\(\int \frac {192 x^2+36 x^3-3 x^4+(-576 x^2-48 x^3+9 x^4) \log (x)+(16-1528 x-47 x^2+24 x^3) \log ^2(x)}{(16+8 x+x^2) \log ^2(x)} \, dx\) [4659]

Optimal result

Integrand size = 70, antiderivative size = 23 \[ \int \frac {192 x^2+36 x^3-3 x^4+\left (-576 x^2-48 x^3+9 x^4\right ) \log (x)+\left (16-1528 x-47 x^2+24 x^3\right ) \log ^2(x)}{\left (16+8 x+x^2\right ) \log ^2(x)} \, dx=x+\frac {3 (-16+x) x^2 \left (4+\frac {x}{\log (x)}\right )}{4+x} \]

[Out]

x+3*(x-16)*x^2/(4+x)*(4+x/ln(x))

Rubi [F]

\[ \int \frac {192 x^2+36 x^3-3 x^4+\left (-576 x^2-48 x^3+9 x^4\right ) \log (x)+\left (16-1528 x-47 x^2+24 x^3\right ) \log ^2(x)}{\left (16+8 x+x^2\right ) \log ^2(x)} \, dx=\int \frac {192 x^2+36 x^3-3 x^4+\left (-576 x^2-48 x^3+9 x^4\right ) \log (x)+\left (16-1528 x-47 x^2+24 x^3\right ) \log ^2(x)}{\left (16+8 x+x^2\right ) \log ^2(x)} \, dx \]

[In]

Int[(192*x^2 + 36*x^3 - 3*x^4 + (-576*x^2 - 48*x^3 + 9*x^4)*Log[x] + (16 - 1528*x - 47*x^2 + 24*x^3)*Log[x]^2)
/((16 + 8*x + x^2)*Log[x]^2),x]

[Out]

-239*x + 12*x^2 - 3840/(4 + x) - 3*Defer[Int][((-16 + x)*x^2)/((4 + x)*Log[x]^2), x] + 3*Defer[Int][(x^2*(-192
- 16*x + 3*x^2))/((4 + x)^2*Log[x]), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {192 x^2+36 x^3-3 x^4+\left (-576 x^2-48 x^3+9 x^4\right ) \log (x)+\left (16-1528 x-47 x^2+24 x^3\right ) \log ^2(x)}{(4+x)^2 \log ^2(x)} \, dx \\ & = \int \left (\frac {16-1528 x-47 x^2+24 x^3}{(4+x)^2}-\frac {3 (-16+x) x^2}{(4+x) \log ^2(x)}+\frac {3 x^2 \left (-192-16 x+3 x^2\right )}{(4+x)^2 \log (x)}\right ) \, dx \\ & = -\left (3 \int \frac {(-16+x) x^2}{(4+x) \log ^2(x)} \, dx\right )+3 \int \frac {x^2 \left (-192-16 x+3 x^2\right )}{(4+x)^2 \log (x)} \, dx+\int \frac {16-1528 x-47 x^2+24 x^3}{(4+x)^2} \, dx \\ & = -\left (3 \int \frac {(-16+x) x^2}{(4+x) \log ^2(x)} \, dx\right )+3 \int \frac {x^2 \left (-192-16 x+3 x^2\right )}{(4+x)^2 \log (x)} \, dx+\int \left (-239+24 x+\frac {3840}{(4+x)^2}\right ) \, dx \\ & = -239 x+12 x^2-\frac {3840}{4+x}-3 \int \frac {(-16+x) x^2}{(4+x) \log ^2(x)} \, dx+3 \int \frac {x^2 \left (-192-16 x+3 x^2\right )}{(4+x)^2 \log (x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43 \[ \int \frac {192 x^2+36 x^3-3 x^4+\left (-576 x^2-48 x^3+9 x^4\right ) \log (x)+\left (16-1528 x-47 x^2+24 x^3\right ) \log ^2(x)}{\left (16+8 x+x^2\right ) \log ^2(x)} \, dx=-239 x+12 x^2-\frac {3840}{4+x}+\frac {3 (-16+x) x^3}{(4+x) \log (x)} \]

[In]

Integrate[(192*x^2 + 36*x^3 - 3*x^4 + (-576*x^2 - 48*x^3 + 9*x^4)*Log[x] + (16 - 1528*x - 47*x^2 + 24*x^3)*Log
[x]^2)/((16 + 8*x + x^2)*Log[x]^2),x]

[Out]

-239*x + 12*x^2 - 3840/(4 + x) + (3*(-16 + x)*x^3)/((4 + x)*Log[x])

Maple [A] (verified)

Time = 0.92 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.74

method	result	size
norman	\(\frac {-16 \ln \left (x \right )-48 x^{3}+3 x^{4}-191 x^{2} \ln \left (x \right )+12 x^{3} \ln \left (x \right )}{\left (4+x \right ) \ln \left (x \right )}\)	\(40\)
risch	\(\frac {12 x^{3}-191 x^{2}-956 x -3840}{4+x}+\frac {3 x^{3} \left (x -16\right )}{\left (4+x \right ) \ln \left (x \right )}\)	\(40\)
parallelrisch	\(\frac {-16 \ln \left (x \right )-48 x^{3}+3 x^{4}-191 x^{2} \ln \left (x \right )+12 x^{3} \ln \left (x \right )}{\left (4+x \right ) \ln \left (x \right )}\)	\(40\)

