Integrand size = 63, antiderivative size = 31 \[ \int \frac {12 x^2+4 x^3-6 x^4-12 x \log \left (16-48 x+36 x^2\right )+(-2+3 x) \log ^2\left (16-48 x+36 x^2\right )}{-2 x^2+3 x^3} \, dx=-x^2+\frac {x+x^2-\left (x-\log \left ((4-6 x)^2\right )\right )^2}{x} \] Output:
(x^2-(x-ln((-6*x+4)^2))^2+x)/x-x^2
Time = 0.06 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.13 \[ \int \frac {12 x^2+4 x^3-6 x^4-12 x \log \left (16-48 x+36 x^2\right )+(-2+3 x) \log ^2\left (16-48 x+36 x^2\right )}{-2 x^2+3 x^3} \, dx=-x^2+2 \log \left (4 (2-3 x)^2\right )-\frac {\log ^2\left (4 (2-3 x)^2\right )}{x} \] Input:
Integrate[(12*x^2 + 4*x^3 - 6*x^4 - 12*x*Log[16 - 48*x + 36*x^2] + (-2 + 3 *x)*Log[16 - 48*x + 36*x^2]^2)/(-2*x^2 + 3*x^3),x]
Output:
-x^2 + 2*Log[4*(2 - 3*x)^2] - Log[4*(2 - 3*x)^2]^2/x
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.87 (sec) , antiderivative size = 104, normalized size of antiderivative = 3.35, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2026, 7293, 2009}
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.
\(\displaystyle \int \frac {-6 x^4+4 x^3+12 x^2+(3 x-2) \log ^2\left (36 x^2-48 x+16\right )-12 x \log \left (36 x^2-48 x+16\right )}{3 x^3-2 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-6 x^4+4 x^3+12 x^2+(3 x-2) \log ^2\left (36 x^2-48 x+16\right )-12 x \log \left (36 x^2-48 x+16\right )}{x^2 (3 x-2)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {2 \left (3 x^2-2 x-6\right )}{3 x-2}+\frac {\log ^2\left (4 (2-3 x)^2\right )}{x^2}-\frac {12 \log \left (4 (2-3 x)^2\right )}{x (3 x-2)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -12 \operatorname {PolyLog}\left (2,\frac {2}{2-3 x}\right )-12 \operatorname {PolyLog}\left (2,1-\frac {3 x}{2}\right )-x^2-\frac {(2-3 x) \log ^2\left (4 (2-3 x)^2\right )}{2 x}+4 \log (2-3 x)+6 \log \left (1-\frac {2}{2-3 x}\right ) \log \left (4 (2-3 x)^2\right )-6 \log \left (4 (2-3 x)^2\right ) \log \left (\frac {3 x}{2}\right )\) |
Input:
Int[(12*x^2 + 4*x^3 - 6*x^4 - 12*x*Log[16 - 48*x + 36*x^2] + (-2 + 3*x)*Lo g[16 - 48*x + 36*x^2]^2)/(-2*x^2 + 3*x^3),x]
Output:
-x^2 + 4*Log[2 - 3*x] + 6*Log[1 - 2/(2 - 3*x)]*Log[4*(2 - 3*x)^2] - ((2 - 3*x)*Log[4*(2 - 3*x)^2]^2)/(2*x) - 6*Log[4*(2 - 3*x)^2]*Log[(3*x)/2] - 12* PolyLog[2, 2/(2 - 3*x)] - 12*PolyLog[2, 1 - (3*x)/2]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Time = 0.64 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06
method | result | size |
risch | \(-\frac {\ln \left (36 x^{2}-48 x +16\right )^{2}}{x}-x^{2}+4 \ln \left (-2+3 x \right )\) | \(33\) |
norman | \(\frac {2 x \ln \left (36 x^{2}-48 x +16\right )-x^{3}-\ln \left (36 x^{2}-48 x +16\right )^{2}}{x}\) | \(40\) |
parallelrisch | \(\frac {-18 x^{3}+36 x \ln \left (36 x^{2}-48 x +16\right )-18 \ln \left (36 x^{2}-48 x +16\right )^{2}+8 x}{18 x}\) | \(44\) |
