Integrand size = 36, antiderivative size = 29 \[ \int \frac {-54 x+(162+54 x) \log \left (\frac {25 x}{4 e^4}\right )-162 \log ^2\left (\frac {25 x}{4 e^4}\right )}{x^3} \, dx=9 \left (x-\frac {-x+x^2+3 \log \left (\frac {25 x}{4 e^4}\right )}{x}\right )^2 \] Output:
3*(x-(3*ln(25/4*x/exp(2)^2)+x^2-x)/x)*(3*x-3*(3*ln(25/4*x/exp(2)^2)+x^2-x) /x)
Time = 0.11 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.90 \[ \int \frac {-54 x+(162+54 x) \log \left (\frac {25 x}{4 e^4}\right )-162 \log ^2\left (\frac {25 x}{4 e^4}\right )}{x^3} \, dx=\frac {27 \left (96+16 x-21 \log \left (\frac {25}{4}\right )-4 x \log \left (\frac {25}{4}\right )+3 \left (-9+\log \left (\frac {625}{16}\right )\right ) \log \left (\frac {25 x}{4}\right )+\log (x) \left (-21-4 x+6 \log \left (\frac {25 x}{4}\right )\right )\right )}{2 x^2} \] Input:
Integrate[(-54*x + (162 + 54*x)*Log[(25*x)/(4*E^4)] - 162*Log[(25*x)/(4*E^ 4)]^2)/x^3,x]
Output:
(27*(96 + 16*x - 21*Log[25/4] - 4*x*Log[25/4] + 3*(-9 + Log[625/16])*Log[( 25*x)/4] + Log[x]*(-21 - 4*x + 6*Log[(25*x)/4])))/(2*x^2)
Leaf count is larger than twice the leaf count of optimal. \(66\) vs. \(2(29)=58\).
Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.28, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {2010, 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.
\(\displaystyle \int \frac {-54 x-162 \log ^2\left (\frac {25 x}{4 e^4}\right )+(54 x+162) \log \left (\frac {25 x}{4 e^4}\right )}{x^3} \, dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (-\frac {162 \left (4 \left (1-\frac {1}{2} \log \left (\frac {5}{2}\right )\right )-\log (x)\right )^2}{x^3}+\frac {54 (x+3) \left (\log (x)-4 \left (1-\frac {1}{2} \log \left (\frac {5}{2}\right )\right )\right )}{x^3}-\frac {54}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {9 (x+3)^2 \left (-\log (x)+4-\log \left (\frac {25}{4}\right )\right )}{x^2}+\frac {81 \left (2 \left (2-\log \left (\frac {5}{2}\right )\right )-\log (x)\right )^2}{x^2}-\frac {81 \left (-\log (x)+4-\log \left (\frac {25}{4}\right )\right )}{x^2}+9 \log (x)\) |
Input:
Int[(-54*x + (162 + 54*x)*Log[(25*x)/(4*E^4)] - 162*Log[(25*x)/(4*E^4)]^2) /x^3,x]
Output:
(81*(2*(2 - Log[5/2]) - Log[x])^2)/x^2 - (81*(4 - Log[25/4] - Log[x]))/x^2 + (9*(3 + x)^2*(4 - Log[25/4] - Log[x]))/x^2 + 9*Log[x]
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Time = 0.41 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90
method | result | size |
risch | \(\frac {81 \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )^{2}}{x^{2}}-\frac {54 \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )}{x}\) | \(26\) |
norman | \(\frac {81 \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )^{2}-54 \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right ) x}{x^{2}}\) | \(29\) |
parallelrisch | \(\frac {81 \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )^{2}-54 \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right ) x}{x^{2}}\) | \(29\) |
parts | \(\frac {81 \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )^{2}}{x^{2}}-\frac {54 \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )}{x}\) | \(30\) |