[In]

int(((24*x^3-47*x^2-1528*x+16)*ln(x)^2+(9*x^4-48*x^3-576*x^2)*ln(x)-3*x^4+36*x^3+192*x^2)/(x^2+8*x+16)/ln(x)^2
,x,method=_RETURNVERBOSE)

[Out]

(-16*ln(x)-48*x^3+3*x^4-191*x^2*ln(x)+12*x^3*ln(x))/(4+x)/ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.70 \[ \int \frac {192 x^2+36 x^3-3 x^4+\left (-576 x^2-48 x^3+9 x^4\right ) \log (x)+\left (16-1528 x-47 x^2+24 x^3\right ) \log ^2(x)}{\left (16+8 x+x^2\right ) \log ^2(x)} \, dx=\frac {3 \, x^{4} - 48 \, x^{3} + {\left (12 \, x^{3} - 191 \, x^{2} - 956 \, x - 3840\right )} \log \left (x\right )}{{\left (x + 4\right )} \log \left (x\right )} \]

[In]

integrate(((24*x^3-47*x^2-1528*x+16)*log(x)^2+(9*x^4-48*x^3-576*x^2)*log(x)-3*x^4+36*x^3+192*x^2)/(x^2+8*x+16)
/log(x)^2,x, algorithm="fricas")

[Out]

(3*x^4 - 48*x^3 + (12*x^3 - 191*x^2 - 956*x - 3840)*log(x))/((x + 4)*log(x))

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {192 x^2+36 x^3-3 x^4+\left (-576 x^2-48 x^3+9 x^4\right ) \log (x)+\left (16-1528 x-47 x^2+24 x^3\right ) \log ^2(x)}{\left (16+8 x+x^2\right ) \log ^2(x)} \, dx=12 x^{2} - 239 x + \frac {3 x^{4} - 48 x^{3}}{\left (x + 4\right ) \log {\left (x \right )}} - \frac {3840}{x + 4} \]

[In]

integrate(((24*x**3-47*x**2-1528*x+16)*ln(x)**2+(9*x**4-48*x**3-576*x**2)*ln(x)-3*x**4+36*x**3+192*x**2)/(x**2
+8*x+16)/ln(x)**2,x)

[Out]

12*x**2 - 239*x + (3*x**4 - 48*x**3)/((x + 4)*log(x)) - 3840/(x + 4)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.70 \[ \int \frac {192 x^2+36 x^3-3 x^4+\left (-576 x^2-48 x^3+9 x^4\right ) \log (x)+\left (16-1528 x-47 x^2+24 x^3\right ) \log ^2(x)}{\left (16+8 x+x^2\right ) \log ^2(x)} \, dx=\frac {3 \, x^{4} - 48 \, x^{3} + {\left (12 \, x^{3} - 191 \, x^{2} - 956 \, x - 3840\right )} \log \left (x\right )}{{\left (x + 4\right )} \log \left (x\right )} \]

[In]

integrate(((24*x^3-47*x^2-1528*x+16)*log(x)^2+(9*x^4-48*x^3-576*x^2)*log(x)-3*x^4+36*x^3+192*x^2)/(x^2+8*x+16)
/log(x)^2,x, algorithm="maxima")

[Out]

(3*x^4 - 48*x^3 + (12*x^3 - 191*x^2 - 956*x - 3840)*log(x))/((x + 4)*log(x))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.65 \[ \int \frac {192 x^2+36 x^3-3 x^4+\left (-576 x^2-48 x^3+9 x^4\right ) \log (x)+\left (16-1528 x-47 x^2+24 x^3\right ) \log ^2(x)}{\left (16+8 x+x^2\right ) \log ^2(x)} \, dx=12 \, x^{2} - 239 \, x + \frac {3 \, {\left (x^{4} - 16 \, x^{3}\right )}}{x \log \left (x\right ) + 4 \, \log \left (x\right )} - \frac {3840}{x + 4} \]

[In]

integrate(((24*x^3-47*x^2-1528*x+16)*log(x)^2+(9*x^4-48*x^3-576*x^2)*log(x)-3*x^4+36*x^3+192*x^2)/(x^2+8*x+16)
/log(x)^2,x, algorithm="giac")

[Out]

12*x^2 - 239*x + 3*(x^4 - 16*x^3)/(x*log(x) + 4*log(x)) - 3840/(x + 4)

Mupad [B] (verification not implemented)

Time = 10.69 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.78 \[ \int \frac {192 x^2+36 x^3-3 x^4+\left (-576 x^2-48 x^3+9 x^4\right ) \log (x)+\left (16-1528 x-47 x^2+24 x^3\right ) \log ^2(x)}{\left (16+8 x+x^2\right ) \log ^2(x)} \, dx=\frac {x\,\left (12\,x^2-191\,x+4\right )}{x+4}-\frac {x\,\left (48\,x^2-3\,x^3\right )}{\ln \left (x\right )\,\left (x+4\right )} \]

[In]

int(-(log(x)*(576*x^2 + 48*x^3 - 9*x^4) + log(x)^2*(1528*x + 47*x^2 - 24*x^3 - 16) - 192*x^2 - 36*x^3 + 3*x^4)
/(log(x)^2*(8*x + x^2 + 16)),x)

[Out]

(x*(12*x^2 - 191*x + 4))/(x + 4) - (x*(48*x^2 - 3*x^3))/(log(x)*(x + 4))