Integrand size = 36, antiderivative size = 22 \[ \int \frac {1-x+e x+2 x^2-3 x^3-x \log \left (5 x+x \log ^2(5)\right )}{x} \, dx=(1-x) \left (-e+x^2+\log \left (x \left (5+\log ^2(5)\right )\right )\right ) \] Output:
(1-x)*(ln((ln(5)^2+5)*x)+x^2-exp(1))
Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {1-x+e x+2 x^2-3 x^3-x \log \left (5 x+x \log ^2(5)\right )}{x} \, dx=e x+x^2-x^3+\log (x)-x \log \left (x \left (5+\log ^2(5)\right )\right ) \] Input:
Integrate[(1 - x + E*x + 2*x^2 - 3*x^3 - x*Log[5*x + x*Log[5]^2])/x,x]
Output:
E*x + x^2 - x^3 + Log[x] - x*Log[x*(5 + Log[5]^2)]
Time = 0.20 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6, 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.
\(\displaystyle \int \frac {-3 x^3+2 x^2+e x-x-x \log \left (5 x+x \log ^2(5)\right )+1}{x} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {-3 x^3+2 x^2+(e-1) x-x \log \left (5 x+x \log ^2(5)\right )+1}{x}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (\frac {-3 x^3+2 x^2-(1-e) x+1}{x}-\log \left (x \left (5+\log ^2(5)\right )\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -x^3+x^2-(1-e) x+x-x \log \left (x \left (5+\log ^2(5)\right )\right )+\log (x)\) |
Input:
Int[(1 - x + E*x + 2*x^2 - 3*x^3 - x*Log[5*x + x*Log[5]^2])/x,x]
Output:
x - (1 - E)*x + x^2 - x^3 + Log[x] - x*Log[x*(5 + Log[5]^2)]
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, 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 = 0.21 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36
method | result | size |
risch | \(-x \ln \left (x \ln \left (5\right )^{2}+5 x \right )-x^{3}+x \,{\mathrm e}+x^{2}+\ln \left (x \right )\) | \(30\) |
norman | \(x^{2}+x \,{\mathrm e}+\ln \left (x \ln \left (5\right )^{2}+5 x \right )-x^{3}-x \ln \left (x \ln \left (5\right )^{2}+5 x \right )\) | \(39\) |
parallelrisch | \(x^{2}+x \,{\mathrm e}+\ln \left (x \ln \left (5\right )^{2}+5 x \right )-x^{3}-x \ln \left (x \ln \left (5\right )^{2}+5 x \right )\) | \(39\) |
parts | \(-x^{3}+x^{2}+x \,{\mathrm e}-x +\ln \left (x \right )-\frac {\left (\ln \left (5\right )^{2}+5\right ) x \ln \left (\left (\ln \left (5\right )^{2}+5\right ) x \right )-\left (\ln \left (5\right )^{2}+5\right ) x}{\ln \left (5\right )^{2}+5}\) | \(56\) |
orering | \(\frac {\left (-1+x \right ) \left (-x \ln \left (x \ln \left (5\right )^{2}+5 x \right )+x \,{\mathrm e}-3 x^{3}+2 x^{2}-x +1\right )}{x}-\frac {\left (2 x^{4}-4 x^{3}+3 x^{2}+1\right ) x \left (\frac {-\ln \left (x \ln \left (5\right )^{2}+5 x \right )-\frac {x \left (\ln \left (5\right )^{2}+5\right )}{x \ln \left (5\right )^{2}+5 x}+{\mathrm e}-9 x^{2}+4 x -1}{x}-\frac {-x \ln \left (x \ln \left (5\right )^{2}+5 x \right )+x \,{\mathrm e}-3 x^{3}+2 x^{2}-x +1}{x^{2}}\right )}{6 x^{3}-2 x^{2}+x +1}\) | \(166\) |
