\(\int \frac {120 x+44 x^2+4 x^3+(-20 x^2-4 x^3) \log (x)+(72-48 x \log (x)+8 x^2 \log ^2(x)) \log (5+x)}{45+9 x+(-30 x-6 x^2) \log (x)+(5 x^2+x^3) \log ^2(x)} \, dx\) [9373]

Optimal result

Integrand size = 83, antiderivative size = 23 \[ \int \frac {120 x+44 x^2+4 x^3+\left (-20 x^2-4 x^3\right ) \log (x)+\left (72-48 x \log (x)+8 x^2 \log ^2(x)\right ) \log (5+x)}{45+9 x+\left (-30 x-6 x^2\right ) \log (x)+\left (5 x^2+x^3\right ) \log ^2(x)} \, dx=\frac {4 x^2}{3-x \log (x)}+4 \log ^2(5+x) \]

[Out]

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

Rubi [F]

\[ \int \frac {120 x+44 x^2+4 x^3+\left (-20 x^2-4 x^3\right ) \log (x)+\left (72-48 x \log (x)+8 x^2 \log ^2(x)\right ) \log (5+x)}{45+9 x+\left (-30 x-6 x^2\right ) \log (x)+\left (5 x^2+x^3\right ) \log ^2(x)} \, dx=\int \frac {120 x+44 x^2+4 x^3+\left (-20 x^2-4 x^3\right ) \log (x)+\left (72-48 x \log (x)+8 x^2 \log ^2(x)\right ) \log (5+x)}{45+9 x+\left (-30 x-6 x^2\right ) \log (x)+\left (5 x^2+x^3\right ) \log ^2(x)} \, dx \]

[In]

Int[(120*x + 44*x^2 + 4*x^3 + (-20*x^2 - 4*x^3)*Log[x] + (72 - 48*x*Log[x] + 8*x^2*Log[x]^2)*Log[5 + x])/(45 +
9*x + (-30*x - 6*x^2)*Log[x] + (5*x^2 + x^3)*Log[x]^2),x]

[Out]

4*Log[5 + x]^2 + 12*Defer[Int][x/(-3 + x*Log[x])^2, x] + 4*Defer[Int][x^2/(-3 + x*Log[x])^2, x] - 4*Defer[Int]
[x/(-3 + x*Log[x]), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {120 x+44 x^2+4 x^3+\left (-20 x^2-4 x^3\right ) \log (x)+\left (72-48 x \log (x)+8 x^2 \log ^2(x)\right ) \log (5+x)}{(5+x) (3-x \log (x))^2} \, dx \\ & = \int \left (\frac {120 x}{(5+x) (-3+x \log (x))^2}+\frac {44 x^2}{(5+x) (-3+x \log (x))^2}+\frac {4 x^3}{(5+x) (-3+x \log (x))^2}-\frac {4 x^2 \log (x)}{(-3+x \log (x))^2}+\frac {8 \log (5+x)}{5+x}\right ) \, dx \\ & = 4 \int \frac {x^3}{(5+x) (-3+x \log (x))^2} \, dx-4 \int \frac {x^2 \log (x)}{(-3+x \log (x))^2} \, dx+8 \int \frac {\log (5+x)}{5+x} \, dx+44 \int \frac {x^2}{(5+x) (-3+x \log (x))^2} \, dx+120 \int \frac {x}{(5+x) (-3+x \log (x))^2} \, dx \\ & = 4 \int \left (\frac {25}{(-3+x \log (x))^2}-\frac {5 x}{(-3+x \log (x))^2}+\frac {x^2}{(-3+x \log (x))^2}-\frac {125}{(5+x) (-3+x \log (x))^2}\right ) \, dx-4 \int \left (\frac {3 x}{(-3+x \log (x))^2}+\frac {x}{-3+x \log (x)}\right ) \, dx+8 \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,5+x\right )+44 \int \left (-\frac {5}{(-3+x \log (x))^2}+\frac {x}{(-3+x \log (x))^2}+\frac {25}{(5+x) (-3+x \log (x))^2}\right ) \, dx+120 \int \left (\frac {1}{(-3+x \log (x))^2}-\frac {5}{(5+x) (-3+x \log (x))^2}\right ) \, dx \\ & = 4 \log ^2(5+x)+4 \int \frac {x^2}{(-3+x \log (x))^2} \, dx-4 \int \frac {x}{-3+x \log (x)} \, dx-12 \int \frac {x}{(-3+x \log (x))^2} \, dx-20 \int \frac {x}{(-3+x \log (x))^2} \, dx+44 \int \frac {x}{(-3+x \log (x))^2} \, dx+100 \int \frac {1}{(-3+x \log (x))^2} \, dx+120 \int \frac {1}{(-3+x \log (x))^2} \, dx-220 \int \frac {1}{(-3+x \log (x))^2} \, dx-500 \int \frac {1}{(5+x) (-3+x \log (x))^2} \, dx-600 \int \frac {1}{(5+x) (-3+x \log (x))^2} \, dx+1100 \int \frac {1}{(5+x) (-3+x \log (x))^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {120 x+44 x^2+4 x^3+\left (-20 x^2-4 x^3\right ) \log (x)+\left (72-48 x \log (x)+8 x^2 \log ^2(x)\right ) \log (5+x)}{45+9 x+\left (-30 x-6 x^2\right ) \log (x)+\left (5 x^2+x^3\right ) \log ^2(x)} \, dx=-\frac {4 x^2}{-3+x \log (x)}+4 \log ^2(5+x) \]

[In]

