Integrand size = 120, antiderivative size = 24 \[ \int \frac {e^{-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (-15-5 \log (x)+e^{\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (4 x^2+6 x^3+\left (2 x^2+3 x^3\right ) \log (x)\right ) \log ^2(2 x+x \log (x))\right )}{(2 x+x \log (x)) \log ^2(2 x+x \log (x))} \, dx=-25+5 e^{-5+\frac {1}{\log (x (2+\log (x)))}}+x \left (x+x^2\right ) \] Output:
x*(x^2+x)+5/exp(5-1/ln((ln(x)+2)*x))-25
Time = 0.06 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {e^{-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (-15-5 \log (x)+e^{\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (4 x^2+6 x^3+\left (2 x^2+3 x^3\right ) \log (x)\right ) \log ^2(2 x+x \log (x))\right )}{(2 x+x \log (x)) \log ^2(2 x+x \log (x))} \, dx=\frac {5 e^{\frac {1}{\log (x (2+\log (x)))}}+e^5 x^2 (1+x)}{e^5} \] Input:
Integrate[(-15 - 5*Log[x] + E^((-1 + 5*Log[2*x + x*Log[x]])/Log[2*x + x*Lo g[x]])*(4*x^2 + 6*x^3 + (2*x^2 + 3*x^3)*Log[x])*Log[2*x + x*Log[x]]^2)/(E^ ((-1 + 5*Log[2*x + x*Log[x]])/Log[2*x + x*Log[x]])*(2*x + x*Log[x])*Log[2* x + x*Log[x]]^2),x]
Output:
(5*E^Log[x*(2 + Log[x])]^(-1) + E^5*x^2*(1 + x))/E^5
Time = 2.54 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {3041, 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.
\(\displaystyle \int \frac {\exp \left (-\frac {5 \log (2 x+x \log (x))-1}{\log (2 x+x \log (x))}\right ) \left (\left (6 x^3+4 x^2+\left (3 x^3+2 x^2\right ) \log (x)\right ) \log ^2(2 x+x \log (x)) \exp \left (\frac {5 \log (2 x+x \log (x))-1}{\log (2 x+x \log (x))}\right )-5 \log (x)-15\right )}{(2 x+x \log (x)) \log ^2(2 x+x \log (x))} \, dx\) |
\(\Big \downarrow \) 3041 |
\(\displaystyle \int \frac {\exp \left (-\frac {5 \log (2 x+x \log (x))-1}{\log (2 x+x \log (x))}\right ) \left (\left (6 x^3+4 x^2+\left (3 x^3+2 x^2\right ) \log (x)\right ) \log ^2(2 x+x \log (x)) \exp \left (\frac {5 \log (2 x+x \log (x))-1}{\log (2 x+x \log (x))}\right )-5 \log (x)-15\right )}{x (\log (x)+2) \log ^2(2 x+x \log (x))}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {5 (-\log (x)-3) \exp \left (-\frac {5 \log (2 x+x \log (x))-1}{\log (2 x+x \log (x))}\right )}{x (\log (x)+2) \log ^2(2 x+x \log (x))}+x (3 x+2) \exp \left (-\frac {5 \log (2 x+x \log (x))-1}{\log (2 x+x \log (x))}-\frac {1}{\log (x (\log (x)+2))}+5\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x^3+x^2+5 e^{\frac {1}{\log (x (\log (x)+2))}-5}\) |
Input:
Int[(-15 - 5*Log[x] + E^((-1 + 5*Log[2*x + x*Log[x]])/Log[2*x + x*Log[x]]) *(4*x^2 + 6*x^3 + (2*x^2 + 3*x^3)*Log[x])*Log[2*x + x*Log[x]]^2)/(E^((-1 + 5*Log[2*x + x*Log[x]])/Log[2*x + x*Log[x]])*(2*x + x*Log[x])*Log[2*x + x* Log[x]]^2),x]
Output:
5*E^(-5 + Log[x*(2 + Log[x])]^(-1)) + x^2 + x^3
Int[(u_.)*((a_.)*(x_)^(m_.) + Log[(c_.)*(x_)^(n_.)]^(q_.)*(b_.)*(x_)^(r_.)) ^(p_.), x_Symbol] :> Int[u*x^(p*r)*(a*x^(m - r) + b*Log[c*x^n]^q)^p, x] /; FreeQ[{a, b, c, m, n, p, q, r}, x] && IntegerQ[p]
Leaf count of result is larger than twice the leaf count of optimal. \(113\) vs. \(2(27)=54\).
