\(\int \frac {288+504 x+16 x^2}{-144 x-504 x^2-437 x^3+16 x^3 \log (x)} \, dx\) [4061]

Optimal result

Integrand size = 34, antiderivative size = 20 \[ \int \frac {288+504 x+16 x^2}{-144 x-504 x^2-437 x^3+16 x^3 \log (x)} \, dx=\log \left (-\frac {1}{4}+\left (-\frac {21}{4}-\frac {3}{x}\right )^2-\log (x)\right ) \]

[Out]

ln((-21/4-3/x)^2-1/4-ln(x))

Rubi [F]

\[ \int \frac {288+504 x+16 x^2}{-144 x-504 x^2-437 x^3+16 x^3 \log (x)} \, dx=\int \frac {288+504 x+16 x^2}{-144 x-504 x^2-437 x^3+16 x^3 \log (x)} \, dx \]

[In]

Int[(288 + 504*x + 16*x^2)/(-144*x - 504*x^2 - 437*x^3 + 16*x^3*Log[x]),x]

[Out]

504*Defer[Int][(-144 - 504*x - 437*x^2 + 16*x^2*Log[x])^(-1), x] + 288*Defer[Int][1/(x*(-144 - 504*x - 437*x^2
+ 16*x^2*Log[x])), x] + 16*Defer[Int][x/(-144 - 504*x - 437*x^2 + 16*x^2*Log[x]), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {504}{-144-504 x-437 x^2+16 x^2 \log (x)}+\frac {288}{x \left (-144-504 x-437 x^2+16 x^2 \log (x)\right )}+\frac {16 x}{-144-504 x-437 x^2+16 x^2 \log (x)}\right ) \, dx \\ & = 16 \int \frac {x}{-144-504 x-437 x^2+16 x^2 \log (x)} \, dx+288 \int \frac {1}{x \left (-144-504 x-437 x^2+16 x^2 \log (x)\right )} \, dx+504 \int \frac {1}{-144-504 x-437 x^2+16 x^2 \log (x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.55 \[ \int \frac {288+504 x+16 x^2}{-144 x-504 x^2-437 x^3+16 x^3 \log (x)} \, dx=8 \left (-\frac {\log (x)}{4}+\frac {1}{8} \log \left (144+504 x+437 x^2-16 x^2 \log (x)\right )\right ) \]

[In]

Integrate[(288 + 504*x + 16*x^2)/(-144*x - 504*x^2 - 437*x^3 + 16*x^3*Log[x]),x]

[Out]

8*(-1/4*Log[x] + Log[144 + 504*x + 437*x^2 - 16*x^2*Log[x]]/8)

Maple [A] (verified)

Time = 2.23 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00

method	result	size
risch	\(\ln \left (\ln \left (x \right )-\frac {437 x^{2}+504 x +144}{16 x^{2}}\right )\)	\(20\)
parallelrisch	\(\ln \left (x^{2} \ln \left (x \right )-\frac {437 x^{2}}{16}-\frac {63 x}{2}-9\right )-2 \ln \left (x \right )\)	\(23\)
default	\(-2 \ln \left (x \right )+\ln \left (16 x^{2} \ln \left (x \right )-437 x^{2}-504 x -144\right )\)	\(24\)
norman	\(-2 \ln \left (x \right )+\ln \left (16 x^{2} \ln \left (x \right )-437 x^{2}-504 x -144\right )\)	\(24\)

[In]

int((16*x^2+504*x+288)/(16*x^3*ln(x)-437*x^3-504*x^2-144*x),x,method=_RETURNVERBOSE)

[Out]

ln(ln(x)-1/16*(437*x^2+504*x+144)/x^2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {288+504 x+16 x^2}{-144 x-504 x^2-437 x^3+16 x^3 \log (x)} \, dx=\log \left (\frac {16 \, x^{2} \log \left (x\right ) - 437 \, x^{2} - 504 \, x - 144}{x^{2}}\right ) \]

[In]

integrate((16*x^2+504*x+288)/(16*x^3*log(x)-437*x^3-504*x^2-144*x),x, algorithm="fricas")

[Out]

log((16*x^2*log(x) - 437*x^2 - 504*x - 144)/x^2)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {288+504 x+16 x^2}{-144 x-504 x^2-437 x^3+16 x^3 \log (x)} \, dx=\log {\left (\log {\left (x \right )} + \frac {- 437 x^{2} - 504 x - 144}{16 x^{2}} \right )} \]

[In]

integrate((16*x**2+504*x+288)/(16*x**3*ln(x)-437*x**3-504*x**2-144*x),x)

[Out]

log(log(x) + (-437*x**2 - 504*x - 144)/(16*x**2))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {288+504 x+16 x^2}{-144 x-504 x^2-437 x^3+16 x^3 \log (x)} \, dx=\log \left (\frac {16 \, x^{2} \log \left (x\right ) - 437 \, x^{2} - 504 \, x - 144}{16 \, x^{2}}\right ) \]

[In]

integrate((16*x^2+504*x+288)/(16*x^3*log(x)-437*x^3-504*x^2-144*x),x, algorithm="maxima")

[Out]

log(1/16*(16*x^2*log(x) - 437*x^2 - 504*x - 144)/x^2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {288+504 x+16 x^2}{-144 x-504 x^2-437 x^3+16 x^3 \log (x)} \, dx=\log \left (16 \, x^{2} \log \left (x\right ) - 437 \, x^{2} - 504 \, x - 144\right ) - 2 \, \log \left (x\right ) \]

[In]

integrate((16*x^2+504*x+288)/(16*x^3*log(x)-437*x^3-504*x^2-144*x),x, algorithm="giac")

[Out]

log(16*x^2*log(x) - 437*x^2 - 504*x - 144) - 2*log(x)

Mupad [B] (verification not implemented)

Time = 10.51 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {288+504 x+16 x^2}{-144 x-504 x^2-437 x^3+16 x^3 \log (x)} \, dx=\ln \left (504\,x-16\,x^2\,\ln \left (x\right )+437\,x^2+144\right )-2\,\ln \left (x\right ) \]

[In]

int(-(504*x + 16*x^2 + 288)/(144*x - 16*x^3*log(x) + 504*x^2 + 437*x^3),x)

[Out]

log(504*x - 16*x^2*log(x) + 437*x^2 + 144) - 2*log(x)