Integrate[(120*x + 44*x^2 + 4*x^3 + (-20*x^2 - 4*x^3)*Log[x] + (72 - 48*x*Log[x] + 8*x^2*Log[x]^2)*Log[5 + x])
/(45 + 9*x + (-30*x - 6*x^2)*Log[x] + (5*x^2 + x^3)*Log[x]^2),x]

[Out]

(-4*x^2)/(-3 + x*Log[x]) + 4*Log[5 + x]^2

Maple [A] (verified)

Time = 1.36 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00

method	result	size
default	\(4 \ln \left (5+x \right )^{2}-\frac {4 x^{2}}{x \ln \left (x \right )-3}\)	\(23\)
risch	\(4 \ln \left (5+x \right )^{2}-\frac {4 x^{2}}{x \ln \left (x \right )-3}\)	\(23\)
parallelrisch	\(-\frac {-1200 \ln \left (5+x \right )^{2} \ln \left (x \right ) x +1200 x^{2}+3600 \ln \left (5+x \right )^{2}}{300 \left (x \ln \left (x \right )-3\right )}\)	\(36\)

[In]

int(((8*x^2*ln(x)^2-48*x*ln(x)+72)*ln(5+x)+(-4*x^3-20*x^2)*ln(x)+4*x^3+44*x^2+120*x)/((x^3+5*x^2)*ln(x)^2+(-6*
x^2-30*x)*ln(x)+9*x+45),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {120 x+44 x^2+4 x^3+\left (-20 x^2-4 x^3\right ) \log (x)+\left (72-48 x \log (x)+8 x^2 \log ^2(x)\right ) \log (5+x)}{45+9 x+\left (-30 x-6 x^2\right ) \log (x)+\left (5 x^2+x^3\right ) \log ^2(x)} \, dx=\frac {4 \, {\left ({\left (x \log \left (x\right ) - 3\right )} \log \left (x + 5\right )^{2} - x^{2}\right )}}{x \log \left (x\right ) - 3} \]

[In]

integrate(((8*x^2*log(x)^2-48*x*log(x)+72)*log(5+x)+(-4*x^3-20*x^2)*log(x)+4*x^3+44*x^2+120*x)/((x^3+5*x^2)*lo
g(x)^2+(-6*x^2-30*x)*log(x)+9*x+45),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {120 x+44 x^2+4 x^3+\left (-20 x^2-4 x^3\right ) \log (x)+\left (72-48 x \log (x)+8 x^2 \log ^2(x)\right ) \log (5+x)}{45+9 x+\left (-30 x-6 x^2\right ) \log (x)+\left (5 x^2+x^3\right ) \log ^2(x)} \, dx=- \frac {4 x^{2}}{x \log {\left (x \right )} - 3} + 4 \log {\left (x + 5 \right )}^{2} \]

[In]

integrate(((8*x**2*ln(x)**2-48*x*ln(x)+72)*ln(5+x)+(-4*x**3-20*x**2)*ln(x)+4*x**3+44*x**2+120*x)/((x**3+5*x**2
)*ln(x)**2+(-6*x**2-30*x)*ln(x)+9*x+45),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {120 x+44 x^2+4 x^3+\left (-20 x^2-4 x^3\right ) \log (x)+\left (72-48 x \log (x)+8 x^2 \log ^2(x)\right ) \log (5+x)}{45+9 x+\left (-30 x-6 x^2\right ) \log (x)+\left (5 x^2+x^3\right ) \log ^2(x)} \, dx=\frac {4 \, {\left ({\left (x \log \left (x\right ) - 3\right )} \log \left (x + 5\right )^{2} - x^{2}\right )}}{x \log \left (x\right ) - 3} \]

[In]

integrate(((8*x^2*log(x)^2-48*x*log(x)+72)*log(5+x)+(-4*x^3-20*x^2)*log(x)+4*x^3+44*x^2+120*x)/((x^3+5*x^2)*lo
g(x)^2+(-6*x^2-30*x)*log(x)+9*x+45),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {120 x+44 x^2+4 x^3+\left (-20 x^2-4 x^3\right ) \log (x)+\left (72-48 x \log (x)+8 x^2 \log ^2(x)\right ) \log (5+x)}{45+9 x+\left (-30 x-6 x^2\right ) \log (x)+\left (5 x^2+x^3\right ) \log ^2(x)} \, dx=4 \, \log \left (x + 5\right )^{2} - \frac {4 \, x^{2}}{x \log \left (x\right ) - 3} \]

[In]

integrate(((8*x^2*log(x)^2-48*x*log(x)+72)*log(5+x)+(-4*x^3-20*x^2)*log(x)+4*x^3+44*x^2+120*x)/((x^3+5*x^2)*lo
g(x)^2+(-6*x^2-30*x)*log(x)+9*x+45),x, algorithm="giac")

[Out]

4*log(x + 5)^2 - 4*x^2/(x*log(x) - 3)

Mupad [B] (verification not implemented)

Time = 14.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {120 x+44 x^2+4 x^3+\left (-20 x^2-4 x^3\right ) \log (x)+\left (72-48 x \log (x)+8 x^2 \log ^2(x)\right ) \log (5+x)}{45+9 x+\left (-30 x-6 x^2\right ) \log (x)+\left (5 x^2+x^3\right ) \log ^2(x)} \, dx=4\,{\ln \left (x+5\right )}^2-\frac {4\,x^2}{x\,\ln \left (x\right )-3} \]

[In]

int((120*x - log(x)*(20*x^2 + 4*x^3) + log(x + 5)*(8*x^2*log(x)^2 - 48*x*log(x) + 72) + 44*x^2 + 4*x^3)/(9*x +
log(x)^2*(5*x^2 + x^3) - log(x)*(30*x + 6*x^2) + 45),x)

[Out]

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