\(\int \frac {-2 \log (x)+(2+(2+x) \log (x)) \log (x+9 e^8 x)}{x \log (x) \log (x+9 e^8 x)} \, dx\) [7266]

Optimal result

Integrand size = 42, antiderivative size = 25 \[ \int \frac {-2 \log (x)+(2+(2+x) \log (x)) \log \left (x+9 e^8 x\right )}{x \log (x) \log \left (x+9 e^8 x\right )} \, dx=x-\log \left (\frac {36 \log ^2\left (x+9 e^8 x\right )}{x^2 \log ^2(x)}\right ) \]

[Out]

x-ln(36/ln(x)^2*ln(9*x*exp(4)^2+x)^2/x^2)

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96, number of steps used = 11, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.119, Rules used = {2494, 6874, 45, 2339, 29} \[ \int \frac {-2 \log (x)+(2+(2+x) \log (x)) \log \left (x+9 e^8 x\right )}{x \log (x) \log \left (x+9 e^8 x\right )} \, dx=x+2 \log (x)+2 \log (\log (x))-2 \log \left (\log \left (\left (1+9 e^8\right ) x\right )\right ) \]

[In]

Int[(-2*Log[x] + (2 + (2 + x)*Log[x])*Log[x + 9*E^8*x])/(x*Log[x]*Log[x + 9*E^8*x]),x]

[Out]

x + 2*Log[x] + 2*Log[Log[x]] - 2*Log[Log[(1 + 9*E^8)*x]]

Rule 29

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 2494

Int[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Int[u*(a + b*Log[c*ExpandToSum[v, x]^n])^p
, x] /; FreeQ[{a, b, c, n, p}, x] && LinearQ[v, x] &&  !LinearMatchQ[v, x] &&  !(EqQ[n, 1] && MatchQ[c*v, (e_.
)*((f_) + (g_.)*x) /; FreeQ[{e, f, g}, x]])

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {-2 \log (x)+(2+(2+x) \log (x)) \log \left (x+9 e^8 x\right )}{x \log (x) \log \left (\left (1+9 e^8\right ) x\right )} \, dx \\ & = \int \left (\frac {2+2 \log (x)+x \log (x)}{x \log (x)}-\frac {2}{x \log \left (\left (1+9 e^8\right ) x\right )}\right ) \, dx \\ & = -\left (2 \int \frac {1}{x \log \left (\left (1+9 e^8\right ) x\right )} \, dx\right )+\int \frac {2+2 \log (x)+x \log (x)}{x \log (x)} \, dx \\ & = -\left (2 \text {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (\left (1+9 e^8\right ) x\right )\right )\right )+\int \left (\frac {2+x}{x}+\frac {2}{x \log (x)}\right ) \, dx \\ & = -2 \log \left (\log \left (\left (1+9 e^8\right ) x\right )\right )+2 \int \frac {1}{x \log (x)} \, dx+\int \frac {2+x}{x} \, dx \\ & = -2 \log \left (\log \left (\left (1+9 e^8\right ) x\right )\right )+2 \text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )+\int \left (1+\frac {2}{x}\right ) \, dx \\ & = x+2 \log (x)+2 \log (\log (x))-2 \log \left (\log \left (\left (1+9 e^8\right ) x\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {-2 \log (x)+(2+(2+x) \log (x)) \log \left (x+9 e^8 x\right )}{x \log (x) \log \left (x+9 e^8 x\right )} \, dx=x+2 \log (x)+2 \log (\log (x))-2 \log \left (\log \left (x+9 e^8 x\right )\right ) \]

[In]

Integrate[(-2*Log[x] + (2 + (2 + x)*Log[x])*Log[x + 9*E^8*x])/(x*Log[x]*Log[x + 9*E^8*x]),x]

[Out]

x + 2*Log[x] + 2*Log[Log[x]] - 2*Log[Log[x + 9*E^8*x]]

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.28

method	result	size
risch	\(2 \ln \left (x \right )+x\)	\(7\)
default	\(2 \ln \left (x \right )+x +2 \ln \left (\ln \left (x \right )\right )-2 \ln \left (\ln \left (9 x \,{\mathrm e}^{8}+x \right )\right )\)	\(25\)
parts	\(2 \ln \left (x \right )+x +2 \ln \left (\ln \left (x \right )\right )-2 \ln \left (\ln \left (9 x \,{\mathrm e}^{8}+x \right )\right )\)	\(25\)
norman	\(x +2 \ln \left (9 x \,{\mathrm e}^{8}+x \right )+2 \ln \left (\ln \left (x \right )\right )-2 \ln \left (\ln \left (9 x \,{\mathrm e}^{8}+x \right )\right )\)	\(33\)
parallelrisch	\(x +2 \ln \left (9 x \,{\mathrm e}^{8}+x \right )+2 \ln \left (\ln \left (x \right )\right )-2 \ln \left (\ln \left (9 x \,{\mathrm e}^{8}+x \right )\right )\)	\(33\)

