Integrand size = 34, antiderivative size = 33 \[ \int \frac {-5+2 x^2+\left (x+4 x^2\right ) \log \left (\frac {-4 e x+5 x \log (3)}{e}\right )}{x} \, dx=3-x-\left (5-x-2 x^2\right ) \log \left (x+5 \left (-x+\frac {x \log (3)}{e}\right )\right ) \] Output:
3-(-2*x^2-x+5)*ln(-4*x+5*x/exp(1)*ln(3))-x
Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18 \[ \int \frac {-5+2 x^2+\left (x+4 x^2\right ) \log \left (\frac {-4 e x+5 x \log (3)}{e}\right )}{x} \, dx=-x-5 \log (x)+x \log \left (x \left (-4+\frac {5 \log (3)}{e}\right )\right )+2 x^2 \log \left (x \left (-4+\frac {5 \log (3)}{e}\right )\right ) \] Input:
Integrate[(-5 + 2*x^2 + (x + 4*x^2)*Log[(-4*E*x + 5*x*Log[3])/E])/x,x]
Output:
-x - 5*Log[x] + x*Log[x*(-4 + (5*Log[3])/E)] + 2*x^2*Log[x*(-4 + (5*Log[3] )/E)]
Time = 0.21 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, 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.
\(\displaystyle \int \frac {2 x^2+\left (4 x^2+x\right ) \log \left (\frac {5 x \log (3)-4 e x}{e}\right )-5}{x} \, dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (\frac {2 x^2-5}{x}+(4 x+1) \log \left (-x \left (4-\frac {5 \log (3)}{e}\right )\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -x+\frac {1}{8} (4 x+1)^2 \log \left (-x \left (4-\frac {5 \log (3)}{e}\right )\right )-\frac {41 \log (x)}{8}\) |
Input:
Int[(-5 + 2*x^2 + (x + 4*x^2)*Log[(-4*E*x + 5*x*Log[3])/E])/x,x]
Output:
-x - (41*Log[x])/8 + ((1 + 4*x)^2*Log[-(x*(4 - (5*Log[3])/E))])/8
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 = 0.35 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97
method | result | size |
risch | \(\left (2 x^{2}+x \right ) \ln \left (\left (5 x \ln \left (3\right )-4 x \,{\mathrm e}\right ) {\mathrm e}^{-1}\right )-x -5 \ln \left (x \right )\) | \(32\) |
norman | \(\ln \left (\left (5 x \ln \left (3\right )-4 x \,{\mathrm e}\right ) {\mathrm e}^{-1}\right ) x -5 \ln \left (\left (5 x \ln \left (3\right )-4 x \,{\mathrm e}\right ) {\mathrm e}^{-1}\right )-x +2 \ln \left (\left (5 x \ln \left (3\right )-4 x \,{\mathrm e}\right ) {\mathrm e}^{-1}\right ) x^{2}\) | \(65\) |
parallelrisch | \(2 \ln \left (-x \left (4 \,{\mathrm e}-5 \ln \left (3\right )\right ) {\mathrm e}^{-1}\right ) x^{2}+\ln \left (-x \left (4 \,{\mathrm e}-5 \ln \left (3\right )\right ) {\mathrm e}^{-1}\right ) x -x -5 \ln \left (-x \left (4 \,{\mathrm e}-5 \ln \left (3\right )\right ) {\mathrm e}^{-1}\right )\) | \(65\) |
orering | \(\frac {\left (12 x^{4}+12 x^{3}-25\right ) \left (\left (4 x^{2}+x \right ) \ln \left (\left (5 x \ln \left (3\right )-4 x \,{\mathrm e}\right ) {\mathrm e}^{-1}\right )+2 x^{2}-5\right )}{\left (16 x^{3}+10 x^{2}+41 x +5\right ) x}-\frac {\left (16 x^{4}+24 x^{3}-x +100\right ) x \left (\frac {\left (8 x +1\right ) \ln \left (\left (5 x \ln \left (3\right )-4 x \,{\mathrm e}\right ) {\mathrm e}^{-1}\right )+\frac {\left (4 x^{2}+x \right ) \left (5 \ln \left (3\right )-4 \,{\mathrm e}\right )}{5 x \ln \left (3\right )-4 x \,{\mathrm e}}+4 x}{x}-\frac {\left (4 x^{2}+x \right ) \ln \left (\left (5 x \ln \left (3\right )-4 x \,{\mathrm e}\right ) {\mathrm e}^{-1}\right )+2 x^{2}-5}{x^{2}}\right )}{4 \left (8 x +1\right ) \left (2 x^{2}+x +5\right )}\) | \(201\) |
