\(\int \frac {-5 x+(10 x+4 x^2) \log (\frac {x}{5+2 x})+(25-4 x^2) \log ^2(\frac {x}{5+2 x})}{(5+2 x) \log ^2(\frac {x}{5+2 x})} \, dx\) [1950]

Optimal result

Integrand size = 64, antiderivative size = 28 \[ \int \frac {-5 x+\left (10 x+4 x^2\right ) \log \left (\frac {x}{5+2 x}\right )+\left (25-4 x^2\right ) \log ^2\left (\frac {x}{5+2 x}\right )}{(5+2 x) \log ^2\left (\frac {x}{5+2 x}\right )} \, dx=-4-(2-x)^2+x+\frac {x^2}{\log \left (\frac {x}{5+2 x}\right )} \]

[Out]

x-4-(2-x)^2+x^2/ln(x/(5+2*x))

Rubi [F]

\[ \int \frac {-5 x+\left (10 x+4 x^2\right ) \log \left (\frac {x}{5+2 x}\right )+\left (25-4 x^2\right ) \log ^2\left (\frac {x}{5+2 x}\right )}{(5+2 x) \log ^2\left (\frac {x}{5+2 x}\right )} \, dx=\int \frac {-5 x+\left (10 x+4 x^2\right ) \log \left (\frac {x}{5+2 x}\right )+\left (25-4 x^2\right ) \log ^2\left (\frac {x}{5+2 x}\right )}{(5+2 x) \log ^2\left (\frac {x}{5+2 x}\right )} \, dx \]

[In]

Int[(-5*x + (10*x + 4*x^2)*Log[x/(5 + 2*x)] + (25 - 4*x^2)*Log[x/(5 + 2*x)]^2)/((5 + 2*x)*Log[x/(5 + 2*x)]^2),
x]

[Out]

5*x - x^2 + 2*Defer[Int][x/Log[x/(5 + 2*x)], x] - 25*Defer[Subst][Defer[Int][x/((1 - 2*x)^2*Log[x]^2), x], x,
x/(5 + 2*x)]

Rubi steps \begin{align*} \text {integral}& = \int \left (5-2 x-\frac {5 x}{(5+2 x) \log ^2\left (\frac {x}{5+2 x}\right )}+\frac {2 x}{\log \left (\frac {x}{5+2 x}\right )}\right ) \, dx \\ & = 5 x-x^2+2 \int \frac {x}{\log \left (\frac {x}{5+2 x}\right )} \, dx-5 \int \frac {x}{(5+2 x) \log ^2\left (\frac {x}{5+2 x}\right )} \, dx \\ & = 5 x-x^2+2 \int \frac {x}{\log \left (\frac {x}{5+2 x}\right )} \, dx-25 \text {Subst}\left (\int \frac {x}{(1-2 x)^2 \log ^2(x)} \, dx,x,\frac {x}{5+2 x}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 2.14 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {-5 x+\left (10 x+4 x^2\right ) \log \left (\frac {x}{5+2 x}\right )+\left (25-4 x^2\right ) \log ^2\left (\frac {x}{5+2 x}\right )}{(5+2 x) \log ^2\left (\frac {x}{5+2 x}\right )} \, dx=5 x-x^2+\frac {x^2}{\log \left (\frac {x}{5+2 x}\right )} \]

[In]

Integrate[(-5*x + (10*x + 4*x^2)*Log[x/(5 + 2*x)] + (25 - 4*x^2)*Log[x/(5 + 2*x)]^2)/((5 + 2*x)*Log[x/(5 + 2*x
)]^2),x]

[Out]

5*x - x^2 + x^2/Log[x/(5 + 2*x)]

Maple [A] (verified)

