Integrand size = 145, antiderivative size = 25 \[ \int \frac {-36+e^{-6+2 x} (-9+18 x)}{\left (16 x+4 e^{-6+2 x} x\right ) \log \left (\frac {x}{4+e^{-6+2 x}}\right )+\left (-16 x-4 e^{-6+2 x} x\right ) \log \left (\frac {x}{4+e^{-6+2 x}}\right ) \log \left (\frac {1}{5} \log \left (\frac {x}{4+e^{-6+2 x}}\right )\right )+\left (4 x+e^{-6+2 x} x\right ) \log \left (\frac {x}{4+e^{-6+2 x}}\right ) \log ^2\left (\frac {1}{5} \log \left (\frac {x}{4+e^{-6+2 x}}\right )\right )} \, dx=\frac {9}{-2+\log \left (\frac {1}{5} \log \left (\frac {x}{4+e^{-6+2 x}}\right )\right )} \] Output:
9/(ln(1/5*ln(x/(exp(-3+x)^2+4)))-2)
Time = 0.05 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {-36+e^{-6+2 x} (-9+18 x)}{\left (16 x+4 e^{-6+2 x} x\right ) \log \left (\frac {x}{4+e^{-6+2 x}}\right )+\left (-16 x-4 e^{-6+2 x} x\right ) \log \left (\frac {x}{4+e^{-6+2 x}}\right ) \log \left (\frac {1}{5} \log \left (\frac {x}{4+e^{-6+2 x}}\right )\right )+\left (4 x+e^{-6+2 x} x\right ) \log \left (\frac {x}{4+e^{-6+2 x}}\right ) \log ^2\left (\frac {1}{5} \log \left (\frac {x}{4+e^{-6+2 x}}\right )\right )} \, dx=\frac {9}{-2+\log \left (\frac {1}{5} \log \left (\frac {x}{4+e^{-6+2 x}}\right )\right )} \] Input:
Integrate[(-36 + E^(-6 + 2*x)*(-9 + 18*x))/((16*x + 4*E^(-6 + 2*x)*x)*Log[ x/(4 + E^(-6 + 2*x))] + (-16*x - 4*E^(-6 + 2*x)*x)*Log[x/(4 + E^(-6 + 2*x) )]*Log[Log[x/(4 + E^(-6 + 2*x))]/5] + (4*x + E^(-6 + 2*x)*x)*Log[x/(4 + E^ (-6 + 2*x))]*Log[Log[x/(4 + E^(-6 + 2*x))]/5]^2),x]
Output:
9/(-2 + Log[Log[x/(4 + E^(-6 + 2*x))]/5])
Time = 0.86 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {7239, 27, 25, 7237}
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.
\(\displaystyle \int \frac {e^{2 x-6} (18 x-9)-36}{\left (e^{2 x-6} x+4 x\right ) \log \left (\frac {x}{e^{2 x-6}+4}\right ) \log ^2\left (\frac {1}{5} \log \left (\frac {x}{e^{2 x-6}+4}\right )\right )+\left (-4 e^{2 x-6} x-16 x\right ) \log \left (\frac {x}{e^{2 x-6}+4}\right ) \log \left (\frac {1}{5} \log \left (\frac {x}{e^{2 x-6}+4}\right )\right )+\left (4 e^{2 x-6} x+16 x\right ) \log \left (\frac {x}{e^{2 x-6}+4}\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {9 \left (e^{2 x} (2 x-1)-4 e^6\right )}{\left (e^{2 x}+4 e^6\right ) x \log \left (\frac {x}{e^{2 x-6}+4}\right ) \left (2-\log \left (\frac {1}{5} \log \left (\frac {x}{e^{2 x-6}+4}\right )\right )\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 9 \int -\frac {e^{2 x} (1-2 x)+4 e^6}{\left (4 e^6+e^{2 x}\right ) x \log \left (\frac {x}{4+e^{2 x-6}}\right ) \left (2-\log \left (\frac {1}{5} \log \left (\frac {x}{4+e^{2 x-6}}\right )\right )\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -9 \int \frac {e^{2 x} (1-2 x)+4 e^6}{\left (4 e^6+e^{2 x}\right ) x \log \left (\frac {x}{4+e^{2 x-6}}\right ) \left (2-\log \left (\frac {1}{5} \log \left (\frac {x}{4+e^{2 x-6}}\right )\right )\right )^2}dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle -\frac {9}{2-\log \left (\frac {1}{5} \log \left (\frac {x}{e^{2 x-6}+4}\right )\right )}\) |
Input:
Int[(-36 + E^(-6 + 2*x)*(-9 + 18*x))/((16*x + 4*E^(-6 + 2*x)*x)*Log[x/(4 + E^(-6 + 2*x))] + (-16*x - 4*E^(-6 + 2*x)*x)*Log[x/(4 + E^(-6 + 2*x))]*Log [Log[x/(4 + E^(-6 + 2*x))]/5] + (4*x + E^(-6 + 2*x)*x)*Log[x/(4 + E^(-6 + 2*x))]*Log[Log[x/(4 + E^(-6 + 2*x))]/5]^2),x]
Output:
-9/(2 - Log[Log[x/(4 + E^(-6 + 2*x))]/5])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 12.57 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92
method | result | size |
parallelrisch | \(\frac {9}{\ln \left (\frac {\ln \left (\frac {x}{{\mathrm e}^{2 x -6}+4}\right )}{5}\right )-2}\) | \(23\) |
risch | \(\frac {9}{\ln \left (\frac {\ln \left (x \right )}{5}-\frac {\ln \left ({\mathrm e}^{2 x -6}+4\right )}{5}-\frac {i \pi \,\operatorname {csgn}\left (\frac {i x}{{\mathrm e}^{2 x -6}+4}\right ) \left (-\operatorname {csgn}\left (\frac {i x}{{\mathrm e}^{2 x -6}+4}\right )+\operatorname {csgn}\left (i x \right )\right ) \left (-\operatorname {csgn}\left (\frac {i x}{{\mathrm e}^{2 x -6}+4}\right )+\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 x -6}+4}\right )\right )}{10}\right )-2}\) | \(98\) |
Input:
int(((18*x-9)*exp(-3+x)^2-36)/((x*exp(-3+x)^2+4*x)*ln(x/(exp(-3+x)^2+4))*l n(1/5*ln(x/(exp(-3+x)^2+4)))^2+(-4*x*exp(-3+x)^2-16*x)*ln(x/(exp(-3+x)^2+4 ))*ln(1/5*ln(x/(exp(-3+x)^2+4)))+(4*x*exp(-3+x)^2+16*x)*ln(x/(exp(-3+x)^2+ 4))),x,method=_RETURNVERBOSE)
Output:
9/(ln(1/5*ln(x/(exp(-3+x)^2+4)))-2)
Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {-36+e^{-6+2 x} (-9+18 x)}{\left (16 x+4 e^{-6+2 x} x\right ) \log \left (\frac {x}{4+e^{-6+2 x}}\right )+\left (-16 x-4 e^{-6+2 x} x\right ) \log \left (\frac {x}{4+e^{-6+2 x}}\right ) \log \left (\frac {1}{5} \log \left (\frac {x}{4+e^{-6+2 x}}\right )\right )+\left (4 x+e^{-6+2 x} x\right ) \log \left (\frac {x}{4+e^{-6+2 x}}\right ) \log ^2\left (\frac {1}{5} \log \left (\frac {x}{4+e^{-6+2 x}}\right )\right )} \, dx=\frac {9}{\log \left (\frac {1}{5} \, \log \left (\frac {x}{e^{\left (2 \, x - 6\right )} + 4}\right )\right ) - 2} \] Input:
integrate(((18*x-9)*exp(-3+x)^2-36)/((x*exp(-3+x)^2+4*x)*log(x/(exp(-3+x)^ 2+4))*log(1/5*log(x/(exp(-3+x)^2+4)))^2+(-4*x*exp(-3+x)^2-16*x)*log(x/(exp (-3+x)^2+4))*log(1/5*log(x/(exp(-3+x)^2+4)))+(4*x*exp(-3+x)^2+16*x)*log(x/ (exp(-3+x)^2+4))),x, algorithm="fricas")
Output:
9/(log(1/5*log(x/(e^(2*x - 6) + 4))) - 2)
Time = 0.49 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {-36+e^{-6+2 x} (-9+18 x)}{\left (16 x+4 e^{-6+2 x} x\right ) \log \left (\frac {x}{4+e^{-6+2 x}}\right )+\left (-16 x-4 e^{-6+2 x} x\right ) \log \left (\frac {x}{4+e^{-6+2 x}}\right ) \log \left (\frac {1}{5} \log \left (\frac {x}{4+e^{-6+2 x}}\right )\right )+\left (4 x+e^{-6+2 x} x\right ) \log \left (\frac {x}{4+e^{-6+2 x}}\right ) \log ^2\left (\frac {1}{5} \log \left (\frac {x}{4+e^{-6+2 x}}\right )\right )} \, dx=\frac {9}{\log {\left (\frac {\log {\left (\frac {x}{e^{2 x - 6} + 4} \right )}}{5} \right )} - 2} \] Input:
integrate(((18*x-9)*exp(-3+x)**2-36)/((x*exp(-3+x)**2+4*x)*ln(x/(exp(-3+x) **2+4))*ln(1/5*ln(x/(exp(-3+x)**2+4)))**2+(-4*x*exp(-3+x)**2-16*x)*ln(x/(e xp(-3+x)**2+4))*ln(1/5*ln(x/(exp(-3+x)**2+4)))+(4*x*exp(-3+x)**2+16*x)*ln( x/(exp(-3+x)**2+4))),x)
Output:
9/(log(log(x/(exp(2*x - 6) + 4))/5) - 2)
Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {-36+e^{-6+2 x} (-9+18 x)}{\left (16 x+4 e^{-6+2 x} x\right ) \log \left (\frac {x}{4+e^{-6+2 x}}\right )+\left (-16 x-4 e^{-6+2 x} x\right ) \log \left (\frac {x}{4+e^{-6+2 x}}\right ) \log \left (\frac {1}{5} \log \left (\frac {x}{4+e^{-6+2 x}}\right )\right )+\left (4 x+e^{-6+2 x} x\right ) \log \left (\frac {x}{4+e^{-6+2 x}}\right ) \log ^2\left (\frac {1}{5} \log \left (\frac {x}{4+e^{-6+2 x}}\right )\right )} \, dx=-\frac {9}{\log \left (5\right ) - \log \left (\log \left (x\right ) - \log \left (4 \, e^{6} + e^{\left (2 \, x\right )}\right ) + 6\right ) + 2} \] Input:
integrate(((18*x-9)*exp(-3+x)^2-36)/((x*exp(-3+x)^2+4*x)*log(x/(exp(-3+x)^ 2+4))*log(1/5*log(x/(exp(-3+x)^2+4)))^2+(-4*x*exp(-3+x)^2-16*x)*log(x/(exp (-3+x)^2+4))*log(1/5*log(x/(exp(-3+x)^2+4)))+(4*x*exp(-3+x)^2+16*x)*log(x/ (exp(-3+x)^2+4))),x, algorithm="maxima")
Output:
-9/(log(5) - log(log(x) - log(4*e^6 + e^(2*x)) + 6) + 2)
Time = 0.29 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {-36+e^{-6+2 x} (-9+18 x)}{\left (16 x+4 e^{-6+2 x} x\right ) \log \left (\frac {x}{4+e^{-6+2 x}}\right )+\left (-16 x-4 e^{-6+2 x} x\right ) \log \left (\frac {x}{4+e^{-6+2 x}}\right ) \log \left (\frac {1}{5} \log \left (\frac {x}{4+e^{-6+2 x}}\right )\right )+\left (4 x+e^{-6+2 x} x\right ) \log \left (\frac {x}{4+e^{-6+2 x}}\right ) \log ^2\left (\frac {1}{5} \log \left (\frac {x}{4+e^{-6+2 x}}\right )\right )} \, dx=\frac {9}{\log \left (\frac {1}{5} \, \log \left (\frac {x}{e^{\left (2 \, x - 6\right )} + 4}\right )\right ) - 2} \] Input:
