Integrand size = 121, antiderivative size = 24 \begin {dmath*} \int \frac {130 x^3+e^{2 x} \left (78 x+52 x^2\right )+e^x \left (208 x^2+52 x^3\right )+26 \log (2)}{12 e^{4 x} x^5+48 e^{3 x} x^6+12 x^9+12 x^6 \log (2)+3 x^3 \log ^2(2)+e^{2 x} \left (72 x^7+12 x^4 \log (2)\right )+e^x \left (48 x^8+24 x^5 \log (2)\right )} \, dx=5-\frac {13}{3 x^2 \left (2 x \left (e^x+x\right )^2+\log (2)\right )} \end {dmath*}
Time = 5.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \begin {dmath*} \int \frac {130 x^3+e^{2 x} \left (78 x+52 x^2\right )+e^x \left (208 x^2+52 x^3\right )+26 \log (2)}{12 e^{4 x} x^5+48 e^{3 x} x^6+12 x^9+12 x^6 \log (2)+3 x^3 \log ^2(2)+e^{2 x} \left (72 x^7+12 x^4 \log (2)\right )+e^x \left (48 x^8+24 x^5 \log (2)\right )} \, dx=-\frac {13}{3 x^2 \left (2 e^{2 x} x+4 e^x x^2+2 x^3+\log (2)\right )} \end {dmath*}
Integrate[(130*x^3 + E^(2*x)*(78*x + 52*x^2) + E^x*(208*x^2 + 52*x^3) + 26 *Log[2])/(12*E^(4*x)*x^5 + 48*E^(3*x)*x^6 + 12*x^9 + 12*x^6*Log[2] + 3*x^3 *Log[2]^2 + E^(2*x)*(72*x^7 + 12*x^4*Log[2]) + E^x*(48*x^8 + 24*x^5*Log[2] )),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.
\(\displaystyle \int \frac {130 x^3+e^{2 x} \left (52 x^2+78 x\right )+e^x \left (52 x^3+208 x^2\right )+26 \log (2)}{12 x^9+48 e^{3 x} x^6+12 x^6 \log (2)+12 e^{4 x} x^5+3 x^3 \log ^2(2)+e^x \left (48 x^8+24 x^5 \log (2)\right )+e^{2 x} \left (72 x^7+12 x^4 \log (2)\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {26 \left (5 x^3+2 e^x (x+4) x^2+e^{2 x} (2 x+3) x+\log (2)\right )}{3 x^3 \left (2 x^3+4 e^x x^2+2 e^{2 x} x+\log (2)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {26}{3} \int \frac {5 x^3+2 e^x (x+4) x^2+e^{2 x} (2 x+3) x+\log (2)}{x^3 \left (2 x^3+4 e^x x^2+2 e^{2 x} x+\log (2)\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {26}{3} \int \left (\frac {2 x+3}{x^3 \left (4 x^3+8 e^x x^2+4 e^{2 x} x+\log (4)\right )}-\frac {4 x^4+4 e^x x^3-4 x^3-4 e^x x^2+\log (4) x+\log (2)}{2 x^3 \left (2 x^3+4 e^x x^2+2 e^{2 x} x+\log (2)\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {26}{3} \left (2 \int \frac {1}{\left (2 x^3+4 e^x x^2+2 e^{2 x} x+\log (2)\right )^2}dx-2 \int \frac {e^x}{\left (2 x^3+4 e^x x^2+2 e^{2 x} x+\log (2)\right )^2}dx-\frac {1}{2} \log (2) \int \frac {1}{x^3 \left (2 x^3+4 e^x x^2+2 e^{2 x} x+\log (2)\right )^2}dx-\frac {1}{2} \log (4) \int \frac {1}{x^2 \left (2 x^3+4 e^x x^2+2 e^{2 x} x+\log (2)\right )^2}dx+2 \int \frac {e^x}{x \left (2 x^3+4 e^x x^2+2 e^{2 x} x+\log (2)\right )^2}dx-2 \int \frac {x}{\left (2 x^3+4 e^x x^2+2 e^{2 x} x+\log (2)\right )^2}dx+3 \int \frac {1}{x^3 \left (4 x^3+8 e^x x^2+4 e^{2 x} x+\log (4)\right )}dx+2 \int \frac {1}{x^2 \left (4 x^3+8 e^x x^2+4 e^{2 x} x+\log (4)\right )}dx\right )\) |
