Integrand size = 83, antiderivative size = 32 \[ \int \frac {e^{-x} \left (768 x^2-128 x^3+52 x^4-2 x^5+e^x \left (-32 x-98 x^3+4 x^4\right )+e^x \left (-768+32 x+48 x^2-2 x^3\right ) \log \left (\frac {1}{3} (-24+x)\right )\right )}{-24 x^2+x^3} \, dx=\log \left (e^{\frac {32}{x}+2 x}\right ) \left (x+e^{-x} x-\log \left (-8+\frac {x}{3}\right )\right ) \]
Time = 0.39 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.59 \[ \int \frac {e^{-x} \left (768 x^2-128 x^3+52 x^4-2 x^5+e^x \left (-32 x-98 x^3+4 x^4\right )+e^x \left (-768+32 x+48 x^2-2 x^3\right ) \log \left (\frac {1}{3} (-24+x)\right )\right )}{-24 x^2+x^3} \, dx=2 \left (x^2+e^{-x} \left (16+x^2\right )-24 \log (24-x)-(-24+x) \log \left (-8+\frac {x}{3}\right )-\frac {16 \log \left (-8+\frac {x}{3}\right )}{x}\right ) \]
Integrate[(768*x^2 - 128*x^3 + 52*x^4 - 2*x^5 + E^x*(-32*x - 98*x^3 + 4*x^ 4) + E^x*(-768 + 32*x + 48*x^2 - 2*x^3)*Log[(-24 + x)/3])/(E^x*(-24*x^2 + x^3)),x]
Time = 1.97 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.47, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {2026, 7293, 2009}
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.
\(\displaystyle \int \frac {e^{-x} \left (-2 x^5+52 x^4-128 x^3+768 x^2+e^x \left (4 x^4-98 x^3-32 x\right )+e^x \left (-2 x^3+48 x^2+32 x-768\right ) \log \left (\frac {x-24}{3}\right )\right )}{x^3-24 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{-x} \left (-2 x^5+52 x^4-128 x^3+768 x^2+e^x \left (4 x^4-98 x^3-32 x\right )+e^x \left (-2 x^3+48 x^2+32 x-768\right ) \log \left (\frac {x-24}{3}\right )\right )}{(x-24) x^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 \left (2 x^4-49 x^3-x^3 \log \left (\frac {x}{3}-8\right )+24 x^2 \log \left (\frac {x}{3}-8\right )-16 x+16 x \log \left (\frac {x}{3}-8\right )-384 \log \left (\frac {x}{3}-8\right )\right )}{(x-24) x^2}-2 e^{-x} \left (x^2-2 x+16\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 e^{-x} x^2+2 x^2+32 e^{-x}-2 x \log \left (\frac {x}{3}-8\right )-\frac {32 \log \left (\frac {x}{3}-8\right )}{x}\) |
Int[(768*x^2 - 128*x^3 + 52*x^4 - 2*x^5 + E^x*(-32*x - 98*x^3 + 4*x^4) + E ^x*(-768 + 32*x + 48*x^2 - 2*x^3)*Log[(-24 + x)/3])/(E^x*(-24*x^2 + x^3)), x]
3.9.63.3.1 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])
Time = 2.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09
method | result | size |
risch | \(-\frac {2 \left (x^{2}+16\right ) \ln \left (\frac {x}{3}-8\right )}{x}+2 \left ({\mathrm e}^{x} x^{2}+x^{2}+16\right ) {\mathrm e}^{-x}\) | \(35\) |
default | \(32 \,{\mathrm e}^{-x}+2 x^{2} {\mathrm e}^{-x}+2 x^{2}+\frac {4 \ln \left (x \right )}{3}-\frac {148 \ln \left (x -24\right )}{3}-6 \ln \left (\frac {x}{3}-8\right ) \left (\frac {x}{3}-8\right )-48-\frac {4 \ln \left (\frac {x}{3}\right )}{3}+\frac {4 \ln \left (\frac {x}{3}-8\right ) \left (\frac {x}{3}-8\right )}{x}\) | \(68\) |
parts | \(32 \,{\mathrm e}^{-x}+2 x^{2} {\mathrm e}^{-x}+2 x^{2}+\frac {4 \ln \left (x \right )}{3}-\frac {148 \ln \left (x -24\right )}{3}-6 \ln \left (\frac {x}{3}-8\right ) \left (\frac {x}{3}-8\right )-48-\frac {4 \ln \left (\frac {x}{3}\right )}{3}+\frac {4 \ln \left (\frac {x}{3}-8\right ) \left (\frac {x}{3}-8\right )}{x}\) | \(68\) |
parallelrisch | \(-\frac {\left (-96 \,{\mathrm e}^{x} x^{3}+96 \,{\mathrm e}^{x} \ln \left (\frac {x}{3}-8\right ) x^{2}+4608 \ln \left ({\mathrm e}^{x}\right ) x \,{\mathrm e}^{x}-4608 \ln \left (x -24\right ) {\mathrm e}^{x} x -96 x^{3}-4608 \,{\mathrm e}^{x} x^{2}+4608 \,{\mathrm e}^{x} \ln \left (\frac {x}{3}-8\right ) x -55296 \,{\mathrm e}^{x} x +1536 \,{\mathrm e}^{x} \ln \left (\frac {x}{3}-8\right )-1536 x \right ) {\mathrm e}^{-x}}{48 x}\) | \(89\) |