parts | \(\frac {{\mathrm e} \left (-\frac {4 \,{\mathrm e} \left (\frac {\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right )^{2} {\mathrm e}^{-2} x^{2} \ln \left (\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right ) {\mathrm e}^{-1} x \right )}{2}-\frac {\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right )^{2} {\mathrm e}^{-2} x^{2}}{4}\right )}{4 \,{\mathrm e}-5 \ln \left (3\right )}+\frac {4 \,{\mathrm e} \left (\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right ) {\mathrm e}^{-1} x \ln \left (\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right ) {\mathrm e}^{-1} x \right )-\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right ) {\mathrm e}^{-1} x \right )}{4 \,{\mathrm e}-5 \ln \left (3\right )}-\frac {5 \ln \left (3\right ) \left (\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right ) {\mathrm e}^{-1} x \ln \left (\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right ) {\mathrm e}^{-1} x \right )-\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right ) {\mathrm e}^{-1} x \right )}{4 \,{\mathrm e}-5 \ln \left (3\right )}\right )}{5 \ln \left (3\right )-4 \,{\mathrm e}}+x^{2}-5 \ln \left (x \right )\) | \(222\) |
derivativedivides | \(\frac {4 \,{\mathrm e}^{2} \left (\frac {\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right )^{2} {\mathrm e}^{-2} x^{2} \ln \left (\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right ) {\mathrm e}^{-1} x \right )}{2}-\frac {\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right )^{2} {\mathrm e}^{-2} x^{2}}{4}\right )}{\left (4 \,{\mathrm e}-5 \ln \left (3\right )\right )^{2}}-\frac {4 \,{\mathrm e}^{2} \left (\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right ) {\mathrm e}^{-1} x \ln \left (\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right ) {\mathrm e}^{-1} x \right )-\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right ) {\mathrm e}^{-1} x \right )}{\left (4 \,{\mathrm e}-5 \ln \left (3\right )\right )^{2}}+\frac {\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right )^{2} x^{2}}{\left (4 \,{\mathrm e}-5 \ln \left (3\right )\right )^{2}}+\frac {5 \,{\mathrm e} \ln \left (3\right ) \left (\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right ) {\mathrm e}^{-1} x \ln \left (\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right ) {\mathrm e}^{-1} x \right )-\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right ) {\mathrm e}^{-1} x \right )}{\left (4 \,{\mathrm e}-5 \ln \left (3\right )\right )^{2}}-\frac {80 \,{\mathrm e}^{2} \ln \left (\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right ) {\mathrm e}^{-1} x \right )}{\left (4 \,{\mathrm e}-5 \ln \left (3\right )\right )^{2}}+\frac {200 \,{\mathrm e} \ln \left (\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right ) {\mathrm e}^{-1} x \right ) \ln \left (3\right )}{\left (4 \,{\mathrm e}-5 \ln \left (3\right )\right )^{2}}-\frac {125 \ln \left (3\right )^{2} \ln \left (\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right ) {\mathrm e}^{-1} x \right )}{\left (4 \,{\mathrm e}-5 \ln \left (3\right )\right )^{2}}\) | \(331\) |
