Integrand size = 172, antiderivative size = 33 \[ \int \frac {15+60 e^{1-4 x}}{-100+75 e^2-75 x+\left (-40+30 e^2-30 x\right ) \log (5)+\left (-4+3 e^2-3 x\right ) \log ^2(5)+e^{1-4 x} \left (75+30 \log (5)+3 \log ^2(5)\right )+\left (-40+30 e^2-30 x+\left (-8+6 e^2-6 x\right ) \log (5)+e^{1-4 x} (30+6 \log (5))\right ) \log \left (-4+3 e^2+3 e^{1-4 x}-3 x\right )+\left (-4+3 e^2+3 e^{1-4 x}-3 x\right ) \log ^2\left (-4+3 e^2+3 e^{1-4 x}-3 x\right )} \, dx=\frac {5}{5+\log (5)+\log \left (-4+3 \left (e^2+e^{\frac {x-4 x^2}{x}}-x\right )\right )} \]
Time = 0.09 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {15+60 e^{1-4 x}}{-100+75 e^2-75 x+\left (-40+30 e^2-30 x\right ) \log (5)+\left (-4+3 e^2-3 x\right ) \log ^2(5)+e^{1-4 x} \left (75+30 \log (5)+3 \log ^2(5)\right )+\left (-40+30 e^2-30 x+\left (-8+6 e^2-6 x\right ) \log (5)+e^{1-4 x} (30+6 \log (5))\right ) \log \left (-4+3 e^2+3 e^{1-4 x}-3 x\right )+\left (-4+3 e^2+3 e^{1-4 x}-3 x\right ) \log ^2\left (-4+3 e^2+3 e^{1-4 x}-3 x\right )} \, dx=\frac {5}{5+\log (5)+\log \left (-4+3 e^2+3 e^{1-4 x}-3 x\right )} \]
Integrate[(15 + 60*E^(1 - 4*x))/(-100 + 75*E^2 - 75*x + (-40 + 30*E^2 - 30 *x)*Log[5] + (-4 + 3*E^2 - 3*x)*Log[5]^2 + E^(1 - 4*x)*(75 + 30*Log[5] + 3 *Log[5]^2) + (-40 + 30*E^2 - 30*x + (-8 + 6*E^2 - 6*x)*Log[5] + E^(1 - 4*x )*(30 + 6*Log[5]))*Log[-4 + 3*E^2 + 3*E^(1 - 4*x) - 3*x] + (-4 + 3*E^2 + 3 *E^(1 - 4*x) - 3*x)*Log[-4 + 3*E^2 + 3*E^(1 - 4*x) - 3*x]^2),x]
Time = 0.69 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {7239, 27, 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.
\(\displaystyle \int \frac {60 e^{1-4 x}+15}{-75 x+\left (-3 x+3 e^{1-4 x}+3 e^2-4\right ) \log ^2\left (-3 x+3 e^{1-4 x}+3 e^2-4\right )+e^{1-4 x} \left (75+3 \log ^2(5)+30 \log (5)\right )+\left (-3 x+3 e^2-4\right ) \log ^2(5)+\left (-30 x+\left (-6 x+6 e^2-8\right ) \log (5)+e^{1-4 x} (30+6 \log (5))+30 e^2-40\right ) \log \left (-3 x+3 e^{1-4 x}+3 e^2-4\right )+\left (-30 x+30 e^2-40\right ) \log (5)+75 e^2-100} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {15 \left (e^{4 x}+4 e\right )}{\left (-e^{4 x} (3 x+4)+3 e^{4 x+2}+3 e\right ) \left (\log \left (-3 x+3 e^{1-4 x}+3 e^2-4\right )+5 \left (1+\frac {\log (5)}{5}\right )\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 15 \int \frac {4 e+e^{4 x}}{\left (-e^{4 x} (3 x+4)+3 e^{4 x+2}+3 e\right ) \left (\log \left (-3 x+3 e^{1-4 x}+3 e^2-4\right )+\log (5)+5\right )^2}dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle \frac {5}{\log \left (-3 x+3 e^{1-4 x}+3 e^2-4\right )+5+\log (5)}\) |
Int[(15 + 60*E^(1 - 4*x))/(-100 + 75*E^2 - 75*x + (-40 + 30*E^2 - 30*x)*Lo g[5] + (-4 + 3*E^2 - 3*x)*Log[5]^2 + E^(1 - 4*x)*(75 + 30*Log[5] + 3*Log[5 ]^2) + (-40 + 30*E^2 - 30*x + (-8 + 6*E^2 - 6*x)*Log[5] + E^(1 - 4*x)*(30 + 6*Log[5]))*Log[-4 + 3*E^2 + 3*E^(1 - 4*x) - 3*x] + (-4 + 3*E^2 + 3*E^(1 - 4*x) - 3*x)*Log[-4 + 3*E^2 + 3*E^(1 - 4*x) - 3*x]^2),x]
