\(\int \frac {-2880-1620 x-225 x^2+(-2880-1610 x-225 x^2) \log (\frac {32 x+9 x^2}{36+10 x})}{(576 x^2+322 x^3+45 x^4) \log ^2(\frac {32 x+9 x^2}{36+10 x})} \, dx\) [623]

Optimal result

Integrand size = 78, antiderivative size = 26 \[ \int \frac {-2880-1620 x-225 x^2+\left (-2880-1610 x-225 x^2\right ) \log \left (\frac {32 x+9 x^2}{36+10 x}\right )}{\left (576 x^2+322 x^3+45 x^4\right ) \log ^2\left (\frac {32 x+9 x^2}{36+10 x}\right )} \, dx=4+\frac {5}{x \log \left (x+\frac {x}{-9-\frac {x}{4+x}}\right )} \]

[Out]

5/ln(x+x/(-x/(4+x)-9))/x+4

Rubi [F]

\[ \int \frac {-2880-1620 x-225 x^2+\left (-2880-1610 x-225 x^2\right ) \log \left (\frac {32 x+9 x^2}{36+10 x}\right )}{\left (576 x^2+322 x^3+45 x^4\right ) \log ^2\left (\frac {32 x+9 x^2}{36+10 x}\right )} \, dx=\int \frac {-2880-1620 x-225 x^2+\left (-2880-1610 x-225 x^2\right ) \log \left (\frac {32 x+9 x^2}{36+10 x}\right )}{\left (576 x^2+322 x^3+45 x^4\right ) \log ^2\left (\frac {32 x+9 x^2}{36+10 x}\right )} \, dx \]

[In]

Int[(-2880 - 1620*x - 225*x^2 + (-2880 - 1610*x - 225*x^2)*Log[(32*x + 9*x^2)/(36 + 10*x)])/((576*x^2 + 322*x^
3 + 45*x^4)*Log[(32*x + 9*x^2)/(36 + 10*x)]^2),x]

[Out]

-5*Defer[Int][1/(x^2*Log[(x*(32 + 9*x))/(36 + 10*x)]^2), x] - (5*Defer[Int][1/(x*Log[(x*(32 + 9*x))/(36 + 10*x
)]^2), x])/288 - (125*Defer[Int][1/((18 + 5*x)*Log[(x*(32 + 9*x))/(36 + 10*x)]^2), x])/18 + (405*Defer[Int][1/
((32 + 9*x)*Log[(x*(32 + 9*x))/(36 + 10*x)]^2), x])/32 - 5*Defer[Int][1/(x^2*Log[(x*(32 + 9*x))/(36 + 10*x)]),
x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-2880-1620 x-225 x^2+\left (-2880-1610 x-225 x^2\right ) \log \left (\frac {32 x+9 x^2}{36+10 x}\right )}{x^2 \left (576+322 x+45 x^2\right ) \log ^2\left (\frac {32 x+9 x^2}{36+10 x}\right )} \, dx \\ & = \int \left (-\frac {45 \left (64+36 x+5 x^2\right )}{x^2 (18+5 x) (32+9 x) \log ^2\left (\frac {x (32+9 x)}{36+10 x}\right )}-\frac {5}{x^2 \log \left (\frac {x (32+9 x)}{36+10 x}\right )}\right ) \, dx \\ & = -\left (5 \int \frac {1}{x^2 \log \left (\frac {x (32+9 x)}{36+10 x}\right )} \, dx\right )-45 \int \frac {64+36 x+5 x^2}{x^2 (18+5 x) (32+9 x) \log ^2\left (\frac {x (32+9 x)}{36+10 x}\right )} \, dx \\ & = -\left (5 \int \frac {1}{x^2 \log \left (\frac {x (32+9 x)}{36+10 x}\right )} \, dx\right )-45 \int \left (\frac {1}{9 x^2 \log ^2\left (\frac {x (32+9 x)}{36+10 x}\right )}+\frac {1}{2592 x \log ^2\left (\frac {x (32+9 x)}{36+10 x}\right )}+\frac {25}{162 (18+5 x) \log ^2\left (\frac {x (32+9 x)}{36+10 x}\right )}-\frac {9}{32 (32+9 x) \log ^2\left (\frac {x (32+9 x)}{36+10 x}\right )}\right ) \, dx \\ & = -\left (\frac {5}{288} \int \frac {1}{x \log ^2\left (\frac {x (32+9 x)}{36+10 x}\right )} \, dx\right )-5 \int \frac {1}{x^2 \log ^2\left (\frac {x (32+9 x)}{36+10 x}\right )} \, dx-5 \int \frac {1}{x^2 \log \left (\frac {x (32+9 x)}{36+10 x}\right )} \, dx-\frac {125}{18} \int \frac {1}{(18+5 x) \log ^2\left (\frac {x (32+9 x)}{36+10 x}\right )} \, dx+\frac {405}{32} \int \frac {1}{(32+9 x) \log ^2\left (\frac {x (32+9 x)}{36+10 x}\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {-2880-1620 x-225 x^2+\left (-2880-1610 x-225 x^2\right ) \log \left (\frac {32 x+9 x^2}{36+10 x}\right )}{\left (576 x^2+322 x^3+45 x^4\right ) \log ^2\left (\frac {32 x+9 x^2}{36+10 x}\right )} \, dx=\frac {5}{x \log \left (\frac {x (32+9 x)}{36+10 x}\right )} \]

[In]

