\(\int \frac {(-16+4 x+(4+4 x) \log (5+5 x)) \log (\log (3))}{e^5 (-100-75 x+25 x^2) \log (5+5 x)+e^5 (-40-30 x+10 x^2) \log (5+5 x) \log (\frac {2}{(-4+x) \log (5+5 x)})+e^5 (-4-3 x+x^2) \log (5+5 x) \log ^2(\frac {2}{(-4+x) \log (5+5 x)})} \, dx\) [127]

Optimal result

Integrand size = 116, antiderivative size = 28 \[ \int \frac {(-16+4 x+(4+4 x) \log (5+5 x)) \log (\log (3))}{e^5 \left (-100-75 x+25 x^2\right ) \log (5+5 x)+e^5 \left (-40-30 x+10 x^2\right ) \log (5+5 x) \log \left (\frac {2}{(-4+x) \log (5+5 x)}\right )+e^5 \left (-4-3 x+x^2\right ) \log (5+5 x) \log ^2\left (\frac {2}{(-4+x) \log (5+5 x)}\right )} \, dx=\frac {4 \log (\log (3))}{e^5 \left (5+\log \left (\frac {2}{(-4+x) \log (5 (1+x))}\right )\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {12, 6820, 6843, 32} \[ \int \frac {(-16+4 x+(4+4 x) \log (5+5 x)) \log (\log (3))}{e^5 \left (-100-75 x+25 x^2\right ) \log (5+5 x)+e^5 \left (-40-30 x+10 x^2\right ) \log (5+5 x) \log \left (\frac {2}{(-4+x) \log (5+5 x)}\right )+e^5 \left (-4-3 x+x^2\right ) \log (5+5 x) \log ^2\left (\frac {2}{(-4+x) \log (5+5 x)}\right )} \, dx=-\frac {4 \log (\log (3))}{5 e^5 \left (\frac {5}{\log \left (-\frac {2}{(4-x) \log (5 (x+1))}\right )}+1\right )} \]

[In]

Int[((-16 + 4*x + (4 + 4*x)*Log[5 + 5*x])*Log[Log[3]])/(E^5*(-100 - 75*x + 25*x^2)*Log[5 + 5*x] + E^5*(-40 - 3
0*x + 10*x^2)*Log[5 + 5*x]*Log[2/((-4 + x)*Log[5 + 5*x])] + E^5*(-4 - 3*x + x^2)*Log[5 + 5*x]*Log[2/((-4 + x)*
Log[5 + 5*x])]^2),x]

[Out]

(-4*Log[Log[3]])/(5*E^5*(1 + 5/Log[-2/((4 - x)*Log[5*(1 + 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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 6820

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

Rule 6843

Int[(u_)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(p*w*D[v, x] - q*v*D[w
, x])]}, Dist[c*p, Subst[Int[(b + a*x^p)^m, x], x, v*w^(m*q + 1)], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p, q}
, x] && EqQ[p + q*(m*p + 1), 0] && IntegerQ[p] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = \log (\log (3)) \int \frac {-16+4 x+(4+4 x) \log (5+5 x)}{e^5 \left (-100-75 x+25 x^2\right ) \log (5+5 x)+e^5 \left (-40-30 x+10 x^2\right ) \log (5+5 x) \log \left (\frac {2}{(-4+x) \log (5+5 x)}\right )+e^5 \left (-4-3 x+x^2\right ) \log (5+5 x) \log ^2\left (\frac {2}{(-4+x) \log (5+5 x)}\right )} \, dx \\ & = \log (\log (3)) \int \frac {4 (4-x-(1+x) \log (5 (1+x)))}{e^5 \left (4+3 x-x^2\right ) \log (5+5 x) \left (5+\log \left (\frac {2}{(-4+x) \log (5 (1+x))}\right )\right )^2} \, dx \\ & = \frac {(4 \log (\log (3))) \int \frac {4-x-(1+x) \log (5 (1+x))}{\left (4+3 x-x^2\right ) \log (5+5 x) \left (5+\log \left (\frac {2}{(-4+x) \log (5 (1+x))}\right )\right )^2} \, dx}{e^5} \\ & = \frac {(4 \log (\log (3))) \text {Subst}\left (\int \frac {1}{(1+x)^2} \, dx,x,\frac {5}{\log \left (\frac {2}{(-4+x) \log (5 (1+x))}\right )}\right )}{5 e^5} \\ & = -\frac {4 \log (\log (3))}{5 e^5 \left (1+\frac {5}{\log \left (-\frac {2}{(4-x) \log (5 (1+x))}\right )}\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {(-16+4 x+(4+4 x) \log (5+5 x)) \log (\log (3))}{e^5 \left (-100-75 x+25 x^2\right ) \log (5+5 x)+e^5 \left (-40-30 x+10 x^2\right ) \log (5+5 x) \log \left (\frac {2}{(-4+x) \log (5+5 x)}\right )+e^5 \left (-4-3 x+x^2\right ) \log (5+5 x) \log ^2\left (\frac {2}{(-4+x) \log (5+5 x)}\right )} \, dx=\frac {4 \log (\log (3))}{e^5 \left (5+\log \left (\frac {2}{(-4+x) \log (5 (1+x))}\right )\right )} \]

