Integrand size = 51, antiderivative size = 24 \[ \int \frac {e^{-4 e^{3 e^9}} \left (10 x+6 x^2+\left (50+50 x+16 x^2\right ) \log (x)+\left (20 x+8 x^2\right ) \log ^2(x)\right )}{x} \, dx=e^{-4 e^{3 e^9}} (x+(5+2 x) \log (x))^2 \] Output:
(x+ln(x)*(5+2*x))^2/exp(2*exp(3*exp(9)))^2
Leaf count is larger than twice the leaf count of optimal. \(57\) vs. \(2(24)=48\).
Time = 0.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.38 \[ \int \frac {e^{-4 e^{3 e^9}} \left (10 x+6 x^2+\left (50+50 x+16 x^2\right ) \log (x)+\left (20 x+8 x^2\right ) \log ^2(x)\right )}{x} \, dx=2 e^{-4 e^{3 e^9}} \left (\frac {x^2}{2}+5 x \log (x)+2 x^2 \log (x)+\frac {25 \log ^2(x)}{2}+10 x \log ^2(x)+2 x^2 \log ^2(x)\right ) \] Input:
Integrate[(10*x + 6*x^2 + (50 + 50*x + 16*x^2)*Log[x] + (20*x + 8*x^2)*Log [x]^2)/(E^(4*E^(3*E^9))*x),x]
Output:
(2*(x^2/2 + 5*x*Log[x] + 2*x^2*Log[x] + (25*Log[x]^2)/2 + 10*x*Log[x]^2 + 2*x^2*Log[x]^2))/E^(4*E^(3*E^9))
Leaf count is larger than twice the leaf count of optimal. \(57\) vs. \(2(24)=48\).
Time = 0.29 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.38, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.078, Rules used = {27, 27, 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.
\(\displaystyle \int \frac {e^{-4 e^{3 e^9}} \left (6 x^2+\left (8 x^2+20 x\right ) \log ^2(x)+\left (16 x^2+50 x+50\right ) \log (x)+10 x\right )}{x} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle e^{-4 e^{3 e^9}} \int \frac {2 \left (3 x^2+5 x+2 \left (2 x^2+5 x\right ) \log ^2(x)+\left (8 x^2+25 x+25\right ) \log (x)\right )}{x}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 e^{-4 e^{3 e^9}} \int \frac {3 x^2+5 x+2 \left (2 x^2+5 x\right ) \log ^2(x)+\left (8 x^2+25 x+25\right ) \log (x)}{x}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle 2 e^{-4 e^{3 e^9}} \int \left (2 (2 x+5) \log ^2(x)+\frac {\left (8 x^2+25 x+25\right ) \log (x)}{x}+3 x+5\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 e^{-4 e^{3 e^9}} \left (\frac {x^2}{2}+2 x^2 \log ^2(x)+2 x^2 \log (x)+10 x \log ^2(x)+\frac {25 \log ^2(x)}{2}+5 x \log (x)\right )\) |
Input:
Int[(10*x + 6*x^2 + (50 + 50*x + 16*x^2)*Log[x] + (20*x + 8*x^2)*Log[x]^2) /(E^(4*E^(3*E^9))*x),x]
Output:
(2*(x^2/2 + 5*x*Log[x] + 2*x^2*Log[x] + (25*Log[x]^2)/2 + 10*x*Log[x]^2 + 2*x^2*Log[x]^2))/E^(4*E^(3*E^9))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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]]
Leaf count of result is larger than twice the leaf count of optimal. \(49\) vs. \(2(23)=46\).