Time = 3.30 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93

method	result	size
risch	\(-x^{2}+5 x +\frac {x^{2}}{\ln \left (\frac {x}{5+2 x}\right )}\)	\(26\)
norman	\(\frac {x^{2}+5 \ln \left (\frac {x}{5+2 x}\right ) x -\ln \left (\frac {x}{5+2 x}\right ) x^{2}}{\ln \left (\frac {x}{5+2 x}\right )}\)	\(46\)
parallelrisch	\(-\frac {4 \ln \left (\frac {x}{5+2 x}\right ) x^{2}-4 x^{2}-20 \ln \left (\frac {x}{5+2 x}\right ) x +75 \ln \left (\frac {x}{5+2 x}\right )}{4 \ln \left (\frac {x}{5+2 x}\right )}\)	\(61\)
derivativedivides	\(25+10 x -\frac {\left (5+2 x \right )^{2}}{4}+\frac {25}{4 \ln \left (\frac {1}{2}-\frac {5}{2 \left (5+2 x \right )}\right )}-\frac {5 \left (5+2 x \right )}{4 \ln \left (\frac {1}{2}-\frac {5}{2 \left (5+2 x \right )}\right )^{2}}+\frac {\left (4 \left (\frac {1}{2}-\frac {5}{2 \left (5+2 x \right )}\right ) \ln \left (\frac {1}{2}-\frac {5}{2 \left (5+2 x \right )}\right )-\ln \left (\frac {1}{2}-\frac {5}{2 \left (5+2 x \right )}\right )+\frac {5}{5+2 x}\right ) \left (5+2 x \right )^{2}}{4 \ln \left (\frac {1}{2}-\frac {5}{2 \left (5+2 x \right )}\right )^{2}}\)	\(124\)
default	\(25+10 x -\frac {\left (5+2 x \right )^{2}}{4}+\frac {25}{4 \ln \left (\frac {1}{2}-\frac {5}{2 \left (5+2 x \right )}\right )}-\frac {5 \left (5+2 x \right )}{4 \ln \left (\frac {1}{2}-\frac {5}{2 \left (5+2 x \right )}\right )^{2}}+\frac {\left (4 \left (\frac {1}{2}-\frac {5}{2 \left (5+2 x \right )}\right ) \ln \left (\frac {1}{2}-\frac {5}{2 \left (5+2 x \right )}\right )-\ln \left (\frac {1}{2}-\frac {5}{2 \left (5+2 x \right )}\right )+\frac {5}{5+2 x}\right ) \left (5+2 x \right )^{2}}{4 \ln \left (\frac {1}{2}-\frac {5}{2 \left (5+2 x \right )}\right )^{2}}\)	\(124\)
parts	\(-x^{2}+5 x +\frac {\frac {25 x \ln \left (\frac {x}{5+2 x}\right )}{5+2 x}-\frac {25 \ln \left (\frac {x}{5+2 x}\right )}{4}-\frac {25 x}{2 \left (5+2 x \right )}+\frac {25}{4}}{\left (\frac {2 x}{5+2 x}-1\right )^{2} \ln \left (\frac {x}{5+2 x}\right )^{2}}+\frac {25}{4 \ln \left (\frac {x}{5+2 x}\right )}+\frac {25}{4 \ln \left (\frac {x}{5+2 x}\right )^{2} \left (\frac {2 x}{5+2 x}-1\right )}\)	\(124\)

[In]

int(((-4*x^2+25)*ln(x/(5+2*x))^2+(4*x^2+10*x)*ln(x/(5+2*x))-5*x)/(5+2*x)/ln(x/(5+2*x))^2,x,method=_RETURNVERBO
SE)

[Out]

-x^2+5*x+x^2/ln(x/(5+2*x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {-5 x+\left (10 x+4 x^2\right ) \log \left (\frac {x}{5+2 x}\right )+\left (25-4 x^2\right ) \log ^2\left (\frac {x}{5+2 x}\right )}{(5+2 x) \log ^2\left (\frac {x}{5+2 x}\right )} \, dx=\frac {x^{2} - {\left (x^{2} - 5 \, x\right )} \log \left (\frac {x}{2 \, x + 5}\right )}{\log \left (\frac {x}{2 \, x + 5}\right )} \]

[In]

integrate(((-4*x^2+25)*log(x/(5+2*x))^2+(4*x^2+10*x)*log(x/(5+2*x))-5*x)/(5+2*x)/log(x/(5+2*x))^2,x, algorithm
="fricas")

[Out]

(x^2 - (x^2 - 5*x)*log(x/(2*x + 5)))/log(x/(2*x + 5))