derivativedivides | \(\frac {\ln \left (5\right )^{6} \ln \left (\left (\ln \left (5\right )^{2}+5\right ) x \right )}{\left (\ln \left (5\right )^{2}+5\right )^{3}}-\frac {\ln \left (5\right )^{4} \left (\left (\ln \left (5\right )^{2}+5\right ) x \ln \left (\left (\ln \left (5\right )^{2}+5\right ) x \right )-\left (\ln \left (5\right )^{2}+5\right ) x \right )}{\left (\ln \left (5\right )^{2}+5\right )^{3}}+\frac {\ln \left (5\right )^{4} {\mathrm e} x}{\left (\ln \left (5\right )^{2}+5\right )^{2}}-\frac {\ln \left (5\right )^{4} x}{\left (\ln \left (5\right )^{2}+5\right )^{2}}+\frac {15 \ln \left (5\right )^{4} \ln \left (\left (\ln \left (5\right )^{2}+5\right ) x \right )}{\left (\ln \left (5\right )^{2}+5\right )^{3}}-\frac {10 \ln \left (5\right )^{2} \left (\left (\ln \left (5\right )^{2}+5\right ) x \ln \left (\left (\ln \left (5\right )^{2}+5\right ) x \right )-\left (\ln \left (5\right )^{2}+5\right ) x \right )}{\left (\ln \left (5\right )^{2}+5\right )^{3}}+\frac {10 \ln \left (5\right )^{2} {\mathrm e} x}{\left (\ln \left (5\right )^{2}+5\right )^{2}}+\frac {\ln \left (5\right )^{2} x^{2}}{\ln \left (5\right )^{2}+5}-\frac {10 x \ln \left (5\right )^{2}}{\left (\ln \left (5\right )^{2}+5\right )^{2}}-x^{3}+\frac {75 \ln \left (5\right )^{2} \ln \left (\left (\ln \left (5\right )^{2}+5\right ) x \right )}{\left (\ln \left (5\right )^{2}+5\right )^{3}}-\frac {25 \left (\left (\ln \left (5\right )^{2}+5\right ) x \ln \left (\left (\ln \left (5\right )^{2}+5\right ) x \right )-\left (\ln \left (5\right )^{2}+5\right ) x \right )}{\left (\ln \left (5\right )^{2}+5\right )^{3}}+\frac {25 \,{\mathrm e} x}{\left (\ln \left (5\right )^{2}+5\right )^{2}}+\frac {5 x^{2}}{\ln \left (5\right )^{2}+5}-\frac {25 x}{\left (\ln \left (5\right )^{2}+5\right )^{2}}+\frac {125 \ln \left (\left (\ln \left (5\right )^{2}+5\right ) x \right )}{\left (\ln \left (5\right )^{2}+5\right )^{3}}\) | \(329\) |
default | \(\frac {\ln \left (5\right )^{6} \ln \left (\left (\ln \left (5\right )^{2}+5\right ) x \right )}{\left (\ln \left (5\right )^{2}+5\right )^{3}}-\frac {\ln \left (5\right )^{4} \left (\left (\ln \left (5\right )^{2}+5\right ) x \ln \left (\left (\ln \left (5\right )^{2}+5\right ) x \right )-\left (\ln \left (5\right )^{2}+5\right ) x \right )}{\left (\ln \left (5\right )^{2}+5\right )^{3}}+\frac {\ln \left (5\right )^{4} {\mathrm e} x}{\left (\ln \left (5\right )^{2}+5\right )^{2}}-\frac {\ln \left (5\right )^{4} x}{\left (\ln \left (5\right )^{2}+5\right )^{2}}+\frac {15 \ln \left (5\right )^{4} \ln \left (\left (\ln \left (5\right )^{2}+5\right ) x \right )}{\left (\ln \left (5\right )^{2}+5\right )^{3}}-\frac {10 \ln \left (5\right )^{2} \left (\left (\ln \left (5\right )^{2}+5\right ) x \ln \left (\left (\ln \left (5\right )^{2}+5\right ) x \right )-\left (\ln \left (5\right )^{2}+5\right ) x \right )}{\left (\ln \left (5\right )^{2}+5\right )^{3}}+\frac {10 \ln \left (5\right )^{2} {\mathrm e} x}{\left (\ln \left (5\right )^{2}+5\right )^{2}}+\frac {\ln \left (5\right )^{2} x^{2}}{\ln \left (5\right )^{2}+5}-\frac {10 x \ln \left (5\right )^{2}}{\left (\ln \left (5\right )^{2}+5\right )^{2}}-x^{3}+\frac {75 \ln \left (5\right )^{2} \ln \left (\left (\ln \left (5\right )^{2}+5\right ) x \right )}{\left (\ln \left (5\right )^{2}+5\right )^{3}}-\frac {25 \left (\left (\ln \left (5\right )^{2}+5\right ) x \ln \left (\left (\ln \left (5\right )^{2}+5\right ) x \right )-\left (\ln \left (5\right )^{2}+5\right ) x \right )}{\left (\ln \left (5\right )^{2}+5\right )^{3}}+\frac {25 \,{\mathrm e} x}{\left (\ln \left (5\right )^{2}+5\right )^{2}}+\frac {5 x^{2}}{\ln \left (5\right )^{2}+5}-\frac {25 x}{\left (\ln \left (5\right )^{2}+5\right )^{2}}+\frac {125 \ln \left (\left (\ln \left (5\right )^{2}+5\right ) x \right )}{\left (\ln \left (5\right )^{2}+5\right )^{3}}\) | \(329\) |
