\(\int \frac {(-2-2 x^2+4 x^3) \log (\frac {1+3 x-x^2+x^3}{x})}{x+3 x^2-x^3+x^4} \, dx\) [7664]

Optimal result

Integrand size = 48, antiderivative size = 19 \[ \int \frac {\left (-2-2 x^2+4 x^3\right ) \log \left (\frac {1+3 x-x^2+x^3}{x}\right )}{x+3 x^2-x^3+x^4} \, dx=-1+e^2+\log ^2\left (3+\frac {1}{x}-x+x^2\right ) \]

[Out]

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

Rubi [F]

\[ \int \frac {\left (-2-2 x^2+4 x^3\right ) \log \left (\frac {1+3 x-x^2+x^3}{x}\right )}{x+3 x^2-x^3+x^4} \, dx=\int \frac {\left (-2-2 x^2+4 x^3\right ) \log \left (\frac {1+3 x-x^2+x^3}{x}\right )}{x+3 x^2-x^3+x^4} \, dx \]

[In]

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

[Out]

-Log[x]^2 - 2*Log[x]*Log[(1 + 3*x - x^2 + x^3)/x] + 6*Defer[Int][Log[x]/(1 + 3*x - x^2 + x^3), x] - 4*Defer[In
t][(x*Log[x])/(1 + 3*x - x^2 + x^3), x] + 6*Defer[Int][(x^2*Log[x])/(1 + 3*x - x^2 + x^3), x] + 6*Defer[Int][L
og[3 + x^(-1) - x + x^2]/(1 + 3*x - x^2 + x^3), x] - 4*Defer[Int][(x*Log[3 + x^(-1) - x + x^2])/(1 + 3*x - x^2
+ x^3), x] + 6*Defer[Int][(x^2*Log[3 + x^(-1) - x + x^2])/(1 + 3*x - x^2 + x^3), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {2 \log \left (\frac {1+3 x-x^2+x^3}{x}\right )}{x}+\frac {2 \left (3-2 x+3 x^2\right ) \log \left (\frac {1+3 x-x^2+x^3}{x}\right )}{1+3 x-x^2+x^3}\right ) \, dx \\ & = -\left (2 \int \frac {\log \left (\frac {1+3 x-x^2+x^3}{x}\right )}{x} \, dx\right )+2 \int \frac {\left (3-2 x+3 x^2\right ) \log \left (\frac {1+3 x-x^2+x^3}{x}\right )}{1+3 x-x^2+x^3} \, dx \\ & = -2 \log (x) \log \left (\frac {1+3 x-x^2+x^3}{x}\right )+2 \int \frac {x \left (\frac {3-2 x+3 x^2}{x}-\frac {1+3 x-x^2+x^3}{x^2}\right ) \log (x)}{1+3 x-x^2+x^3} \, dx+2 \int \left (\frac {3 \log \left (\frac {1+3 x-x^2+x^3}{x}\right )}{1+3 x-x^2+x^3}-\frac {2 x \log \left (\frac {1+3 x-x^2+x^3}{x}\right )}{1+3 x-x^2+x^3}+\frac {3 x^2 \log \left (\frac {1+3 x-x^2+x^3}{x}\right )}{1+3 x-x^2+x^3}\right ) \, dx \\ & = -2 \log (x) \log \left (\frac {1+3 x-x^2+x^3}{x}\right )+2 \int \left (-\frac {\log (x)}{x}+\frac {\left (3-2 x+3 x^2\right ) \log (x)}{1+3 x-x^2+x^3}\right ) \, dx-4 \int \frac {x \log \left (\frac {1+3 x-x^2+x^3}{x}\right )}{1+3 x-x^2+x^3} \, dx+6 \int \frac {\log \left (\frac {1+3 x-x^2+x^3}{x}\right )}{1+3 x-x^2+x^3} \, dx+6 \int \frac {x^2 \log \left (\frac {1+3 x-x^2+x^3}{x}\right )}{1+3 x-x^2+x^3} \, dx \\ & = -2 \log (x) \log \left (\frac {1+3 x-x^2+x^3}{x}\right )-2 \int \frac {\log (x)}{x} \, dx+2 \int \frac {\left (3-2 x+3 x^2\right ) \log (x)}{1+3 x-x^2+x^3} \, dx-4 \int \frac {x \log \left (3+\frac {1}{x}-x+x^2\right )}{1+3 x-x^2+x^3} \, dx+6 \int \frac {\log \left (3+\frac {1}{x}-x+x^2\right )}{1+3 x-x^2+x^3} \, dx+6 \int \frac {x^2 \log \left (3+\frac {1}{x}-x+x^2\right )}{1+3 x-x^2+x^3} \, dx \\ & = -\log ^2(x)-2 \log (x) \log \left (\frac {1+3 x-x^2+x^3}{x}\right )+2 \int \left (\frac {3 \log (x)}{1+3 x-x^2+x^3}-\frac {2 x \log (x)}{1+3 x-x^2+x^3}+\frac {3 x^2 \log (x)}{1+3 x-x^2+x^3}\right ) \, dx-4 \int \frac {x \log \left (3+\frac {1}{x}-x+x^2\right )}{1+3 x-x^2+x^3} \, dx+6 \int \frac {\log \left (3+\frac {1}{x}-x+x^2\right )}{1+3 x-x^2+x^3} \, dx+6 \int \frac {x^2 \log \left (3+\frac {1}{x}-x+x^2\right )}{1+3 x-x^2+x^3} \, dx \\ & = -\log ^2(x)-2 \log (x) \log \left (\frac {1+3 x-x^2+x^3}{x}\right )-4 \int \frac {x \log (x)}{1+3 x-x^2+x^3} \, dx-4 \int \frac {x \log \left (3+\frac {1}{x}-x+x^2\right )}{1+3 x-x^2+x^3} \, dx+6 \int \frac {\log (x)}{1+3 x-x^2+x^3} \, dx+6 \int \frac {x^2 \log (x)}{1+3 x-x^2+x^3} \, dx+6 \int \frac {\log \left (3+\frac {1}{x}-x+x^2\right )}{1+3 x-x^2+x^3} \, dx+6 \int \frac {x^2 \log \left (3+\frac {1}{x}-x+x^2\right )}{1+3 x-x^2+x^3} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {\left (-2-2 x^2+4 x^3\right ) \log \left (\frac {1+3 x-x^2+x^3}{x}\right )}{x+3 x^2-x^3+x^4} \, dx=\log ^2\left (3+\frac {1}{x}-x+x^2\right ) \]