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61 \[ \int \frac {-5 x+\left (10 x+4 x^2\right ) \log \left (\frac {x}{5+2 x}\right )+\left (25-4 x^2\right ) \log ^2\left (\frac {x}{5+2 x}\right )}{(5+2 x) \log ^2\left (\frac {x}{5+2 x}\right )} \, dx=- x^{2} + \frac {x^{2}}{\log {\left (\frac {x}{2 x + 5} \right )}} + 5 x \]

[In]

integrate(((-4*x**2+25)*ln(x/(5+2*x))**2+(4*x**2+10*x)*ln(x/(5+2*x))-5*x)/(5+2*x)/ln(x/(5+2*x))**2,x)

[Out]

-x**2 + x**2/log(x/(2*x + 5)) + 5*x

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.57 \[ \int \frac {-5 x+\left (10 x+4 x^2\right ) \log \left (\frac {x}{5+2 x}\right )+\left (25-4 x^2\right ) \log ^2\left (\frac {x}{5+2 x}\right )}{(5+2 x) \log ^2\left (\frac {x}{5+2 x}\right )} \, dx=-\frac {x^{2} + {\left (x^{2} - 5 \, x\right )} \log \left (2 \, x + 5\right ) - {\left (x^{2} - 5 \, x\right )} \log \left (x\right )}{\log \left (2 \, x + 5\right ) - \log \left (x\right )} \]

[In]

integrate(((-4*x^2+25)*log(x/(5+2*x))^2+(4*x^2+10*x)*log(x/(5+2*x))-5*x)/(5+2*x)/log(x/(5+2*x))^2,x, algorithm
="maxima")

[Out]

-(x^2 + (x^2 - 5*x)*log(2*x + 5) - (x^2 - 5*x)*log(x))/(log(2*x + 5) - log(x))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 110 vs. \(2 (26) = 52\).

Time = 0.27 (sec) , antiderivative size = 110, normalized size of antiderivative = 3.93 \[ \int \frac {-5 x+\left (10 x+4 x^2\right ) \log \left (\frac {x}{5+2 x}\right )+\left (25-4 x^2\right ) \log ^2\left (\frac {x}{5+2 x}\right )}{(5+2 x) \log ^2\left (\frac {x}{5+2 x}\right )} \, dx=\frac {25 \, {\left (\frac {8 \, x}{2 \, x + 5} - 3\right )}}{4 \, {\left (\frac {4 \, x}{2 \, x + 5} - \frac {4 \, x^{2}}{{\left (2 \, x + 5\right )}^{2}} - 1\right )}} - \frac {25 \, x^{2}}{{\left (2 \, x + 5\right )}^{2} {\left (\frac {4 \, x \log \left (\frac {x}{2 \, x + 5}\right )}{2 \, x + 5} - \frac {4 \, x^{2} \log \left (\frac {x}{2 \, x + 5}\right )}{{\left (2 \, x + 5\right )}^{2}} - \log \left (\frac {x}{2 \, x + 5}\right )\right )}} \]

[In]

integrate(((-4*x^2+25)*log(x/(5+2*x))^2+(4*x^2+10*x)*log(x/(5+2*x))-5*x)/(5+2*x)/log(x/(5+2*x))^2,x, algorithm
="giac")

[Out]

25/4*(8*x/(2*x + 5) - 3)/(4*x/(2*x + 5) - 4*x^2/(2*x + 5)^2 - 1) - 25*x^2/((2*x + 5)^2*(4*x*log(x/(2*x + 5))/(
2*x + 5) - 4*x^2*log(x/(2*x + 5))/(2*x + 5)^2 - log(x/(2*x + 5))))

Mupad [B] (verification not implemented)

Time = 9.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {-5 x+\left (10 x+4 x^2\right ) \log \left (\frac {x}{5+2 x}\right )+\left (25-4 x^2\right ) \log ^2\left (\frac {x}{5+2 x}\right )}{(5+2 x) \log ^2\left (\frac {x}{5+2 x}\right )} \, dx=5\,x+\frac {x^2}{\ln \left (\frac {x}{2\,x+5}\right )}-x^2 \]

[In]

int(-(5*x + log(x/(2*x + 5))^2*(4*x^2 - 25) - log(x/(2*x + 5))*(10*x + 4*x^2))/(log(x/(2*x + 5))^2*(2*x + 5)),
x)

[Out]

5*x + x^2/log(x/(2*x + 5)) - x^2