Input:
int((-x*ln(x*ln(5)^2+5*x)+x*exp(1)-3*x^3+2*x^2-x+1)/x,x,method=_RETURNVERB OSE)
Output:
-x*ln(x*ln(5)^2+5*x)-x^3+x*exp(1)+x^2+ln(x)
Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {1-x+e x+2 x^2-3 x^3-x \log \left (5 x+x \log ^2(5)\right )}{x} \, dx=-x^{3} + x^{2} + x e - {\left (x - 1\right )} \log \left (x \log \left (5\right )^{2} + 5 \, x\right ) \] Input:
integrate((-x*log(x*log(5)^2+5*x)+exp(1)*x-3*x^3+2*x^2-x+1)/x,x, algorithm ="fricas")
Output:
-x^3 + x^2 + x*e - (x - 1)*log(x*log(5)^2 + 5*x)
Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {1-x+e x+2 x^2-3 x^3-x \log \left (5 x+x \log ^2(5)\right )}{x} \, dx=- x^{3} + x^{2} - x \log {\left (x \log {\left (5 \right )}^{2} + 5 x \right )} + e x + \log {\left (x \right )} \] Input:
integrate((-x*ln(x*ln(5)**2+5*x)+exp(1)*x-3*x**3+2*x**2-x+1)/x,x)
Output:
-x**3 + x**2 - x*log(x*log(5)**2 + 5*x) + E*x + log(x)
Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (22) = 44\).
Time = 0.03 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.73 \[ \int \frac {1-x+e x+2 x^2-3 x^3-x \log \left (5 x+x \log ^2(5)\right )}{x} \, dx=-x^{3} + x^{2} + x e - x + \frac {x \log \left (5\right )^{2} - {\left (x \log \left (5\right )^{2} + 5 \, x\right )} \log \left (x \log \left (5\right )^{2} + 5 \, x\right ) + 5 \, x}{\log \left (5\right )^{2} + 5} + \log \left (x\right ) \] Input:
integrate((-x*log(x*log(5)^2+5*x)+exp(1)*x-3*x^3+2*x^2-x+1)/x,x, algorithm ="maxima")
Output:
-x^3 + x^2 + x*e - x + (x*log(5)^2 - (x*log(5)^2 + 5*x)*log(x*log(5)^2 + 5 *x) + 5*x)/(log(5)^2 + 5) + log(x)
Time = 0.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {1-x+e x+2 x^2-3 x^3-x \log \left (5 x+x \log ^2(5)\right )}{x} \, dx=-x^{3} + x^{2} + x e - x \log \left (x \log \left (5\right )^{2} + 5 \, x\right ) + \log \left (x\right ) \] Input:
integrate((-x*log(x*log(5)^2+5*x)+exp(1)*x-3*x^3+2*x^2-x+1)/x,x, algorithm ="giac")
Output:
-x^3 + x^2 + x*e - x*log(x*log(5)^2 + 5*x) + log(x)
Time = 3.76 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {1-x+e x+2 x^2-3 x^3-x \log \left (5 x+x \log ^2(5)\right )}{x} \, dx=\ln \left (x\right )-x\,\ln \left (5\,x+x\,{\ln \left (5\right )}^2\right )+x\,\mathrm {e}+x^2-x^3 \] Input:
int(-(x + x*log(5*x + x*log(5)^2) - x*exp(1) - 2*x^2 + 3*x^3 - 1)/x,x)
Output:
log(x) - x*log(5*x + x*log(5)^2) + x*exp(1) + x^2 - x^3
Time = 0.20 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {1-x+e x+2 x^2-3 x^3-x \log \left (5 x+x \log ^2(5)\right )}{x} \, dx=-\mathrm {log}\left (\mathrm {log}\left (5\right )^{2} x +5 x \right ) x +\mathrm {log}\left (x \right )+e x -x^{3}+x^{2} \] Input:
int((-x*log(x*log(5)^2+5*x)+exp(1)*x-3*x^3+2*x^2-x+1)/x,x)
Output:
- log(log(5)**2*x + 5*x)*x + log(x) + e*x - x**3 + x**2