Time = 1.05 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.08
method | result | size |
default | \({\mathrm e}^{-4 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} \left (4 x^{2} \ln \left (x \right )^{2}+4 x^{2} \ln \left (x \right )+x^{2}+20 x \ln \left (x \right )^{2}+10 x \ln \left (x \right )+25 \ln \left (x \right )^{2}\right )\) | \(50\) |
parallelrisch | \({\mathrm e}^{-4 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} \left (4 x^{2} \ln \left (x \right )^{2}+4 x^{2} \ln \left (x \right )+x^{2}+20 x \ln \left (x \right )^{2}+10 x \ln \left (x \right )+25 \ln \left (x \right )^{2}\right )\) | \(50\) |
risch | \({\mathrm e}^{-4 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} \left (4 x^{2}+20 x +25\right ) \ln \left (x \right )^{2}+{\mathrm e}^{-4 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} \left (4 x^{2}+10 x \right ) \ln \left (x \right )+{\mathrm e}^{-4 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} x^{2}\) | \(57\) |
norman | \(\left ({\mathrm e}^{-2 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} x^{2}+25 \,{\mathrm e}^{-2 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} \ln \left (x \right )^{2}+10 \,{\mathrm e}^{-2 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} x \ln \left (x \right )+20 \,{\mathrm e}^{-2 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} x \ln \left (x \right )^{2}+4 \,{\mathrm e}^{-2 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} x^{2} \ln \left (x \right )+4 \,{\mathrm e}^{-2 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} x^{2} \ln \left (x \right )^{2}\right ) {\mathrm e}^{-2 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}}\) | \(99\) |
parts | \(2 \,{\mathrm e}^{-4 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} \left (5 x +\frac {3}{2} x^{2}\right )+4 \,{\mathrm e}^{-4 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} \left (5 x \ln \left (x \right )^{2}-10 x \ln \left (x \right )+10 x +x^{2} \ln \left (x \right )^{2}-x^{2} \ln \left (x \right )+\frac {x^{2}}{2}\right )+2 \,{\mathrm e}^{-4 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}} \left (4 x^{2} \ln \left (x \right )-2 x^{2}+25 x \ln \left (x \right )-25 x +\frac {25 \ln \left (x \right )^{2}}{2}\right )\) | \(110\) |
orering | \(\frac {\left (448 x^{6}+3760 x^{5}-5500 x^{4}-16250 x^{3}-143125 x^{2}-40625 x +15625\right ) \left (\left (8 x^{2}+20 x \right ) \ln \left (x \right )^{2}+\left (16 x^{2}+50 x +50\right ) \ln \left (x \right )+6 x^{2}+10 x \right ) {\mathrm e}^{-4 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}}}{4 \left (128 x^{5}+600 x^{4}-1850 x^{3}-9500 x^{2}-5625 x +3125\right ) x}-\frac {3 \left (128 x^{6}+1600 x^{5}-36250 x^{2}-153125 x -31250\right ) x \left (\frac {\left (\left (16 x +20\right ) \ln \left (x \right )^{2}+\frac {2 \left (8 x^{2}+20 x \right ) \ln \left (x \right )}{x}+\left (32 x +50\right ) \ln \left (x \right )+\frac {16 x^{2}+50 x +50}{x}+12 x +10\right ) {\mathrm e}^{-4 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}}}{x}-\frac {\left (\left (8 x^{2}+20 x \right ) \ln \left (x \right )^{2}+\left (16 x^{2}+50 x +50\right ) \ln \left (x \right )+6 x^{2}+10 x \right ) {\mathrm e}^{-4 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}}}{x^{2}}\right )}{8 \left (128 x^{5}+600 x^{4}-1850 x^{3}-9500 x^{2}-5625 x +3125\right )}+\frac {\left (256 x^{6}+4800 x^{5}-29750 x^{3}+217500 x^{2}+459375 x +62500\right ) x^{2} \left (\frac {\left (16 \ln \left (x \right )^{2}+\frac {4 \left (16 x +20\right ) \ln \left (x \right )}{x}+\frac {16 x^{2}+40 x}{x^{2}}-\frac {2 \left (8 x^{2}+20 x \right ) \ln \left (x \right )}{x^{2}}+32 \ln \left (x \right )+\frac {64 x +100}{x}-\frac {16 x^{2}+50 x +50}{x^{2}}+12\right ) {\mathrm e}^{-4 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}}}{x}-\frac {2 \left (\left (16 x +20\right ) \ln \left (x \right )^{2}+\frac {2 \left (8 x^{2}+20 x \right ) \ln \left (x \right )}{x}+\left (32 x +50\right ) \ln \left (x \right )+\frac {16 x^{2}+50 x +50}{x}+12 x +10\right ) {\mathrm e}^{-4 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}}}{x^{2}}+\frac {2 \left (\left (8 x^{2}+20 x \right ) \ln \left (x \right )^{2}+\left (16 x^{2}+50 x +50\right ) \ln \left (x \right )+6 x^{2}+10 x \right ) {\mathrm e}^{-4 \,{\mathrm e}^{3 \,{\mathrm e}^{9}}}}{x^{3}}\right )}{16 \left (5+x \right ) \left (32 x^{3}-50 x^{2}-350 x +125\right ) \left (4 x +5\right )}\) | \(551\) |
Input:
int(((8*x^2+20*x)*ln(x)^2+(16*x^2+50*x+50)*ln(x)+6*x^2+10*x)/x/exp(2*exp(3 *exp(9)))^2,x,method=_RETURNVERBOSE)
Output:
1/exp(2*exp(3*exp(9)))^2*(4*x^2*ln(x)^2+4*x^2*ln(x)+x^2+20*x*ln(x)^2+10*x* ln(x)+25*ln(x)^2)
Time = 0.10 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.71 \[ \int \frac {e^{-4 e^{3 e^9}} \left (10 x+6 x^2+\left (50+50 x+16 x^2\right ) \log (x)+\left (20 x+8 x^2\right ) \log ^2(x)\right )}{x} \, dx={\left ({\left (4 \, x^{2} + 20 \, x + 25\right )} \log \left (x\right )^{2} + x^{2} + 2 \, {\left (2 \, x^{2} + 5 \, x\right )} \log \left (x\right )\right )} e^{\left (-4 \, e^{\left (3 \, e^{9}\right )}\right )} \] Input:
integrate(((8*x^2+20*x)*log(x)^2+(16*x^2+50*x+50)*log(x)+6*x^2+10*x)/x/exp (2*exp(3*exp(9)))^2,x, algorithm="fricas")
Output:
((4*x^2 + 20*x + 25)*log(x)^2 + x^2 + 2*(2*x^2 + 5*x)*log(x))*e^(-4*e^(3*e ^9))
Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (20) = 40\).
Time = 0.12 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.50 \[ \int \frac {e^{-4 e^{3 e^9}} \left (10 x+6 x^2+\left (50+50 x+16 x^2\right ) \log (x)+\left (20 x+8 x^2\right ) \log ^2(x)\right )}{x} \, dx=\frac {x^{2}}{e^{4 e^{3 e^{9}}}} + \frac {\left (4 x^{2} + 10 x\right ) \log {\left (x \right )}}{e^{4 e^{3 e^{9}}}} + \frac {\left (4 x^{2} + 20 x + 25\right ) \log {\left (x \right )}^{2}}{e^{4 e^{3 e^{9}}}} \] Input:
integrate(((8*x**2+20*x)*ln(x)**2+(16*x**2+50*x+50)*ln(x)+6*x**2+10*x)/x/e xp(2*exp(3*exp(9)))**2,x)
Output:
x**2*exp(-4*exp(3*exp(9))) + (4*x**2 + 10*x)*exp(-4*exp(3*exp(9)))*log(x) + (4*x**2 + 20*x + 25)*exp(-4*exp(3*exp(9)))*log(x)**2
Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (21) = 42\).
