\(\int \frac {-x+30 x^3-6 x^4+(10-2 x-30 x^3+6 x^4) \log (5-x)+(15 x^3-3 x^4) \log ^2(5-x)}{(-15 x^3+3 x^4) \log ^2(5-x)} \, dx\) [6505]

Optimal result

Integrand size = 78, antiderivative size = 24 \[ \int \frac {-x+30 x^3-6 x^4+\left (10-2 x-30 x^3+6 x^4\right ) \log (5-x)+\left (15 x^3-3 x^4\right ) \log ^2(5-x)}{\left (-15 x^3+3 x^4\right ) \log ^2(5-x)} \, dx=-x+\frac {2 \left (-5+\frac {1}{6 x^2}+x\right )}{\log (5-x)} \]

[Out]

2*(x-5+1/6/x^2)/ln(5-x)-x

Rubi [F]

\[ \int \frac {-x+30 x^3-6 x^4+\left (10-2 x-30 x^3+6 x^4\right ) \log (5-x)+\left (15 x^3-3 x^4\right ) \log ^2(5-x)}{\left (-15 x^3+3 x^4\right ) \log ^2(5-x)} \, dx=\int \frac {-x+30 x^3-6 x^4+\left (10-2 x-30 x^3+6 x^4\right ) \log (5-x)+\left (15 x^3-3 x^4\right ) \log ^2(5-x)}{\left (-15 x^3+3 x^4\right ) \log ^2(5-x)} \, dx \]

[In]

Int[(-x + 30*x^3 - 6*x^4 + (10 - 2*x - 30*x^3 + 6*x^4)*Log[5 - x] + (15*x^3 - 3*x^4)*Log[5 - x]^2)/((-15*x^3 +
3*x^4)*Log[5 - x]^2),x]

[Out]

-x + 1/(75*Log[5 - x]) - (2*(5 - x))/Log[5 - x] + Defer[Int][1/(x^2*Log[5 - x]^2), x]/15 + Defer[Int][1/(x*Log
[5 - x]^2), x]/75 - (2*Defer[Int][1/(x^3*Log[5 - x]), x])/3

Rubi steps \begin{align*} \text {integral}& = \int \frac {-x+30 x^3-6 x^4+\left (10-2 x-30 x^3+6 x^4\right ) \log (5-x)+\left (15 x^3-3 x^4\right ) \log ^2(5-x)}{x^3 (-15+3 x) \log ^2(5-x)} \, dx \\ & = \int \left (-1+\frac {-1+30 x^2-6 x^3}{3 (-5+x) x^2 \log ^2(5-x)}+\frac {2 \left (-1+3 x^3\right )}{3 x^3 \log (5-x)}\right ) \, dx \\ & = -x+\frac {1}{3} \int \frac {-1+30 x^2-6 x^3}{(-5+x) x^2 \log ^2(5-x)} \, dx+\frac {2}{3} \int \frac {-1+3 x^3}{x^3 \log (5-x)} \, dx \\ & = -x+\frac {1}{3} \int \left (-\frac {6}{\log ^2(5-x)}-\frac {1}{25 (-5+x) \log ^2(5-x)}+\frac {1}{5 x^2 \log ^2(5-x)}+\frac {1}{25 x \log ^2(5-x)}\right ) \, dx+\frac {2}{3} \int \left (\frac {3}{\log (5-x)}-\frac {1}{x^3 \log (5-x)}\right ) \, dx \\ & = -x-\frac {1}{75} \int \frac {1}{(-5+x) \log ^2(5-x)} \, dx+\frac {1}{75} \int \frac {1}{x \log ^2(5-x)} \, dx+\frac {1}{15} \int \frac {1}{x^2 \log ^2(5-x)} \, dx-\frac {2}{3} \int \frac {1}{x^3 \log (5-x)} \, dx-2 \int \frac {1}{\log ^2(5-x)} \, dx+2 \int \frac {1}{\log (5-x)} \, dx \\ & = -x+\frac {1}{75} \int \frac {1}{x \log ^2(5-x)} \, dx-\frac {1}{75} \text {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,5-x\right )+\frac {1}{15} \int \frac {1}{x^2 \log ^2(5-x)} \, dx-\frac {2}{3} \int \frac {1}{x^3 \log (5-x)} \, dx+2 \text {Subst}\left (\int \frac {1}{\log ^2(x)} \, dx,x,5-x\right )-2 \text {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,5-x\right ) \\ & = -x-\frac {2 (5-x)}{\log (5-x)}-2 \text {li}(5-x)+\frac {1}{75} \int \frac {1}{x \log ^2(5-x)} \, dx-\frac {1}{75} \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (5-x)\right )+\frac {1}{15} \int \frac {1}{x^2 \log ^2(5-x)} \, dx-\frac {2}{3} \int \frac {1}{x^3 \log (5-x)} \, dx+2 \text {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,5-x\right ) \\ & = -x+\frac {1}{75 \log (5-x)}-\frac {2 (5-x)}{\log (5-x)}+\frac {1}{75} \int \frac {1}{x \log ^2(5-x)} \, dx+\frac {1}{15} \int \frac {1}{x^2 \log ^2(5-x)} \, dx-\frac {2}{3} \int \frac {1}{x^3 \log (5-x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {-x+30 x^3-6 x^4+\left (10-2 x-30 x^3+6 x^4\right ) \log (5-x)+\left (15 x^3-3 x^4\right ) \log ^2(5-x)}{\left (-15 x^3+3 x^4\right ) \log ^2(5-x)} \, dx=\frac {1}{3} \left (-3 x+\frac {-30+\frac {1}{x^2}+6 x}{\log (5-x)}\right ) \]

