Integrand size = 89, antiderivative size = 30 \[ \int \frac {e^x (130+80 x)+e^x \left (50 x^2+80 x^3+e^5 \left (169 x^2+208 x^3+64 x^4\right )\right ) \log (x)+e^x \left (-130-30 x+80 x^2\right ) \log (x) \log (\log (x))}{\left (169 x^2+208 x^3+64 x^4\right ) \log (x)} \, dx=e^x \left (e^5+\frac {x+\log (\log (x))}{\frac {x}{2}+\frac {4}{5} x (1+x)}\right ) \]
Time = 3.50 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {e^x (130+80 x)+e^x \left (50 x^2+80 x^3+e^5 \left (169 x^2+208 x^3+64 x^4\right )\right ) \log (x)+e^x \left (-130-30 x+80 x^2\right ) \log (x) \log (\log (x))}{\left (169 x^2+208 x^3+64 x^4\right ) \log (x)} \, dx=\frac {e^x \left (x \left (10+e^5 (13+8 x)\right )+10 \log (\log (x))\right )}{x (13+8 x)} \]
Integrate[(E^x*(130 + 80*x) + E^x*(50*x^2 + 80*x^3 + E^5*(169*x^2 + 208*x^ 3 + 64*x^4))*Log[x] + E^x*(-130 - 30*x + 80*x^2)*Log[x]*Log[Log[x]])/((169 *x^2 + 208*x^3 + 64*x^4)*Log[x]),x]
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.
| \(\displaystyle \int \frac {e^x \left (80 x^2-30 x-130\right ) \log (x) \log (\log (x))+e^x \left (80 x^3+50 x^2+e^5 \left (64 x^4+208 x^3+169 x^2\right )\right ) \log (x)+e^x (80 x+130)}{\left (64 x^4+208 x^3+169 x^2\right ) \log (x)} \, dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {e^x \left (80 x^2-30 x-130\right ) \log (x) \log (\log (x))+e^x \left (80 x^3+50 x^2+e^5 \left (64 x^4+208 x^3+169 x^2\right )\right ) \log (x)+e^x (80 x+130)}{x^2 \left (64 x^2+208 x+169\right ) \log (x)}dx\) |
| \(\Big \downarrow \) 2007 |
| \(\displaystyle \int \frac {e^x \left (80 x^2-30 x-130\right ) \log (x) \log (\log (x))+e^x \left (80 x^3+50 x^2+e^5 \left (64 x^4+208 x^3+169 x^2\right )\right ) \log (x)+e^x (80 x+130)}{x^2 (8 x+13)^2 \log (x)}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {16 e^x \left (-64 e^5 x^4 \log (x)-80 \left (1+\frac {13 e^5}{5}\right ) x^3 \log (x)-50 \left (1+\frac {169 e^5}{50}\right ) x^2 \log (x)-80 x^2 \log (x) \log (\log (x))-80 x+30 x \log (x) \log (\log (x))+130 \log (x) \log (\log (x))-130\right )}{2197 x \log (x)}+\frac {128 e^x \left (64 e^5 x^4 \log (x)+80 \left (1+\frac {13 e^5}{5}\right ) x^3 \log (x)+50 \left (1+\frac {169 e^5}{50}\right ) x^2 \log (x)+80 x^2 \log (x) \log (\log (x))+80 x-30 x \log (x) \log (\log (x))-130 \log (x) \log (\log (x))+130\right )}{2197 (8 x+13) \log (x)}+\frac {e^x \left (64 e^5 x^4 \log (x)+80 \left (1+\frac {13 e^5}{5}\right ) x^3 \log (x)+50 \left (1+\frac {169 e^5}{50}\right ) x^2 \log (x)+80 x^2 \log (x) \log (\log (x))+80 x-30 x \log (x) \log (\log (x))-130 \log (x) \log (\log (x))+130\right )}{169 x^2 \log (x)}+\frac {64 e^x \left (64 e^5 x^4 \log (x)+80 \left (1+\frac {13 e^5}{5}\right ) x^3 \log (x)+50 \left (1+\frac {169 e^5}{50}\right ) x^2 \log (x)+80 x^2 \log (x) \log (\log (x))+80 x-30 x \log (x) \log (\log (x))-130 \log (x) \log (\log (x))+130\right )}{169 (8 x+13)^2 \log (x)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {10}{13} \int \frac {e^x}{x^2 \log (x)}dx-\frac {10}{13} \int \frac {e^x \log (\log (x))}{x^2}dx-\frac {80}{169} \int \frac {e^x}{x \log (x)}dx+\frac {640}{169} \int \frac {e^x}{(8 x+13) \log (x)}dx+\frac {10}{13} \int \frac {e^x \log (\log (x))}{x}dx+\frac {640}{13} \int \frac {e^x \log (\log (x))}{(8 x+13)^2}dx-\frac {80}{13} \int \frac {e^x \log (\log (x))}{8 x+13}dx-\frac {3 \left (50+169 e^5\right ) \operatorname {ExpIntegralEi}\left (\frac {1}{8} (8 x+13)\right )}{832 e^{13/8}}+\frac {11 \left (5+13 e^5\right ) \operatorname {ExpIntegralEi}\left (\frac {1}{8} (8 x+13)\right )}{32 e^{13/8}}-\frac {247}{64} e^{27/8} \operatorname {ExpIntegralEi}\left (\frac {1}{8} (8 x+13)\right )-\frac {20 \operatorname {ExpIntegralEi}\left (\frac {1}{8} (8 x+13)\right )}{13 e^{13/8}}+\frac {128}{169} e^{x+5} x^2+\frac {128 \left (10+13 e^5\right ) e^x x^2}{2197}-\frac {256 \left (5+13 e^5\right ) e^x x^2}{2197}-\frac {1280 e^x x}{2197}-\frac {464}{169} e^{x+5} x-\frac {16 \left (50+169 e^5\right ) e^x x}{2197}-\frac {256 \left (10+13 e^5\right ) e^x x}{2197}+\frac {928 \left (5+13 e^5\right ) e^x x}{2197}+\frac {3360 e^x}{2197}+\frac {971 e^{x+5}}{169}-\frac {169 e^{x+5}}{8 (8 x+13)}-\frac {\left (50+169 e^5\right ) e^x}{8 (8 x+13)}+\frac {13 \left (5+13 e^5\right ) e^x}{4 (8 x+13)}+\frac {42 \left (50+169 e^5\right ) e^x}{2197}+\frac {256 \left (10+13 e^5\right ) e^x}{2197}-\frac {1604 \left (5+13 e^5\right ) e^x}{2197}\) |
Int[(E^x*(130 + 80*x) + E^x*(50*x^2 + 80*x^3 + E^5*(169*x^2 + 208*x^3 + 64 *x^4))*Log[x] + E^x*(-130 - 30*x + 80*x^2)*Log[x]*Log[Log[x]])/((169*x^2 + 208*x^3 + 64*x^4)*Log[x]),x]
3.14.8.3.1 Defintions of rubi rules used
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[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Time = 9.00 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.33
| method | result | size |
| risch | \(\frac {10 \,{\mathrm e}^{x} \ln \left (\ln \left (x \right )\right )}{\left (8 x +13\right ) x}+\frac {\left (8 x \,{\mathrm e}^{5}+13 \,{\mathrm e}^{5}+10\right ) {\mathrm e}^{x}}{8 x +13}\) | \(40\) |
| parallelrisch | \(\frac {64 x^{2} {\mathrm e}^{5} {\mathrm e}^{x}+104 x \,{\mathrm e}^{5} {\mathrm e}^{x}+80 \,{\mathrm e}^{x} x +80 \,{\mathrm e}^{x} \ln \left (\ln \left (x \right )\right )}{8 x \left (8 x +13\right )}\) | \(42\) |
int(((80*x^2-30*x-130)*exp(x)*ln(x)*ln(ln(x))+((64*x^4+208*x^3+169*x^2)*ex p(5)+80*x^3+50*x^2)*exp(x)*ln(x)+(80*x+130)*exp(x))/(64*x^4+208*x^3+169*x^ 2)/ln(x),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.30 \[ \int \frac {e^x (130+80 x)+e^x \left (50 x^2+80 x^3+e^5 \left (169 x^2+208 x^3+64 x^4\right )\right ) \log (x)+e^x \left (-130-30 x+80 x^2\right ) \log (x) \log (\log (x))}{\left (169 x^2+208 x^3+64 x^4\right ) \log (x)} \, dx=\frac {{\left ({\left (8 \, x^{2} + 13 \, x\right )} e^{5} + 10 \, x\right )} e^{x} + 10 \, e^{x} \log \left (\log \left (x\right )\right )}{8 \, x^{2} + 13 \, x} \]
