Integrand size = 135, antiderivative size = 23 \[ \int \frac {-45 e^4+e^x (-9-9 x)+45 \log (3)-9 \log (5)}{4+20 e^4 x+25 e^8 x^2+e^{2 x} x^2+\left (-20 x-50 e^4 x^2\right ) \log (3)+25 x^2 \log ^2(3)+\left (4 x+10 e^4 x^2-10 x^2 \log (3)\right ) \log (5)+x^2 \log ^2(5)+e^x \left (4 x+10 e^4 x^2-10 x^2 \log (3)+2 x^2 \log (5)\right )} \, dx=\frac {9}{2+x \left (5 e^4+e^x-5 \log (3)+\log (5)\right )} \]
Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {-45 e^4+e^x (-9-9 x)+45 \log (3)-9 \log (5)}{4+20 e^4 x+25 e^8 x^2+e^{2 x} x^2+\left (-20 x-50 e^4 x^2\right ) \log (3)+25 x^2 \log ^2(3)+\left (4 x+10 e^4 x^2-10 x^2 \log (3)\right ) \log (5)+x^2 \log ^2(5)+e^x \left (4 x+10 e^4 x^2-10 x^2 \log (3)+2 x^2 \log (5)\right )} \, dx=\frac {9}{2+5 e^4 x+e^x x-x \log \left (\frac {243}{5}\right )} \]
Integrate[(-45*E^4 + E^x*(-9 - 9*x) + 45*Log[3] - 9*Log[5])/(4 + 20*E^4*x + 25*E^8*x^2 + E^(2*x)*x^2 + (-20*x - 50*E^4*x^2)*Log[3] + 25*x^2*Log[3]^2 + (4*x + 10*E^4*x^2 - 10*x^2*Log[3])*Log[5] + x^2*Log[5]^2 + E^x*(4*x + 1 0*E^4*x^2 - 10*x^2*Log[3] + 2*x^2*Log[5])),x]
Time = 0.54 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {6, 6, 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.
\(\displaystyle \int \frac {e^x (-9 x-9)-45 e^4-9 \log (5)+45 \log (3)}{e^{2 x} x^2+25 e^8 x^2+x^2 \log ^2(5)+25 x^2 \log ^2(3)+e^x \left (10 e^4 x^2+2 x^2 \log (5)-10 x^2 \log (3)+4 x\right )+\log (5) \left (10 e^4 x^2-10 x^2 \log (3)+4 x\right )+\left (-50 e^4 x^2-20 x\right ) \log (3)+20 e^4 x+4} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {e^x (-9 x-9)-45 e^4-9 \log (5)+45 \log (3)}{e^{2 x} x^2+x^2 \log ^2(5)+25 x^2 \left (e^8+\log ^2(3)\right )+e^x \left (10 e^4 x^2+2 x^2 \log (5)-10 x^2 \log (3)+4 x\right )+\log (5) \left (10 e^4 x^2-10 x^2 \log (3)+4 x\right )+\left (-50 e^4 x^2-20 x\right ) \log (3)+20 e^4 x+4}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {e^x (-9 x-9)-45 e^4-9 \log (5)+45 \log (3)}{e^{2 x} x^2+x^2 \left (25 \left (e^8+\log ^2(3)\right )+\log ^2(5)\right )+e^x \left (10 e^4 x^2+2 x^2 \log (5)-10 x^2 \log (3)+4 x\right )+\log (5) \left (10 e^4 x^2-10 x^2 \log (3)+4 x\right )+\left (-50 e^4 x^2-20 x\right ) \log (3)+20 e^4 x+4}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {9 \left (-e^x (x+1)-5 e^4 \left (1-\frac {\log (3) \log \left (\frac {243}{5}\right )}{e^4 \log (243)}\right )\right )}{\left (e^x x+5 e^4 x \left (1-\frac {\log \left (\frac {243}{5}\right )}{5 e^4}\right )+2\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 9 \int -\frac {e^x (x+1)-\log \left (\frac {243}{5}\right )+5 e^4}{\left (e^x x+\left (5 e^4-\log \left (\frac {243}{5}\right )\right ) x+2\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -9 \int \frac {e^x (x+1)-\log \left (\frac {243}{5}\right )+5 e^4}{\left (e^x x+\left (5 e^4-\log \left (\frac {243}{5}\right )\right ) x+2\right )^2}dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle \frac {9}{e^x x+x \left (5 e^4-\log \left (\frac {243}{5}\right )\right )+2}\) |