[In]

Integrate[((-16 + 4*x + (4 + 4*x)*Log[5 + 5*x])*Log[Log[3]])/(E^5*(-100 - 75*x + 25*x^2)*Log[5 + 5*x] + E^5*(-
40 - 30*x + 10*x^2)*Log[5 + 5*x]*Log[2/((-4 + x)*Log[5 + 5*x])] + E^5*(-4 - 3*x + x^2)*Log[5 + 5*x]*Log[2/((-4
+ x)*Log[5 + 5*x])]^2),x]

[Out]

(4*Log[Log[3]])/(E^5*(5 + Log[2/((-4 + x)*Log[5*(1 + x)])]))

Maple [A] (verified)

Time = 2.80 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07

method	result	size
parallelrisch	\(\frac {4 \,{\mathrm e}^{-5} \ln \left (\ln \left (3\right )\right )}{\ln \left (\frac {2}{\left (x -4\right ) \ln \left (5 x +5\right )}\right )+5}\)	\(30\)
default	\(\frac {4 \ln \left (\ln \left (3\right )\right ) {\mathrm e}^{-5}}{\ln \left (2\right )+\ln \left (\frac {1}{\left (x -4\right ) \left (\ln \left (5\right )+\ln \left (1+x \right )\right )}\right )+5}\)	\(32\)
risch	\(\frac {8 \ln \left (\ln \left (3\right )\right ) {\mathrm e}^{-5}}{10-i \pi \,\operatorname {csgn}\left (\frac {i}{x -4}\right ) \operatorname {csgn}\left (\frac {i}{\ln \left (5 x +5\right )}\right ) \operatorname {csgn}\left (\frac {i}{\ln \left (5 x +5\right ) \left (x -4\right )}\right )+i \pi \,\operatorname {csgn}\left (\frac {i}{x -4}\right ) \operatorname {csgn}\left (\frac {i}{\ln \left (5 x +5\right ) \left (x -4\right )}\right )^{2}+i \pi \,\operatorname {csgn}\left (\frac {i}{\ln \left (5 x +5\right )}\right ) \operatorname {csgn}\left (\frac {i}{\ln \left (5 x +5\right ) \left (x -4\right )}\right )^{2}-i \pi \operatorname {csgn}\left (\frac {i}{\ln \left (5 x +5\right ) \left (x -4\right )}\right )^{3}+2 \ln \left (2\right )-2 \ln \left (x -4\right )-2 \ln \left (\ln \left (5 x +5\right )\right )}\)	\(163\)

[In]

int(((4+4*x)*ln(5*x+5)+4*x-16)*ln(ln(3))/((x^2-3*x-4)*exp(5)*ln(5*x+5)*ln(2/(x-4)/ln(5*x+5))^2+(10*x^2-30*x-40
)*exp(5)*ln(5*x+5)*ln(2/(x-4)/ln(5*x+5))+(25*x^2-75*x-100)*exp(5)*ln(5*x+5)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {(-16+4 x+(4+4 x) \log (5+5 x)) \log (\log (3))}{e^5 \left (-100-75 x+25 x^2\right ) \log (5+5 x)+e^5 \left (-40-30 x+10 x^2\right ) \log (5+5 x) \log \left (\frac {2}{(-4+x) \log (5+5 x)}\right )+e^5 \left (-4-3 x+x^2\right ) \log (5+5 x) \log ^2\left (\frac {2}{(-4+x) \log (5+5 x)}\right )} \, dx=\frac {4 \, \log \left (\log \left (3\right )\right )}{e^{5} \log \left (\frac {2}{{\left (x - 4\right )} \log \left (5 \, x + 5\right )}\right ) + 5 \, e^{5}} \]

