\(\int \frac {6 x+3 x^2+e^4 (6 x+3 x^2)+(6 x+3 x^2) \log (3)+(-170-414 x-241 x^2-53 x^3-4 x^4+e^4 (-170-244 x-82 x^2-8 x^3)+(-170-244 x-82 x^2-8 x^3) \log (3)) \log (\frac {17+4 x}{5+x})}{(170 x+159 x^2+45 x^3+4 x^4) \log (\frac {17+4 x}{5+x})} \, dx\) [6966]

Optimal result

Integrand size = 140, antiderivative size = 33 \[ \int \frac {6 x+3 x^2+e^4 \left (6 x+3 x^2\right )+\left (6 x+3 x^2\right ) \log (3)+\left (-170-414 x-241 x^2-53 x^3-4 x^4+e^4 \left (-170-244 x-82 x^2-8 x^3\right )+\left (-170-244 x-82 x^2-8 x^3\right ) \log (3)\right ) \log \left (\frac {17+4 x}{5+x}\right )}{\left (170 x+159 x^2+45 x^3+4 x^4\right ) \log \left (\frac {17+4 x}{5+x}\right )} \, dx=-x+\left (1+e^4+\log (3)\right ) \log \left (\frac {5 \log \left (4-\frac {3}{5+x}\right )}{x (2+x)}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.73, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {6, 6820, 6874, 907, 2561, 2339, 29} \[ \int \frac {6 x+3 x^2+e^4 \left (6 x+3 x^2\right )+\left (6 x+3 x^2\right ) \log (3)+\left (-170-414 x-241 x^2-53 x^3-4 x^4+e^4 \left (-170-244 x-82 x^2-8 x^3\right )+\left (-170-244 x-82 x^2-8 x^3\right ) \log (3)\right ) \log \left (\frac {17+4 x}{5+x}\right )}{\left (170 x+159 x^2+45 x^3+4 x^4\right ) \log \left (\frac {17+4 x}{5+x}\right )} \, dx=-x-\frac {1}{2} \left (2+2 e^4+\log (9)\right ) \log (x)-\frac {1}{2} \left (2+2 e^4+\log (9)\right ) \log (x+2)+\left (1+e^4+\log (3)\right ) \log \left (\log \left (\frac {4 x+17}{x+5}\right )\right ) \]

[In]

Int[(6*x + 3*x^2 + E^4*(6*x + 3*x^2) + (6*x + 3*x^2)*Log[3] + (-170 - 414*x - 241*x^2 - 53*x^3 - 4*x^4 + E^4*(
-170 - 244*x - 82*x^2 - 8*x^3) + (-170 - 244*x - 82*x^2 - 8*x^3)*Log[3])*Log[(17 + 4*x)/(5 + x)])/((170*x + 15
9*x^2 + 45*x^3 + 4*x^4)*Log[(17 + 4*x)/(5 + x)]),x]

[Out]

-x - ((2 + 2*E^4 + Log[9])*Log[x])/2 - ((2 + 2*E^4 + Log[9])*Log[2 + x])/2 + (1 + E^4 + Log[3])*Log[Log[(17 +
4*x)/(5 + x)]]

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 29

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

Rule 907

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 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 2561

