\(\int \frac {-2 x+x^4-2 x^5+(4 x+10 x^2-20 x^3) \log (x)+(25-50 x) \log ^2(x)}{x^4+10 x^2 \log (x)+25 \log ^2(x)} \, dx\) [7445]

Optimal result

Integrand size = 59, antiderivative size = 22 \[ \int \frac {-2 x+x^4-2 x^5+\left (4 x+10 x^2-20 x^3\right ) \log (x)+(25-50 x) \log ^2(x)}{x^4+10 x^2 \log (x)+25 \log ^2(x)} \, dx=3+x-x^2-\frac {2}{5+\frac {x^2}{\log (x)}} \]

[Out]

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

Rubi [F]

\[ \int \frac {-2 x+x^4-2 x^5+\left (4 x+10 x^2-20 x^3\right ) \log (x)+(25-50 x) \log ^2(x)}{x^4+10 x^2 \log (x)+25 \log ^2(x)} \, dx=\int \frac {-2 x+x^4-2 x^5+\left (4 x+10 x^2-20 x^3\right ) \log (x)+(25-50 x) \log ^2(x)}{x^4+10 x^2 \log (x)+25 \log ^2(x)} \, dx \]

[In]

Int[(-2*x + x^4 - 2*x^5 + (4*x + 10*x^2 - 20*x^3)*Log[x] + (25 - 50*x)*Log[x]^2)/(x^4 + 10*x^2*Log[x] + 25*Log
[x]^2),x]

[Out]

x - x^2 - 2*Defer[Int][x/(x^2 + 5*Log[x])^2, x] - (4*Defer[Int][x^3/(x^2 + 5*Log[x])^2, x])/5 + (4*Defer[Int][
x/(x^2 + 5*Log[x]), x])/5

Rubi steps \begin{align*} \text {integral}& = \int \frac {-2 x+x^4-2 x^5+\left (4 x+10 x^2-20 x^3\right ) \log (x)+(25-50 x) \log ^2(x)}{\left (x^2+5 \log (x)\right )^2} \, dx \\ & = \int \left (1-2 x-\frac {2 x \left (5+2 x^2\right )}{5 \left (x^2+5 \log (x)\right )^2}+\frac {4 x}{5 \left (x^2+5 \log (x)\right )}\right ) \, dx \\ & = x-x^2-\frac {2}{5} \int \frac {x \left (5+2 x^2\right )}{\left (x^2+5 \log (x)\right )^2} \, dx+\frac {4}{5} \int \frac {x}{x^2+5 \log (x)} \, dx \\ & = x-x^2-\frac {2}{5} \int \left (\frac {5 x}{\left (x^2+5 \log (x)\right )^2}+\frac {2 x^3}{\left (x^2+5 \log (x)\right )^2}\right ) \, dx+\frac {4}{5} \int \frac {x}{x^2+5 \log (x)} \, dx \\ & = x-x^2-\frac {4}{5} \int \frac {x^3}{\left (x^2+5 \log (x)\right )^2} \, dx+\frac {4}{5} \int \frac {x}{x^2+5 \log (x)} \, dx-2 \int \frac {x}{\left (x^2+5 \log (x)\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {-2 x+x^4-2 x^5+\left (4 x+10 x^2-20 x^3\right ) \log (x)+(25-50 x) \log ^2(x)}{x^4+10 x^2 \log (x)+25 \log ^2(x)} \, dx=x-x^2+\frac {2 x^2}{5 \left (x^2+5 \log (x)\right )} \]

[In]

Integrate[(-2*x + x^4 - 2*x^5 + (4*x + 10*x^2 - 20*x^3)*Log[x] + (25 - 50*x)*Log[x]^2)/(x^4 + 10*x^2*Log[x] +
25*Log[x]^2),x]

[Out]

x - x^2 + (2*x^2)/(5*(x^2 + 5*Log[x]))

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05

method	result	size
risch	\(-x^{2}+x +\frac {2 x^{2}}{5 \left (x^{2}+5 \ln \left (x \right )\right )}\)	\(23\)
norman	\(\frac {x^{3}-2 \ln \left (x \right )-x^{4}+5 x \ln \left (x \right )-5 x^{2} \ln \left (x \right )}{x^{2}+5 \ln \left (x \right )}\)	\(37\)
default	\(-\frac {x^{4}-\frac {2 x^{2}}{5}-x^{3}-5 x \ln \left (x \right )+5 x^{2} \ln \left (x \right )}{x^{2}+5 \ln \left (x \right )}\)	\(39\)
parallelrisch	\(\frac {-5 x^{4}+5 x^{3}-25 x^{2} \ln \left (x \right )+2 x^{2}+25 x \ln \left (x \right )}{5 x^{2}+25 \ln \left (x \right )}\)	\(41\)

