Integrand size = 90, antiderivative size = 26 \[ \int \frac {e^{\frac {256-\log \left (x^3\right )}{-5+x}} \left (25 x-10 x^2+x^3+\left (-30-757 x-252 x^2-x^3\right ) \log (6+2 x)+\left (3 x+x^2\right ) \log \left (x^3\right ) \log (6+2 x)\right )}{75 x^2-5 x^3-7 x^4+x^5} \, dx=\frac {e^{\frac {-256+\log \left (x^3\right )}{5-x}} \log (2 (3+x))}{x} \]
Time = 5.15 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {e^{\frac {256-\log \left (x^3\right )}{-5+x}} \left (25 x-10 x^2+x^3+\left (-30-757 x-252 x^2-x^3\right ) \log (6+2 x)+\left (3 x+x^2\right ) \log \left (x^3\right ) \log (6+2 x)\right )}{75 x^2-5 x^3-7 x^4+x^5} \, dx=\frac {e^{\frac {256}{-5+x}} \left (x^3\right )^{-\frac {1}{-5+x}} \log (2 (3+x))}{x} \]
Integrate[(E^((256 - Log[x^3])/(-5 + x))*(25*x - 10*x^2 + x^3 + (-30 - 757 *x - 252*x^2 - x^3)*Log[6 + 2*x] + (3*x + x^2)*Log[x^3]*Log[6 + 2*x]))/(75 *x^2 - 5*x^3 - 7*x^4 + x^5),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.
\(\displaystyle \int \frac {e^{\frac {256-\log \left (x^3\right )}{x-5}} \left (x^3-10 x^2+\left (-x^3-252 x^2-757 x-30\right ) \log (2 x+6)+\left (x^2+3 x\right ) \log \left (x^3\right ) \log (2 x+6)+25 x\right )}{x^5-7 x^4-5 x^3+75 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{\frac {256-\log \left (x^3\right )}{x-5}} \left (x^3-10 x^2+\left (-x^3-252 x^2-757 x-30\right ) \log (2 x+6)+\left (x^2+3 x\right ) \log \left (x^3\right ) \log (2 x+6)+25 x\right )}{x^2 \left (x^3-7 x^2-5 x+75\right )}dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (\frac {e^{\frac {256-\log \left (x^3\right )}{x-5}} \left (x^3-10 x^2+\left (-x^3-252 x^2-757 x-30\right ) \log (2 x+6)+\left (x^2+3 x\right ) \log \left (x^3\right ) \log (2 x+6)+25 x\right )}{64 x^2 (x+3)}-\frac {e^{\frac {256-\log \left (x^3\right )}{x-5}} \left (x^3-10 x^2+\left (-x^3-252 x^2-757 x-30\right ) \log (2 x+6)+\left (x^2+3 x\right ) \log \left (x^3\right ) \log (2 x+6)+25 x\right )}{64 (x-5) x^2}+\frac {e^{\frac {256-\log \left (x^3\right )}{x-5}} \left (x^3-10 x^2+\left (-x^3-252 x^2-757 x-30\right ) \log (2 x+6)+\left (x^2+3 x\right ) \log \left (x^3\right ) \log (2 x+6)+25 x\right )}{8 (x-5)^2 x^2}\right )dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^{\frac {256}{x-5}} \left (x^3\right )^{\frac {1}{5-x}} \left ((x-5)^2 x-(x+3) \left (-x \log \left (x^3\right )+x^2+249 x+10\right ) \log (2 (x+3))\right )}{(5-x)^2 x^2 (x+3)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{\frac {256}{x-5}} \left (x^3\right )^{\frac {1}{5-x}}}{x (x+3)}+\frac {e^{\frac {256}{x-5}} \left (x^3\right )^{\frac {1}{5-x}} \left (x \log \left (x^3\right )-x^2-249 x-10\right ) \log (2 x+6)}{(5-x)^2 x^2}\right )dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \left (\frac {e^{\frac {256}{x-5}} \left (x^3\right )^{\frac {1}{5-x}}}{x (x+3)}+\frac {e^{\frac {256}{x-5}} \left (x^3\right )^{\frac {1}{5-x}} \left (x \log \left (x^3\right )-x^2-249 x-10\right ) \log (2 x+6)}{(5-x)^2 x^2}\right )dx\) |
Int[(E^((256 - Log[x^3])/(-5 + x))*(25*x - 10*x^2 + x^3 + (-30 - 757*x - 2 52*x^2 - x^3)*Log[6 + 2*x] + (3*x + x^2)*Log[x^3]*Log[6 + 2*x]))/(75*x^2 - 5*x^3 - 7*x^4 + x^5),x]
3.21.69.3.1 Defintions of rubi rules used
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])
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u, Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 4.46 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96
method | result | size |
parallelrisch | \(\frac {\ln \left (2 x +6\right ) {\mathrm e}^{-\frac {\ln \left (x^{3}\right )-256}{-5+x}}}{x}\) | \(25\) |