Time = 3.49 (sec) , antiderivative size = 114, normalized size of antiderivative = 4.75
method | result | size |
parallelrisch | \(-\frac {\left (-4 \,{\mathrm e}^{\frac {5 \ln \left (\left (\ln \left (x \right )+2\right ) x \right )-1}{\ln \left (\left (\ln \left (x \right )+2\right ) x \right )}} \ln \left (\left (\ln \left (x \right )+2\right ) x \right ) x^{3}-4 x^{2} \ln \left (\left (\ln \left (x \right )+2\right ) x \right ) {\mathrm e}^{\frac {5 \ln \left (\left (\ln \left (x \right )+2\right ) x \right )-1}{\ln \left (\left (\ln \left (x \right )+2\right ) x \right )}}-20 \ln \left (\left (\ln \left (x \right )+2\right ) x \right )\right ) {\mathrm e}^{-\frac {5 \ln \left (\left (\ln \left (x \right )+2\right ) x \right )-1}{\ln \left (\left (\ln \left (x \right )+2\right ) x \right )}}}{4 \ln \left (\left (\ln \left (x \right )+2\right ) x \right )}\) | \(114\) |
risch | \(x^{3}+x^{2}+5 \,{\mathrm e}^{-\frac {-5 i \pi \operatorname {csgn}\left (i x \left (\ln \left (x \right )+2\right )\right )^{3}+5 i \pi \operatorname {csgn}\left (i x \left (\ln \left (x \right )+2\right )\right )^{2} \operatorname {csgn}\left (i x \right )+5 i \pi \operatorname {csgn}\left (i x \left (\ln \left (x \right )+2\right )\right )^{2} \operatorname {csgn}\left (i \left (\ln \left (x \right )+2\right )\right )-5 i \pi \,\operatorname {csgn}\left (i x \left (\ln \left (x \right )+2\right )\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left (\ln \left (x \right )+2\right )\right )+10 \ln \left (x \right )+10 \ln \left (\ln \left (x \right )+2\right )-2}{-i \pi \operatorname {csgn}\left (i x \left (\ln \left (x \right )+2\right )\right )^{3}+i \pi \operatorname {csgn}\left (i x \left (\ln \left (x \right )+2\right )\right )^{2} \operatorname {csgn}\left (i x \right )+i \pi \operatorname {csgn}\left (i x \left (\ln \left (x \right )+2\right )\right )^{2} \operatorname {csgn}\left (i \left (\ln \left (x \right )+2\right )\right )-i \pi \,\operatorname {csgn}\left (i x \left (\ln \left (x \right )+2\right )\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left (\ln \left (x \right )+2\right )\right )+2 \ln \left (x \right )+2 \ln \left (\ln \left (x \right )+2\right )}}\) | \(208\) |
Input:
int((((3*x^3+2*x^2)*ln(x)+6*x^3+4*x^2)*ln(x*ln(x)+2*x)^2*exp((5*ln(x*ln(x) +2*x)-1)/ln(x*ln(x)+2*x))-5*ln(x)-15)/(x*ln(x)+2*x)/ln(x*ln(x)+2*x)^2/exp( (5*ln(x*ln(x)+2*x)-1)/ln(x*ln(x)+2*x)),x,method=_RETURNVERBOSE)
Output:
-1/4*(-4*exp((5*ln((ln(x)+2)*x)-1)/ln((ln(x)+2)*x))*ln((ln(x)+2)*x)*x^3-4* x^2*ln((ln(x)+2)*x)*exp((5*ln((ln(x)+2)*x)-1)/ln((ln(x)+2)*x))-20*ln((ln(x )+2)*x))/ln((ln(x)+2)*x)/exp((5*ln((ln(x)+2)*x)-1)/ln((ln(x)+2)*x))
Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (23) = 46\).
