Integrand size = 83, antiderivative size = 33 \[ \int \frac {-2+\left (5 x-5 e^x x-40 x^2+\left (x-e^x x-8 x^2\right ) \log \left (\frac {x^2}{16}\right )\right ) \log ^2\left (5+\log \left (\frac {x^2}{16}\right )\right )}{\left (5 x+x \log \left (\frac {x^2}{16}\right )\right ) \log ^2\left (5+\log \left (\frac {x^2}{16}\right )\right )} \, dx=-e^x+x-4 x^2+\frac {1}{3} \left (1+\frac {3}{\log \left (5+\log \left (\frac {x^2}{16}\right )\right )}\right ) \] Output:
x-4*x^2+1/3+1/ln(ln(1/16*x^2)+5)-exp(x)
Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {-2+\left (5 x-5 e^x x-40 x^2+\left (x-e^x x-8 x^2\right ) \log \left (\frac {x^2}{16}\right )\right ) \log ^2\left (5+\log \left (\frac {x^2}{16}\right )\right )}{\left (5 x+x \log \left (\frac {x^2}{16}\right )\right ) \log ^2\left (5+\log \left (\frac {x^2}{16}\right )\right )} \, dx=-e^x+x-4 x^2+\frac {1}{\log \left (5+\log \left (\frac {x^2}{16}\right )\right )} \] Input:
Integrate[(-2 + (5*x - 5*E^x*x - 40*x^2 + (x - E^x*x - 8*x^2)*Log[x^2/16]) *Log[5 + Log[x^2/16]]^2)/((5*x + x*Log[x^2/16])*Log[5 + Log[x^2/16]]^2),x]
Output:
-E^x + x - 4*x^2 + Log[5 + Log[x^2/16]]^(-1)
Time = 1.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {3041, 7239, 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.
\(\displaystyle \int \frac {\left (-40 x^2+\left (-8 x^2-e^x x+x\right ) \log \left (\frac {x^2}{16}\right )-5 e^x x+5 x\right ) \log ^2\left (\log \left (\frac {x^2}{16}\right )+5\right )-2}{\left (x \log \left (\frac {x^2}{16}\right )+5 x\right ) \log ^2\left (\log \left (\frac {x^2}{16}\right )+5\right )} \, dx\) |
\(\Big \downarrow \) 3041 |
\(\displaystyle \int \frac {\left (-40 x^2+\left (-8 x^2-e^x x+x\right ) \log \left (\frac {x^2}{16}\right )-5 e^x x+5 x\right ) \log ^2\left (\log \left (\frac {x^2}{16}\right )+5\right )-2}{x \left (\log \left (\frac {x^2}{16}\right )+5\right ) \log ^2\left (\log \left (\frac {x^2}{16}\right )+5\right )}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (-\frac {2}{x \left (\log \left (\frac {x^2}{16}\right )+5\right ) \log ^2\left (\log \left (\frac {x^2}{16}\right )+5\right )}-8 x-e^x+1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 x^2+\frac {1}{\log \left (\log \left (\frac {x^2}{16}\right )+5\right )}+x-e^x\) |
Input:
Int[(-2 + (5*x - 5*E^x*x - 40*x^2 + (x - E^x*x - 8*x^2)*Log[x^2/16])*Log[5 + Log[x^2/16]]^2)/((5*x + x*Log[x^2/16])*Log[5 + Log[x^2/16]]^2),x]
Output:
-E^x + x - 4*x^2 + Log[5 + Log[x^2/16]]^(-1)
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]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 2.33 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.70
method | result | size |
default | \(-{\mathrm e}^{x}-4 x^{2}+x +\frac {1}{\ln \left (\ln \left (\frac {x^{2}}{16}\right )+5\right )}\) | \(23\) |
parts | \(-{\mathrm e}^{x}-4 x^{2}+x +\frac {1}{\ln \left (\ln \left (\frac {x^{2}}{16}\right )+5\right )}\) | \(23\) |
