3.41.61 $\int \frac{288+504x+16x^2}{-144x-504x^2-437x^3+16x^3\log (x)}dx$

Optimal. Leaf size=20 $$\log (-\frac{1}{4}+{(-\frac{21}{4}-\frac{3}{x})}^2-\log (x))$$

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Rubi [F]  time = 0.25, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.000, Rules used = {} $$\begin{array}{c}\int \frac{288+504x+16x^2}{-144x-504x^2-437x^3+16x^3\log (x)}dx\end{array}$$

Verification is not applicable to the result.

[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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (\frac{504}{-144-504x-437x^2+16x^2\log (x)}+\frac{288}{x(-144-504x-437x^2+16x^2\log (x))}+\frac{16x}{-144-504x-437x^2+16x^2\log (x)})dx\\ & =16\int \frac{x}{-144-504x-437x^2+16x^2\log (x)}dx+288\int \frac{1}{x(-144-504x-437x^2+16x^2\log (x))}dx+504\int \frac{1}{-144-504x-437x^2+16x^2\log (x)}dx\end{array}\end{array}$$

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Mathematica [A]  time = 0.12, size = 31, normalized size = 1.55 $$\begin{array}{c}8(-\frac{\log (x)}{4}+\frac{1}{8}\log (144+504x+437x^2-16x^2\log (x)))\end{array}$$

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.51, size = 22, normalized size = 1.10 $$\begin{array}{c}\log \left(\frac{16x^2\log (x)-437x^2-504x-144}{x^2}\right)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.17, size = 23, normalized size = 1.15 $$\begin{array}{c}\log (16x^2\log (x)-437x^2-504x-144)-2\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.02, size = 20, normalized size = 1.00

method	result	size
risch	$\ln (\ln (x)-\frac{437x^2+504x+144}{16x^2})$	$20$
norman	$-2\ln (x)+\ln (16x^2\ln (x)-437x^2-504x-144)$	$24$

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.40, size = 23, normalized size = 1.15 $$\begin{array}{c}\log \left(\frac{16x^2\log (x)-437x^2-504x-144}{16x^2}\right)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 3.21, size = 23, normalized size = 1.15 $$\begin{array}{c}\ln (504x-16x^2\ln (x)+437x^2+144)-2\ln (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.18, size = 20, normalized size = 1.00 $$\begin{array}{c}\log (\log (x)+\frac{-437x^2-504x-144}{16x^2})\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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))

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