Int[(-45*E^4 + E^x*(-9 - 9*x) + 45*Log[3] - 9*Log[5])/(4 + 20*E^4*x + 25*E ^8*x^2 + E^(2*x)*x^2 + (-20*x - 50*E^4*x^2)*Log[3] + 25*x^2*Log[3]^2 + (4* x + 10*E^4*x^2 - 10*x^2*Log[3])*Log[5] + x^2*Log[5]^2 + E^x*(4*x + 10*E^4* x^2 - 10*x^2*Log[3] + 2*x^2*Log[5])),x]
3.8.66.3.1 Defintions of rubi rules used
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_)*(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 = 1.71 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09
method | result | size |
norman | \(\frac {9}{5 x \,{\mathrm e}^{4}+{\mathrm e}^{x} x -5 x \ln \left (3\right )+x \ln \left (5\right )+2}\) | \(25\) |
risch | \(\frac {9}{5 x \,{\mathrm e}^{4}+{\mathrm e}^{x} x -5 x \ln \left (3\right )+x \ln \left (5\right )+2}\) | \(25\) |
parallelrisch | \(\frac {9}{5 x \,{\mathrm e}^{4}+{\mathrm e}^{x} x -5 x \ln \left (3\right )+x \ln \left (5\right )+2}\) | \(25\) |
int(((-9*x-9)*exp(x)-9*ln(5)+45*ln(3)-45*exp(4))/(exp(x)^2*x^2+(2*x^2*ln(5 )-10*x^2*ln(3)+10*x^2*exp(4)+4*x)*exp(x)+x^2*ln(5)^2+(-10*x^2*ln(3)+10*x^2 *exp(4)+4*x)*ln(5)+25*x^2*ln(3)^2+(-50*x^2*exp(4)-20*x)*ln(3)+25*x^2*exp(4 )^2+20*x*exp(4)+4),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {-45 e^4+e^x (-9-9 x)+45 \log (3)-9 \log (5)}{4+20 e^4 x+25 e^8 x^2+e^{2 x} x^2+\left (-20 x-50 e^4 x^2\right ) \log (3)+25 x^2 \log ^2(3)+\left (4 x+10 e^4 x^2-10 x^2 \log (3)\right ) \log (5)+x^2 \log ^2(5)+e^x \left (4 x+10 e^4 x^2-10 x^2 \log (3)+2 x^2 \log (5)\right )} \, dx=\frac {9}{5 \, x e^{4} + x e^{x} + x \log \left (5\right ) - 5 \, x \log \left (3\right ) + 2} \]
integrate(((-9*x-9)*exp(x)-9*log(5)+45*log(3)-45*exp(4))/(exp(x)^2*x^2+(2* x^2*log(5)-10*x^2*log(3)+10*x^2*exp(4)+4*x)*exp(x)+x^2*log(5)^2+(-10*x^2*l og(3)+10*x^2*exp(4)+4*x)*log(5)+25*x^2*log(3)^2+(-50*x^2*exp(4)-20*x)*log( 3)+25*x^2*exp(4)^2+20*x*exp(4)+4),x, algorithm=\
Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {-45 e^4+e^x (-9-9 x)+45 \log (3)-9 \log (5)}{4+20 e^4 x+25 e^8 x^2+e^{2 x} x^2+\left (-20 x-50 e^4 x^2\right ) \log (3)+25 x^2 \log ^2(3)+\left (4 x+10 e^4 x^2-10 x^2 \log (3)\right ) \log (5)+x^2 \log ^2(5)+e^x \left (4 x+10 e^4 x^2-10 x^2 \log (3)+2 x^2 \log (5)\right )} \, dx=\frac {9}{x e^{x} - 5 x \log {\left (3 \right )} + x \log {\left (5 \right )} + 5 x e^{4} + 2} \]
