Integrand size = 84, antiderivative size = 26 \[ \int -\frac {2 e^{\frac {2 x}{e^{10} (25+5 x)+e^5 (-50-10 x) \log (4)+(25+5 x) \log ^2(4)}}}{e^{10} \left (25+10 x+x^2\right )+e^5 \left (-50-20 x-2 x^2\right ) \log (4)+\left (25+10 x+x^2\right ) \log ^2(4)} \, dx=5-e^{\frac {2 x}{5 (5+x) \left (e^5-\log (4)\right )^2}} \] Output:
5-exp(1/2*x/(exp(5)-2*ln(2))^2/(25/4+5/4*x))
Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int -\frac {2 e^{\frac {2 x}{e^{10} (25+5 x)+e^5 (-50-10 x) \log (4)+(25+5 x) \log ^2(4)}}}{e^{10} \left (25+10 x+x^2\right )+e^5 \left (-50-20 x-2 x^2\right ) \log (4)+\left (25+10 x+x^2\right ) \log ^2(4)} \, dx=-e^{\frac {2 x}{5 (5+x) \left (e^5-\log (4)\right )^2}} \] Input:
Integrate[(-2*E^((2*x)/(E^10*(25 + 5*x) + E^5*(-50 - 10*x)*Log[4] + (25 + 5*x)*Log[4]^2)))/(E^10*(25 + 10*x + x^2) + E^5*(-50 - 20*x - 2*x^2)*Log[4] + (25 + 10*x + x^2)*Log[4]^2),x]
Output:
-E^((2*x)/(5*(5 + x)*(E^5 - Log[4])^2))
Time = 0.33 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {6, 27, 6, 25, 25, 27, 2007, 7257}
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.
| \(\displaystyle \int -\frac {2 \exp \left (\frac {2 x}{e^{10} (5 x+25)+(5 x+25) \log ^2(4)+e^5 (-10 x-50) \log (4)}\right )}{e^{10} \left (x^2+10 x+25\right )+\left (x^2+10 x+25\right ) \log ^2(4)+e^5 \left (-2 x^2-20 x-50\right ) \log (4)} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int -\frac {2 \exp \left (\frac {2 x}{e^{10} (5 x+25)+(5 x+25) \log ^2(4)+e^5 (-10 x-50) \log (4)}\right )}{\left (x^2+10 x+25\right ) \left (e^{10}+\log ^2(4)\right )+e^5 \left (-2 x^2-20 x-50\right ) \log (4)}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -2 \int -\frac {\exp \left (\frac {2 x}{5 \left (\log ^2(4) (x+5)-2 e^5 \log (4) (x+5)+e^{10} (x+5)\right )}\right )}{2 e^5 \left (x^2+10 x+25\right ) \log (4)-\left (x^2+10 x+25\right ) \left (e^{10}+\log ^2(4)\right )}dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle -2 \int -\frac {\exp \left (\frac {2 x}{5 \left (\log ^2(4) (x+5)-2 e^5 \log (4) (x+5)+e^{10} (x+5)\right )}\right )}{\left (x^2+10 x+25\right ) \left (-e^{10}+2 e^5 \log (4)-\log ^2(4)\right )}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle 2 \int -\frac {\exp \left (\frac {2 x}{5 \left (\log ^2(4) (x+5)-2 e^5 \log (4) (x+5)+e^{10} (x+5)\right )}\right )}{\left (x^2+10 x+25\right ) \left (e^5-\log (4)\right )^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -2 \int \frac {\exp \left (\frac {2 x}{5 \left (\log ^2(4) (x+5)-2 e^5 \log (4) (x+5)+e^{10} (x+5)\right )}\right )}{\left (x^2+10 x+25\right ) \left (e^5-\log (4)\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {2 \int \frac {\exp \left (\frac {2 x}{5 \left (\log ^2(4) (x+5)-2 e^5 \log (4) (x+5)+e^{10} (x+5)\right )}\right )}{x^2+10 x+25}dx}{\left (e^5-\log (4)\right )^2}\) |
