\(\int \frac {6+x-5 \log (x)}{64+48 x+12 x^2+x^3+(240+120 x+15 x^2) \log (x)+(300+75 x) \log ^2(x)+125 \log ^3(x)} \, dx\) [7623]

Optimal result

Integrand size = 52, antiderivative size = 12 \[ \int \frac {6+x-5 \log (x)}{64+48 x+12 x^2+x^3+\left (240+120 x+15 x^2\right ) \log (x)+(300+75 x) \log ^2(x)+125 \log ^3(x)} \, dx=-\frac {x}{(4+x+5 \log (x))^2} \]

[Out]

-x/(5*ln(x)+4+x)^2

Rubi [F]

\[ \int \frac {6+x-5 \log (x)}{64+48 x+12 x^2+x^3+\left (240+120 x+15 x^2\right ) \log (x)+(300+75 x) \log ^2(x)+125 \log ^3(x)} \, dx=\int \frac {6+x-5 \log (x)}{64+48 x+12 x^2+x^3+\left (240+120 x+15 x^2\right ) \log (x)+(300+75 x) \log ^2(x)+125 \log ^3(x)} \, dx \]

[In]

Int[(6 + x - 5*Log[x])/(64 + 48*x + 12*x^2 + x^3 + (240 + 120*x + 15*x^2)*Log[x] + (300 + 75*x)*Log[x]^2 + 125
*Log[x]^3),x]

[Out]

10*Defer[Int][(4 + x + 5*Log[x])^(-3), x] + 2*Defer[Int][x/(4 + x + 5*Log[x])^3, x] - Defer[Int][(4 + x + 5*Lo
g[x])^(-2), x]

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {6+x-5 \log (x)}{64+48 x+12 x^2+x^3+\left (240+120 x+15 x^2\right ) \log (x)+(300+75 x) \log ^2(x)+125 \log ^3(x)} \, dx=-\frac {x}{(4+x+5 \log (x))^2} \]

[In]

Integrate[(6 + x - 5*Log[x])/(64 + 48*x + 12*x^2 + x^3 + (240 + 120*x + 15*x^2)*Log[x] + (300 + 75*x)*Log[x]^2
+ 125*Log[x]^3),x]

[Out]

-(x/(4 + x + 5*Log[x])^2)

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.08

method	result	size
default	\(-\frac {x}{\left (5 \ln \left (x \right )+4+x \right )^{2}}\)	\(13\)
norman	\(-\frac {x}{\left (5 \ln \left (x \right )+4+x \right )^{2}}\)	\(13\)
risch	\(-\frac {x}{\left (5 \ln \left (x \right )+4+x \right )^{2}}\)	\(13\)
parallelrisch	\(-\frac {x}{x^{2}+10 x \ln \left (x \right )+25 \ln \left (x \right )^{2}+8 x +40 \ln \left (x \right )+16}\)	\(29\)

[In]

int((-5*ln(x)+x+6)/(125*ln(x)^3+(75*x+300)*ln(x)^2+(15*x^2+120*x+240)*ln(x)+x^3+12*x^2+48*x+64),x,method=_RETU
RNVERBOSE)

[Out]

-x/(5*ln(x)+4+x)^2

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 26 vs. \(2 (12) = 24\).

Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 2.17 \[ \int \frac {6+x-5 \log (x)}{64+48 x+12 x^2+x^3+\left (240+120 x+15 x^2\right ) \log (x)+(300+75 x) \log ^2(x)+125 \log ^3(x)} \, dx=-\frac {x}{x^{2} + 10 \, {\left (x + 4\right )} \log \left (x\right ) + 25 \, \log \left (x\right )^{2} + 8 \, x + 16} \]

[In]

integrate((-5*log(x)+x+6)/(125*log(x)^3+(75*x+300)*log(x)^2+(15*x^2+120*x+240)*log(x)+x^3+12*x^2+48*x+64),x, a
lgorithm="fricas")

[Out]

-x/(x^2 + 10*(x + 4)*log(x) + 25*log(x)^2 + 8*x + 16)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 26 vs. \(2 (12) = 24\).

Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 2.17 \[ \int \frac {6+x-5 \log (x)}{64+48 x+12 x^2+x^3+\left (240+120 x+15 x^2\right ) \log (x)+(300+75 x) \log ^2(x)+125 \log ^3(x)} \, dx=- \frac {x}{x^{2} + 8 x + \left (10 x + 40\right ) \log {\left (x \right )} + 25 \log {\left (x \right )}^{2} + 16} \]

[In]

integrate((-5*ln(x)+x+6)/(125*ln(x)**3+(75*x+300)*ln(x)**2+(15*x**2+120*x+240)*ln(x)+x**3+12*x**2+48*x+64),x)

[Out]

-x/(x**2 + 8*x + (10*x + 40)*log(x) + 25*log(x)**2 + 16)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 26 vs. \(2 (12) = 24\).

Time = 0.22 (sec) , antiderivative size = 26, normalized size of antiderivative = 2.17 \[ \int \frac {6+x-5 \log (x)}{64+48 x+12 x^2+x^3+\left (240+120 x+15 x^2\right ) \log (x)+(300+75 x) \log ^2(x)+125 \log ^3(x)} \, dx=-\frac {x}{x^{2} + 10 \, {\left (x + 4\right )} \log \left (x\right ) + 25 \, \log \left (x\right )^{2} + 8 \, x + 16} \]

[In]

integrate((-5*log(x)+x+6)/(125*log(x)^3+(75*x+300)*log(x)^2+(15*x^2+120*x+240)*log(x)+x^3+12*x^2+48*x+64),x, a
lgorithm="maxima")

[Out]

-x/(x^2 + 10*(x + 4)*log(x) + 25*log(x)^2 + 8*x + 16)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (12) = 24\).

Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 4.42 \[ \int \frac {6+x-5 \log (x)}{64+48 x+12 x^2+x^3+\left (240+120 x+15 x^2\right ) \log (x)+(300+75 x) \log ^2(x)+125 \log ^3(x)} \, dx=-\frac {x^{2} + 5 \, x}{x^{3} + 10 \, x^{2} \log \left (x\right ) + 25 \, x \log \left (x\right )^{2} + 13 \, x^{2} + 90 \, x \log \left (x\right ) + 125 \, \log \left (x\right )^{2} + 56 \, x + 200 \, \log \left (x\right ) + 80} \]

[In]

integrate((-5*log(x)+x+6)/(125*log(x)^3+(75*x+300)*log(x)^2+(15*x^2+120*x+240)*log(x)+x^3+12*x^2+48*x+64),x, a
lgorithm="giac")

[Out]

-(x^2 + 5*x)/(x^3 + 10*x^2*log(x) + 25*x*log(x)^2 + 13*x^2 + 90*x*log(x) + 125*log(x)^2 + 56*x + 200*log(x) +
80)

Mupad [B] (verification not implemented)

Time = 13.94 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {6+x-5 \log (x)}{64+48 x+12 x^2+x^3+\left (240+120 x+15 x^2\right ) \log (x)+(300+75 x) \log ^2(x)+125 \log ^3(x)} \, dx=-\frac {x}{{\left (x+5\,\ln \left (x\right )+4\right )}^2} \]

[In]

int((x - 5*log(x) + 6)/(48*x + 125*log(x)^3 + log(x)*(120*x + 15*x^2 + 240) + 12*x^2 + x^3 + log(x)^2*(75*x +
300) + 64),x)

[Out]

-x/(x + 5*log(x) + 4)^2