Time = 0.09 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.67 \[ \int \frac {e^{-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (-15-5 \log (x)+e^{\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (4 x^2+6 x^3+\left (2 x^2+3 x^3\right ) \log (x)\right ) \log ^2(2 x+x \log (x))\right )}{(2 x+x \log (x)) \log ^2(2 x+x \log (x))} \, dx={\left ({\left (x^{3} + x^{2}\right )} e^{\left (\frac {5 \, \log \left (x \log \left (x\right ) + 2 \, x\right ) - 1}{\log \left (x \log \left (x\right ) + 2 \, x\right )}\right )} + 5\right )} e^{\left (-\frac {5 \, \log \left (x \log \left (x\right ) + 2 \, x\right ) - 1}{\log \left (x \log \left (x\right ) + 2 \, x\right )}\right )} \] Input:
integrate((((3*x^3+2*x^2)*log(x)+6*x^3+4*x^2)*log(x*log(x)+2*x)^2*exp((5*l og(x*log(x)+2*x)-1)/log(x*log(x)+2*x))-5*log(x)-15)/(x*log(x)+2*x)/log(x*l og(x)+2*x)^2/exp((5*log(x*log(x)+2*x)-1)/log(x*log(x)+2*x)),x, algorithm=" fricas")
Output:
((x^3 + x^2)*e^((5*log(x*log(x) + 2*x) - 1)/log(x*log(x) + 2*x)) + 5)*e^(- (5*log(x*log(x) + 2*x) - 1)/log(x*log(x) + 2*x))
Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {e^{-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (-15-5 \log (x)+e^{\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (4 x^2+6 x^3+\left (2 x^2+3 x^3\right ) \log (x)\right ) \log ^2(2 x+x \log (x))\right )}{(2 x+x \log (x)) \log ^2(2 x+x \log (x))} \, dx=x^{3} + x^{2} + 5 e^{- \frac {5 \log {\left (x \log {\left (x \right )} + 2 x \right )} - 1}{\log {\left (x \log {\left (x \right )} + 2 x \right )}}} \] Input:
integrate((((3*x**3+2*x**2)*ln(x)+6*x**3+4*x**2)*ln(x*ln(x)+2*x)**2*exp((5 *ln(x*ln(x)+2*x)-1)/ln(x*ln(x)+2*x))-5*ln(x)-15)/(x*ln(x)+2*x)/ln(x*ln(x)+ 2*x)**2/exp((5*ln(x*ln(x)+2*x)-1)/ln(x*ln(x)+2*x)),x)
Output:
x**3 + x**2 + 5*exp(-(5*log(x*log(x) + 2*x) - 1)/log(x*log(x) + 2*x))
Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (23) = 46\).