[In]

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

[Out]

Log[3 + x^(-1) - x + x^2]^2

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11

method	result	size
default	\(\ln \left (\frac {x^{3}-x^{2}+3 x +1}{x}\right )^{2}\)	\(21\)
norman	\(\ln \left (\frac {x^{3}-x^{2}+3 x +1}{x}\right )^{2}\)	\(21\)
risch	\(\ln \left (\frac {x^{3}-x^{2}+3 x +1}{x}\right )^{2}\)	\(21\)
parts	\(-2 \ln \left (\frac {x^{3}-x^{2}+3 x +1}{x}\right ) \ln \left (x \right )+2 \ln \left (\frac {x^{3}-x^{2}+3 x +1}{x}\right ) \ln \left (x^{3}-x^{2}+3 x +1\right )+2 \ln \left (x \right ) \ln \left (x^{3}-x^{2}+3 x +1\right )-2 \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (\textit {\_Z}^{3}-\textit {\_Z}^{2}+3 \textit {\_Z} +1\right )}{\sum }\left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (x^{3}-x^{2}+3 x +1\right )-\frac {\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right )^{2}}{2}-\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {\underline {\hspace {1.25 ex}}\alpha -1-\sqrt {-3 \underline {\hspace {1.25 ex}}\alpha ^{2}+2 \underline {\hspace {1.25 ex}}\alpha -11}+2 x}{3 \underline {\hspace {1.25 ex}}\alpha -1-\sqrt {-3 \underline {\hspace {1.25 ex}}\alpha ^{2}+2 \underline {\hspace {1.25 ex}}\alpha -11}}\right )-\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {\underline {\hspace {1.25 ex}}\alpha -1+\sqrt {-3 \underline {\hspace {1.25 ex}}\alpha ^{2}+2 \underline {\hspace {1.25 ex}}\alpha -11}+2 x}{3 \underline {\hspace {1.25 ex}}\alpha -1+\sqrt {-3 \underline {\hspace {1.25 ex}}\alpha ^{2}+2 \underline {\hspace {1.25 ex}}\alpha -11}}\right )-\operatorname {dilog}\left (\frac {\underline {\hspace {1.25 ex}}\alpha -1-\sqrt {-3 \underline {\hspace {1.25 ex}}\alpha ^{2}+2 \underline {\hspace {1.25 ex}}\alpha -11}+2 x}{3 \underline {\hspace {1.25 ex}}\alpha -1-\sqrt {-3 \underline {\hspace {1.25 ex}}\alpha ^{2}+2 \underline {\hspace {1.25 ex}}\alpha -11}}\right )-\operatorname {dilog}\left (\frac {\underline {\hspace {1.25 ex}}\alpha -1+\sqrt {-3 \underline {\hspace {1.25 ex}}\alpha ^{2}+2 \underline {\hspace {1.25 ex}}\alpha -11}+2 x}{3 \underline {\hspace {1.25 ex}}\alpha -1+\sqrt {-3 \underline {\hspace {1.25 ex}}\alpha ^{2}+2 \underline {\hspace {1.25 ex}}\alpha -11}}\right )\right )\right )-\ln \left (x \right )^{2}\)	\(317\)

