\(\int \frac {5+5 x+5 x^2+5 x^3+(10 x-10 x^2) \log (x)}{(x+3 x^2+3 x^3+x^4) \log ^2(x)} \, dx\) [3215]

Optimal result

Integrand size = 49, antiderivative size = 17 \[ \int \frac {5+5 x+5 x^2+5 x^3+\left (10 x-10 x^2\right ) \log (x)}{\left (x+3 x^2+3 x^3+x^4\right ) \log ^2(x)} \, dx=\frac {-5-5 x^2}{(1+x)^2 \log (x)} \]

[Out]

1/ln(x)/(1+x)^2*(-5*x^2-5)

Rubi [F]

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

[In]

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

[Out]

5*Defer[Int][(1 + x^2)/(x*(1 + x)^2*Log[x]^2), x] - 10*Defer[Int][(-1 + x)/((1 + x)^3*Log[x]), x]

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {5+5 x+5 x^2+5 x^3+\left (10 x-10 x^2\right ) \log (x)}{\left (x+3 x^2+3 x^3+x^4\right ) \log ^2(x)} \, dx=\frac {5 \left (-1-x^2\right )}{(1+x)^2 \log (x)} \]

[In]

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

[Out]

(5*(-1 - x^2))/((1 + x)^2*Log[x])

Maple [A] (verified)

Time = 16.38 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06

method	result	size
norman	\(\frac {-5 x^{2}-5}{\ln \left (x \right ) \left (1+x \right )^{2}}\)	\(18\)
risch	\(-\frac {5 \left (x^{2}+1\right )}{\left (x^{2}+2 x +1\right ) \ln \left (x \right )}\)	\(22\)
parallelrisch	\(\frac {-5 x^{2}-5}{\ln \left (x \right ) \left (x^{2}+2 x +1\right )}\)	\(23\)
default	\(\frac {10 x \ln \left (x \right )+5 \ln \left (x \right )-5 x -5}{\ln \left (x \right )^{2} \left (1+x \right )^{2}}-\frac {5 \left (1+x +\ln \left (x \right )\right )}{\ln \left (x \right )^{2} \left (1+x \right )^{2}}+\frac {10}{\ln \left (x \right )^{2} \left (1+x \right )}-\frac {5}{\ln \left (x \right )}\)	\(58\)
parts	\(\frac {10 x \ln \left (x \right )+5 \ln \left (x \right )-5 x -5}{\ln \left (x \right )^{2} \left (1+x \right )^{2}}-\frac {5 \left (1+x +\ln \left (x \right )\right )}{\ln \left (x \right )^{2} \left (1+x \right )^{2}}+\frac {10}{\ln \left (x \right )^{2} \left (1+x \right )}-\frac {5}{\ln \left (x \right )}\)	\(58\)

[In]

int(((-10*x^2+10*x)*ln(x)+5*x^3+5*x^2+5*x+5)/(x^4+3*x^3+3*x^2+x)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

1/ln(x)/(1+x)^2*(-5*x^2-5)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.24 \[ \int \frac {5+5 x+5 x^2+5 x^3+\left (10 x-10 x^2\right ) \log (x)}{\left (x+3 x^2+3 x^3+x^4\right ) \log ^2(x)} \, dx=-\frac {5 \, {\left (x^{2} + 1\right )}}{{\left (x^{2} + 2 \, x + 1\right )} \log \left (x\right )} \]

[In]

integrate(((-10*x^2+10*x)*log(x)+5*x^3+5*x^2+5*x+5)/(x^4+3*x^3+3*x^2+x)/log(x)^2,x, algorithm="fricas")

[Out]

-5*(x^2 + 1)/((x^2 + 2*x + 1)*log(x))

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \[ \int \frac {5+5 x+5 x^2+5 x^3+\left (10 x-10 x^2\right ) \log (x)}{\left (x+3 x^2+3 x^3+x^4\right ) \log ^2(x)} \, dx=\frac {- 5 x^{2} - 5}{\left (x^{2} + 2 x + 1\right ) \log {\left (x \right )}} \]

[In]

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

[Out]

(-5*x**2 - 5)/((x**2 + 2*x + 1)*log(x))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.24 \[ \int \frac {5+5 x+5 x^2+5 x^3+\left (10 x-10 x^2\right ) \log (x)}{\left (x+3 x^2+3 x^3+x^4\right ) \log ^2(x)} \, dx=-\frac {5 \, {\left (x^{2} + 1\right )}}{{\left (x^{2} + 2 \, x + 1\right )} \log \left (x\right )} \]

[In]

integrate(((-10*x^2+10*x)*log(x)+5*x^3+5*x^2+5*x+5)/(x^4+3*x^3+3*x^2+x)/log(x)^2,x, algorithm="maxima")

[Out]

-5*(x^2 + 1)/((x^2 + 2*x + 1)*log(x))

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.35 \[ \int \frac {5+5 x+5 x^2+5 x^3+\left (10 x-10 x^2\right ) \log (x)}{\left (x+3 x^2+3 x^3+x^4\right ) \log ^2(x)} \, dx=-\frac {5 \, {\left (x^{2} + 1\right )}}{x^{2} \log \left (x\right ) + 2 \, x \log \left (x\right ) + \log \left (x\right )} \]

[In]

integrate(((-10*x^2+10*x)*log(x)+5*x^3+5*x^2+5*x+5)/(x^4+3*x^3+3*x^2+x)/log(x)^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.46 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {5+5 x+5 x^2+5 x^3+\left (10 x-10 x^2\right ) \log (x)}{\left (x+3 x^2+3 x^3+x^4\right ) \log ^2(x)} \, dx=-\frac {5\,\left (x^2+1\right )}{\ln \left (x\right )\,{\left (x+1\right )}^2} \]

[In]

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

[Out]

-(5*(x^2 + 1))/(log(x)*(x + 1)^2)