integrate(((18*x-9)*exp(-3+x)^2-36)/((x*exp(-3+x)^2+4*x)*log(x/(exp(-3+x)^ 2+4))*log(1/5*log(x/(exp(-3+x)^2+4)))^2+(-4*x*exp(-3+x)^2-16*x)*log(x/(exp (-3+x)^2+4))*log(1/5*log(x/(exp(-3+x)^2+4)))+(4*x*exp(-3+x)^2+16*x)*log(x/ (exp(-3+x)^2+4))),x, algorithm="giac")
Output:
9/(log(1/5*log(x/(e^(2*x - 6) + 4))) - 2)
Time = 4.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {-36+e^{-6+2 x} (-9+18 x)}{\left (16 x+4 e^{-6+2 x} x\right ) \log \left (\frac {x}{4+e^{-6+2 x}}\right )+\left (-16 x-4 e^{-6+2 x} x\right ) \log \left (\frac {x}{4+e^{-6+2 x}}\right ) \log \left (\frac {1}{5} \log \left (\frac {x}{4+e^{-6+2 x}}\right )\right )+\left (4 x+e^{-6+2 x} x\right ) \log \left (\frac {x}{4+e^{-6+2 x}}\right ) \log ^2\left (\frac {1}{5} \log \left (\frac {x}{4+e^{-6+2 x}}\right )\right )} \, dx=\frac {9}{\ln \left (\frac {\ln \left (x\right )}{5}-\frac {\ln \left ({\mathrm {e}}^{2\,x}+4\,{\mathrm {e}}^6\right )}{5}+\frac {6}{5}\right )-2} \] Input:
int((exp(2*x - 6)*(18*x - 9) - 36)/(log(x/(exp(2*x - 6) + 4))*(16*x + 4*x* exp(2*x - 6)) - log(log(x/(exp(2*x - 6) + 4))/5)*log(x/(exp(2*x - 6) + 4)) *(16*x + 4*x*exp(2*x - 6)) + log(log(x/(exp(2*x - 6) + 4))/5)^2*log(x/(exp (2*x - 6) + 4))*(4*x + x*exp(2*x - 6))),x)
Output:
9/(log(log(x)/5 - log(exp(2*x) + 4*exp(6))/5 + 6/5) - 2)
Time = 0.21 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.08 \[ \int \frac {-36+e^{-6+2 x} (-9+18 x)}{\left (16 x+4 e^{-6+2 x} x\right ) \log \left (\frac {x}{4+e^{-6+2 x}}\right )+\left (-16 x-4 e^{-6+2 x} x\right ) \log \left (\frac {x}{4+e^{-6+2 x}}\right ) \log \left (\frac {1}{5} \log \left (\frac {x}{4+e^{-6+2 x}}\right )\right )+\left (4 x+e^{-6+2 x} x\right ) \log \left (\frac {x}{4+e^{-6+2 x}}\right ) \log ^2\left (\frac {1}{5} \log \left (\frac {x}{4+e^{-6+2 x}}\right )\right )} \, dx=\frac {9 \,\mathrm {log}\left (\frac {\mathrm {log}\left (\frac {e^{6} x}{e^{2 x}+4 e^{6}}\right )}{5}\right )}{2 \,\mathrm {log}\left (\frac {\mathrm {log}\left (\frac {e^{6} x}{e^{2 x}+4 e^{6}}\right )}{5}\right )-4} \] Input:
int(((18*x-9)*exp(-3+x)^2-36)/((x*exp(-3+x)^2+4*x)*log(x/(exp(-3+x)^2+4))* log(1/5*log(x/(exp(-3+x)^2+4)))^2+(-4*x*exp(-3+x)^2-16*x)*log(x/(exp(-3+x) ^2+4))*log(1/5*log(x/(exp(-3+x)^2+4)))+(4*x*exp(-3+x)^2+16*x)*log(x/(exp(- 3+x)^2+4))),x)
Output:
(9*log(log((e**6*x)/(e**(2*x) + 4*e**6))/5))/(2*(log(log((e**6*x)/(e**(2*x ) + 4*e**6))/5) - 2))