risch | \(-4 x^{2}+x -{\mathrm e}^{x}+\frac {1}{\ln \left (2 \ln \left (x \right )-4 \ln \left (2\right )-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) {\left (-\operatorname {csgn}\left (i x^{2}\right )+\operatorname {csgn}\left (i x \right )\right )}^{2}}{2}+5\right )}\) | \(53\) |
parallelrisch | \(\frac {20-80 \ln \left (\ln \left (\frac {x^{2}}{16}\right )+5\right ) x^{2}+20 \ln \left (\ln \left (\frac {x^{2}}{16}\right )+5\right ) x -20 \,{\mathrm e}^{x} \ln \left (\ln \left (\frac {x^{2}}{16}\right )+5\right )}{20 \ln \left (\ln \left (\frac {x^{2}}{16}\right )+5\right )}\) | \(55\) |
Input:
int((((-exp(x)*x-8*x^2+x)*ln(1/16*x^2)-5*exp(x)*x-40*x^2+5*x)*ln(ln(1/16*x ^2)+5)^2-2)/(x*ln(1/16*x^2)+5*x)/ln(ln(1/16*x^2)+5)^2,x,method=_RETURNVERB OSE)
Output:
-exp(x)-4*x^2+x+1/ln(ln(1/16*x^2)+5)
Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09 \[ \int \frac {-2+\left (5 x-5 e^x x-40 x^2+\left (x-e^x x-8 x^2\right ) \log \left (\frac {x^2}{16}\right )\right ) \log ^2\left (5+\log \left (\frac {x^2}{16}\right )\right )}{\left (5 x+x \log \left (\frac {x^2}{16}\right )\right ) \log ^2\left (5+\log \left (\frac {x^2}{16}\right )\right )} \, dx=-\frac {{\left (4 \, x^{2} - x + e^{x}\right )} \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) - 1}{\log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right )} \] Input:
integrate((((-exp(x)*x-8*x^2+x)*log(1/16*x^2)-5*exp(x)*x-40*x^2+5*x)*log(l og(1/16*x^2)+5)^2-2)/(x*log(1/16*x^2)+5*x)/log(log(1/16*x^2)+5)^2,x, algor ithm="fricas")
Output:
-((4*x^2 - x + e^x)*log(log(1/16*x^2) + 5) - 1)/log(log(1/16*x^2) + 5)
Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.61 \[ \int \frac {-2+\left (5 x-5 e^x x-40 x^2+\left (x-e^x x-8 x^2\right ) \log \left (\frac {x^2}{16}\right )\right ) \log ^2\left (5+\log \left (\frac {x^2}{16}\right )\right )}{\left (5 x+x \log \left (\frac {x^2}{16}\right )\right ) \log ^2\left (5+\log \left (\frac {x^2}{16}\right )\right )} \, dx=- 4 x^{2} + x - e^{x} + \frac {1}{\log {\left (\log {\left (\frac {x^{2}}{16} \right )} + 5 \right )}} \] Input:
integrate((((-exp(x)*x-8*x**2+x)*ln(1/16*x**2)-5*exp(x)*x-40*x**2+5*x)*ln( ln(1/16*x**2)+5)**2-2)/(x*ln(1/16*x**2)+5*x)/ln(ln(1/16*x**2)+5)**2,x)
Output:
-4*x**2 + x - exp(x) + 1/log(log(x**2/16) + 5)
Time = 0.16 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {-2+\left (5 x-5 e^x x-40 x^2+\left (x-e^x x-8 x^2\right ) \log \left (\frac {x^2}{16}\right )\right ) \log ^2\left (5+\log \left (\frac {x^2}{16}\right )\right )}{\left (5 x+x \log \left (\frac {x^2}{16}\right )\right ) \log ^2\left (5+\log \left (\frac {x^2}{16}\right )\right )} \, dx=-4 \, x^{2} + x + \frac {1}{\log \left (-4 \, \log \left (2\right ) + 2 \, \log \left (x\right ) + 5\right )} - e^{x} \] Input:
integrate((((-exp(x)*x-8*x^2+x)*log(1/16*x^2)-5*exp(x)*x-40*x^2+5*x)*log(l og(1/16*x^2)+5)^2-2)/(x*log(1/16*x^2)+5*x)/log(log(1/16*x^2)+5)^2,x, algor ithm="maxima")
Output:
-4*x^2 + x + 1/log(-4*log(2) + 2*log(x) + 5) - e^x
Leaf count of result is larger than twice the leaf count of optimal. 186 vs. \(2 (23) = 46\).