derivativedivides | \(\frac {25 \,{\mathrm e}^{-8} \left (216 \,{\mathrm e}^{4} \left (-\frac {4 \,{\mathrm e}^{4} \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )}{25 x}-\frac {4 \,{\mathrm e}^{4}}{25 x}\right )+\frac {864 \,{\mathrm e}^{8}}{25 x}+\frac {1296 \,{\mathrm e}^{8} \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )^{2}}{25 x^{2}}\right )}{16}\) | \(69\) |
default | \(\frac {25 \,{\mathrm e}^{-8} \left (216 \,{\mathrm e}^{4} \left (-\frac {4 \,{\mathrm e}^{4} \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )}{25 x}-\frac {4 \,{\mathrm e}^{4}}{25 x}\right )+\frac {864 \,{\mathrm e}^{8}}{25 x}+\frac {1296 \,{\mathrm e}^{8} \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )^{2}}{25 x^{2}}\right )}{16}\) | \(69\) |
orering | \(-\frac {\left (10 x^{2}-123 x +171\right ) \left (-162 \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )^{2}+\left (54 x +162\right ) \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )-54 x \right )}{2 x^{2} \left (x^{2}-15 x +36\right )}-\frac {x^{2} \left (7 x^{2}-72 x +81\right ) \left (\frac {-\frac {324 \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )}{x}+54 \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )+\frac {54 x +162}{x}-54}{x^{3}}-\frac {3 \left (-162 \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )^{2}+\left (54 x +162\right ) \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )-54 x \right )}{x^{4}}\right )}{2 \left (x^{2}-15 x +36\right )}-\frac {\left (x^{2}-9 x +9\right ) x^{3} \left (\frac {-\frac {324}{x^{2}}+\frac {324 \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )}{x^{2}}+\frac {108}{x}-\frac {54 x +162}{x^{2}}}{x^{3}}-\frac {6 \left (-\frac {324 \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )}{x}+54 \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )+\frac {54 x +162}{x}-54\right )}{x^{4}}+\frac {-1944 \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )^{2}+12 \left (54 x +162\right ) \ln \left (\frac {25 x \,{\mathrm e}^{-4}}{4}\right )-648 x}{x^{5}}\right )}{2 \left (-3+x \right ) \left (x -12\right )}\) | \(292\) |
Input:
int((-162*ln(25/4*x/exp(2)^2)^2+(54*x+162)*ln(25/4*x/exp(2)^2)-54*x)/x^3,x ,method=_RETURNVERBOSE)
Output:
81/x^2*ln(25/4*x*exp(-4))^2-54*ln(25/4*x*exp(-4))/x
Time = 0.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {-54 x+(162+54 x) \log \left (\frac {25 x}{4 e^4}\right )-162 \log ^2\left (\frac {25 x}{4 e^4}\right )}{x^3} \, dx=-\frac {27 \, {\left (2 \, x \log \left (\frac {25}{4} \, x e^{\left (-4\right )}\right ) - 3 \, \log \left (\frac {25}{4} \, x e^{\left (-4\right )}\right )^{2}\right )}}{x^{2}} \] Input:
integrate((-162*log(25/4*x/exp(2)^2)^2+(54*x+162)*log(25/4*x/exp(2)^2)-54* x)/x^3,x, algorithm="fricas")
Output:
-27*(2*x*log(25/4*x*e^(-4)) - 3*log(25/4*x*e^(-4))^2)/x^2
Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {-54 x+(162+54 x) \log \left (\frac {25 x}{4 e^4}\right )-162 \log ^2\left (\frac {25 x}{4 e^4}\right )}{x^3} \, dx=- \frac {54 \log {\left (\frac {25 x}{4 e^{4}} \right )}}{x} + \frac {81 \log {\left (\frac {25 x}{4 e^{4}} \right )}^{2}}{x^{2}} \] Input:
integrate((-162*ln(25/4*x/exp(2)**2)**2+(54*x+162)*ln(25/4*x/exp(2)**2)-54 *x)/x**3,x)
Output:
-54*log(25*x*exp(-4)/4)/x + 81*log(25*x*exp(-4)/4)**2/x**2
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (26) = 52\).