3.8.30.3.1 Defintions of rubi rules used
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 = 0.54 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82
method | result | size |
norman | \(\frac {5}{\ln \left (5\right )+\ln \left (3 \,{\mathrm e}^{-4 x +1}+3 \,{\mathrm e}^{2}-3 x -4\right )+5}\) | \(27\) |
risch | \(\frac {5}{\ln \left (5\right )+\ln \left (3 \,{\mathrm e}^{-4 x +1}+3 \,{\mathrm e}^{2}-3 x -4\right )+5}\) | \(27\) |
parallelrisch | \(\frac {5}{\ln \left (5\right )+\ln \left (3 \,{\mathrm e}^{-4 x +1}+3 \,{\mathrm e}^{2}-3 x -4\right )+5}\) | \(27\) |
int((60*exp(-4*x+1)+15)/((3*exp(-4*x+1)+3*exp(2)-3*x-4)*ln(3*exp(-4*x+1)+3 *exp(2)-3*x-4)^2+((6*ln(5)+30)*exp(-4*x+1)+(6*exp(2)-6*x-8)*ln(5)+30*exp(2 )-30*x-40)*ln(3*exp(-4*x+1)+3*exp(2)-3*x-4)+(3*ln(5)^2+30*ln(5)+75)*exp(-4 *x+1)+(3*exp(2)-3*x-4)*ln(5)^2+(30*exp(2)-30*x-40)*ln(5)+75*exp(2)-75*x-10 0),x,method=_RETURNVERBOSE)
Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \frac {15+60 e^{1-4 x}}{-100+75 e^2-75 x+\left (-40+30 e^2-30 x\right ) \log (5)+\left (-4+3 e^2-3 x\right ) \log ^2(5)+e^{1-4 x} \left (75+30 \log (5)+3 \log ^2(5)\right )+\left (-40+30 e^2-30 x+\left (-8+6 e^2-6 x\right ) \log (5)+e^{1-4 x} (30+6 \log (5))\right ) \log \left (-4+3 e^2+3 e^{1-4 x}-3 x\right )+\left (-4+3 e^2+3 e^{1-4 x}-3 x\right ) \log ^2\left (-4+3 e^2+3 e^{1-4 x}-3 x\right )} \, dx=\frac {5}{\log \left (5\right ) + \log \left (-3 \, x + 3 \, e^{2} + 3 \, e^{\left (-4 \, x + 1\right )} - 4\right ) + 5} \]
integrate((60*exp(-4*x+1)+15)/((3*exp(-4*x+1)+3*exp(2)-3*x-4)*log(3*exp(-4 *x+1)+3*exp(2)-3*x-4)^2+((6*log(5)+30)*exp(-4*x+1)+(6*exp(2)-6*x-8)*log(5) +30*exp(2)-30*x-40)*log(3*exp(-4*x+1)+3*exp(2)-3*x-4)+(3*log(5)^2+30*log(5 )+75)*exp(-4*x+1)+(3*exp(2)-3*x-4)*log(5)^2+(30*exp(2)-30*x-40)*log(5)+75* exp(2)-75*x-100),x, algorithm=\
Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \frac {15+60 e^{1-4 x}}{-100+75 e^2-75 x+\left (-40+30 e^2-30 x\right ) \log (5)+\left (-4+3 e^2-3 x\right ) \log ^2(5)+e^{1-4 x} \left (75+30 \log (5)+3 \log ^2(5)\right )+\left (-40+30 e^2-30 x+\left (-8+6 e^2-6 x\right ) \log (5)+e^{1-4 x} (30+6 \log (5))\right ) \log \left (-4+3 e^2+3 e^{1-4 x}-3 x\right )+\left (-4+3 e^2+3 e^{1-4 x}-3 x\right ) \log ^2\left (-4+3 e^2+3 e^{1-4 x}-3 x\right )} \, dx=\frac {5}{\log {\left (- 3 x + 3 e^{1 - 4 x} - 4 + 3 e^{2} \right )} + \log {\left (5 \right )} + 5} \]
integrate((60*exp(-4*x+1)+15)/((3*exp(-4*x+1)+3*exp(2)-3*x-4)*ln(3*exp(-4* x+1)+3*exp(2)-3*x-4)**2+((6*ln(5)+30)*exp(-4*x+1)+(6*exp(2)-6*x-8)*ln(5)+3 0*exp(2)-30*x-40)*ln(3*exp(-4*x+1)+3*exp(2)-3*x-4)+(3*ln(5)**2+30*ln(5)+75 )*exp(-4*x+1)+(3*exp(2)-3*x-4)*ln(5)**2+(30*exp(2)-30*x-40)*ln(5)+75*exp(2 )-75*x-100),x)