integrate(((80*x^2-30*x-130)*exp(x)*log(x)*log(log(x))+((64*x^4+208*x^3+16 9*x^2)*exp(5)+80*x^3+50*x^2)*exp(x)*log(x)+(80*x+130)*exp(x))/(64*x^4+208* x^3+169*x^2)/log(x),x, algorithm=\
Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {e^x (130+80 x)+e^x \left (50 x^2+80 x^3+e^5 \left (169 x^2+208 x^3+64 x^4\right )\right ) \log (x)+e^x \left (-130-30 x+80 x^2\right ) \log (x) \log (\log (x))}{\left (169 x^2+208 x^3+64 x^4\right ) \log (x)} \, dx=\frac {\left (8 x^{2} e^{5} + 10 x + 13 x e^{5} + 10 \log {\left (\log {\left (x \right )} \right )}\right ) e^{x}}{8 x^{2} + 13 x} \]
integrate(((80*x**2-30*x-130)*exp(x)*ln(x)*ln(ln(x))+((64*x**4+208*x**3+16 9*x**2)*exp(5)+80*x**3+50*x**2)*exp(x)*ln(x)+(80*x+130)*exp(x))/(64*x**4+2 08*x**3+169*x**2)/ln(x),x)
Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.30 \[ \int \frac {e^x (130+80 x)+e^x \left (50 x^2+80 x^3+e^5 \left (169 x^2+208 x^3+64 x^4\right )\right ) \log (x)+e^x \left (-130-30 x+80 x^2\right ) \log (x) \log (\log (x))}{\left (169 x^2+208 x^3+64 x^4\right ) \log (x)} \, dx=\frac {{\left (8 \, x^{2} e^{5} + x {\left (13 \, e^{5} + 10\right )}\right )} e^{x} + 10 \, e^{x} \log \left (\log \left (x\right )\right )}{8 \, x^{2} + 13 \, x} \]
integrate(((80*x^2-30*x-130)*exp(x)*log(x)*log(log(x))+((64*x^4+208*x^3+16 9*x^2)*exp(5)+80*x^3+50*x^2)*exp(x)*log(x)+(80*x+130)*exp(x))/(64*x^4+208* x^3+169*x^2)/log(x),x, algorithm=\
Time = 0.30 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.37 \[ \int \frac {e^x (130+80 x)+e^x \left (50 x^2+80 x^3+e^5 \left (169 x^2+208 x^3+64 x^4\right )\right ) \log (x)+e^x \left (-130-30 x+80 x^2\right ) \log (x) \log (\log (x))}{\left (169 x^2+208 x^3+64 x^4\right ) \log (x)} \, dx=\frac {8 \, x^{2} e^{\left (x + 5\right )} + 13 \, x e^{\left (x + 5\right )} + 10 \, x e^{x} + 10 \, e^{x} \log \left (\log \left (x\right )\right )}{8 \, x^{2} + 13 \, x} \]
integrate(((80*x^2-30*x-130)*exp(x)*log(x)*log(log(x))+((64*x^4+208*x^3+16 9*x^2)*exp(5)+80*x^3+50*x^2)*exp(x)*log(x)+(80*x+130)*exp(x))/(64*x^4+208* x^3+169*x^2)/log(x),x, algorithm=\
Timed out. \[ \int \frac {e^x (130+80 x)+e^x \left (50 x^2+80 x^3+e^5 \left (169 x^2+208 x^3+64 x^4\right )\right ) \log (x)+e^x \left (-130-30 x+80 x^2\right ) \log (x) \log (\log (x))}{\left (169 x^2+208 x^3+64 x^4\right ) \log (x)} \, dx=\int \frac {{\mathrm {e}}^x\,\left (80\,x+130\right )+{\mathrm {e}}^x\,\ln \left (x\right )\,\left ({\mathrm {e}}^5\,\left (64\,x^4+208\,x^3+169\,x^2\right )+50\,x^2+80\,x^3\right )-\ln \left (\ln \left (x\right )\right )\,{\mathrm {e}}^x\,\ln \left (x\right )\,\left (-80\,x^2+30\,x+130\right )}{\ln \left (x\right )\,\left (64\,x^4+208\,x^3+169\,x^2\right )} \,d x \]
int((exp(x)*(80*x + 130) + exp(x)*log(x)*(exp(5)*(169*x^2 + 208*x^3 + 64*x ^4) + 50*x^2 + 80*x^3) - log(log(x))*exp(x)*log(x)*(30*x - 80*x^2 + 130))/ (log(x)*(169*x^2 + 208*x^3 + 64*x^4)),x)