Integrand size = 48, antiderivative size = 19 \[ \int \frac {\left (-2-2 x^2+4 x^3\right ) \log \left (\frac {1+3 x-x^2+x^3}{x}\right )}{x+3 x^2-x^3+x^4} \, dx=-1+e^2+\log ^2\left (3+\frac {1}{x}-x+x^2\right ) \] Output:
-1+exp(2)+ln(x^2-x+3+1/x)^2
Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {\left (-2-2 x^2+4 x^3\right ) \log \left (\frac {1+3 x-x^2+x^3}{x}\right )}{x+3 x^2-x^3+x^4} \, dx=\log ^2\left (3+\frac {1}{x}-x+x^2\right ) \] Input:
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]
Output:
Log[3 + x^(-1) - x + x^2]^2
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (4 x^3-2 x^2-2\right ) \log \left (\frac {x^3-x^2+3 x+1}{x}\right )}{x^4-x^3+3 x^2+x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (4 x^3-2 x^2-2\right ) \log \left (\frac {x^3-x^2+3 x+1}{x}\right )}{x \left (x^3-x^2+3 x+1\right )}dx\) |
\(\Big \downarrow \) 3008 |
\(\displaystyle \int \left (\frac {2 \left (3 x^2-2 x+3\right ) \log \left (\frac {x^3-x^2+3 x+1}{x}\right )}{x^3-x^2+3 x+1}-\frac {2 \log \left (\frac {x^3-x^2+3 x+1}{x}\right )}{x}\right )dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \left (\frac {2 \left (3 x^2-2 x+3\right ) \log \left (x^2-x+\frac {1}{x}+3\right )}{x^3-x^2+3 x+1}-\frac {2 \log \left (x^2-x+\frac {1}{x}+3\right )}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 6 \int \frac {\log (x)}{x^3-x^2+3 x+1}dx-4 \int \frac {x \log (x)}{x^3-x^2+3 x+1}dx+6 \int \frac {x^2 \log (x)}{x^3-x^2+3 x+1}dx+6 \int \frac {\log \left (x^2-x+3+\frac {1}{x}\right )}{x^3-x^2+3 x+1}dx-4 \int \frac {x \log \left (x^2-x+3+\frac {1}{x}\right )}{x^3-x^2+3 x+1}dx+6 \int \frac {x^2 \log \left (x^2-x+3+\frac {1}{x}\right )}{x^3-x^2+3 x+1}dx-2 \log \left (x^2-x+\frac {1}{x}+3\right ) \log (x)-\log ^2(x)\) |
Input:
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]
Output:
$Aborted
Time = 0.21 (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 | \(-2 \ln \left (\frac {x^{3}-x^{2}+3 x +1}{x}\right ) \ln \left (x \right )+2 \ln \left (\frac {x^{3}-x^{2}+3 x +1}{x}\right ) \ln \left (x^{3}-x^{2}+3 x +1\right )-2 \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (\textit {\_Z}^{3}-\textit {\_Z}^{2}+3 \textit {\_Z} +1\right )}{\sum }\left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (x^{3}-x^{2}+3 x +1\right )-\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {\underline {\hspace {1.25 ex}}\alpha -1-\sqrt {-3 \underline {\hspace {1.25 ex}}\alpha ^{2}+2 \underline {\hspace {1.25 ex}}\alpha -11}+2 x}{3 \underline {\hspace {1.25 ex}}\alpha -1-\sqrt {-3 \underline {\hspace {1.25 ex}}\alpha ^{2}+2 \underline {\hspace {1.25 ex}}\alpha -11}}\right )-\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {\underline {\hspace {1.25 ex}}\alpha -1+\sqrt {-3 \underline {\hspace {1.25 ex}}\alpha ^{2}+2 \underline {\hspace {1.25 ex}}\alpha -11}+2 x}{3 \underline {\hspace {1.25 ex}}\alpha -1+\sqrt {-3 \underline {\hspace {1.25 ex}}\alpha ^{2}+2 \underline {\hspace {1.25 ex}}\alpha -11}}\right )-\operatorname {dilog}\left (\frac {\underline {\hspace {1.25 ex}}\alpha -1-\sqrt {-3 \underline {\hspace {1.25 ex}}\alpha ^{2}+2 \underline {\hspace {1.25 ex}}\alpha -11}+2 x}{3 \underline {\hspace {1.25 ex}}\alpha -1-\sqrt {-3 \underline {\hspace {1.25 ex}}\alpha ^{2}+2 \underline {\hspace {1.25 ex}}\alpha -11}}\right )-\operatorname {dilog}\left (\frac {\underline {\hspace {1.25 ex}}\alpha -1+\sqrt {-3 \underline {\hspace {1.25 ex}}\alpha ^{2}+2 \underline {\hspace {1.25 ex}}\alpha -11}+2 x}{3 \underline {\hspace {1.25 ex}}\alpha -1+\sqrt {-3 \underline {\hspace {1.25 ex}}\alpha ^{2}+2 \underline {\hspace {1.25 ex}}\alpha -11}}\right )-\frac {\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right )^{2}}{2}\right )\right )+2 \ln \left (x \right ) \ln \left (x^{3}-x^{2}+3 x +1\right )-\ln \left (x \right )^{2}\) | \(317\) |
Input:
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=_RETU RNVERBOSE)
Output:
ln((x^3-x^2+3*x+1)/x)^2
Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {\left (-2-2 x^2+4 x^3\right ) \log \left (\frac {1+3 x-x^2+x^3}{x}\right )}{x+3 x^2-x^3+x^4} \, dx=\log \left (\frac {x^{3} - x^{2} + 3 \, x + 1}{x}\right )^{2} \] Input:
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, algo rithm="fricas")
Output:
log((x^3 - x^2 + 3*x + 1)/x)^2
Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {\left (-2-2 x^2+4 x^3\right ) \log \left (\frac {1+3 x-x^2+x^3}{x}\right )}{x+3 x^2-x^3+x^4} \, dx=\log {\left (\frac {x^{3} - x^{2} + 3 x + 1}{x} \right )}^{2} \] Input:
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 )
Output:
log((x**3 - x**2 + 3*x + 1)/x)**2
Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (18) = 36\).
Time = 0.03 (sec) , antiderivative size = 82, normalized size of antiderivative = 4.32 \[ \int \frac {\left (-2-2 x^2+4 x^3\right ) \log \left (\frac {1+3 x-x^2+x^3}{x}\right )}{x+3 x^2-x^3+x^4} \, dx=-\log \left (x^{3} - x^{2} + 3 \, x + 1\right )^{2} + 2 \, \log \left (x^{3} - x^{2} + 3 \, x + 1\right ) \log \left (x\right ) - \log \left (x\right )^{2} + 2 \, {\left (\log \left (x^{3} - x^{2} + 3 \, x + 1\right ) - \log \left (x\right )\right )} \log \left (\frac {x^{3} - x^{2} + 3 \, x + 1}{x}\right ) \] Input:
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, algo rithm="maxima")
Output:
-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)
\[ \int \frac {\left (-2-2 x^2+4 x^3\right ) \log \left (\frac {1+3 x-x^2+x^3}{x}\right )}{x+3 x^2-x^3+x^4} \, dx=\int { \frac {2 \, {\left (2 \, x^{3} - x^{2} - 1\right )} \log \left (\frac {x^{3} - x^{2} + 3 \, x + 1}{x}\right )}{x^{4} - x^{3} + 3 \, x^{2} + x} \,d x } \] Input:
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, algo rithm="giac")
Output:
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)
Time = 3.29 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {\left (-2-2 x^2+4 x^3\right ) \log \left (\frac {1+3 x-x^2+x^3}{x}\right )}{x+3 x^2-x^3+x^4} \, dx={\ln \left (\frac {1}{x}-x+x^2+3\right )}^2 \] Input:
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)
Output:
log(1/x - x + x^2 + 3)^2
Time = 0.20 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {\left (-2-2 x^2+4 x^3\right ) \log \left (\frac {1+3 x-x^2+x^3}{x}\right )}{x+3 x^2-x^3+x^4} \, dx=\mathrm {log}\left (\frac {x^{3}-x^{2}+3 x +1}{x}\right )^{2} \] Input:
int((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)
Output:
log((x**3 - x**2 + 3*x + 1)/x)**2