orering | \(\frac {\left (81 x^{5}-513 x^{4}+1008 x^{3}-882 x^{2}+1152 x -512\right ) \left (\left (-2+3 x \right ) \ln \left (36 x^{2}-48 x +16\right )^{2}-12 x \ln \left (36 x^{2}-48 x +16\right )-6 x^{4}+4 x^{3}+12 x^{2}\right )}{3 \left (81 x^{3}-126 x^{2}+42 x -32\right ) \left (3 x^{3}-2 x^{2}\right )}-\frac {\left (-2+3 x \right ) \left (162 x^{5}-1161 x^{4}+1296 x^{3}-216 x^{2}-1536 x +512\right ) \left (\frac {3 \ln \left (36 x^{2}-48 x +16\right )^{2}+\frac {2 \left (-2+3 x \right ) \ln \left (36 x^{2}-48 x +16\right ) \left (72 x -48\right )}{36 x^{2}-48 x +16}-12 \ln \left (36 x^{2}-48 x +16\right )-\frac {12 x \left (72 x -48\right )}{36 x^{2}-48 x +16}-24 x^{3}+12 x^{2}+24 x}{3 x^{3}-2 x^{2}}-\frac {\left (\left (-2+3 x \right ) \ln \left (36 x^{2}-48 x +16\right )^{2}-12 x \ln \left (36 x^{2}-48 x +16\right )-6 x^{4}+4 x^{3}+12 x^{2}\right ) \left (9 x^{2}-4 x \right )}{\left (3 x^{3}-2 x^{2}\right )^{2}}\right )}{18 \left (81 x^{3}-126 x^{2}+42 x -32\right )}-\frac {\left (81 x^{4}-405 x^{3}+324 x^{2}-256\right ) x \left (-2+3 x \right )^{2} \left (\frac {\frac {12 \ln \left (36 x^{2}-48 x +16\right ) \left (72 x -48\right )}{36 x^{2}-48 x +16}+\frac {2 \left (-2+3 x \right ) \left (72 x -48\right )^{2}}{\left (36 x^{2}-48 x +16\right )^{2}}+\frac {144 \left (-2+3 x \right ) \ln \left (36 x^{2}-48 x +16\right )}{36 x^{2}-48 x +16}-\frac {2 \left (-2+3 x \right ) \ln \left (36 x^{2}-48 x +16\right ) \left (72 x -48\right )^{2}}{\left (36 x^{2}-48 x +16\right )^{2}}-\frac {24 \left (72 x -48\right )}{36 x^{2}-48 x +16}-\frac {864 x}{36 x^{2}-48 x +16}+\frac {12 x \left (72 x -48\right )^{2}}{\left (36 x^{2}-48 x +16\right )^{2}}-72 x^{2}+24 x +24}{3 x^{3}-2 x^{2}}-\frac {2 \left (3 \ln \left (36 x^{2}-48 x +16\right )^{2}+\frac {2 \left (-2+3 x \right ) \ln \left (36 x^{2}-48 x +16\right ) \left (72 x -48\right )}{36 x^{2}-48 x +16}-12 \ln \left (36 x^{2}-48 x +16\right )-\frac {12 x \left (72 x -48\right )}{36 x^{2}-48 x +16}-24 x^{3}+12 x^{2}+24 x \right ) \left (9 x^{2}-4 x \right )}{\left (3 x^{3}-2 x^{2}\right )^{2}}+\frac {2 \left (\left (-2+3 x \right ) \ln \left (36 x^{2}-48 x +16\right )^{2}-12 x \ln \left (36 x^{2}-48 x +16\right )-6 x^{4}+4 x^{3}+12 x^{2}\right ) \left (9 x^{2}-4 x \right )^{2}}{\left (3 x^{3}-2 x^{2}\right )^{3}}-\frac {\left (\left (-2+3 x \right ) \ln \left (36 x^{2}-48 x +16\right )^{2}-12 x \ln \left (36 x^{2}-48 x +16\right )-6 x^{4}+4 x^{3}+12 x^{2}\right ) \left (18 x -4\right )}{\left (3 x^{3}-2 x^{2}\right )^{2}}\right )}{54 \left (81 x^{3}-126 x^{2}+42 x -32\right )}\) | \(855\) |
Input:
int(((-2+3*x)*ln(36*x^2-48*x+16)^2-12*x*ln(36*x^2-48*x+16)-6*x^4+4*x^3+12* x^2)/(3*x^3-2*x^2),x,method=_RETURNVERBOSE)
Output:
-1/x*ln(36*x^2-48*x+16)^2-x^2+4*ln(-2+3*x)
Time = 0.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {12 x^2+4 x^3-6 x^4-12 x \log \left (16-48 x+36 x^2\right )+(-2+3 x) \log ^2\left (16-48 x+36 x^2\right )}{-2 x^2+3 x^3} \, dx=-\frac {x^{3} - 2 \, x \log \left (36 \, x^{2} - 48 \, x + 16\right ) + \log \left (36 \, x^{2} - 48 \, x + 16\right )^{2}}{x} \] Input:
integrate(((-2+3*x)*log(36*x^2-48*x+16)^2-12*x*log(36*x^2-48*x+16)-6*x^4+4 *x^3+12*x^2)/(3*x^3-2*x^2),x, algorithm="fricas")
Output:
-(x^3 - 2*x*log(36*x^2 - 48*x + 16) + log(36*x^2 - 48*x + 16)^2)/x
Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {12 x^2+4 x^3-6 x^4-12 x \log \left (16-48 x+36 x^2\right )+(-2+3 x) \log ^2\left (16-48 x+36 x^2\right )}{-2 x^2+3 x^3} \, dx=- x^{2} + 4 \log {\left (3 x - 2 \right )} - \frac {\log {\left (36 x^{2} - 48 x + 16 \right )}^{2}}{x} \] Input:
integrate(((-2+3*x)*ln(36*x**2-48*x+16)**2-12*x*ln(36*x**2-48*x+16)-6*x**4 +4*x**3+12*x**2)/(3*x**3-2*x**2),x)
Output:
-x**2 + 4*log(3*x - 2) - log(36*x**2 - 48*x + 16)**2/x
Time = 0.14 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \frac {12 x^2+4 x^3-6 x^4-12 x \log \left (16-48 x+36 x^2\right )+(-2+3 x) \log ^2\left (16-48 x+36 x^2\right )}{-2 x^2+3 x^3} \, dx=-x^{2} - \frac {4 \, {\left (\log \left (2\right )^{2} + 2 \, \log \left (2\right ) \log \left (3 \, x - 2\right ) + \log \left (3 \, x - 2\right )^{2}\right )}}{x} + 4 \, \log \left (3 \, x - 2\right ) \] Input:
integrate(((-2+3*x)*log(36*x^2-48*x+16)^2-12*x*log(36*x^2-48*x+16)-6*x^4+4 *x^3+12*x^2)/(3*x^3-2*x^2),x, algorithm="maxima")
Output:
-x^2 - 4*(log(2)^2 + 2*log(2)*log(3*x - 2) + log(3*x - 2)^2)/x + 4*log(3*x - 2)
Time = 0.15 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {12 x^2+4 x^3-6 x^4-12 x \log \left (16-48 x+36 x^2\right )+(-2+3 x) \log ^2\left (16-48 x+36 x^2\right )}{-2 x^2+3 x^3} \, dx=-x^{2} - \frac {\log \left (36 \, x^{2} - 48 \, x + 16\right )^{2}}{x} + 4 \, \log \left (3 \, x - 2\right ) \] Input:
integrate(((-2+3*x)*log(36*x^2-48*x+16)^2-12*x*log(36*x^2-48*x+16)-6*x^4+4 *x^3+12*x^2)/(3*x^3-2*x^2),x, algorithm="giac")
Output:
-x^2 - log(36*x^2 - 48*x + 16)^2/x + 4*log(3*x - 2)
Time = 0.11 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {12 x^2+4 x^3-6 x^4-12 x \log \left (16-48 x+36 x^2\right )+(-2+3 x) \log ^2\left (16-48 x+36 x^2\right )}{-2 x^2+3 x^3} \, dx=2\,\ln \left ({\left (3\,x-2\right )}^2\right )-\frac {{\ln \left (36\,x^2-48\,x+16\right )}^2}{x}-x^2 \] Input:
int(-(log(36*x^2 - 48*x + 16)^2*(3*x - 2) - 12*x*log(36*x^2 - 48*x + 16) + 12*x^2 + 4*x^3 - 6*x^4)/(2*x^2 - 3*x^3),x)
Output:
2*log((3*x - 2)^2) - log(36*x^2 - 48*x + 16)^2/x - x^2
Time = 0.16 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {12 x^2+4 x^3-6 x^4-12 x \log \left (16-48 x+36 x^2\right )+(-2+3 x) \log ^2\left (16-48 x+36 x^2\right )}{-2 x^2+3 x^3} \, dx=\frac {-\mathrm {log}\left (36 x^{2}-48 x +16\right )^{2}+4 \,\mathrm {log}\left (3 x -2\right ) x -x^{3}}{x} \] Input:
int(((-2+3*x)*log(36*x^2-48*x+16)^2-12*x*log(36*x^2-48*x+16)-6*x^4+4*x^3+1 2*x^2)/(3*x^3-2*x^2),x)
Output:
( - log(36*x**2 - 48*x + 16)**2 + 4*log(3*x - 2)*x - x**3)/x