| \(\Big \downarrow \) 2007 |
| \(\displaystyle -\frac {2 \int \frac {\exp \left (\frac {2 x}{5 \left (\log ^2(4) (x+5)-2 e^5 \log (4) (x+5)+e^{10} (x+5)\right )}\right )}{(x+5)^2}dx}{\left (e^5-\log (4)\right )^2}\) |
| \(\Big \downarrow \) 7257 |
| \(\displaystyle -\exp \left (\frac {2 x}{5 \left (e^{10} (x+5)+(x+5) \log ^2(4)-2 e^5 (x+5) \log (4)\right )}\right )\) |
Input:
Int[(-2*E^((2*x)/(E^10*(25 + 5*x) + E^5*(-50 - 10*x)*Log[4] + (25 + 5*x)*L og[4]^2)))/(E^10*(25 + 10*x + x^2) + E^5*(-50 - 20*x - 2*x^2)*Log[4] + (25 + 10*x + x^2)*Log[4]^2),x]
Output:
-E^((2*x)/(5*(E^10*(5 + x) - 2*E^5*(5 + x)*Log[4] + (5 + x)*Log[4]^2)))
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Sim p[q*(F^v/Log[F]), x] /; !FalseQ[q]] /; FreeQ[F, x]
Time = 0.57 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12
| method | result | size |
| risch | \(-{\mathrm e}^{\frac {2 x}{5 \left (5+x \right ) \left (4 \ln \left (2\right )^{2}-4 \,{\mathrm e}^{5} \ln \left (2\right )+{\mathrm e}^{10}\right )}}\) | \(29\) |
| gosper | \(-{\mathrm e}^{\frac {2 x}{5 \left (4 x \ln \left (2\right )^{2}-4 \,{\mathrm e}^{5} \ln \left (2\right ) x +{\mathrm e}^{10} x +20 \ln \left (2\right )^{2}-20 \,{\mathrm e}^{5} \ln \left (2\right )+5 \,{\mathrm e}^{10}\right )}}\) | \(48\) |
| orering | \(-\frac {\left (5+x \right )^{2} \left (4 \ln \left (2\right )^{2}-4 \,{\mathrm e}^{5} \ln \left (2\right )+{\mathrm e}^{10}\right ) {\mathrm e}^{\frac {2 x}{4 \left (25+5 x \right ) \ln \left (2\right )^{2}+2 \left (-10 x -50\right ) {\mathrm e}^{5} \ln \left (2\right )+\left (25+5 x \right ) {\mathrm e}^{10}}}}{4 \left (x^{2}+10 x +25\right ) \ln \left (2\right )^{2}+2 \left (-2 x^{2}-20 x -50\right ) {\mathrm e}^{5} \ln \left (2\right )+\left (x^{2}+10 x +25\right ) {\mathrm e}^{10}}\) | \(108\) |
| derivativedivides | \(\frac {25 \left (4 \,{\mathrm e}^{5} \ln \left (2\right )-4 \ln \left (2\right )^{2}-{\mathrm e}^{10}\right ) \left (\frac {4 \ln \left (2\right )^{2}}{25}-\frac {4 \,{\mathrm e}^{5} \ln \left (2\right )}{25}+\frac {{\mathrm e}^{10}}{25}\right ) {\mathrm e}^{\frac {2}{5 \left (4 \ln \left (2\right )^{2}-4 \,{\mathrm e}^{5} \ln \left (2\right )+{\mathrm e}^{10}\right )}-\frac {2}{\left (5+x \right ) \left (4 \ln \left (2\right )^{2}-4 \,{\mathrm e}^{5} \ln \left (2\right )+{\mathrm e}^{10}\right )}}}{\left (4 \ln \left (2\right )^{2}-4 \,{\mathrm e}^{5} \ln \left (2\right )+{\mathrm e}^{10}\right )^{2}}\) | \(109\) |