[In]

int((4*x^3-2*x^2-2)*ln((x^3-x^2+3*x+1)/x)/(x^4-x^3+3*x^2+x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {\left (-2-2 x^2+4 x^3\right ) \log \left (\frac {1+3 x-x^2+x^3}{x}\right )}{x+3 x^2-x^3+x^4} \, dx=\log \left (\frac {x^{3} - x^{2} + 3 \, x + 1}{x}\right )^{2} \]

[In]

integrate((4*x^3-2*x^2-2)*log((x^3-x^2+3*x+1)/x)/(x^4-x^3+3*x^2+x),x, algorithm="fricas")

[Out]

log((x^3 - x^2 + 3*x + 1)/x)^2

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {\left (-2-2 x^2+4 x^3\right ) \log \left (\frac {1+3 x-x^2+x^3}{x}\right )}{x+3 x^2-x^3+x^4} \, dx=\log {\left (\frac {x^{3} - x^{2} + 3 x + 1}{x} \right )}^{2} \]

[In]

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

[Out]

log((x**3 - x**2 + 3*x + 1)/x)**2

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (18) = 36\).

Time = 0.21 (sec) , antiderivative size = 82, normalized size of antiderivative = 4.32 \[ \int \frac {\left (-2-2 x^2+4 x^3\right ) \log \left (\frac {1+3 x-x^2+x^3}{x}\right )}{x+3 x^2-x^3+x^4} \, dx=-\log \left (x^{3} - x^{2} + 3 \, x + 1\right )^{2} + 2 \, \log \left (x^{3} - x^{2} + 3 \, x + 1\right ) \log \left (x\right ) - \log \left (x\right )^{2} + 2 \, {\left (\log \left (x^{3} - x^{2} + 3 \, x + 1\right ) - \log \left (x\right )\right )} \log \left (\frac {x^{3} - x^{2} + 3 \, x + 1}{x}\right ) \]

[In]

integrate((4*x^3-2*x^2-2)*log((x^3-x^2+3*x+1)/x)/(x^4-x^3+3*x^2+x),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {\left (-2-2 x^2+4 x^3\right ) \log \left (\frac {1+3 x-x^2+x^3}{x}\right )}{x+3 x^2-x^3+x^4} \, dx=\int { \frac {2 \, {\left (2 \, x^{3} - x^{2} - 1\right )} \log \left (\frac {x^{3} - x^{2} + 3 \, x + 1}{x}\right )}{x^{4} - x^{3} + 3 \, x^{2} + x} \,d x } \]

[In]

integrate((4*x^3-2*x^2-2)*log((x^3-x^2+3*x+1)/x)/(x^4-x^3+3*x^2+x),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.70 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {\left (-2-2 x^2+4 x^3\right ) \log \left (\frac {1+3 x-x^2+x^3}{x}\right )}{x+3 x^2-x^3+x^4} \, dx={\ln \left (\frac {1}{x}-x+x^2+3\right )}^2 \]

[In]

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

[Out]

log(1/x - x + x^2 + 3)^2