\(\int \frac {-300-150 x+55 x^2-x^3+(100-25 x) \log (4-x)}{-300 x-13 x^2+26 x^3-x^4+(100 x-29 x^2+x^3) \log (4-x)} \, dx\) [7880]

Optimal result

Integrand size = 68, antiderivative size = 20 \[ \int \frac {-300-150 x+55 x^2-x^3+(100-25 x) \log (4-x)}{-300 x-13 x^2+26 x^3-x^4+\left (100 x-29 x^2+x^3\right ) \log (4-x)} \, dx=\log \left (\frac {4 x (-3-x+\log (4-x))}{-25+x}\right ) \]

[Out]

ln(4*(ln(-x+4)-3-x)/(x-25)*x)

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {6873, 6860, 36, 31, 29, 6816} \[ \int \frac {-300-150 x+55 x^2-x^3+(100-25 x) \log (4-x)}{-300 x-13 x^2+26 x^3-x^4+\left (100 x-29 x^2+x^3\right ) \log (4-x)} \, dx=-\log (25-x)+\log (x)+\log (x-\log (4-x)+3) \]

[In]

Int[(-300 - 150*x + 55*x^2 - x^3 + (100 - 25*x)*Log[4 - x])/(-300*x - 13*x^2 + 26*x^3 - x^4 + (100*x - 29*x^2
+ x^3)*Log[4 - x]),x]

[Out]

-Log[25 - x] + Log[x] + Log[3 + x - Log[4 - x]]

Rule 29

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6860

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rule 6873

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {300+150 x-55 x^2+x^3-(100-25 x) \log (4-x)}{x \left (100-29 x+x^2\right ) (3+x-\log (4-x))} \, dx \\ & = \int \left (-\frac {25}{(-25+x) x}+\frac {-5+x}{(-4+x) (3+x-\log (4-x))}\right ) \, dx \\ & = -\left (25 \int \frac {1}{(-25+x) x} \, dx\right )+\int \frac {-5+x}{(-4+x) (3+x-\log (4-x))} \, dx \\ & = \log (3+x-\log (4-x))-\int \frac {1}{-25+x} \, dx+\int \frac {1}{x} \, dx \\ & = -\log (25-x)+\log (x)+\log (3+x-\log (4-x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {-300-150 x+55 x^2-x^3+(100-25 x) \log (4-x)}{-300 x-13 x^2+26 x^3-x^4+\left (100 x-29 x^2+x^3\right ) \log (4-x)} \, dx=-\log (25-x)+\log (x)+\log (3+x-\log (4-x)) \]

[In]

Integrate[(-300 - 150*x + 55*x^2 - x^3 + (100 - 25*x)*Log[4 - x])/(-300*x - 13*x^2 + 26*x^3 - x^4 + (100*x - 2
9*x^2 + x^3)*Log[4 - x]),x]

[Out]

-Log[25 - x] + Log[x] + Log[3 + x - Log[4 - x]]

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10

method	result	size
norman	\(-\ln \left (x -25\right )+\ln \left (x \right )+\ln \left (x -\ln \left (-x +4\right )+3\right )\)	\(22\)
risch	\(-\ln \left (x -25\right )+\ln \left (x \right )+\ln \left (\ln \left (-x +4\right )-3-x \right )\)	\(22\)
parallelrisch	\(-\ln \left (x -25\right )+\ln \left (x \right )+\ln \left (x -\ln \left (-x +4\right )+3\right )\)	\(22\)
derivativedivides	\(-\ln \left (-x +25\right )+\ln \left (-x \right )+\ln \left (\ln \left (-x +4\right )-3-x \right )\)	\(26\)
default	\(-\ln \left (-x +25\right )+\ln \left (-x \right )+\ln \left (\ln \left (-x +4\right )-3-x \right )\)	\(26\)