[In]

integrate(((4+4*x)*log(5*x+5)+4*x-16)*log(log(3))/((x^2-3*x-4)*exp(5)*log(5*x+5)*log(2/(x-4)/log(5*x+5))^2+(10
*x^2-30*x-40)*exp(5)*log(5*x+5)*log(2/(x-4)/log(5*x+5))+(25*x^2-75*x-100)*exp(5)*log(5*x+5)),x, algorithm="fri
cas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {(-16+4 x+(4+4 x) \log (5+5 x)) \log (\log (3))}{e^5 \left (-100-75 x+25 x^2\right ) \log (5+5 x)+e^5 \left (-40-30 x+10 x^2\right ) \log (5+5 x) \log \left (\frac {2}{(-4+x) \log (5+5 x)}\right )+e^5 \left (-4-3 x+x^2\right ) \log (5+5 x) \log ^2\left (\frac {2}{(-4+x) \log (5+5 x)}\right )} \, dx=\frac {4 \log {\left (\log {\left (3 \right )} \right )}}{e^{5} \log {\left (\frac {2}{\left (x - 4\right ) \log {\left (5 x + 5 \right )}} \right )} + 5 e^{5}} \]

[In]

integrate(((4+4*x)*ln(5*x+5)+4*x-16)*ln(ln(3))/((x**2-3*x-4)*exp(5)*ln(5*x+5)*ln(2/(x-4)/ln(5*x+5))**2+(10*x**
2-30*x-40)*exp(5)*ln(5*x+5)*ln(2/(x-4)/ln(5*x+5))+(25*x**2-75*x-100)*exp(5)*ln(5*x+5)),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \frac {(-16+4 x+(4+4 x) \log (5+5 x)) \log (\log (3))}{e^5 \left (-100-75 x+25 x^2\right ) \log (5+5 x)+e^5 \left (-40-30 x+10 x^2\right ) \log (5+5 x) \log \left (\frac {2}{(-4+x) \log (5+5 x)}\right )+e^5 \left (-4-3 x+x^2\right ) \log (5+5 x) \log ^2\left (\frac {2}{(-4+x) \log (5+5 x)}\right )} \, dx=\frac {4 \, \log \left (\log \left (3\right )\right )}{{\left (\log \left (2\right ) + 5\right )} e^{5} - e^{5} \log \left (x - 4\right ) - e^{5} \log \left (\log \left (5\right ) + \log \left (x + 1\right )\right )} \]

[In]

integrate(((4+4*x)*log(5*x+5)+4*x-16)*log(log(3))/((x^2-3*x-4)*exp(5)*log(5*x+5)*log(2/(x-4)/log(5*x+5))^2+(10
*x^2-30*x-40)*exp(5)*log(5*x+5)*log(2/(x-4)/log(5*x+5))+(25*x^2-75*x-100)*exp(5)*log(5*x+5)),x, algorithm="max
ima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 735 vs. \(2 (27) = 54\).

Time = 1.09 (sec) , antiderivative size = 735, normalized size of antiderivative = 26.25 \[ \int \frac {(-16+4 x+(4+4 x) \log (5+5 x)) \log (\log (3))}{e^5 \left (-100-75 x+25 x^2\right ) \log (5+5 x)+e^5 \left (-40-30 x+10 x^2\right ) \log (5+5 x) \log \left (\frac {2}{(-4+x) \log (5+5 x)}\right )+e^5 \left (-4-3 x+x^2\right ) \log (5+5 x) \log ^2\left (\frac {2}{(-4+x) \log (5+5 x)}\right )} \, dx=\text {Too large to display} \]

[In]

integrate(((4+4*x)*log(5*x+5)+4*x-16)*log(log(3))/((x^2-3*x-4)*exp(5)*log(5*x+5)*log(2/(x-4)/log(5*x+5))^2+(10
*x^2-30*x-40)*exp(5)*log(5*x+5)*log(2/(x-4)/log(5*x+5))+(25*x^2-75*x-100)*exp(5)*log(5*x+5)),x, algorithm="gia
c")

[Out]