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*(B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m
_.)*((h_.) + (i_.)*(x_))^(q_.), x_Symbol] :> Dist[(b*c - a*d)^(m + q + 1)*(g/b)^m*(i/d)^q, Subst[Int[x^m*((A +
B*Log[e*x^n])^p/(b - d*x)^(m + q + 2)), x], x, (a + b*x)/(c + d*x)], x] /; FreeQ[{a, b, c, d, e, f, g, h, i,
A, B, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[b*f - a*g, 0] && EqQ[d*h - c*i, 0] && IntegersQ[m, q]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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 {6 x+3 x^2+\left (6 x+3 x^2\right ) \left (e^4+\log (3)\right )+\left (-170-414 x-241 x^2-53 x^3-4 x^4+e^4 \left (-170-244 x-82 x^2-8 x^3\right )+\left (-170-244 x-82 x^2-8 x^3\right ) \log (3)\right ) \log \left (\frac {17+4 x}{5+x}\right )}{\left (170 x+159 x^2+45 x^3+4 x^4\right ) \log \left (\frac {17+4 x}{5+x}\right )} \, dx \\ & = \int \frac {6 x+3 x^2+3 x (2+x) \left (e^4+\log (3)\right )-\left (85+37 x+4 x^2\right ) \left (2+x^2+2 e^4 (1+x)+\log (9)+x (4+\log (9))\right ) \log \left (\frac {17+4 x}{5+x}\right )}{x \left (170+159 x+45 x^2+4 x^3\right ) \log \left (\frac {17+4 x}{5+x}\right )} \, dx \\ & = \int \left (\frac {-2-2 e^4-x^2-\log (9)-x \left (4+2 e^4+\log (9)\right )}{x (2+x)}+\frac {3 \left (1+e^4+\log (3)\right )}{(5+x) (17+4 x) \log \left (\frac {17+4 x}{5+x}\right )}\right ) \, dx \\ & = \left (3 \left (1+e^4+\log (3)\right )\right ) \int \frac {1}{(5+x) (17+4 x) \log \left (\frac {17+4 x}{5+x}\right )} \, dx+\int \frac {-2-2 e^4-x^2-\log (9)-x \left (4+2 e^4+\log (9)\right )}{x (2+x)} \, dx \\ & = \left (1+e^4+\log (3)\right ) \text {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,\frac {17+4 x}{5+x}\right )+\int \left (-1+\frac {-2-2 e^4-\log (9)}{2 x}+\frac {-2-2 e^4-\log (9)}{2 (2+x)}\right ) \, dx \\ & = -x-\frac {1}{2} \left (2+2 e^4+\log (9)\right ) \log (x)-\frac {1}{2} \left (2+2 e^4+\log (9)\right ) \log (2+x)+\left (1+e^4+\log (3)\right ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (\frac {17+4 x}{5+x}\right )\right ) \\ & = -x-\frac {1}{2} \left (2+2 e^4+\log (9)\right ) \log (x)-\frac {1}{2} \left (2+2 e^4+\log (9)\right ) \log (2+x)+\left (1+e^4+\log (3)\right ) \log \left (\log \left (\frac {17+4 x}{5+x}\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.73 \[ \int \frac {6 x+3 x^2+e^4 \left (6 x+3 x^2\right )+\left (6 x+3 x^2\right ) \log (3)+\left (-170-414 x-241 x^2-53 x^3-4 x^4+e^4 \left (-170-244 x-82 x^2-8 x^3\right )+\left (-170-244 x-82 x^2-8 x^3\right ) \log (3)\right ) \log \left (\frac {17+4 x}{5+x}\right )}{\left (170 x+159 x^2+45 x^3+4 x^4\right ) \log \left (\frac {17+4 x}{5+x}\right )} \, dx=-x-\frac {1}{2} \left (2+2 e^4+\log (9)\right ) \log (x)-\frac {1}{2} \left (2+2 e^4+\log (9)\right ) \log (2+x)+\left (1+e^4+\log (3)\right ) \log \left (\log \left (\frac {17+4 x}{5+x}\right )\right ) \]

[In]

Integrate[(6*x + 3*x^2 + E^4*(6*x + 3*x^2) + (6*x + 3*x^2)*Log[3] + (-170 - 414*x - 241*x^2 - 53*x^3 - 4*x^4 +
E^4*(-170 - 244*x - 82*x^2 - 8*x^3) + (-170 - 244*x - 82*x^2 - 8*x^3)*Log[3])*Log[(17 + 4*x)/(5 + x)])/((170*
x + 159*x^2 + 45*x^3 + 4*x^4)*Log[(17 + 4*x)/(5 + x)]),x]

[Out]

-x - ((2 + 2*E^4 + Log[9])*Log[x])/2 - ((2 + 2*E^4 + Log[9])*Log[2 + x])/2 + (1 + E^4 + Log[3])*Log[Log[(17 +
4*x)/(5 + x)]]