Time = 0.04 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.83 \[ \int \frac {-54 x+(162+54 x) \log \left (\frac {25 x}{4 e^4}\right )-162 \log ^2\left (\frac {25 x}{4 e^4}\right )}{x^3} \, dx=-\frac {54 \, \log \left (\frac {25}{4} \, x e^{\left (-4\right )}\right )}{x} + \frac {81 \, {\left (2 \, \log \left (\frac {25}{4} \, x e^{\left (-4\right )}\right )^{2} + 2 \, \log \left (\frac {25}{4} \, x e^{\left (-4\right )}\right ) + 1\right )}}{2 \, x^{2}} - \frac {81 \, \log \left (\frac {25}{4} \, x e^{\left (-4\right )}\right )}{x^{2}} - \frac {81}{2 \, x^{2}} \] Input:
integrate((-162*log(25/4*x/exp(2)^2)^2+(54*x+162)*log(25/4*x/exp(2)^2)-54* x)/x^3,x, algorithm="maxima")
Output:
-54*log(25/4*x*e^(-4))/x + 81/2*(2*log(25/4*x*e^(-4))^2 + 2*log(25/4*x*e^( -4)) + 1)/x^2 - 81*log(25/4*x*e^(-4))/x^2 - 81/2/x^2
Time = 0.11 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {-54 x+(162+54 x) \log \left (\frac {25 x}{4 e^4}\right )-162 \log ^2\left (\frac {25 x}{4 e^4}\right )}{x^3} \, dx=-\frac {54 \, {\left (x + 12\right )} \log \left (\frac {25}{4} \, x\right )}{x^{2}} + \frac {81 \, \log \left (\frac {25}{4} \, x\right )^{2}}{x^{2}} + \frac {216 \, {\left (x + 6\right )}}{x^{2}} \] Input:
integrate((-162*log(25/4*x/exp(2)^2)^2+(54*x+162)*log(25/4*x/exp(2)^2)-54* x)/x^3,x, algorithm="giac")
Output:
-54*(x + 12)*log(25/4*x)/x^2 + 81*log(25/4*x)^2/x^2 + 216*(x + 6)/x^2
Time = 3.06 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {-54 x+(162+54 x) \log \left (\frac {25 x}{4 e^4}\right )-162 \log ^2\left (\frac {25 x}{4 e^4}\right )}{x^3} \, dx=-\frac {27\,\ln \left (\frac {25\,x\,{\mathrm {e}}^{-4}}{4}\right )\,\left (2\,x-3\,\ln \left (\frac {25\,x\,{\mathrm {e}}^{-4}}{4}\right )\right )}{x^2} \] Input:
int(-(54*x + 162*log((25*x*exp(-4))/4)^2 - log((25*x*exp(-4))/4)*(54*x + 1 62))/x^3,x)
Output:
-(27*log((25*x*exp(-4))/4)*(2*x - 3*log((25*x*exp(-4))/4)))/x^2
Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {-54 x+(162+54 x) \log \left (\frac {25 x}{4 e^4}\right )-162 \log ^2\left (\frac {25 x}{4 e^4}\right )}{x^3} \, dx=\frac {27 \,\mathrm {log}\left (\frac {25 x}{4 e^{4}}\right ) \left (3 \,\mathrm {log}\left (\frac {25 x}{4 e^{4}}\right )-2 x \right )}{x^{2}} \] Input:
int((-162*log(25/4*x/exp(2)^2)^2+(54*x+162)*log(25/4*x/exp(2)^2)-54*x)/x^3 ,x)
Output:
(27*log((25*x)/(4*e**4))*(3*log((25*x)/(4*e**4)) - 2*x))/x**2