[In]

int((((2+x)*ln(x)+2)*ln(9*x*exp(4)^2+x)-2*ln(x))/x/ln(x)/ln(9*x*exp(4)^2+x),x,method=_RETURNVERBOSE)

[Out]

2*ln(x)+x

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {-2 \log (x)+(2+(2+x) \log (x)) \log \left (x+9 e^8 x\right )}{x \log (x) \log \left (x+9 e^8 x\right )} \, dx=x + 2 \, \log \left (x\right ) - 2 \, \log \left (\log \left (x\right ) + \log \left (9 \, e^{8} + 1\right )\right ) + 2 \, \log \left (\log \left (x\right )\right ) \]

[In]

integrate((((2+x)*log(x)+2)*log(9*x*exp(4)^2+x)-2*log(x))/x/log(x)/log(9*x*exp(4)^2+x),x, algorithm="fricas")

[Out]

x + 2*log(x) - 2*log(log(x) + log(9*e^8 + 1)) + 2*log(log(x))

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {-2 \log (x)+(2+(2+x) \log (x)) \log \left (x+9 e^8 x\right )}{x \log (x) \log \left (x+9 e^8 x\right )} \, dx=x + 2 \log {\left (x \right )} - 2 \log {\left (\log {\left (x \right )} + \log {\left (1 + 9 e^{8} \right )} \right )} + 2 \log {\left (\log {\left (x \right )} \right )} \]

[In]

integrate((((2+x)*ln(x)+2)*ln(9*x*exp(4)**2+x)-2*ln(x))/x/ln(x)/ln(9*x*exp(4)**2+x),x)

[Out]

x + 2*log(x) - 2*log(log(x) + log(1 + 9*exp(8))) + 2*log(log(x))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {-2 \log (x)+(2+(2+x) \log (x)) \log \left (x+9 e^8 x\right )}{x \log (x) \log \left (x+9 e^8 x\right )} \, dx=x + 2 \, \log \left (x\right ) - 2 \, \log \left (\log \left (9 \, x e^{8} + x\right )\right ) + 2 \, \log \left (\log \left (x\right )\right ) \]

[In]

integrate((((2+x)*log(x)+2)*log(9*x*exp(4)^2+x)-2*log(x))/x/log(x)/log(9*x*exp(4)^2+x),x, algorithm="maxima")

[Out]

x + 2*log(x) - 2*log(log(9*x*e^8 + x)) + 2*log(log(x))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {-2 \log (x)+(2+(2+x) \log (x)) \log \left (x+9 e^8 x\right )}{x \log (x) \log \left (x+9 e^8 x\right )} \, dx=x + 2 \, \log \left (x\right ) - 2 \, \log \left (-\log \left (x\right ) - \log \left (9 \, e^{8} + 1\right )\right ) + 2 \, \log \left (\log \left (x\right )\right ) \]

[In]

integrate((((2+x)*log(x)+2)*log(9*x*exp(4)^2+x)-2*log(x))/x/log(x)/log(9*x*exp(4)^2+x),x, algorithm="giac")

[Out]

x + 2*log(x) - 2*log(-log(x) - log(9*e^8 + 1)) + 2*log(log(x))

Mupad [B] (verification not implemented)

Time = 11.15 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {-2 \log (x)+(2+(2+x) \log (x)) \log \left (x+9 e^8 x\right )}{x \log (x) \log \left (x+9 e^8 x\right )} \, dx=x+2\,\ln \left (\ln \left (x\right )\right )-2\,\ln \left (\ln \left (x\,\left (9\,{\mathrm {e}}^8+1\right )\right )\right )+2\,\ln \left (x\right ) \]

[In]

int(-(2*log(x) - log(x + 9*x*exp(8))*(log(x)*(x + 2) + 2))/(x*log(x + 9*x*exp(8))*log(x)),x)

[Out]

x + 2*log(log(x)) - 2*log(log(x*(9*exp(8) + 1))) + 2*log(x)