Int[(130*x^3 + E^(2*x)*(78*x + 52*x^2) + E^x*(208*x^2 + 52*x^3) + 26*Log[2 ])/(12*E^(4*x)*x^5 + 48*E^(3*x)*x^6 + 12*x^9 + 12*x^6*Log[2] + 3*x^3*Log[2 ]^2 + E^(2*x)*(72*x^7 + 12*x^4*Log[2]) + E^x*(48*x^8 + 24*x^5*Log[2])),x]
3.1.56.3.1 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_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25
method | result | size |
risch | \(-\frac {13}{3 x^{2} \left (2 x \,{\mathrm e}^{2 x}+4 \,{\mathrm e}^{x} x^{2}+2 x^{3}+\ln \left (2\right )\right )}\) | \(30\) |
parallelrisch | \(-\frac {13}{3 x^{2} \left (2 x \,{\mathrm e}^{2 x}+4 \,{\mathrm e}^{x} x^{2}+2 x^{3}+\ln \left (2\right )\right )}\) | \(30\) |
int(((52*x^2+78*x)*exp(x)^2+(52*x^3+208*x^2)*exp(x)+26*ln(2)+130*x^3)/(12* x^5*exp(x)^4+48*x^6*exp(x)^3+(12*x^4*ln(2)+72*x^7)*exp(x)^2+(24*x^5*ln(2)+ 48*x^8)*exp(x)+3*x^3*ln(2)^2+12*x^6*ln(2)+12*x^9),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \begin {dmath*} \int \frac {130 x^3+e^{2 x} \left (78 x+52 x^2\right )+e^x \left (208 x^2+52 x^3\right )+26 \log (2)}{12 e^{4 x} x^5+48 e^{3 x} x^6+12 x^9+12 x^6 \log (2)+3 x^3 \log ^2(2)+e^{2 x} \left (72 x^7+12 x^4 \log (2)\right )+e^x \left (48 x^8+24 x^5 \log (2)\right )} \, dx=-\frac {13}{3 \, {\left (2 \, x^{5} + 4 \, x^{4} e^{x} + 2 \, x^{3} e^{\left (2 \, x\right )} + x^{2} \log \left (2\right )\right )}} \end {dmath*}
integrate(((52*x^2+78*x)*exp(x)^2+(52*x^3+208*x^2)*exp(x)+26*log(2)+130*x^ 3)/(12*x^5*exp(x)^4+48*x^6*exp(x)^3+(12*x^4*log(2)+72*x^7)*exp(x)^2+(24*x^ 5*log(2)+48*x^8)*exp(x)+3*x^3*log(2)^2+12*x^6*log(2)+12*x^9),x, algorithm= \
Time = 0.09 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \begin {dmath*} \int \frac {130 x^3+e^{2 x} \left (78 x+52 x^2\right )+e^x \left (208 x^2+52 x^3\right )+26 \log (2)}{12 e^{4 x} x^5+48 e^{3 x} x^6+12 x^9+12 x^6 \log (2)+3 x^3 \log ^2(2)+e^{2 x} \left (72 x^7+12 x^4 \log (2)\right )+e^x \left (48 x^8+24 x^5 \log (2)\right )} \, dx=- \frac {13}{6 x^{5} + 12 x^{4} e^{x} + 6 x^{3} e^{2 x} + 3 x^{2} \log {\left (2 \right )}} \end {dmath*}
integrate(((52*x**2+78*x)*exp(x)**2+(52*x**3+208*x**2)*exp(x)+26*ln(2)+130 *x**3)/(12*x**5*exp(x)**4+48*x**6*exp(x)**3+(12*x**4*ln(2)+72*x**7)*exp(x) **2+(24*x**5*ln(2)+48*x**8)*exp(x)+3*x**3*ln(2)**2+12*x**6*ln(2)+12*x**9), x)
