 0.35 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.04 \[ \int \frac {e^{\frac {3}{\log (1+\log (3 x))}} (12+3 x)+\left (x+12 x^3+3 x^4+\left (x+12 x^3+3 x^4\right ) \log (3 x)\right ) \log ^2(1+\log (3 x))}{\left (4 x+x^2+\left (4 x+x^2\right ) \log (3 x)\right ) \log ^2(1+\log (3 x))} \, dx=x^{3} - \frac {x e^{\left (\frac {3}{\log \left (\log \left (3\right ) + \log \left (x\right ) + 1\right )}\right )}}{x + 4} - \frac {4 \, e^{\left (\frac {3}{\log \left (\log \left (3\right ) + \log \left (x\right ) + 1\right )}\right )}}{x + 4} + \log \left (x + 4\right ) \]

[In]

integrate(((3*x+12)*exp(3/log(log(3*x)+1))+((3*x^4+12*x^3+x)*log(3*x)+3*x^4+12*x^3+x)*log(log(3*x)+1)^2)/((x^2
+4*x)*log(3*x)+x^2+4*x)/log(log(3*x)+1)^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 1.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {3}{\log (1+\log (3 x))}} (12+3 x)+\left (x+12 x^3+3 x^4+\left (x+12 x^3+3 x^4\right ) \log (3 x)\right ) \log ^2(1+\log (3 x))}{\left (4 x+x^2+\left (4 x+x^2\right ) \log (3 x)\right ) \log ^2(1+\log (3 x))} \, dx=x^{3} - e^{\left (\frac {3}{\log \left (\log \left (3 \, x\right ) + 1\right )}\right )} + \log \left (x + 4\right ) \]

[In]

integrate(((3*x+12)*exp(3/log(log(3*x)+1))+((3*x^4+12*x^3+x)*log(3*x)+3*x^4+12*x^3+x)*log(log(3*x)+1)^2)/((x^2
+4*x)*log(3*x)+x^2+4*x)/log(log(3*x)+1)^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.20 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {3}{\log (1+\log (3 x))}} (12+3 x)+\left (x+12 x^3+3 x^4+\left (x+12 x^3+3 x^4\right ) \log (3 x)\right ) \log ^2(1+\log (3 x))}{\left (4 x+x^2+\left (4 x+x^2\right ) \log (3 x)\right ) \log ^2(1+\log (3 x))} \, dx=\ln \left (x+4\right )-{\mathrm {e}}^{\frac {3}{\ln \left (\ln \left (3\,x\right )+1\right )}}+x^3 \]

[In]

int((exp(3/log(log(3*x) + 1))*(3*x + 12) + log(log(3*x) + 1)^2*(x + 12*x^3 + 3*x^4 + log(3*x)*(x + 12*x^3 + 3*
x^4)))/(log(log(3*x) + 1)^2*(4*x + log(3*x)*(4*x + x^2) + x^2)),x)

[Out]

log(x + 4) - exp(3/log(log(3*x) + 1)) + x^3