risch | \(\frac {\ln \left (2 x +6\right ) x^{-\frac {3}{-5+x}} {\mathrm e}^{\frac {i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}-2 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )-i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )^{2}-i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right )^{2}+i \pi \operatorname {csgn}\left (i x^{3}\right )^{3}+512}{2 x -10}}}{x}\) | \(153\) |
int(((x^2+3*x)*ln(2*x+6)*ln(x^3)+(-x^3-252*x^2-757*x-30)*ln(2*x+6)+x^3-10* x^2+25*x)*exp((-ln(x^3)+256)/(-5+x))/(x^5-7*x^4-5*x^3+75*x^2),x,method=_RE TURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {256-\log \left (x^3\right )}{-5+x}} \left (25 x-10 x^2+x^3+\left (-30-757 x-252 x^2-x^3\right ) \log (6+2 x)+\left (3 x+x^2\right ) \log \left (x^3\right ) \log (6+2 x)\right )}{75 x^2-5 x^3-7 x^4+x^5} \, dx=\frac {e^{\left (-\frac {\log \left (x^{3}\right ) - 256}{x - 5}\right )} \log \left (2 \, x + 6\right )}{x} \]
integrate(((x^2+3*x)*log(2*x+6)*log(x^3)+(-x^3-252*x^2-757*x-30)*log(2*x+6 )+x^3-10*x^2+25*x)*exp((-log(x^3)+256)/(-5+x))/(x^5-7*x^4-5*x^3+75*x^2),x, algorithm=\
Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {e^{\frac {256-\log \left (x^3\right )}{-5+x}} \left (25 x-10 x^2+x^3+\left (-30-757 x-252 x^2-x^3\right ) \log (6+2 x)+\left (3 x+x^2\right ) \log \left (x^3\right ) \log (6+2 x)\right )}{75 x^2-5 x^3-7 x^4+x^5} \, dx=\frac {e^{\frac {256 - \log {\left (x^{3} \right )}}{x - 5}} \log {\left (2 x + 6 \right )}}{x} \]
integrate(((x**2+3*x)*ln(2*x+6)*ln(x**3)+(-x**3-252*x**2-757*x-30)*ln(2*x+ 6)+x**3-10*x**2+25*x)*exp((-ln(x**3)+256)/(-5+x))/(x**5-7*x**4-5*x**3+75*x **2),x)
Time = 0.36 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {e^{\frac {256-\log \left (x^3\right )}{-5+x}} \left (25 x-10 x^2+x^3+\left (-30-757 x-252 x^2-x^3\right ) \log (6+2 x)+\left (3 x+x^2\right ) \log \left (x^3\right ) \log (6+2 x)\right )}{75 x^2-5 x^3-7 x^4+x^5} \, dx=\frac {{\left (\log \left (2\right ) + \log \left (x + 3\right )\right )} e^{\left (-\frac {3 \, \log \left (x\right )}{x - 5} + \frac {256}{x - 5}\right )}}{x} \]
integrate(((x^2+3*x)*log(2*x+6)*log(x^3)+(-x^3-252*x^2-757*x-30)*log(2*x+6 )+x^3-10*x^2+25*x)*exp((-log(x^3)+256)/(-5+x))/(x^5-7*x^4-5*x^3+75*x^2),x, algorithm=\
\[ \int \frac {e^{\frac {256-\log \left (x^3\right )}{-5+x}} \left (25 x-10 x^2+x^3+\left (-30-757 x-252 x^2-x^3\right ) \log (6+2 x)+\left (3 x+x^2\right ) \log \left (x^3\right ) \log (6+2 x)\right )}{75 x^2-5 x^3-7 x^4+x^5} \, dx=\int { \frac {{\left (x^{3} + {\left (x^{2} + 3 \, x\right )} \log \left (x^{3}\right ) \log \left (2 \, x + 6\right ) - 10 \, x^{2} - {\left (x^{3} + 252 \, x^{2} + 757 \, x + 30\right )} \log \left (2 \, x + 6\right ) + 25 \, x\right )} e^{\left (-\frac {\log \left (x^{3}\right ) - 256}{x - 5}\right )}}{x^{5} - 7 \, x^{4} - 5 \, x^{3} + 75 \, x^{2}} \,d x } \]
integrate(((x^2+3*x)*log(2*x+6)*log(x^3)+(-x^3-252*x^2-757*x-30)*log(2*x+6 )+x^3-10*x^2+25*x)*exp((-log(x^3)+256)/(-5+x))/(x^5-7*x^4-5*x^3+75*x^2),x, algorithm=\
integrate((x^3 + (x^2 + 3*x)*log(x^3)*log(2*x + 6) - 10*x^2 - (x^3 + 252*x ^2 + 757*x + 30)*log(2*x + 6) + 25*x)*e^(-(log(x^3) - 256)/(x - 5))/(x^5 - 7*x^4 - 5*x^3 + 75*x^2), x)
Timed out. \[ \int \frac {e^{\frac {256-\log \left (x^3\right )}{-5+x}} \left (25 x-10 x^2+x^3+\left (-30-757 x-252 x^2-x^3\right ) \log (6+2 x)+\left (3 x+x^2\right ) \log \left (x^3\right ) \log (6+2 x)\right )}{75 x^2-5 x^3-7 x^4+x^5} \, dx=\int \frac {{\mathrm {e}}^{-\frac {\ln \left (x^3\right )-256}{x-5}}\,\left (25\,x-\ln \left (2\,x+6\right )\,\left (x^3+252\,x^2+757\,x+30\right )-10\,x^2+x^3+\ln \left (x^3\right )\,\ln \left (2\,x+6\right )\,\left (x^2+3\,x\right )\right )}{x^5-7\,x^4-5\,x^3+75\,x^2} \,d x \]
int((exp(-(log(x^3) - 256)/(x - 5))*(25*x - log(2*x + 6)*(757*x + 252*x^2 + x^3 + 30) - 10*x^2 + x^3 + log(x^3)*log(2*x + 6)*(3*x + x^2)))/(75*x^2 - 5*x^3 - 7*x^4 + x^5),x)