[In]

Integrate[(-x + 30*x^3 - 6*x^4 + (10 - 2*x - 30*x^3 + 6*x^4)*Log[5 - x] + (15*x^3 - 3*x^4)*Log[5 - x]^2)/((-15
*x^3 + 3*x^4)*Log[5 - x]^2),x]

[Out]

(-3*x + (-30 + x^(-2) + 6*x)/Log[5 - x])/3

Maple [A] (verified)

Time = 8.11 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25

method	result	size
risch	\(-x +\frac {6 x^{3}-30 x^{2}+1}{3 x^{2} \ln \left (5-x \right )}\)	\(30\)
norman	\(\frac {\frac {1}{3}-10 x^{2}+2 x^{3}-x^{3} \ln \left (5-x \right )}{\ln \left (5-x \right ) x^{2}}\)	\(36\)
parallelrisch	\(\frac {1-3 x^{3} \ln \left (5-x \right )+6 x^{3}-30 \ln \left (5-x \right ) x^{2}-30 x^{2}}{3 \ln \left (5-x \right ) x^{2}}\)	\(48\)
derivativedivides	\(5-x +\frac {5 \ln \left (5-x \right )+x}{15 x^{2} \ln \left (5-x \right )^{2}}-\frac {2 \left (5-x \right )}{\ln \left (5-x \right )}-\frac {1}{15 \ln \left (5-x \right )^{2} x}\)	\(57\)
default	\(5-x +\frac {5 \ln \left (5-x \right )+x}{15 x^{2} \ln \left (5-x \right )^{2}}-\frac {2 \left (5-x \right )}{\ln \left (5-x \right )}-\frac {1}{15 \ln \left (5-x \right )^{2} x}\)	\(57\)

[In]

int(((-3*x^4+15*x^3)*ln(5-x)^2+(6*x^4-30*x^3-2*x+10)*ln(5-x)-6*x^4+30*x^3-x)/(3*x^4-15*x^3)/ln(5-x)^2,x,method
=_RETURNVERBOSE)

[Out]

-x+1/3/x^2*(6*x^3-30*x^2+1)/ln(5-x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {-x+30 x^3-6 x^4+\left (10-2 x-30 x^3+6 x^4\right ) \log (5-x)+\left (15 x^3-3 x^4\right ) \log ^2(5-x)}{\left (-15 x^3+3 x^4\right ) \log ^2(5-x)} \, dx=-\frac {3 \, x^{3} \log \left (-x + 5\right ) - 6 \, x^{3} + 30 \, x^{2} - 1}{3 \, x^{2} \log \left (-x + 5\right )} \]