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.52

method	result	size
parts	\(\frac {\left (3+3 \,{\mathrm e}^{4}+3 \ln \left (3\right )\right ) \ln \left (\ln \left (4-\frac {3}{5+x}\right )\right )}{3}-x -\left (1+{\mathrm e}^{4}+\ln \left (3\right )\right ) \ln \left (x \right )-\left (1+{\mathrm e}^{4}+\ln \left (3\right )\right ) \ln \left (2+x \right )\)	\(50\)
norman	\(-x +\left (1+{\mathrm e}^{4}+\ln \left (3\right )\right ) \ln \left (\ln \left (\frac {4 x +17}{5+x}\right )\right )+\left (-{\mathrm e}^{4}-\ln \left (3\right )-1\right ) \ln \left (x \right )+\left (-{\mathrm e}^{4}-\ln \left (3\right )-1\right ) \ln \left (2+x \right )\)	\(53\)
risch	\(-\ln \left (x^{2}+2 x \right ) {\mathrm e}^{4}+\ln \left (\ln \left (\frac {4 x +17}{5+x}\right )\right ) {\mathrm e}^{4}-\ln \left (x^{2}+2 x \right ) \ln \left (3\right )+\ln \left (\ln \left (\frac {4 x +17}{5+x}\right )\right ) \ln \left (3\right )-\ln \left (x^{2}+2 x \right )+\ln \left (\ln \left (\frac {4 x +17}{5+x}\right )\right )-x\)	\(84\)
parallelrisch	\(-{\mathrm e}^{4} \ln \left (x \right )+\ln \left (\ln \left (\frac {4 x +17}{5+x}\right )\right ) {\mathrm e}^{4}-{\mathrm e}^{4} \ln \left (2+x \right )-\ln \left (3\right ) \ln \left (x \right )+\ln \left (\ln \left (\frac {4 x +17}{5+x}\right )\right ) \ln \left (3\right )-\ln \left (3\right ) \ln \left (2+x \right )+\frac {37}{2}-\ln \left (x \right )+\ln \left (\ln \left (\frac {4 x +17}{5+x}\right )\right )-\ln \left (2+x \right )-x\)	\(89\)
derivativedivides	\(\ln \left (3\right ) \left (\ln \left (\ln \left (4-\frac {3}{5+x}\right )\right )-\ln \left (1-\frac {3}{5+x}\right )+2 \ln \left (-\frac {3}{5+x}\right )-\ln \left (3-\frac {15}{5+x}\right )\right )+3 \left (\frac {{\mathrm e}^{4}}{3}+\frac {1}{3}\right ) \ln \left (\ln \left (4-\frac {3}{5+x}\right )\right )+\left (-{\mathrm e}^{4}-1\right ) \ln \left (1-\frac {3}{5+x}\right )+\left (2 \,{\mathrm e}^{4}+2\right ) \ln \left (-\frac {3}{5+x}\right )-5-x +\left (-{\mathrm e}^{4}-1\right ) \ln \left (3-\frac {15}{5+x}\right )\)	\(123\)
default	\(\ln \left (3\right ) \left (\ln \left (\ln \left (4-\frac {3}{5+x}\right )\right )-\ln \left (1-\frac {3}{5+x}\right )+2 \ln \left (-\frac {3}{5+x}\right )-\ln \left (3-\frac {15}{5+x}\right )\right )+3 \left (\frac {{\mathrm e}^{4}}{3}+\frac {1}{3}\right ) \ln \left (\ln \left (4-\frac {3}{5+x}\right )\right )+\left (-{\mathrm e}^{4}-1\right ) \ln \left (1-\frac {3}{5+x}\right )+\left (2 \,{\mathrm e}^{4}+2\right ) \ln \left (-\frac {3}{5+x}\right )-5-x +\left (-{\mathrm e}^{4}-1\right ) \ln \left (3-\frac {15}{5+x}\right )\)	\(123\)

[In]

int((((-8*x^3-82*x^2-244*x-170)*ln(3)+(-8*x^3-82*x^2-244*x-170)*exp(4)-4*x^4-53*x^3-241*x^2-414*x-170)*ln((4*x
+17)/(5+x))+(3*x^2+6*x)*ln(3)+(3*x^2+6*x)*exp(4)+3*x^2+6*x)/(4*x^4+45*x^3+159*x^2+170*x)/ln((4*x+17)/(5+x)),x,
method=_RETURNVERBOSE)

[Out]