| default | \(\frac {25 \left (4 \,{\mathrm e}^{5} \ln \left (2\right )-4 \ln \left (2\right )^{2}-{\mathrm e}^{10}\right ) \left (\frac {4 \ln \left (2\right )^{2}}{25}-\frac {4 \,{\mathrm e}^{5} \ln \left (2\right )}{25}+\frac {{\mathrm e}^{10}}{25}\right ) {\mathrm e}^{\frac {2}{5 \left (4 \ln \left (2\right )^{2}-4 \,{\mathrm e}^{5} \ln \left (2\right )+{\mathrm e}^{10}\right )}-\frac {2}{\left (5+x \right ) \left (4 \ln \left (2\right )^{2}-4 \,{\mathrm e}^{5} \ln \left (2\right )+{\mathrm e}^{10}\right )}}}{\left (4 \ln \left (2\right )^{2}-4 \,{\mathrm e}^{5} \ln \left (2\right )+{\mathrm e}^{10}\right )^{2}}\) | \(109\) |
| norman | \(\frac {\left (5 \,{\mathrm e}^{5}-10 \ln \left (2\right )\right ) {\mathrm e}^{\frac {2 x}{4 \left (25+5 x \right ) \ln \left (2\right )^{2}+2 \left (-10 x -50\right ) {\mathrm e}^{5} \ln \left (2\right )+\left (25+5 x \right ) {\mathrm e}^{10}}}+\left ({\mathrm e}^{5}-2 \ln \left (2\right )\right ) x \,{\mathrm e}^{\frac {2 x}{4 \left (25+5 x \right ) \ln \left (2\right )^{2}+2 \left (-10 x -50\right ) {\mathrm e}^{5} \ln \left (2\right )+\left (25+5 x \right ) {\mathrm e}^{10}}}}{\left (5+x \right ) \left (-{\mathrm e}^{5}+2 \ln \left (2\right )\right )}\) | \(116\) |
| parallelrisch | \(-\frac {4 \ln \left (2\right )^{2} {\mathrm e}^{\frac {2 x}{5 \left (4 x \ln \left (2\right )^{2}-4 \,{\mathrm e}^{5} \ln \left (2\right ) x +{\mathrm e}^{10} x +20 \ln \left (2\right )^{2}-20 \,{\mathrm e}^{5} \ln \left (2\right )+5 \,{\mathrm e}^{10}\right )}}-4 \ln \left (2\right ) {\mathrm e}^{5} {\mathrm e}^{\frac {2 x}{5 \left (4 x \ln \left (2\right )^{2}-4 \,{\mathrm e}^{5} \ln \left (2\right ) x +{\mathrm e}^{10} x +20 \ln \left (2\right )^{2}-20 \,{\mathrm e}^{5} \ln \left (2\right )+5 \,{\mathrm e}^{10}\right )}}+{\mathrm e}^{10} {\mathrm e}^{\frac {2 x}{5 \left (4 x \ln \left (2\right )^{2}-4 \,{\mathrm e}^{5} \ln \left (2\right ) x +{\mathrm e}^{10} x +20 \ln \left (2\right )^{2}-20 \,{\mathrm e}^{5} \ln \left (2\right )+5 \,{\mathrm e}^{10}\right )}}}{4 \ln \left (2\right )^{2}-4 \,{\mathrm e}^{5} \ln \left (2\right )+{\mathrm e}^{10}}\) | \(175\) |
Input:
int(-2*exp(2*x/(4*(25+5*x)*ln(2)^2+2*(-10*x-50)*exp(5)*ln(2)+(25+5*x)*exp( 5)^2))/(4*(x^2+10*x+25)*ln(2)^2+2*(-2*x^2-20*x-50)*exp(5)*ln(2)+(x^2+10*x+ 25)*exp(5)^2),x,method=_RETURNVERBOSE)
Output:
-exp(2/5*x/(5+x)/(4*ln(2)^2-4*exp(5)*ln(2)+exp(10)))
Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int -\frac {2 e^{\frac {2 x}{e^{10} (25+5 x)+e^5 (-50-10 x) \log (4)+(25+5 x) \log ^2(4)}}}{e^{10} \left (25+10 x+x^2\right )+e^5 \left (-50-20 x-2 x^2\right ) \log (4)+\left (25+10 x+x^2\right ) \log ^2(4)} \, dx=-e^{\left (-\frac {2 \, x}{5 \, {\left (4 \, {\left (x + 5\right )} e^{5} \log \left (2\right ) - 4 \, {\left (x + 5\right )} \log \left (2\right )^{2} - {\left (x + 5\right )} e^{10}\right )}}\right )} \] Input:
integrate(-2*exp(2*x/(4*(25+5*x)*log(2)^2+2*(-10*x-50)*exp(5)*log(2)+(25+5 *x)*exp(5)^2))/(4*(x^2+10*x+25)*log(2)^2+2*(-2*x^2-20*x-50)*exp(5)*log(2)+ (x^2+10*x+25)*exp(5)^2),x, algorithm="fricas")
Output:
-e^(-2/5*x/(4*(x + 5)*e^5*log(2) - 4*(x + 5)*log(2)^2 - (x + 5)*e^10))
Time = 0.20 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int -\frac {2 e^{\frac {2 x}{e^{10} (25+5 x)+e^5 (-50-10 x) \log (4)+(25+5 x) \log ^2(4)}}}{e^{10} \left (25+10 x+x^2\right )+e^5 \left (-50-20 x-2 x^2\right ) \log (4)+\left (25+10 x+x^2\right ) \log ^2(4)} \, dx=- e^{\frac {2 x}{\left (- 20 x - 100\right ) e^{5} \log {\left (2 \right )} + \left (5 x + 25\right ) e^{10} + \left (20 x + 100\right ) \log {\left (2 \right )}^{2}}} \] Input:
integrate(-2*exp(2*x/(4*(25+5*x)*ln(2)**2+2*(-10*x-50)*exp(5)*ln(2)+(25+5* x)*exp(5)**2))/(4*(x**2+10*x+25)*ln(2)**2+2*(-2*x**2-20*x-50)*exp(5)*ln(2) +(x**2+10*x+25)*exp(5)**2),x)
Output:
-exp(2*x/((-20*x - 100)*exp(5)*log(2) + (5*x + 25)*exp(10) + (20*x + 100)* log(2)**2))
Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (22) = 44\).
Time = 0.15 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.50 \[ \int -\frac {2 e^{\frac {2 x}{e^{10} (25+5 x)+e^5 (-50-10 x) \log (4)+(25+5 x) \log ^2(4)}}}{e^{10} \left (25+10 x+x^2\right )+e^5 \left (-50-20 x-2 x^2\right ) \log (4)+\left (25+10 x+x^2\right ) \log ^2(4)} \, dx=-e^{\left (\frac {2}{{\left (4 \, e^{5} \log \left (2\right ) - 4 \, \log \left (2\right )^{2} - e^{10}\right )} x + 20 \, e^{5} \log \left (2\right ) - 20 \, \log \left (2\right )^{2} - 5 \, e^{10}} - \frac {2}{5 \, {\left (4 \, e^{5} \log \left (2\right ) - 4 \, \log \left (2\right )^{2} - e^{10}\right )}}\right )} \] Input:
integrate(-2*exp(2*x/(4*(25+5*x)*log(2)^2+2*(-10*x-50)*exp(5)*log(2)+(25+5 *x)*exp(5)^2))/(4*(x^2+10*x+25)*log(2)^2+2*(-2*x^2-20*x-50)*exp(5)*log(2)+ (x^2+10*x+25)*exp(5)^2),x, algorithm="maxima")
Output:
-e^(2/((4*e^5*log(2) - 4*log(2)^2 - e^10)*x + 20*e^5*log(2) - 20*log(2)^2 - 5*e^10) - 2/5/(4*e^5*log(2) - 4*log(2)^2 - e^10))
