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

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

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Rubi [A]  time = 0.37, antiderivative size = 36, normalized size of antiderivative = 1.29, number of steps used = 5, number of rules used = 4, integrand size = 116, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {12, 6688, 6711, 32} \begin {gather*} -\frac {4 \log (\log (3))}{5 e^5 \left (\frac {5}{\log \left (-\frac {2}{(4-x) \log (5 (x+1))}\right )}+1\right )} \end {gather*}

Antiderivative was successfully verified.

[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 6688

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

Rule 6711

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 {gather*} \begin {aligned} \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))) \operatorname {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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 28, normalized size = 1.00 \begin {gather*} \frac {4 \log (\log (3))}{e^5 \left (5+\log \left (\frac {2}{(-4+x) \log (5 (1+x))}\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[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)])]))

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fricas [A]  time = 0.87, size = 31, normalized size = 1.11 \begin {gather*} \frac {4 \, \log \left (\log \relax (3)\right )}{e^{5} \log \left (\frac {2}{{\left (x - 4\right )} \log \left (5 \, x + 5\right )}\right ) + 5 \, e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+4)*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)

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giac [B]  time = 3.36, size = 735, normalized size = 26.25 result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+4)*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)

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maple [C]  time = 0.16, size = 162, normalized size = 5.79

method	result	size
risch	\(\frac {8 i \ln \left (\ln \relax (3)\right ) {\mathrm e}^{-5}}{\pi \,\mathrm {csgn}\left (\frac {i}{x -4}\right ) \mathrm {csgn}\left (\frac {i}{\ln \left (5 x +5\right )}\right ) \mathrm {csgn}\left (\frac {i}{\ln \left (5 x +5\right ) \left (x -4\right )}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{x -4}\right ) \mathrm {csgn}\left (\frac {i}{\ln \left (5 x +5\right ) \left (x -4\right )}\right )^{2}-\pi \,\mathrm {csgn}\left (\frac {i}{\ln \left (5 x +5\right )}\right ) \mathrm {csgn}\left (\frac {i}{\ln \left (5 x +5\right ) \left (x -4\right )}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i}{\ln \left (5 x +5\right ) \left (x -4\right )}\right )^{3}+2 i \ln \relax (2)-2 i \ln \left (x -4\right )-2 i \ln \left (\ln \left (5 x +5\right )\right )+10 i}\)	\(162\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x+4)*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]

8*I*ln(ln(3))*exp(-5)/(Pi*csgn(I/(x-4))*csgn(I/ln(5*x+5))*csgn(I/ln(5*x+5)/(x-4))-Pi*csgn(I/(x-4))*csgn(I/ln(5
*x+5)/(x-4))^2-Pi*csgn(I/ln(5*x+5))*csgn(I/ln(5*x+5)/(x-4))^2+Pi*csgn(I/ln(5*x+5)/(x-4))^3+2*I*ln(2)-2*I*ln(x-
4)-2*I*ln(ln(5*x+5))+10*I)

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maxima [A]  time = 0.63, size = 35, normalized size = 1.25 \begin {gather*} \frac {4 \, \log \left (\log \relax (3)\right )}{{\left (\log \relax (2) + 5\right )} e^{5} - e^{5} \log \left (x - 4\right ) - e^{5} \log \left (\log \relax (5) + \log \left (x + 1\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+4)*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)))

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mupad [B]  time = 2.13, size = 27, normalized size = 0.96 \begin {gather*} \frac {4\,{\mathrm {e}}^{-5}\,\ln \left (\ln \relax (3)\right )}{\ln \left (\frac {2}{\ln \left (5\,x+5\right )\,\left (x-4\right )}\right )+5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.43, size = 27, normalized size = 0.96 \begin {gather*} \frac {4 \log {\left (\log {\relax (3 )} \right )}}{e^{5} \log {\left (\frac {2}{\left (x - 4\right ) \log {\left (5 x + 5 \right )}} \right )} + 5 e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+4)*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))

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