Time = 0.35 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \begin {dmath*} \int \frac {130 x^3+e^{2 x} \left (78 x+52 x^2\right )+e^x \left (208 x^2+52 x^3\right )+26 \log (2)}{12 e^{4 x} x^5+48 e^{3 x} x^6+12 x^9+12 x^6 \log (2)+3 x^3 \log ^2(2)+e^{2 x} \left (72 x^7+12 x^4 \log (2)\right )+e^x \left (48 x^8+24 x^5 \log (2)\right )} \, dx=-\frac {13}{3 \, {\left (2 \, x^{5} + 4 \, x^{4} e^{x} + 2 \, x^{3} e^{\left (2 \, x\right )} + x^{2} \log \left (2\right )\right )}} \end {dmath*}
integrate(((52*x^2+78*x)*exp(x)^2+(52*x^3+208*x^2)*exp(x)+26*log(2)+130*x^ 3)/(12*x^5*exp(x)^4+48*x^6*exp(x)^3+(12*x^4*log(2)+72*x^7)*exp(x)^2+(24*x^ 5*log(2)+48*x^8)*exp(x)+3*x^3*log(2)^2+12*x^6*log(2)+12*x^9),x, algorithm= \
Time = 0.72 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \begin {dmath*} \int \frac {130 x^3+e^{2 x} \left (78 x+52 x^2\right )+e^x \left (208 x^2+52 x^3\right )+26 \log (2)}{12 e^{4 x} x^5+48 e^{3 x} x^6+12 x^9+12 x^6 \log (2)+3 x^3 \log ^2(2)+e^{2 x} \left (72 x^7+12 x^4 \log (2)\right )+e^x \left (48 x^8+24 x^5 \log (2)\right )} \, dx=-\frac {26}{3 \, {\left (2 \, x^{5} + 4 \, x^{4} e^{x} + 2 \, x^{3} e^{\left (2 \, x\right )} + x^{2} \log \left (2\right )\right )}} \end {dmath*}
integrate(((52*x^2+78*x)*exp(x)^2+(52*x^3+208*x^2)*exp(x)+26*log(2)+130*x^ 3)/(12*x^5*exp(x)^4+48*x^6*exp(x)^3+(12*x^4*log(2)+72*x^7)*exp(x)^2+(24*x^ 5*log(2)+48*x^8)*exp(x)+3*x^3*log(2)^2+12*x^6*log(2)+12*x^9),x, algorithm= \
Timed out. \begin {dmath*} \int \frac {130 x^3+e^{2 x} \left (78 x+52 x^2\right )+e^x \left (208 x^2+52 x^3\right )+26 \log (2)}{12 e^{4 x} x^5+48 e^{3 x} x^6+12 x^9+12 x^6 \log (2)+3 x^3 \log ^2(2)+e^{2 x} \left (72 x^7+12 x^4 \log (2)\right )+e^x \left (48 x^8+24 x^5 \log (2)\right )} \, dx=\int \frac {26\,\ln \left (2\right )+{\mathrm {e}}^{2\,x}\,\left (52\,x^2+78\,x\right )+{\mathrm {e}}^x\,\left (52\,x^3+208\,x^2\right )+130\,x^3}{3\,x^3\,{\ln \left (2\right )}^2+12\,x^5\,{\mathrm {e}}^{4\,x}+48\,x^6\,{\mathrm {e}}^{3\,x}+{\mathrm {e}}^x\,\left (48\,x^8+24\,\ln \left (2\right )\,x^5\right )+12\,x^6\,\ln \left (2\right )+{\mathrm {e}}^{2\,x}\,\left (72\,x^7+12\,\ln \left (2\right )\,x^4\right )+12\,x^9} \,d x \end {dmath*}
int((26*log(2) + exp(2*x)*(78*x + 52*x^2) + exp(x)*(208*x^2 + 52*x^3) + 13 0*x^3)/(3*x^3*log(2)^2 + 12*x^5*exp(4*x) + 48*x^6*exp(3*x) + exp(x)*(24*x^ 5*log(2) + 48*x^8) + 12*x^6*log(2) + exp(2*x)*(12*x^4*log(2) + 72*x^7) + 1 2*x^9),x)