default | \(\frac {4 \,{\mathrm e}^{2} \left (\frac {\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right )^{2} {\mathrm e}^{-2} x^{2} \ln \left (\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right ) {\mathrm e}^{-1} x \right )}{2}-\frac {\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right )^{2} {\mathrm e}^{-2} x^{2}}{4}\right )}{\left (4 \,{\mathrm e}-5 \ln \left (3\right )\right )^{2}}-\frac {4 \,{\mathrm e}^{2} \left (\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right ) {\mathrm e}^{-1} x \ln \left (\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right ) {\mathrm e}^{-1} x \right )-\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right ) {\mathrm e}^{-1} x \right )}{\left (4 \,{\mathrm e}-5 \ln \left (3\right )\right )^{2}}+\frac {\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right )^{2} x^{2}}{\left (4 \,{\mathrm e}-5 \ln \left (3\right )\right )^{2}}+\frac {5 \,{\mathrm e} \ln \left (3\right ) \left (\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right ) {\mathrm e}^{-1} x \ln \left (\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right ) {\mathrm e}^{-1} x \right )-\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right ) {\mathrm e}^{-1} x \right )}{\left (4 \,{\mathrm e}-5 \ln \left (3\right )\right )^{2}}-\frac {80 \,{\mathrm e}^{2} \ln \left (\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right ) {\mathrm e}^{-1} x \right )}{\left (4 \,{\mathrm e}-5 \ln \left (3\right )\right )^{2}}+\frac {200 \,{\mathrm e} \ln \left (\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right ) {\mathrm e}^{-1} x \right ) \ln \left (3\right )}{\left (4 \,{\mathrm e}-5 \ln \left (3\right )\right )^{2}}-\frac {125 \ln \left (3\right )^{2} \ln \left (\left (5 \ln \left (3\right )-4 \,{\mathrm e}\right ) {\mathrm e}^{-1} x \right )}{\left (4 \,{\mathrm e}-5 \ln \left (3\right )\right )^{2}}\) | \(331\) |
Input:
int(((4*x^2+x)*ln((5*x*ln(3)-4*x*exp(1))/exp(1))+2*x^2-5)/x,x,method=_RETU RNVERBOSE)
Output:
(2*x^2+x)*ln((5*x*ln(3)-4*x*exp(1))*exp(-1))-x-5*ln(x)
Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {-5+2 x^2+\left (x+4 x^2\right ) \log \left (\frac {-4 e x+5 x \log (3)}{e}\right )}{x} \, dx={\left (2 \, x^{2} + x - 5\right )} \log \left (-{\left (4 \, x e - 5 \, x \log \left (3\right )\right )} e^{\left (-1\right )}\right ) - x \] Input:
integrate(((4*x^2+x)*log((5*x*log(3)-4*exp(1)*x)/exp(1))+2*x^2-5)/x,x, alg orithm="fricas")
Output:
(2*x^2 + x - 5)*log(-(4*x*e - 5*x*log(3))*e^(-1)) - x
Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {-5+2 x^2+\left (x+4 x^2\right ) \log \left (\frac {-4 e x+5 x \log (3)}{e}\right )}{x} \, dx=- x + \left (2 x^{2} + x\right ) \log {\left (\frac {- 4 e x + 5 x \log {\left (3 \right )}}{e} \right )} - 5 \log {\left (x \right )} \] Input:
integrate(((4*x**2+x)*ln((5*x*ln(3)-4*exp(1)*x)/exp(1))+2*x**2-5)/x,x)
Output:
-x + (2*x**2 + x)*log((-4*E*x + 5*x*log(3))*exp(-1)) - 5*log(x)
Leaf count of result is larger than twice the leaf count of optimal. 98 vs. \(2 (26) = 52\).