1/3*(3+3*exp(4)+3*ln(3))*ln(ln(4-3/(5+x)))-x-(1+exp(4)+ln(3))*ln(x)-(1+exp(4)+ln(3))*ln(2+x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.21 \[ \int \frac {6 x+3 x^2+e^4 \left (6 x+3 x^2\right )+\left (6 x+3 x^2\right ) \log (3)+\left (-170-414 x-241 x^2-53 x^3-4 x^4+e^4 \left (-170-244 x-82 x^2-8 x^3\right )+\left (-170-244 x-82 x^2-8 x^3\right ) \log (3)\right ) \log \left (\frac {17+4 x}{5+x}\right )}{\left (170 x+159 x^2+45 x^3+4 x^4\right ) \log \left (\frac {17+4 x}{5+x}\right )} \, dx=-{\left (e^{4} + \log \left (3\right ) + 1\right )} \log \left (x^{2} + 2 \, x\right ) + {\left (e^{4} + \log \left (3\right ) + 1\right )} \log \left (\log \left (\frac {4 \, x + 17}{x + 5}\right )\right ) - x \]

[In]

integrate((((-8*x^3-82*x^2-244*x-170)*log(3)+(-8*x^3-82*x^2-244*x-170)*exp(4)-4*x^4-53*x^3-241*x^2-414*x-170)*
log((4*x+17)/(5+x))+(3*x^2+6*x)*log(3)+(3*x^2+6*x)*exp(4)+3*x^2+6*x)/(4*x^4+45*x^3+159*x^2+170*x)/log((4*x+17)
/(5+x)),x, algorithm="fricas")

[Out]

-(e^4 + log(3) + 1)*log(x^2 + 2*x) + (e^4 + log(3) + 1)*log(log((4*x + 17)/(x + 5))) - x

Sympy [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.12 \[ \int \frac {6 x+3 x^2+e^4 \left (6 x+3 x^2\right )+\left (6 x+3 x^2\right ) \log (3)+\left (-170-414 x-241 x^2-53 x^3-4 x^4+e^4 \left (-170-244 x-82 x^2-8 x^3\right )+\left (-170-244 x-82 x^2-8 x^3\right ) \log (3)\right ) \log \left (\frac {17+4 x}{5+x}\right )}{\left (170 x+159 x^2+45 x^3+4 x^4\right ) \log \left (\frac {17+4 x}{5+x}\right )} \, dx=- x - \left (1 + \log {\left (3 \right )} + e^{4}\right ) \log {\left (x^{2} + 2 x \right )} + \left (1 + \log {\left (3 \right )} + e^{4}\right ) \log {\left (\log {\left (\frac {4 x + 17}{x + 5} \right )} \right )} \]

[In]

integrate((((-8*x**3-82*x**2-244*x-170)*ln(3)+(-8*x**3-82*x**2-244*x-170)*exp(4)-4*x**4-53*x**3-241*x**2-414*x
-170)*ln((4*x+17)/(5+x))+(3*x**2+6*x)*ln(3)+(3*x**2+6*x)*exp(4)+3*x**2+6*x)/(4*x**4+45*x**3+159*x**2+170*x)/ln
((4*x+17)/(5+x)),x)

[Out]

-x - (1 + log(3) + exp(4))*log(x**2 + 2*x) + (1 + log(3) + exp(4))*log(log((4*x + 17)/(x + 5)))

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.42 \[ \int \frac {6 x+3 x^2+e^4 \left (6 x+3 x^2\right )+\left (6 x+3 x^2\right ) \log (3)+\left (-170-414 x-241 x^2-53 x^3-4 x^4+e^4 \left (-170-244 x-82 x^2-8 x^3\right )+\left (-170-244 x-82 x^2-8 x^3\right ) \log (3)\right ) \log \left (\frac {17+4 x}{5+x}\right )}{\left (170 x+159 x^2+45 x^3+4 x^4\right ) \log \left (\frac {17+4 x}{5+x}\right )} \, dx=-{\left (e^{4} + \log \left (3\right ) + 1\right )} \log \left (x + 2\right ) - {\left (e^{4} + \log \left (3\right ) + 1\right )} \log \left (x\right ) + {\left (e^{4} + \log \left (3\right ) + 1\right )} \log \left (\log \left (4 \, x + 17\right ) - \log \left (x + 5\right )\right ) - x \]

[In]

integrate((((-8*x^3-82*x^2-244*x-170)*log(3)+(-8*x^3-82*x^2-244*x-170)*exp(4)-4*x^4-53*x^3-241*x^2-414*x-170)*
log((4*x+17)/(5+x))+(3*x^2+6*x)*log(3)+(3*x^2+6*x)*exp(4)+3*x^2+6*x)/(4*x^4+45*x^3+159*x^2+170*x)/log((4*x+17)
/(5+x)),x, algorithm="maxima")

[Out]

-(e^4 + log(3) + 1)*log(x + 2) - (e^4 + log(3) + 1)*log(x) + (e^4 + log(3) + 1)*log(log(4*x + 17) - log(x + 5)
) - x

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 502 vs. \(2 (32) = 64\).