[In]

int(((-25*x+100)*ln(-x+4)-x^3+55*x^2-150*x-300)/((x^3-29*x^2+100*x)*ln(-x+4)-x^4+26*x^3-13*x^2-300*x),x,method
=_RETURNVERBOSE)

[Out]

-ln(x-25)+ln(x)+ln(x-ln(-x+4)+3)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {-300-150 x+55 x^2-x^3+(100-25 x) \log (4-x)}{-300 x-13 x^2+26 x^3-x^4+\left (100 x-29 x^2+x^3\right ) \log (4-x)} \, dx=-\log \left (x - 25\right ) + \log \left (x\right ) + \log \left (-x + \log \left (-x + 4\right ) - 3\right ) \]

[In]

integrate(((-25*x+100)*log(-x+4)-x^3+55*x^2-150*x-300)/((x^3-29*x^2+100*x)*log(-x+4)-x^4+26*x^3-13*x^2-300*x),
x, algorithm="fricas")

[Out]

-log(x - 25) + log(x) + log(-x + log(-x + 4) - 3)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {-300-150 x+55 x^2-x^3+(100-25 x) \log (4-x)}{-300 x-13 x^2+26 x^3-x^4+\left (100 x-29 x^2+x^3\right ) \log (4-x)} \, dx=\log {\left (x \right )} - \log {\left (x - 25 \right )} + \log {\left (- x + \log {\left (4 - x \right )} - 3 \right )} \]

[In]

integrate(((-25*x+100)*ln(-x+4)-x**3+55*x**2-150*x-300)/((x**3-29*x**2+100*x)*ln(-x+4)-x**4+26*x**3-13*x**2-30
0*x),x)

[Out]

log(x) - log(x - 25) + log(-x + log(4 - x) - 3)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {-300-150 x+55 x^2-x^3+(100-25 x) \log (4-x)}{-300 x-13 x^2+26 x^3-x^4+\left (100 x-29 x^2+x^3\right ) \log (4-x)} \, dx=-\log \left (x - 25\right ) + \log \left (x\right ) + \log \left (-x + \log \left (-x + 4\right ) - 3\right ) \]

[In]

integrate(((-25*x+100)*log(-x+4)-x^3+55*x^2-150*x-300)/((x^3-29*x^2+100*x)*log(-x+4)-x^4+26*x^3-13*x^2-300*x),
x, algorithm="maxima")

[Out]

-log(x - 25) + log(x) + log(-x + log(-x + 4) - 3)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25 \[ \int \frac {-300-150 x+55 x^2-x^3+(100-25 x) \log (4-x)}{-300 x-13 x^2+26 x^3-x^4+\left (100 x-29 x^2+x^3\right ) \log (4-x)} \, dx=\log \left (-x\right ) + \log \left (x - \log \left (-x + 4\right ) + 3\right ) - \log \left (-x + 25\right ) \]

[In]

integrate(((-25*x+100)*log(-x+4)-x^3+55*x^2-150*x-300)/((x^3-29*x^2+100*x)*log(-x+4)-x^4+26*x^3-13*x^2-300*x),
x, algorithm="giac")

[Out]

log(-x) + log(x - log(-x + 4) + 3) - log(-x + 25)

Mupad [B] (verification not implemented)

Time = 11.64 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {-300-150 x+55 x^2-x^3+(100-25 x) \log (4-x)}{-300 x-13 x^2+26 x^3-x^4+\left (100 x-29 x^2+x^3\right ) \log (4-x)} \, dx=\ln \left (x-\ln \left (4-x\right )+3\right )+2\,\mathrm {atanh}\left (\frac {2\,x}{25}-1\right ) \]

[In]

int((150*x + log(4 - x)*(25*x - 100) - 55*x^2 + x^3 + 300)/(300*x - log(4 - x)*(100*x - 29*x^2 + x^3) + 13*x^2
- 26*x^3 + x^4),x)

[Out]

log(x - log(4 - x) + 3) + 2*atanh((2*x)/25 - 1)