Integrate[(-2880 - 1620*x - 225*x^2 + (-2880 - 1610*x - 225*x^2)*Log[(32*x + 9*x^2)/(36 + 10*x)])/((576*x^2 +
322*x^3 + 45*x^4)*Log[(32*x + 9*x^2)/(36 + 10*x)]^2),x]

[Out]

5/(x*Log[(x*(32 + 9*x))/(36 + 10*x)])

Maple [A] (verified)

Time = 0.77 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00

method	result	size
norman	\(\frac {5}{x \ln \left (\frac {9 x^{2}+32 x}{10 x +36}\right )}\)	\(26\)
risch	\(\frac {5}{x \ln \left (\frac {9 x^{2}+32 x}{10 x +36}\right )}\)	\(26\)

[In]

int(((-225*x^2-1610*x-2880)*ln((9*x^2+32*x)/(10*x+36))-225*x^2-1620*x-2880)/(45*x^4+322*x^3+576*x^2)/ln((9*x^2
+32*x)/(10*x+36))^2,x,method=_RETURNVERBOSE)

[Out]

5/x/ln((9*x^2+32*x)/(10*x+36))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {-2880-1620 x-225 x^2+\left (-2880-1610 x-225 x^2\right ) \log \left (\frac {32 x+9 x^2}{36+10 x}\right )}{\left (576 x^2+322 x^3+45 x^4\right ) \log ^2\left (\frac {32 x+9 x^2}{36+10 x}\right )} \, dx=\frac {5}{x \log \left (\frac {9 \, x^{2} + 32 \, x}{2 \, {\left (5 \, x + 18\right )}}\right )} \]

[In]

integrate(((-225*x^2-1610*x-2880)*log((9*x^2+32*x)/(10*x+36))-225*x^2-1620*x-2880)/(45*x^4+322*x^3+576*x^2)/lo
g((9*x^2+32*x)/(10*x+36))^2,x, algorithm="fricas")

[Out]

5/(x*log(1/2*(9*x^2 + 32*x)/(5*x + 18)))

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65 \[ \int \frac {-2880-1620 x-225 x^2+\left (-2880-1610 x-225 x^2\right ) \log \left (\frac {32 x+9 x^2}{36+10 x}\right )}{\left (576 x^2+322 x^3+45 x^4\right ) \log ^2\left (\frac {32 x+9 x^2}{36+10 x}\right )} \, dx=\frac {5}{x \log {\left (\frac {9 x^{2} + 32 x}{10 x + 36} \right )}} \]

[In]

integrate(((-225*x**2-1610*x-2880)*ln((9*x**2+32*x)/(10*x+36))-225*x**2-1620*x-2880)/(45*x**4+322*x**3+576*x**
2)/ln((9*x**2+32*x)/(10*x+36))**2,x)

[Out]

5/(x*log((9*x**2 + 32*x)/(10*x + 36)))

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {-2880-1620 x-225 x^2+\left (-2880-1610 x-225 x^2\right ) \log \left (\frac {32 x+9 x^2}{36+10 x}\right )}{\left (576 x^2+322 x^3+45 x^4\right ) \log ^2\left (\frac {32 x+9 x^2}{36+10 x}\right )} \, dx=-\frac {5}{x \log \left (2\right ) - x \log \left (9 \, x + 32\right ) + x \log \left (5 \, x + 18\right ) - x \log \left (x\right )} \]

[In]

integrate(((-225*x^2-1610*x-2880)*log((9*x^2+32*x)/(10*x+36))-225*x^2-1620*x-2880)/(45*x^4+322*x^3+576*x^2)/lo
g((9*x^2+32*x)/(10*x+36))^2,x, algorithm="maxima")

[Out]

-5/(x*log(2) - x*log(9*x + 32) + x*log(5*x + 18) - x*log(x))

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {-2880-1620 x-225 x^2+\left (-2880-1610 x-225 x^2\right ) \log \left (\frac {32 x+9 x^2}{36+10 x}\right )}{\left (576 x^2+322 x^3+45 x^4\right ) \log ^2\left (\frac {32 x+9 x^2}{36+10 x}\right )} \, dx=\frac {5}{x \log \left (\frac {9 \, x^{2} + 32 \, x}{2 \, {\left (5 \, x + 18\right )}}\right )} \]

[In]

integrate(((-225*x^2-1610*x-2880)*log((9*x^2+32*x)/(10*x+36))-225*x^2-1620*x-2880)/(45*x^4+322*x^3+576*x^2)/lo
g((9*x^2+32*x)/(10*x+36))^2,x, algorithm="giac")

[Out]

5/(x*log(1/2*(9*x^2 + 32*x)/(5*x + 18)))

Mupad [B] (verification not implemented)

Time = 9.53 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {-2880-1620 x-225 x^2+\left (-2880-1610 x-225 x^2\right ) \log \left (\frac {32 x+9 x^2}{36+10 x}\right )}{\left (576 x^2+322 x^3+45 x^4\right ) \log ^2\left (\frac {32 x+9 x^2}{36+10 x}\right )} \, dx=\frac {5}{x\,\ln \left (\frac {9\,x^2+32\,x}{10\,x+36}\right )} \]

[In]

int(-(1620*x + log((32*x + 9*x^2)/(10*x + 36))*(1610*x + 225*x^2 + 2880) + 225*x^2 + 2880)/(log((32*x + 9*x^2)
/(10*x + 36))^2*(576*x^2 + 322*x^3 + 45*x^4)),x)

[Out]

5/(x*log((32*x + 9*x^2)/(10*x + 36)))