Time = 0.35 (sec) , antiderivative size = 502, normalized size of antiderivative = 15.21 \[ \int \frac {6 x+3 x^2+e^4 \left (6 x+3 x^2\right )+\left (6 x+3 x^2\right ) \log (3)+\left (-170-414 x-241 x^2-53 x^3-4 x^4+e^4 \left (-170-244 x-82 x^2-8 x^3\right )+\left (-170-244 x-82 x^2-8 x^3\right ) \log (3)\right ) \log \left (\frac {17+4 x}{5+x}\right )}{\left (170 x+159 x^2+45 x^3+4 x^4\right ) \log \left (\frac {17+4 x}{5+x}\right )} \, dx=-\frac {\frac {{\left (4 \, x + 17\right )} e^{4} \log \left (\frac {5 \, {\left (4 \, x + 17\right )}^{2}}{{\left (x + 5\right )}^{2}} - \frac {32 \, {\left (4 \, x + 17\right )}}{x + 5} + 51\right )}{x + 5} - 4 \, e^{4} \log \left (\frac {5 \, {\left (4 \, x + 17\right )}^{2}}{{\left (x + 5\right )}^{2}} - \frac {32 \, {\left (4 \, x + 17\right )}}{x + 5} + 51\right ) + \frac {{\left (4 \, x + 17\right )} \log \left (3\right ) \log \left (\frac {5 \, {\left (4 \, x + 17\right )}^{2}}{{\left (x + 5\right )}^{2}} - \frac {32 \, {\left (4 \, x + 17\right )}}{x + 5} + 51\right )}{x + 5} - 4 \, \log \left (3\right ) \log \left (\frac {5 \, {\left (4 \, x + 17\right )}^{2}}{{\left (x + 5\right )}^{2}} - \frac {32 \, {\left (4 \, x + 17\right )}}{x + 5} + 51\right ) - \frac {2 \, {\left (4 \, x + 17\right )} e^{4} \log \left (\frac {4 \, x + 17}{x + 5} - 4\right )}{x + 5} + 8 \, e^{4} \log \left (\frac {4 \, x + 17}{x + 5} - 4\right ) - \frac {2 \, {\left (4 \, x + 17\right )} \log \left (3\right ) \log \left (\frac {4 \, x + 17}{x + 5} - 4\right )}{x + 5} + 8 \, \log \left (3\right ) \log \left (\frac {4 \, x + 17}{x + 5} - 4\right ) - \frac {{\left (4 \, x + 17\right )} e^{4} \log \left (\log \left (\frac {4 \, x + 17}{x + 5}\right )\right )}{x + 5} + 4 \, e^{4} \log \left (\log \left (\frac {4 \, x + 17}{x + 5}\right )\right ) - \frac {{\left (4 \, x + 17\right )} \log \left (3\right ) \log \left (\log \left (\frac {4 \, x + 17}{x + 5}\right )\right )}{x + 5} + 4 \, \log \left (3\right ) \log \left (\log \left (\frac {4 \, x + 17}{x + 5}\right )\right ) + \frac {{\left (4 \, x + 17\right )} \log \left (\frac {5 \, {\left (4 \, x + 17\right )}^{2}}{{\left (x + 5\right )}^{2}} - \frac {32 \, {\left (4 \, x + 17\right )}}{x + 5} + 51\right )}{x + 5} - \frac {2 \, {\left (4 \, x + 17\right )} \log \left (\frac {4 \, x + 17}{x + 5} - 4\right )}{x + 5} - \frac {{\left (4 \, x + 17\right )} \log \left (\log \left (\frac {4 \, x + 17}{x + 5}\right )\right )}{x + 5} - 4 \, \log \left (\frac {5 \, {\left (4 \, x + 17\right )}^{2}}{{\left (x + 5\right )}^{2}} - \frac {32 \, {\left (4 \, x + 17\right )}}{x + 5} + 51\right ) + 8 \, \log \left (\frac {4 \, x + 17}{x + 5} - 4\right ) + 4 \, \log \left (\log \left (\frac {4 \, x + 17}{x + 5}\right )\right ) - 3}{\frac {4 \, x + 17}{x + 5} - 4} \]