Time = 0.03 (sec) , antiderivative size = 98, normalized size of antiderivative = 2.97 \[ \int \frac {-5+2 x^2+\left (x+4 x^2\right ) \log \left (\frac {-4 e x+5 x \log (3)}{e}\right )}{x} \, dx=\frac {{\left (5 \, e^{\left (-1\right )} \log \left (3\right ) - 4\right )} x^{2} e}{4 \, e - 5 \, \log \left (3\right )} + 2 \, x^{2} \log \left (5 \, x e^{\left (-1\right )} \log \left (3\right ) - 4 \, x\right ) + x^{2} - \frac {5 \, x e^{\left (-1\right )} \log \left (3\right ) - {\left (5 \, x e^{\left (-1\right )} \log \left (3\right ) - 4 \, x\right )} \log \left (5 \, x e^{\left (-1\right )} \log \left (3\right ) - 4 \, x\right ) - 4 \, x}{5 \, e^{\left (-1\right )} \log \left (3\right ) - 4} - 5 \, \log \left (x\right ) \] Input:
integrate(((4*x^2+x)*log((5*x*log(3)-4*exp(1)*x)/exp(1))+2*x^2-5)/x,x, alg orithm="maxima")
Output:
(5*e^(-1)*log(3) - 4)*x^2*e/(4*e - 5*log(3)) + 2*x^2*log(5*x*e^(-1)*log(3) - 4*x) + x^2 - (5*x*e^(-1)*log(3) - (5*x*e^(-1)*log(3) - 4*x)*log(5*x*e^( -1)*log(3) - 4*x) - 4*x)/(5*e^(-1)*log(3) - 4) - 5*log(x)
Time = 0.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.33 \[ \int \frac {-5+2 x^2+\left (x+4 x^2\right ) \log \left (\frac {-4 e x+5 x \log (3)}{e}\right )}{x} \, dx=2 \, x^{2} \log \left (-4 \, x e + 5 \, x \log \left (3\right )\right ) - 2 \, x^{2} + x \log \left (-4 \, x e + 5 \, x \log \left (3\right )\right ) - 2 \, x - 5 \, \log \left (x\right ) \] Input:
integrate(((4*x^2+x)*log((5*x*log(3)-4*exp(1)*x)/exp(1))+2*x^2-5)/x,x, alg orithm="giac")
Output:
2*x^2*log(-4*x*e + 5*x*log(3)) - 2*x^2 + x*log(-4*x*e + 5*x*log(3)) - 2*x - 5*log(x)
Time = 1.61 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18 \[ \int \frac {-5+2 x^2+\left (x+4 x^2\right ) \log \left (\frac {-4 e x+5 x \log (3)}{e}\right )}{x} \, dx=x\,\ln \left (5\,x\,{\mathrm {e}}^{-1}\,\ln \left (3\right )-4\,x\right )-5\,\ln \left (x\right )-x+2\,x^2\,\ln \left (5\,x\,{\mathrm {e}}^{-1}\,\ln \left (3\right )-4\,x\right ) \] Input:
int((log(-exp(-1)*(4*x*exp(1) - 5*x*log(3)))*(x + 4*x^2) + 2*x^2 - 5)/x,x)
Output:
x*log(5*x*exp(-1)*log(3) - 4*x) - 5*log(x) - x + 2*x^2*log(5*x*exp(-1)*log (3) - 4*x)
Time = 0.20 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.36 \[ \int \frac {-5+2 x^2+\left (x+4 x^2\right ) \log \left (\frac {-4 e x+5 x \log (3)}{e}\right )}{x} \, dx=2 \,\mathrm {log}\left (\frac {5 \,\mathrm {log}\left (3\right ) x -4 e x}{e}\right ) x^{2}+\mathrm {log}\left (\frac {5 \,\mathrm {log}\left (3\right ) x -4 e x}{e}\right ) x -5 \,\mathrm {log}\left (x \right )-x \] Input:
int(((4*x^2+x)*log((5*x*log(3)-4*exp(1)*x)/exp(1))+2*x^2-5)/x,x)
Output:
2*log((5*log(3)*x - 4*e*x)/e)*x**2 + log((5*log(3)*x - 4*e*x)/e)*x - 5*log (x) - x