\[ \int -\frac {2 e^{\frac {2 x}{e^{10} (25+5 x)+e^5 (-50-10 x) \log (4)+(25+5 x) \log ^2(4)}}}{e^{10} \left (25+10 x+x^2\right )+e^5 \left (-50-20 x-2 x^2\right ) \log (4)+\left (25+10 x+x^2\right ) \log ^2(4)} \, dx=\int { \frac {2 \, e^{\left (-\frac {2 \, x}{5 \, {\left (4 \, {\left (x + 5\right )} e^{5} \log \left (2\right ) - 4 \, {\left (x + 5\right )} \log \left (2\right )^{2} - {\left (x + 5\right )} e^{10}\right )}}\right )}}{4 \, {\left (x^{2} + 10 \, x + 25\right )} e^{5} \log \left (2\right ) - 4 \, {\left (x^{2} + 10 \, x + 25\right )} \log \left (2\right )^{2} - {\left (x^{2} + 10 \, x + 25\right )} e^{10}} \,d x } \] Input:
integrate(-2*exp(2*x/(4*(25+5*x)*log(2)^2+2*(-10*x-50)*exp(5)*log(2)+(25+5 *x)*exp(5)^2))/(4*(x^2+10*x+25)*log(2)^2+2*(-2*x^2-20*x-50)*exp(5)*log(2)+ (x^2+10*x+25)*exp(5)^2),x, algorithm="giac")
Output:
integrate(2*e^(-2/5*x/(4*(x + 5)*e^5*log(2) - 4*(x + 5)*log(2)^2 - (x + 5) *e^10))/(4*(x^2 + 10*x + 25)*e^5*log(2) - 4*(x^2 + 10*x + 25)*log(2)^2 - ( x^2 + 10*x + 25)*e^10), x)
Time = 0.70 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69 \[ \int -\frac {2 e^{\frac {2 x}{e^{10} (25+5 x)+e^5 (-50-10 x) \log (4)+(25+5 x) \log ^2(4)}}}{e^{10} \left (25+10 x+x^2\right )+e^5 \left (-50-20 x-2 x^2\right ) \log (4)+\left (25+10 x+x^2\right ) \log ^2(4)} \, dx=-{\mathrm {e}}^{\frac {2\,x}{25\,{\mathrm {e}}^{10}-100\,{\mathrm {e}}^5\,\ln \left (2\right )+5\,x\,{\mathrm {e}}^{10}+20\,x\,{\ln \left (2\right )}^2+100\,{\ln \left (2\right )}^2-20\,x\,{\mathrm {e}}^5\,\ln \left (2\right )}} \] Input:
int(-(2*exp((2*x)/(4*log(2)^2*(5*x + 25) + exp(10)*(5*x + 25) - 2*exp(5)*l og(2)*(10*x + 50))))/(4*log(2)^2*(10*x + x^2 + 25) + exp(10)*(10*x + x^2 + 25) - 2*exp(5)*log(2)*(20*x + 2*x^2 + 50)),x)
Output:
-exp((2*x)/(25*exp(10) - 100*exp(5)*log(2) + 5*x*exp(10) + 20*x*log(2)^2 + 100*log(2)^2 - 20*x*exp(5)*log(2)))
Time = 0.18 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.88 \[ \int -\frac {2 e^{\frac {2 x}{e^{10} (25+5 x)+e^5 (-50-10 x) \log (4)+(25+5 x) \log ^2(4)}}}{e^{10} \left (25+10 x+x^2\right )+e^5 \left (-50-20 x-2 x^2\right ) \log (4)+\left (25+10 x+x^2\right ) \log ^2(4)} \, dx=-e^{\frac {2 x}{20 \mathrm {log}\left (2\right )^{2} x +100 \mathrm {log}\left (2\right )^{2}-20 \,\mathrm {log}\left (2\right ) e^{5} x -100 \,\mathrm {log}\left (2\right ) e^{5}+5 e^{10} x +25 e^{10}}} \] Input:
int(-2*exp(2*x/(4*(25+5*x)*log(2)^2+2*(-10*x-50)*exp(5)*log(2)+(25+5*x)*ex p(5)^2))/(4*(x^2+10*x+25)*log(2)^2+2*(-2*x^2-20*x-50)*exp(5)*log(2)+(x^2+1 0*x+25)*exp(5)^2),x)
Output:
- e**((2*x)/(20*log(2)**2*x + 100*log(2)**2 - 20*log(2)*e**5*x - 100*log( 2)*e**5 + 5*e**10*x + 25*e**10))