[In]

integrate(((-3*x^4+15*x^3)*log(5-x)^2+(6*x^4-30*x^3-2*x+10)*log(5-x)-6*x^4+30*x^3-x)/(3*x^4-15*x^3)/log(5-x)^2
,x, algorithm="fricas")

[Out]

-1/3*(3*x^3*log(-x + 5) - 6*x^3 + 30*x^2 - 1)/(x^2*log(-x + 5))

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {-x+30 x^3-6 x^4+\left (10-2 x-30 x^3+6 x^4\right ) \log (5-x)+\left (15 x^3-3 x^4\right ) \log ^2(5-x)}{\left (-15 x^3+3 x^4\right ) \log ^2(5-x)} \, dx=- x + \frac {6 x^{3} - 30 x^{2} + 1}{3 x^{2} \log {\left (5 - x \right )}} \]

[In]

integrate(((-3*x**4+15*x**3)*ln(5-x)**2+(6*x**4-30*x**3-2*x+10)*ln(5-x)-6*x**4+30*x**3-x)/(3*x**4-15*x**3)/ln(
5-x)**2,x)

[Out]

-x + (6*x**3 - 30*x**2 + 1)/(3*x**2*log(5 - x))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {-x+30 x^3-6 x^4+\left (10-2 x-30 x^3+6 x^4\right ) \log (5-x)+\left (15 x^3-3 x^4\right ) \log ^2(5-x)}{\left (-15 x^3+3 x^4\right ) \log ^2(5-x)} \, dx=-\frac {3 \, x^{3} \log \left (-x + 5\right ) - 6 \, x^{3} + 30 \, x^{2} - 1}{3 \, x^{2} \log \left (-x + 5\right )} \]

[In]

integrate(((-3*x^4+15*x^3)*log(5-x)^2+(6*x^4-30*x^3-2*x+10)*log(5-x)-6*x^4+30*x^3-x)/(3*x^4-15*x^3)/log(5-x)^2
,x, algorithm="maxima")

[Out]

-1/3*(3*x^3*log(-x + 5) - 6*x^3 + 30*x^2 - 1)/(x^2*log(-x + 5))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (22) = 44\).

Time = 0.30 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.50 \[ \int \frac {-x+30 x^3-6 x^4+\left (10-2 x-30 x^3+6 x^4\right ) \log (5-x)+\left (15 x^3-3 x^4\right ) \log ^2(5-x)}{\left (-15 x^3+3 x^4\right ) \log ^2(5-x)} \, dx=-x + \frac {6 \, {\left (x - 5\right )}^{3} + 60 \, {\left (x - 5\right )}^{2} + 150 \, x - 749}{3 \, {\left ({\left (x - 5\right )}^{2} \log \left (-x + 5\right ) + 10 \, {\left (x - 5\right )} \log \left (-x + 5\right ) + 25 \, \log \left (-x + 5\right )\right )}} + 5 \]

[In]

integrate(((-3*x^4+15*x^3)*log(5-x)^2+(6*x^4-30*x^3-2*x+10)*log(5-x)-6*x^4+30*x^3-x)/(3*x^4-15*x^3)/log(5-x)^2
,x, algorithm="giac")

[Out]

-x + 1/3*(6*(x - 5)^3 + 60*(x - 5)^2 + 150*x - 749)/((x - 5)^2*log(-x + 5) + 10*(x - 5)*log(-x + 5) + 25*log(-
x + 5)) + 5

Mupad [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {-x+30 x^3-6 x^4+\left (10-2 x-30 x^3+6 x^4\right ) \log (5-x)+\left (15 x^3-3 x^4\right ) \log ^2(5-x)}{\left (-15 x^3+3 x^4\right ) \log ^2(5-x)} \, dx=\frac {2\,x^3-10\,x^2+\frac {1}{3}}{x^2\,\ln \left (5-x\right )}-x \]

[In]

int((x - log(5 - x)^2*(15*x^3 - 3*x^4) + log(5 - x)*(2*x + 30*x^3 - 6*x^4 - 10) - 30*x^3 + 6*x^4)/(log(5 - x)^
2*(15*x^3 - 3*x^4)),x)

[Out]

(2*x^3 - 10*x^2 + 1/3)/(x^2*log(5 - x)) - x