Time = 0.25 (sec) , antiderivative size = 186, normalized size of antiderivative = 5.64 \[ \int \frac {-2+\left (5 x-5 e^x x-40 x^2+\left (x-e^x x-8 x^2\right ) \log \left (\frac {x^2}{16}\right )\right ) \log ^2\left (5+\log \left (\frac {x^2}{16}\right )\right )}{\left (5 x+x \log \left (\frac {x^2}{16}\right )\right ) \log ^2\left (5+\log \left (\frac {x^2}{16}\right )\right )} \, dx=-\frac {16 \, x^{2} \log \left (2\right ) \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) - 4 \, x^{2} \log \left (x^{2}\right ) \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) - 20 \, x^{2} \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) - 4 \, x \log \left (2\right ) \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) + 4 \, e^{x} \log \left (2\right ) \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) + x \log \left (x^{2}\right ) \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) - e^{x} \log \left (x^{2}\right ) \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) + 5 \, x \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) - 5 \, e^{x} \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) + \log \left (\frac {1}{16} \, x^{2}\right ) + 5}{4 \, \log \left (2\right ) \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) - \log \left (x^{2}\right ) \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) - 5 \, \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right )} \] Input:
integrate((((-exp(x)*x-8*x^2+x)*log(1/16*x^2)-5*exp(x)*x-40*x^2+5*x)*log(l og(1/16*x^2)+5)^2-2)/(x*log(1/16*x^2)+5*x)/log(log(1/16*x^2)+5)^2,x, algor ithm="giac")
Output:
-(16*x^2*log(2)*log(log(1/16*x^2) + 5) - 4*x^2*log(x^2)*log(log(1/16*x^2) + 5) - 20*x^2*log(log(1/16*x^2) + 5) - 4*x*log(2)*log(log(1/16*x^2) + 5) + 4*e^x*log(2)*log(log(1/16*x^2) + 5) + x*log(x^2)*log(log(1/16*x^2) + 5) - e^x*log(x^2)*log(log(1/16*x^2) + 5) + 5*x*log(log(1/16*x^2) + 5) - 5*e^x* log(log(1/16*x^2) + 5) + log(1/16*x^2) + 5)/(4*log(2)*log(log(1/16*x^2) + 5) - log(x^2)*log(log(1/16*x^2) + 5) - 5*log(log(1/16*x^2) + 5))
Time = 1.62 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.67 \[ \int \frac {-2+\left (5 x-5 e^x x-40 x^2+\left (x-e^x x-8 x^2\right ) \log \left (\frac {x^2}{16}\right )\right ) \log ^2\left (5+\log \left (\frac {x^2}{16}\right )\right )}{\left (5 x+x \log \left (\frac {x^2}{16}\right )\right ) \log ^2\left (5+\log \left (\frac {x^2}{16}\right )\right )} \, dx=x-{\mathrm {e}}^x+\frac {1}{\ln \left (\ln \left (\frac {x^2}{16}\right )+5\right )}-4\,x^2 \] Input:
int(-(log(log(x^2/16) + 5)^2*(log(x^2/16)*(x*exp(x) - x + 8*x^2) - 5*x + 5 *x*exp(x) + 40*x^2) + 2)/(log(log(x^2/16) + 5)^2*(5*x + x*log(x^2/16))),x)
Output:
x - exp(x) + 1/log(log(x^2/16) + 5) - 4*x^2
Time = 0.19 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.61 \[ \int \frac {-2+\left (5 x-5 e^x x-40 x^2+\left (x-e^x x-8 x^2\right ) \log \left (\frac {x^2}{16}\right )\right ) \log ^2\left (5+\log \left (\frac {x^2}{16}\right )\right )}{\left (5 x+x \log \left (\frac {x^2}{16}\right )\right ) \log ^2\left (5+\log \left (\frac {x^2}{16}\right )\right )} \, dx=\frac {-e^{x} \mathrm {log}\left (\mathrm {log}\left (\frac {x^{2}}{16}\right )+5\right )-4 \,\mathrm {log}\left (\mathrm {log}\left (\frac {x^{2}}{16}\right )+5\right ) x^{2}+\mathrm {log}\left (\mathrm {log}\left (\frac {x^{2}}{16}\right )+5\right ) x +1}{\mathrm {log}\left (\mathrm {log}\left (\frac {x^{2}}{16}\right )+5\right )} \] Input:
int((((-exp(x)*x-8*x^2+x)*log(1/16*x^2)-5*exp(x)*x-40*x^2+5*x)*log(log(1/1 6*x^2)+5)^2-2)/(x*log(1/16*x^2)+5*x)/log(log(1/16*x^2)+5)^2,x)
Output:
( - e**x*log(log(x**2/16) + 5) - 4*log(log(x**2/16) + 5)*x**2 + log(log(x* *2/16) + 5)*x + 1)/log(log(x**2/16) + 5)