\(\int \frac {-4 x-2 x^2+(-12-4 x) \log (3+x)+(3+x+3 x^2-5 x^3-2 x^4) \log ^2(3+x)}{(3 x^2+x^3) \log ^2(3+x)} \, dx\) [2568]

Optimal result

Integrand size = 62, antiderivative size = 26 \[ \int \frac {-4 x-2 x^2+(-12-4 x) \log (3+x)+\left (3+x+3 x^2-5 x^3-2 x^4\right ) \log ^2(3+x)}{\left (3 x^2+x^3\right ) \log ^2(3+x)} \, dx=x+x \left (-x+\frac {-1+\frac {4+2 x}{\log (3+x)}}{x^2}\right ) \]

[Out]

x+(((4+2*x)/ln(3+x)-1)/x^2-x)*x

Rubi [F]

\[ \int \frac {-4 x-2 x^2+(-12-4 x) \log (3+x)+\left (3+x+3 x^2-5 x^3-2 x^4\right ) \log ^2(3+x)}{\left (3 x^2+x^3\right ) \log ^2(3+x)} \, dx=\int \frac {-4 x-2 x^2+(-12-4 x) \log (3+x)+\left (3+x+3 x^2-5 x^3-2 x^4\right ) \log ^2(3+x)}{\left (3 x^2+x^3\right ) \log ^2(3+x)} \, dx \]

[In]

Int[(-4*x - 2*x^2 + (-12 - 4*x)*Log[3 + x] + (3 + x + 3*x^2 - 5*x^3 - 2*x^4)*Log[3 + x]^2)/((3*x^2 + x^3)*Log[
3 + x]^2),x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {-4 x-2 x^2+(-12-4 x) \log (3+x)+\left (3+x+3 x^2-5 x^3-2 x^4\right ) \log ^2(3+x)}{x^2 (3+x) \log ^2(3+x)} \, dx \\ & = \int \left (\frac {1+x^2-2 x^3}{x^2}-\frac {2 (2+x)}{x (3+x) \log ^2(3+x)}-\frac {4}{x^2 \log (3+x)}\right ) \, dx \\ & = -\left (2 \int \frac {2+x}{x (3+x) \log ^2(3+x)} \, dx\right )-4 \int \frac {1}{x^2 \log (3+x)} \, dx+\int \frac {1+x^2-2 x^3}{x^2} \, dx \\ & = -\left (2 \int \left (\frac {2}{3 x \log ^2(3+x)}+\frac {1}{3 (3+x) \log ^2(3+x)}\right ) \, dx\right )-4 \int \frac {1}{x^2 \log (3+x)} \, dx+\int \left (1+\frac {1}{x^2}-2 x\right ) \, dx \\ & = -\frac {1}{x}+x-x^2-\frac {2}{3} \int \frac {1}{(3+x) \log ^2(3+x)} \, dx-\frac {4}{3} \int \frac {1}{x \log ^2(3+x)} \, dx-4 \int \frac {1}{x^2 \log (3+x)} \, dx \\ & = -\frac {1}{x}+x-x^2-\frac {2}{3} \text {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,3+x\right )-\frac {4}{3} \int \frac {1}{x \log ^2(3+x)} \, dx-4 \int \frac {1}{x^2 \log (3+x)} \, dx \\ & = -\frac {1}{x}+x-x^2-\frac {2}{3} \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (3+x)\right )-\frac {4}{3} \int \frac {1}{x \log ^2(3+x)} \, dx-4 \int \frac {1}{x^2 \log (3+x)} \, dx \\ & = -\frac {1}{x}+x-x^2+\frac {2}{3 \log (3+x)}-\frac {4}{3} \int \frac {1}{x \log ^2(3+x)} \, dx-4 \int \frac {1}{x^2 \log (3+x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {-4 x-2 x^2+(-12-4 x) \log (3+x)+\left (3+x+3 x^2-5 x^3-2 x^4\right ) \log ^2(3+x)}{\left (3 x^2+x^3\right ) \log ^2(3+x)} \, dx=-\frac {1}{x}+x-x^2+\frac {2 (2+x)}{x \log (3+x)} \]

[In]

Integrate[(-4*x - 2*x^2 + (-12 - 4*x)*Log[3 + x] + (3 + x + 3*x^2 - 5*x^3 - 2*x^4)*Log[3 + x]^2)/((3*x^2 + x^3
)*Log[3 + x]^2),x]

[Out]

-x^(-1) + x - x^2 + (2*(2 + x))/(x*Log[3 + x])

Maple [A] (verified)