Time = 0.05 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.75 \[ \int \frac {e^{-4 e^{3 e^9}} \left (10 x+6 x^2+\left (50+50 x+16 x^2\right ) \log (x)+\left (20 x+8 x^2\right ) \log ^2(x)\right )}{x} \, dx={\left (2 \, {\left (2 \, \log \left (x\right )^{2} - 2 \, \log \left (x\right ) + 1\right )} x^{2} + 8 \, x^{2} \log \left (x\right ) + 20 \, {\left (\log \left (x\right )^{2} - 2 \, \log \left (x\right ) + 2\right )} x - x^{2} + 50 \, x \log \left (x\right ) + 25 \, \log \left (x\right )^{2} - 40 \, x\right )} e^{\left (-4 \, e^{\left (3 \, e^{9}\right )}\right )} \] Input:
integrate(((8*x^2+20*x)*log(x)^2+(16*x^2+50*x+50)*log(x)+6*x^2+10*x)/x/exp (2*exp(3*exp(9)))^2,x, algorithm="maxima")
Output:
(2*(2*log(x)^2 - 2*log(x) + 1)*x^2 + 8*x^2*log(x) + 20*(log(x)^2 - 2*log(x ) + 2)*x - x^2 + 50*x*log(x) + 25*log(x)^2 - 40*x)*e^(-4*e^(3*e^9))
Time = 0.12 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.71 \[ \int \frac {e^{-4 e^{3 e^9}} \left (10 x+6 x^2+\left (50+50 x+16 x^2\right ) \log (x)+\left (20 x+8 x^2\right ) \log ^2(x)\right )}{x} \, dx={\left ({\left (4 \, x^{2} + 20 \, x + 25\right )} \log \left (x\right )^{2} + x^{2} + 2 \, {\left (2 \, x^{2} + 5 \, x\right )} \log \left (x\right )\right )} e^{\left (-4 \, e^{\left (3 \, e^{9}\right )}\right )} \] Input:
integrate(((8*x^2+20*x)*log(x)^2+(16*x^2+50*x+50)*log(x)+6*x^2+10*x)/x/exp (2*exp(3*exp(9)))^2,x, algorithm="giac")
Output:
((4*x^2 + 20*x + 25)*log(x)^2 + x^2 + 2*(2*x^2 + 5*x)*log(x))*e^(-4*e^(3*e ^9))
Time = 2.85 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {e^{-4 e^{3 e^9}} \left (10 x+6 x^2+\left (50+50 x+16 x^2\right ) \log (x)+\left (20 x+8 x^2\right ) \log ^2(x)\right )}{x} \, dx={\mathrm {e}}^{-4\,{\mathrm {e}}^{3\,{\mathrm {e}}^9}}\,{\left (x+5\,\ln \left (x\right )+2\,x\,\ln \left (x\right )\right )}^2 \] Input:
int((exp(-4*exp(3*exp(9)))*(10*x + log(x)^2*(20*x + 8*x^2) + log(x)*(50*x + 16*x^2 + 50) + 6*x^2))/x,x)
Output:
exp(-4*exp(3*exp(9)))*(x + 5*log(x) + 2*x*log(x))^2
Time = 0.16 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.17 \[ \int \frac {e^{-4 e^{3 e^9}} \left (10 x+6 x^2+\left (50+50 x+16 x^2\right ) \log (x)+\left (20 x+8 x^2\right ) \log ^2(x)\right )}{x} \, dx=\frac {4 \mathrm {log}\left (x \right )^{2} x^{2}+20 \mathrm {log}\left (x \right )^{2} x +25 \mathrm {log}\left (x \right )^{2}+4 \,\mathrm {log}\left (x \right ) x^{2}+10 \,\mathrm {log}\left (x \right ) x +x^{2}}{e^{4 e^{3 e^{9}}}} \] Input:
int(((8*x^2+20*x)*log(x)^2+(16*x^2+50*x+50)*log(x)+6*x^2+10*x)/x/exp(2*exp (3*exp(9)))^2,x)
Output:
(4*log(x)**2*x**2 + 20*log(x)**2*x + 25*log(x)**2 + 4*log(x)*x**2 + 10*log (x)*x + x**2)/e**(4*e**(3*e**9))