[In]

integrate((((-8*x^3-82*x^2-244*x-170)*log(3)+(-8*x^3-82*x^2-244*x-170)*exp(4)-4*x^4-53*x^3-241*x^2-414*x-170)*
log((4*x+17)/(5+x))+(3*x^2+6*x)*log(3)+(3*x^2+6*x)*exp(4)+3*x^2+6*x)/(4*x^4+45*x^3+159*x^2+170*x)/log((4*x+17)
/(5+x)),x, algorithm="giac")

[Out]

-((4*x + 17)*e^4*log(5*(4*x + 17)^2/(x + 5)^2 - 32*(4*x + 17)/(x + 5) + 51)/(x + 5) - 4*e^4*log(5*(4*x + 17)^2
/(x + 5)^2 - 32*(4*x + 17)/(x + 5) + 51) + (4*x + 17)*log(3)*log(5*(4*x + 17)^2/(x + 5)^2 - 32*(4*x + 17)/(x +
5) + 51)/(x + 5) - 4*log(3)*log(5*(4*x + 17)^2/(x + 5)^2 - 32*(4*x + 17)/(x + 5) + 51) - 2*(4*x + 17)*e^4*log
((4*x + 17)/(x + 5) - 4)/(x + 5) + 8*e^4*log((4*x + 17)/(x + 5) - 4) - 2*(4*x + 17)*log(3)*log((4*x + 17)/(x +
5) - 4)/(x + 5) + 8*log(3)*log((4*x + 17)/(x + 5) - 4) - (4*x + 17)*e^4*log(log((4*x + 17)/(x + 5)))/(x + 5)
+ 4*e^4*log(log((4*x + 17)/(x + 5))) - (4*x + 17)*log(3)*log(log((4*x + 17)/(x + 5)))/(x + 5) + 4*log(3)*log(l
og((4*x + 17)/(x + 5))) + (4*x + 17)*log(5*(4*x + 17)^2/(x + 5)^2 - 32*(4*x + 17)/(x + 5) + 51)/(x + 5) - 2*(4
*x + 17)*log((4*x + 17)/(x + 5) - 4)/(x + 5) - (4*x + 17)*log(log((4*x + 17)/(x + 5)))/(x + 5) - 4*log(5*(4*x
+ 17)^2/(x + 5)^2 - 32*(4*x + 17)/(x + 5) + 51) + 8*log((4*x + 17)/(x + 5) - 4) + 4*log(log((4*x + 17)/(x + 5)
)) - 3)/((4*x + 17)/(x + 5) - 4)

Mupad [B] (verification not implemented)

Time = 0.73 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.15 \[ \int \frac {6 x+3 x^2+e^4 \left (6 x+3 x^2\right )+\left (6 x+3 x^2\right ) \log (3)+\left (-170-414 x-241 x^2-53 x^3-4 x^4+e^4 \left (-170-244 x-82 x^2-8 x^3\right )+\left (-170-244 x-82 x^2-8 x^3\right ) \log (3)\right ) \log \left (\frac {17+4 x}{5+x}\right )}{\left (170 x+159 x^2+45 x^3+4 x^4\right ) \log \left (\frac {17+4 x}{5+x}\right )} \, dx=\ln \left (\ln \left (\frac {4\,x+17}{x+5}\right )\right )\,\left ({\mathrm {e}}^4+\ln \left (3\right )+1\right )-x-\ln \left (x\,\left (x+2\right )\right )\,\left ({\mathrm {e}}^4+\ln \left (3\right )+1\right ) \]

[In]

int((6*x + exp(4)*(6*x + 3*x^2) + log(3)*(6*x + 3*x^2) - log((4*x + 17)/(x + 5))*(414*x + exp(4)*(244*x + 82*x
^2 + 8*x^3 + 170) + log(3)*(244*x + 82*x^2 + 8*x^3 + 170) + 241*x^2 + 53*x^3 + 4*x^4 + 170) + 3*x^2)/(log((4*x
+ 17)/(x + 5))*(170*x + 159*x^2 + 45*x^3 + 4*x^4)),x)

[Out]

log(log((4*x + 17)/(x + 5)))*(exp(4) + log(3) + 1) - x - log(x*(x + 2))*(exp(4) + log(3) + 1)