Time = 3.40 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19

method	result	size
risch	\(-\frac {x^{3}-x^{2}+1}{x}+\frac {4+2 x}{x \ln \left (3+x \right )}\)	\(31\)
parts	\(\frac {4}{\ln \left (3+x \right ) x}-x^{2}+x -\frac {1}{x}+\frac {2}{\ln \left (3+x \right )}\)	\(32\)
derivativedivides	\(\frac {4}{\ln \left (3+x \right ) x}-\left (3+x \right )^{2}+21+7 x -\frac {1}{x}+\frac {2}{\ln \left (3+x \right )}\)	\(37\)
default	\(\frac {4}{\ln \left (3+x \right ) x}-\left (3+x \right )^{2}+21+7 x -\frac {1}{x}+\frac {2}{\ln \left (3+x \right )}\)	\(37\)
norman	\(\frac {4+\ln \left (3+x \right ) x^{2}+2 x -\ln \left (3+x \right ) x^{3}-\ln \left (3+x \right )}{x \ln \left (3+x \right )}\)	\(39\)
parallelrisch	\(\frac {-\ln \left (3+x \right ) x^{3}+\ln \left (3+x \right ) x^{2}+4+3 x \ln \left (3+x \right )+2 x -\ln \left (3+x \right )}{x \ln \left (3+x \right )}\)	\(46\)

[In]

int(((-2*x^4-5*x^3+3*x^2+x+3)*ln(3+x)^2+(-4*x-12)*ln(3+x)-2*x^2-4*x)/(x^3+3*x^2)/ln(3+x)^2,x,method=_RETURNVER
BOSE)

[Out]

-(x^3-x^2+1)/x+2*(2+x)/x/ln(3+x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {-4 x-2 x^2+(-12-4 x) \log (3+x)+\left (3+x+3 x^2-5 x^3-2 x^4\right ) \log ^2(3+x)}{\left (3 x^2+x^3\right ) \log ^2(3+x)} \, dx=-\frac {{\left (x^{3} - x^{2} + 1\right )} \log \left (x + 3\right ) - 2 \, x - 4}{x \log \left (x + 3\right )} \]

[In]

integrate(((-2*x^4-5*x^3+3*x^2+x+3)*log(3+x)^2+(-4*x-12)*log(3+x)-2*x^2-4*x)/(x^3+3*x^2)/log(3+x)^2,x, algorit
hm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {-4 x-2 x^2+(-12-4 x) \log (3+x)+\left (3+x+3 x^2-5 x^3-2 x^4\right ) \log ^2(3+x)}{\left (3 x^2+x^3\right ) \log ^2(3+x)} \, dx=- x^{2} + x + \frac {2 x + 4}{x \log {\left (x + 3 \right )}} - \frac {1}{x} \]

[In]

integrate(((-2*x**4-5*x**3+3*x**2+x+3)*ln(3+x)**2+(-4*x-12)*ln(3+x)-2*x**2-4*x)/(x**3+3*x**2)/ln(3+x)**2,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {-4 x-2 x^2+(-12-4 x) \log (3+x)+\left (3+x+3 x^2-5 x^3-2 x^4\right ) \log ^2(3+x)}{\left (3 x^2+x^3\right ) \log ^2(3+x)} \, dx=-\frac {{\left (x^{3} - x^{2} + 1\right )} \log \left (x + 3\right ) - 2 \, x - 4}{x \log \left (x + 3\right )} \]

[In]

integrate(((-2*x^4-5*x^3+3*x^2+x+3)*log(3+x)^2+(-4*x-12)*log(3+x)-2*x^2-4*x)/(x^3+3*x^2)/log(3+x)^2,x, algorit
hm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {-4 x-2 x^2+(-12-4 x) \log (3+x)+\left (3+x+3 x^2-5 x^3-2 x^4\right ) \log ^2(3+x)}{\left (3 x^2+x^3\right ) \log ^2(3+x)} \, dx=-x^{2} + x - \frac {1}{x} + \frac {2 \, {\left (x + 2\right )}}{x \log \left (x + 3\right )} \]

[In]

integrate(((-2*x^4-5*x^3+3*x^2+x+3)*log(3+x)^2+(-4*x-12)*log(3+x)-2*x^2-4*x)/(x^3+3*x^2)/log(3+x)^2,x, algorit
hm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {-4 x-2 x^2+(-12-4 x) \log (3+x)+\left (3+x+3 x^2-5 x^3-2 x^4\right ) \log ^2(3+x)}{\left (3 x^2+x^3\right ) \log ^2(3+x)} \, dx=x+\frac {2}{\ln \left (x+3\right )}-\frac {1}{x}-x^2+\frac {4}{x\,\ln \left (x+3\right )} \]

[In]

int(-(4*x - log(x + 3)^2*(x + 3*x^2 - 5*x^3 - 2*x^4 + 3) + 2*x^2 + log(x + 3)*(4*x + 12))/(log(x + 3)^2*(3*x^2
+ x^3)),x)

[Out]

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