-8*(2*log(2) - log(-1/2*pi^2*x^2*sgn(5*x + 5) + 1/2*pi^2*x^2 + x^2*log(abs(5*x + 5))^2 + 4*pi^2*x*sgn(5*x + 5)
- 4*pi^2*x - 8*x*log(abs(5*x + 5))^2 - 8*pi^2*sgn(5*x + 5) + 8*pi^2 + 16*log(abs(5*x + 5))^2) + 10)*log(log(3
))/(8*pi^2*e^5*sgn(4*pi + pi*x*sgn(5*x + 5) - pi*x - 4*pi*sgn(5*x + 5))*sgn(x*log(abs(5*x + 5)) - 4*log(abs(5*
x + 5))) + 4*pi*arctan(-1/2*(pi - pi*sgn(5*x + 5))/log(abs(5*x + 5)))*e^5*sgn(4*pi + pi*x*sgn(5*x + 5) - pi*x
- 4*pi*sgn(5*x + 5))*sgn(x*log(abs(5*x + 5)) - 4*log(abs(5*x + 5))) - 8*pi^2*e^5*sgn(4*pi + pi*x*sgn(5*x + 5)
- pi*x - 4*pi*sgn(5*x + 5)) - 4*pi*arctan(-1/2*(pi - pi*sgn(5*x + 5))/log(abs(5*x + 5)))*e^5*sgn(4*pi + pi*x*s
gn(5*x + 5) - pi*x - 4*pi*sgn(5*x + 5)) + 2*pi^2*e^5*sgn(x*log(abs(5*x + 5)) - 4*log(abs(5*x + 5))) - 18*pi^2*
e^5 - 16*pi*arctan(-1/2*(pi - pi*sgn(5*x + 5))/log(abs(5*x + 5)))*e^5 - 4*arctan(-1/2*(pi - pi*sgn(5*x + 5))/l
og(abs(5*x + 5)))^2*e^5 - 4*e^5*log(2)^2 + 4*e^5*log(2)*log(-1/2*pi^2*x^2*sgn(5*x + 5) + 1/2*pi^2*x^2 + x^2*lo
g(abs(5*x + 5))^2 + 4*pi^2*x*sgn(5*x + 5) - 4*pi^2*x - 8*x*log(abs(5*x + 5))^2 - 8*pi^2*sgn(5*x + 5) + 8*pi^2
+ 16*log(abs(5*x + 5))^2) - e^5*log(-1/2*pi^2*x^2*sgn(5*x + 5) + 1/2*pi^2*x^2 + x^2*log(abs(5*x + 5))^2 + 4*pi
^2*x*sgn(5*x + 5) - 4*pi^2*x - 8*x*log(abs(5*x + 5))^2 - 8*pi^2*sgn(5*x + 5) + 8*pi^2 + 16*log(abs(5*x + 5))^2
)^2 - 40*e^5*log(2) + 20*e^5*log(-1/2*pi^2*x^2*sgn(5*x + 5) + 1/2*pi^2*x^2 + x^2*log(abs(5*x + 5))^2 + 4*pi^2*
x*sgn(5*x + 5) - 4*pi^2*x - 8*x*log(abs(5*x + 5))^2 - 8*pi^2*sgn(5*x + 5) + 8*pi^2 + 16*log(abs(5*x + 5))^2) -
100*e^5)

Mupad [B] (verification not implemented)

Time = 9.81 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {(-16+4 x+(4+4 x) \log (5+5 x)) \log (\log (3))}{e^5 \left (-100-75 x+25 x^2\right ) \log (5+5 x)+e^5 \left (-40-30 x+10 x^2\right ) \log (5+5 x) \log \left (\frac {2}{(-4+x) \log (5+5 x)}\right )+e^5 \left (-4-3 x+x^2\right ) \log (5+5 x) \log ^2\left (\frac {2}{(-4+x) \log (5+5 x)}\right )} \, dx=\frac {4\,{\mathrm {e}}^{-5}\,\ln \left (\ln \left (3\right )\right )}{\ln \left (\frac {2}{\ln \left (5\,x+5\right )\,\left (x-4\right )}\right )+5} \]

[In]

int(-(log(log(3))*(4*x + log(5*x + 5)*(4*x + 4) - 16))/(exp(5)*log(5*x + 5)*(75*x - 25*x^2 + 100) + log(2/(log
(5*x + 5)*(x - 4)))^2*exp(5)*log(5*x + 5)*(3*x - x^2 + 4) + log(2/(log(5*x + 5)*(x - 4)))*exp(5)*log(5*x + 5)*
(30*x - 10*x^2 + 40)),x)

[Out]

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