Time = 0.49 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09 \[ \int \frac {15+60 e^{1-4 x}}{-100+75 e^2-75 x+\left (-40+30 e^2-30 x\right ) \log (5)+\left (-4+3 e^2-3 x\right ) \log ^2(5)+e^{1-4 x} \left (75+30 \log (5)+3 \log ^2(5)\right )+\left (-40+30 e^2-30 x+\left (-8+6 e^2-6 x\right ) \log (5)+e^{1-4 x} (30+6 \log (5))\right ) \log \left (-4+3 e^2+3 e^{1-4 x}-3 x\right )+\left (-4+3 e^2+3 e^{1-4 x}-3 x\right ) \log ^2\left (-4+3 e^2+3 e^{1-4 x}-3 x\right )} \, dx=-\frac {5}{4 \, x - \log \left (5\right ) - \log \left (-{\left (3 \, x - 3 \, e^{2} + 4\right )} e^{\left (4 \, x\right )} + 3 \, e\right ) - 5} \]
integrate((60*exp(-4*x+1)+15)/((3*exp(-4*x+1)+3*exp(2)-3*x-4)*log(3*exp(-4 *x+1)+3*exp(2)-3*x-4)^2+((6*log(5)+30)*exp(-4*x+1)+(6*exp(2)-6*x-8)*log(5) +30*exp(2)-30*x-40)*log(3*exp(-4*x+1)+3*exp(2)-3*x-4)+(3*log(5)^2+30*log(5 )+75)*exp(-4*x+1)+(3*exp(2)-3*x-4)*log(5)^2+(30*exp(2)-30*x-40)*log(5)+75* exp(2)-75*x-100),x, algorithm=\
Time = 0.84 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \frac {15+60 e^{1-4 x}}{-100+75 e^2-75 x+\left (-40+30 e^2-30 x\right ) \log (5)+\left (-4+3 e^2-3 x\right ) \log ^2(5)+e^{1-4 x} \left (75+30 \log (5)+3 \log ^2(5)\right )+\left (-40+30 e^2-30 x+\left (-8+6 e^2-6 x\right ) \log (5)+e^{1-4 x} (30+6 \log (5))\right ) \log \left (-4+3 e^2+3 e^{1-4 x}-3 x\right )+\left (-4+3 e^2+3 e^{1-4 x}-3 x\right ) \log ^2\left (-4+3 e^2+3 e^{1-4 x}-3 x\right )} \, dx=\frac {5}{\log \left (5\right ) + \log \left (-3 \, x + 3 \, e^{2} + 3 \, e^{\left (-4 \, x + 1\right )} - 4\right ) + 5} \]
integrate((60*exp(-4*x+1)+15)/((3*exp(-4*x+1)+3*exp(2)-3*x-4)*log(3*exp(-4 *x+1)+3*exp(2)-3*x-4)^2+((6*log(5)+30)*exp(-4*x+1)+(6*exp(2)-6*x-8)*log(5) +30*exp(2)-30*x-40)*log(3*exp(-4*x+1)+3*exp(2)-3*x-4)+(3*log(5)^2+30*log(5 )+75)*exp(-4*x+1)+(3*exp(2)-3*x-4)*log(5)^2+(30*exp(2)-30*x-40)*log(5)+75* exp(2)-75*x-100),x, algorithm=\
Time = 21.55 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {15+60 e^{1-4 x}}{-100+75 e^2-75 x+\left (-40+30 e^2-30 x\right ) \log (5)+\left (-4+3 e^2-3 x\right ) \log ^2(5)+e^{1-4 x} \left (75+30 \log (5)+3 \log ^2(5)\right )+\left (-40+30 e^2-30 x+\left (-8+6 e^2-6 x\right ) \log (5)+e^{1-4 x} (30+6 \log (5))\right ) \log \left (-4+3 e^2+3 e^{1-4 x}-3 x\right )+\left (-4+3 e^2+3 e^{1-4 x}-3 x\right ) \log ^2\left (-4+3 e^2+3 e^{1-4 x}-3 x\right )} \, dx=\frac {5}{\ln \left (15\,{\mathrm {e}}^2-15\,x+15\,{\mathrm {e}}^{-4\,x}\,\mathrm {e}-20\right )+5} \]
int(-(60*exp(1 - 4*x) + 15)/(75*x - 75*exp(2) + log(3*exp(2) - 3*x + 3*exp (1 - 4*x) - 4)^2*(3*x - 3*exp(2) - 3*exp(1 - 4*x) + 4) + log(5)^2*(3*x - 3 *exp(2) + 4) - exp(1 - 4*x)*(30*log(5) + 3*log(5)^2 + 75) + log(3*exp(2) - 3*x + 3*exp(1 - 4*x) - 4)*(30*x - 30*exp(2) - exp(1 - 4*x)*(6*log(5) + 30 ) + log(5)*(6*x - 6*exp(2) + 8) + 40) + log(5)*(30*x - 30*exp(2) + 40) + 1 00),x)