integrate(((-9*x-9)*exp(x)-9*ln(5)+45*ln(3)-45*exp(4))/(exp(x)**2*x**2+(2* x**2*ln(5)-10*x**2*ln(3)+10*x**2*exp(4)+4*x)*exp(x)+x**2*ln(5)**2+(-10*x** 2*ln(3)+10*x**2*exp(4)+4*x)*ln(5)+25*x**2*ln(3)**2+(-50*x**2*exp(4)-20*x)* ln(3)+25*x**2*exp(4)**2+20*x*exp(4)+4),x)
Time = 0.32 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {-45 e^4+e^x (-9-9 x)+45 \log (3)-9 \log (5)}{4+20 e^4 x+25 e^8 x^2+e^{2 x} x^2+\left (-20 x-50 e^4 x^2\right ) \log (3)+25 x^2 \log ^2(3)+\left (4 x+10 e^4 x^2-10 x^2 \log (3)\right ) \log (5)+x^2 \log ^2(5)+e^x \left (4 x+10 e^4 x^2-10 x^2 \log (3)+2 x^2 \log (5)\right )} \, dx=\frac {9}{x {\left (5 \, e^{4} + \log \left (5\right ) - 5 \, \log \left (3\right )\right )} + x e^{x} + 2} \]
integrate(((-9*x-9)*exp(x)-9*log(5)+45*log(3)-45*exp(4))/(exp(x)^2*x^2+(2* x^2*log(5)-10*x^2*log(3)+10*x^2*exp(4)+4*x)*exp(x)+x^2*log(5)^2+(-10*x^2*l og(3)+10*x^2*exp(4)+4*x)*log(5)+25*x^2*log(3)^2+(-50*x^2*exp(4)-20*x)*log( 3)+25*x^2*exp(4)^2+20*x*exp(4)+4),x, algorithm=\
Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {-45 e^4+e^x (-9-9 x)+45 \log (3)-9 \log (5)}{4+20 e^4 x+25 e^8 x^2+e^{2 x} x^2+\left (-20 x-50 e^4 x^2\right ) \log (3)+25 x^2 \log ^2(3)+\left (4 x+10 e^4 x^2-10 x^2 \log (3)\right ) \log (5)+x^2 \log ^2(5)+e^x \left (4 x+10 e^4 x^2-10 x^2 \log (3)+2 x^2 \log (5)\right )} \, dx=\frac {9}{5 \, x e^{4} + x e^{x} + x \log \left (5\right ) - 5 \, x \log \left (3\right ) + 2} \]
integrate(((-9*x-9)*exp(x)-9*log(5)+45*log(3)-45*exp(4))/(exp(x)^2*x^2+(2* x^2*log(5)-10*x^2*log(3)+10*x^2*exp(4)+4*x)*exp(x)+x^2*log(5)^2+(-10*x^2*l og(3)+10*x^2*exp(4)+4*x)*log(5)+25*x^2*log(3)^2+(-50*x^2*exp(4)-20*x)*log( 3)+25*x^2*exp(4)^2+20*x*exp(4)+4),x, algorithm=\
Time = 9.11 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.96 \[ \int \frac {-45 e^4+e^x (-9-9 x)+45 \log (3)-9 \log (5)}{4+20 e^4 x+25 e^8 x^2+e^{2 x} x^2+\left (-20 x-50 e^4 x^2\right ) \log (3)+25 x^2 \log ^2(3)+\left (4 x+10 e^4 x^2-10 x^2 \log (3)\right ) \log (5)+x^2 \log ^2(5)+e^x \left (4 x+10 e^4 x^2-10 x^2 \log (3)+2 x^2 \log (5)\right )} \, dx=-\frac {\frac {9\,x\,{\mathrm {e}}^x}{2}+x\,\left (\frac {45\,{\mathrm {e}}^4}{2}-\ln \left (\frac {31381059609\,\sqrt {15}}{3125}\right )\right )}{5\,x\,{\mathrm {e}}^4-5\,x\,\ln \left (3\right )+x\,\ln \left (5\right )+x\,{\mathrm {e}}^x+2} \]
int(-(45*exp(4) - 45*log(3) + 9*log(5) + exp(x)*(9*x + 9))/(25*x^2*log(3)^ 2 + x^2*log(5)^2 + 20*x*exp(4) + log(5)*(4*x + 10*x^2*exp(4) - 10*x^2*log( 3)) + exp(x)*(4*x + 10*x^2*exp(4) - 10*x^2*log(3) + 2*x^2*log(5)) + x^2*ex p(2*x) + 25*x^2*exp(8) - log(3)*(20*x + 50*x^2*exp(4)) + 4),x)