[In]

int(((-50*x+25)*ln(x)^2+(-20*x^3+10*x^2+4*x)*ln(x)-2*x^5+x^4-2*x)/(25*ln(x)^2+10*x^2*ln(x)+x^4),x,method=_RETU
RNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.77 \[ \int \frac {-2 x+x^4-2 x^5+\left (4 x+10 x^2-20 x^3\right ) \log (x)+(25-50 x) \log ^2(x)}{x^4+10 x^2 \log (x)+25 \log ^2(x)} \, dx=-\frac {5 \, x^{4} - 5 \, x^{3} - 2 \, x^{2} + 25 \, {\left (x^{2} - x\right )} \log \left (x\right )}{5 \, {\left (x^{2} + 5 \, \log \left (x\right )\right )}} \]

[In]

integrate(((-50*x+25)*log(x)^2+(-20*x^3+10*x^2+4*x)*log(x)-2*x^5+x^4-2*x)/(25*log(x)^2+10*x^2*log(x)+x^4),x, a
lgorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {-2 x+x^4-2 x^5+\left (4 x+10 x^2-20 x^3\right ) \log (x)+(25-50 x) \log ^2(x)}{x^4+10 x^2 \log (x)+25 \log ^2(x)} \, dx=- x^{2} + \frac {2 x^{2}}{5 x^{2} + 25 \log {\left (x \right )}} + x \]

[In]

integrate(((-50*x+25)*ln(x)**2+(-20*x**3+10*x**2+4*x)*ln(x)-2*x**5+x**4-2*x)/(25*ln(x)**2+10*x**2*ln(x)+x**4),
x)

[Out]

-x**2 + 2*x**2/(5*x**2 + 25*log(x)) + x

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.77 \[ \int \frac {-2 x+x^4-2 x^5+\left (4 x+10 x^2-20 x^3\right ) \log (x)+(25-50 x) \log ^2(x)}{x^4+10 x^2 \log (x)+25 \log ^2(x)} \, dx=-\frac {5 \, x^{4} - 5 \, x^{3} - 2 \, x^{2} + 25 \, {\left (x^{2} - x\right )} \log \left (x\right )}{5 \, {\left (x^{2} + 5 \, \log \left (x\right )\right )}} \]

[In]

integrate(((-50*x+25)*log(x)^2+(-20*x^3+10*x^2+4*x)*log(x)-2*x^5+x^4-2*x)/(25*log(x)^2+10*x^2*log(x)+x^4),x, a
lgorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-2 x+x^4-2 x^5+\left (4 x+10 x^2-20 x^3\right ) \log (x)+(25-50 x) \log ^2(x)}{x^4+10 x^2 \log (x)+25 \log ^2(x)} \, dx=-x^{2} + x + \frac {2 \, x^{2}}{5 \, {\left (x^{2} + 5 \, \log \left (x\right )\right )}} \]

[In]

integrate(((-50*x+25)*log(x)^2+(-20*x^3+10*x^2+4*x)*log(x)-2*x^5+x^4-2*x)/(25*log(x)^2+10*x^2*log(x)+x^4),x, a
lgorithm="giac")

[Out]

-x^2 + x + 2/5*x^2/(x^2 + 5*log(x))

Mupad [B] (verification not implemented)

Time = 12.31 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {-2 x+x^4-2 x^5+\left (4 x+10 x^2-20 x^3\right ) \log (x)+(25-50 x) \log ^2(x)}{x^4+10 x^2 \log (x)+25 \log ^2(x)} \, dx=x+\frac {2\,x^2}{5\,\left (5\,\ln \left (x\right )+x^2\right )}-x^2 \]

[In]

int(-(2*x - x^4 + 2*x^5 + log(x)^2*(50*x - 25) - log(x)*(4*x + 10*x^2 - 20*x^3))/(10*x^2*log(x) + 25*log(x)^2
+ x^4),x)

[Out]

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