\(\int \frac {24+2 x-2 x^2-7 x \log (\frac {81 x^2}{2})}{(16 x-8 x^2+x^3) \log ^2(\frac {81 x^2}{2})} \, dx\) [2733]

Optimal result

Integrand size = 46, antiderivative size = 19 \[ \int \frac {24+2 x-2 x^2-7 x \log \left (\frac {81 x^2}{2}\right )}{\left (16 x-8 x^2+x^3\right ) \log ^2\left (\frac {81 x^2}{2}\right )} \, dx=\frac {3+x}{(-4+x) \log \left (\frac {81 x^2}{2}\right )} \]

[Out]

(3+x)/(x-4)/ln(81/2*x^2)

Rubi [F]

\[ \int \frac {24+2 x-2 x^2-7 x \log \left (\frac {81 x^2}{2}\right )}{\left (16 x-8 x^2+x^3\right ) \log ^2\left (\frac {81 x^2}{2}\right )} \, dx=\int \frac {24+2 x-2 x^2-7 x \log \left (\frac {81 x^2}{2}\right )}{\left (16 x-8 x^2+x^3\right ) \log ^2\left (\frac {81 x^2}{2}\right )} \, dx \]

[In]

Int[(24 + 2*x - 2*x^2 - 7*x*Log[(81*x^2)/2])/((16*x - 8*x^2 + x^3)*Log[(81*x^2)/2]^2),x]

[Out]

-2*Defer[Int][(3 + x)/((-4 + x)*x*Log[(81*x^2)/2]^2), x] - 7*Defer[Int][1/((-4 + x)^2*Log[(81*x^2)/2]), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {24+2 x-2 x^2-7 x \log \left (\frac {81 x^2}{2}\right )}{x \left (16-8 x+x^2\right ) \log ^2\left (\frac {81 x^2}{2}\right )} \, dx \\ & = \int \frac {24+2 x-2 x^2-7 x \log \left (\frac {81 x^2}{2}\right )}{(-4+x)^2 x \log ^2\left (\frac {81 x^2}{2}\right )} \, dx \\ & = \int \left (-\frac {2 (3+x)}{(-4+x) x \log ^2\left (\frac {81 x^2}{2}\right )}-\frac {7}{(-4+x)^2 \log \left (\frac {81 x^2}{2}\right )}\right ) \, dx \\ & = -\left (2 \int \frac {3+x}{(-4+x) x \log ^2\left (\frac {81 x^2}{2}\right )} \, dx\right )-7 \int \frac {1}{(-4+x)^2 \log \left (\frac {81 x^2}{2}\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {24+2 x-2 x^2-7 x \log \left (\frac {81 x^2}{2}\right )}{\left (16 x-8 x^2+x^3\right ) \log ^2\left (\frac {81 x^2}{2}\right )} \, dx=\frac {3+x}{(-4+x) \log \left (\frac {81 x^2}{2}\right )} \]

[In]

Integrate[(24 + 2*x - 2*x^2 - 7*x*Log[(81*x^2)/2])/((16*x - 8*x^2 + x^3)*Log[(81*x^2)/2]^2),x]

[Out]

(3 + x)/((-4 + x)*Log[(81*x^2)/2])

Maple [A] (verified)

Time = 1.38 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95

method	result	size
norman	\(\frac {3+x}{\left (x -4\right ) \ln \left (\frac {81 x^{2}}{2}\right )}\)	\(18\)
risch	\(\frac {3+x}{\left (x -4\right ) \ln \left (\frac {81 x^{2}}{2}\right )}\)	\(18\)
parallelrisch	\(\frac {2 x +6}{2 \left (x -4\right ) \ln \left (\frac {81 x^{2}}{2}\right )}\)	\(21\)

[In]

int((-7*x*ln(81/2*x^2)-2*x^2+2*x+24)/(x^3-8*x^2+16*x)/ln(81/2*x^2)^2,x,method=_RETURNVERBOSE)

[Out]

(3+x)/(x-4)/ln(81/2*x^2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {24+2 x-2 x^2-7 x \log \left (\frac {81 x^2}{2}\right )}{\left (16 x-8 x^2+x^3\right ) \log ^2\left (\frac {81 x^2}{2}\right )} \, dx=\frac {x + 3}{{\left (x - 4\right )} \log \left (\frac {81}{2} \, x^{2}\right )} \]

[In]

integrate((-7*x*log(81/2*x^2)-2*x^2+2*x+24)/(x^3-8*x^2+16*x)/log(81/2*x^2)^2,x, algorithm="fricas")

[Out]

(x + 3)/((x - 4)*log(81/2*x^2))

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {24+2 x-2 x^2-7 x \log \left (\frac {81 x^2}{2}\right )}{\left (16 x-8 x^2+x^3\right ) \log ^2\left (\frac {81 x^2}{2}\right )} \, dx=\frac {x + 3}{\left (x - 4\right ) \log {\left (\frac {81 x^{2}}{2} \right )}} \]

[In]

integrate((-7*x*ln(81/2*x**2)-2*x**2+2*x+24)/(x**3-8*x**2+16*x)/ln(81/2*x**2)**2,x)

[Out]

(x + 3)/((x - 4)*log(81*x**2/2))

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.74 \[ \int \frac {24+2 x-2 x^2-7 x \log \left (\frac {81 x^2}{2}\right )}{\left (16 x-8 x^2+x^3\right ) \log ^2\left (\frac {81 x^2}{2}\right )} \, dx=\frac {x + 3}{x {\left (4 \, \log \left (3\right ) - \log \left (2\right )\right )} + 2 \, {\left (x - 4\right )} \log \left (x\right ) - 16 \, \log \left (3\right ) + 4 \, \log \left (2\right )} \]

[In]

integrate((-7*x*log(81/2*x^2)-2*x^2+2*x+24)/(x^3-8*x^2+16*x)/log(81/2*x^2)^2,x, algorithm="maxima")

[Out]

(x + 3)/(x*(4*log(3) - log(2)) + 2*(x - 4)*log(x) - 16*log(3) + 4*log(2))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {24+2 x-2 x^2-7 x \log \left (\frac {81 x^2}{2}\right )}{\left (16 x-8 x^2+x^3\right ) \log ^2\left (\frac {81 x^2}{2}\right )} \, dx=\frac {x + 3}{x \log \left (\frac {81}{2} \, x^{2}\right ) - 4 \, \log \left (\frac {81}{2} \, x^{2}\right )} \]

[In]

integrate((-7*x*log(81/2*x^2)-2*x^2+2*x+24)/(x^3-8*x^2+16*x)/log(81/2*x^2)^2,x, algorithm="giac")

[Out]

(x + 3)/(x*log(81/2*x^2) - 4*log(81/2*x^2))

Mupad [B] (verification not implemented)

Time = 9.60 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {24+2 x-2 x^2-7 x \log \left (\frac {81 x^2}{2}\right )}{\left (16 x-8 x^2+x^3\right ) \log ^2\left (\frac {81 x^2}{2}\right )} \, dx=\frac {x+3}{\ln \left (\frac {81\,x^2}{2}\right )\,\left (x-4\right )} \]

[In]

int((2*x - 7*x*log((81*x^2)/2) - 2*x^2 + 24)/(log((81*x^2)/2)^2*(16*x - 8*x^2 + x^3)),x)

[Out]

(x + 3)/(log((81*x^2)/2)*(x - 4))