Time = 0.31 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.46 \[ \int \frac {e^{-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (-15-5 \log (x)+e^{\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (4 x^2+6 x^3+\left (2 x^2+3 x^3\right ) \log (x)\right ) \log ^2(2 x+x \log (x))\right )}{(2 x+x \log (x)) \log ^2(2 x+x \log (x))} \, dx=x^{3} + x^{2} + \frac {5 \, e^{\left (\frac {1}{\log \left (x\right ) + \log \left (\log \left (x\right ) + 2\right )}\right )} \log \left (x\right )}{e^{5} \log \left (x\right ) + 3 \, e^{5}} + \frac {15 \, e^{\left (\frac {1}{\log \left (x\right ) + \log \left (\log \left (x\right ) + 2\right )}\right )}}{e^{5} \log \left (x\right ) + 3 \, e^{5}} \] Input:
integrate((((3*x^3+2*x^2)*log(x)+6*x^3+4*x^2)*log(x*log(x)+2*x)^2*exp((5*l og(x*log(x)+2*x)-1)/log(x*log(x)+2*x))-5*log(x)-15)/(x*log(x)+2*x)/log(x*l og(x)+2*x)^2/exp((5*log(x*log(x)+2*x)-1)/log(x*log(x)+2*x)),x, algorithm=" maxima")
Output:
x^3 + x^2 + 5*e^(1/(log(x) + log(log(x) + 2)))*log(x)/(e^5*log(x) + 3*e^5) + 15*e^(1/(log(x) + log(log(x) + 2)))/(e^5*log(x) + 3*e^5)
Time = 0.58 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (-15-5 \log (x)+e^{\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (4 x^2+6 x^3+\left (2 x^2+3 x^3\right ) \log (x)\right ) \log ^2(2 x+x \log (x))\right )}{(2 x+x \log (x)) \log ^2(2 x+x \log (x))} \, dx=x^{3} + x^{2} + 5 \, e^{\left (\frac {1}{\log \left (x \log \left (x\right ) + 2 \, x\right )} - 5\right )} \] Input:
integrate((((3*x^3+2*x^2)*log(x)+6*x^3+4*x^2)*log(x*log(x)+2*x)^2*exp((5*l og(x*log(x)+2*x)-1)/log(x*log(x)+2*x))-5*log(x)-15)/(x*log(x)+2*x)/log(x*l og(x)+2*x)^2/exp((5*log(x*log(x)+2*x)-1)/log(x*log(x)+2*x)),x, algorithm=" giac")
Output:
x^3 + x^2 + 5*e^(1/log(x*log(x) + 2*x) - 5)
Time = 7.64 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (-15-5 \log (x)+e^{\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (4 x^2+6 x^3+\left (2 x^2+3 x^3\right ) \log (x)\right ) \log ^2(2 x+x \log (x))\right )}{(2 x+x \log (x)) \log ^2(2 x+x \log (x))} \, dx=x^2+x^3+5\,{\mathrm {e}}^{\frac {1}{\ln \left (2\,x+x\,\ln \left (x\right )\right )}}\,{\mathrm {e}}^{-5} \] Input:
int(-(exp(-(5*log(2*x + x*log(x)) - 1)/log(2*x + x*log(x)))*(5*log(x) - lo g(2*x + x*log(x))^2*exp((5*log(2*x + x*log(x)) - 1)/log(2*x + x*log(x)))*( log(x)*(2*x^2 + 3*x^3) + 4*x^2 + 6*x^3) + 15))/(log(2*x + x*log(x))^2*(2*x + x*log(x))),x)
Output:
x^2 + x^3 + 5*exp(1/log(2*x + x*log(x)))*exp(-5)
Time = 0.19 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {e^{-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (-15-5 \log (x)+e^{\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (4 x^2+6 x^3+\left (2 x^2+3 x^3\right ) \log (x)\right ) \log ^2(2 x+x \log (x))\right )}{(2 x+x \log (x)) \log ^2(2 x+x \log (x))} \, dx=\frac {5 e^{\frac {1}{\mathrm {log}\left (\mathrm {log}\left (x \right ) x +2 x \right )}}+e^{5} x^{3}+e^{5} x^{2}}{e^{5}} \] Input:
int((((3*x^3+2*x^2)*log(x)+6*x^3+4*x^2)*log(x*log(x)+2*x)^2*exp((5*log(x*l og(x)+2*x)-1)/log(x*log(x)+2*x))-5*log(x)-15)/(x*log(x)+2*x)/log(x*log(x)+ 2*x)^2/exp((5*log(x*log(x)+2*x)-1)/log(x*log(x)+2*x)),x)
Output:
(5*e**(1/log(log(x)*x + 2*x)) + e**5*x**3 + e**5*x**2)/e**5