int(((-2*x^3+48*x^2+32*x-768)*exp(x)*ln(1/3*x-8)+(4*x^4-98*x^3-32*x)*exp(x )-2*x^5+52*x^4-128*x^3+768*x^2)/(x^3-24*x^2)/exp(x),x,method=_RETURNVERBOS E)
Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16 \[ \int \frac {e^{-x} \left (768 x^2-128 x^3+52 x^4-2 x^5+e^x \left (-32 x-98 x^3+4 x^4\right )+e^x \left (-768+32 x+48 x^2-2 x^3\right ) \log \left (\frac {1}{3} (-24+x)\right )\right )}{-24 x^2+x^3} \, dx=\frac {2 \, {\left (x^{3} e^{x} + x^{3} - {\left (x^{2} + 16\right )} e^{x} \log \left (\frac {1}{3} \, x - 8\right ) + 16 \, x\right )} e^{\left (-x\right )}}{x} \]
integrate(((-2*x^3+48*x^2+32*x-768)*exp(x)*log(1/3*x-8)+(4*x^4-98*x^3-32*x )*exp(x)-2*x^5+52*x^4-128*x^3+768*x^2)/(x^3-24*x^2)/exp(x),x, algorithm=\
Time = 0.16 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-x} \left (768 x^2-128 x^3+52 x^4-2 x^5+e^x \left (-32 x-98 x^3+4 x^4\right )+e^x \left (-768+32 x+48 x^2-2 x^3\right ) \log \left (\frac {1}{3} (-24+x)\right )\right )}{-24 x^2+x^3} \, dx=2 x^{2} + \left (2 x^{2} + 32\right ) e^{- x} + \frac {\left (- 2 x^{2} - 32\right ) \log {\left (\frac {x}{3} - 8 \right )}}{x} \]
integrate(((-2*x**3+48*x**2+32*x-768)*exp(x)*ln(1/3*x-8)+(4*x**4-98*x**3-3 2*x)*exp(x)-2*x**5+52*x**4-128*x**3+768*x**2)/(x**3-24*x**2)/exp(x),x)
\[ \int \frac {e^{-x} \left (768 x^2-128 x^3+52 x^4-2 x^5+e^x \left (-32 x-98 x^3+4 x^4\right )+e^x \left (-768+32 x+48 x^2-2 x^3\right ) \log \left (\frac {1}{3} (-24+x)\right )\right )}{-24 x^2+x^3} \, dx=\int { -\frac {2 \, {\left (x^{5} - 26 \, x^{4} + 64 \, x^{3} + {\left (x^{3} - 24 \, x^{2} - 16 \, x + 384\right )} e^{x} \log \left (\frac {1}{3} \, x - 8\right ) - 384 \, x^{2} - {\left (2 \, x^{4} - 49 \, x^{3} - 16 \, x\right )} e^{x}\right )} e^{\left (-x\right )}}{x^{3} - 24 \, x^{2}} \,d x } \]
integrate(((-2*x^3+48*x^2+32*x-768)*exp(x)*log(1/3*x-8)+(4*x^4-98*x^3-32*x )*exp(x)-2*x^5+52*x^4-128*x^3+768*x^2)/(x^3-24*x^2)/exp(x),x, algorithm=\
-768*e^(-24)*exp_integral_e(1, x - 24) + 2*(x^4 + x^3*(log(3) - 24) - 24*x ^2*log(3) + (x^4 - 24*x^3 + 16*x^2)*e^(-x) + 16*x*log(3) - (x^3 - 24*x^2 + 16*x - 384)*log(x - 24) - 384*log(3))/(x^2 - 24*x) + 768*integrate(e^(-x) /(x^2 - 48*x + 576), x)
Leaf count of result is larger than twice the leaf count of optimal. 122 vs. \(2 (28) = 56\).
Time = 0.27 (sec) , antiderivative size = 122, normalized size of antiderivative = 3.81 \[ \int \frac {e^{-x} \left (768 x^2-128 x^3+52 x^4-2 x^5+e^x \left (-32 x-98 x^3+4 x^4\right )+e^x \left (-768+32 x+48 x^2-2 x^3\right ) \log \left (\frac {1}{3} (-24+x)\right )\right )}{-24 x^2+x^3} \, dx=\frac {2 \, {\left ({\left (x - 24\right )}^{3} e^{24} + {\left (x - 24\right )}^{3} e^{\left (-x + 24\right )} - {\left (x - 24\right )}^{2} e^{24} \log \left (\frac {1}{3} \, x - 8\right ) + 72 \, {\left (x - 24\right )}^{2} e^{24} + 72 \, {\left (x - 24\right )}^{2} e^{\left (-x + 24\right )} - 48 \, {\left (x - 24\right )} e^{24} \log \left (\frac {1}{3} \, x - 8\right ) + 1152 \, {\left (x - 24\right )} e^{24} + 1744 \, {\left (x - 24\right )} e^{\left (-x + 24\right )} - 592 \, e^{24} \log \left (\frac {1}{3} \, x - 8\right ) + 14208 \, e^{\left (-x + 24\right )}\right )}}{{\left (x - 24\right )} e^{24} + 24 \, e^{24}} \]
integrate(((-2*x^3+48*x^2+32*x-768)*exp(x)*log(1/3*x-8)+(4*x^4-98*x^3-32*x )*exp(x)-2*x^5+52*x^4-128*x^3+768*x^2)/(x^3-24*x^2)/exp(x),x, algorithm=\
2*((x - 24)^3*e^24 + (x - 24)^3*e^(-x + 24) - (x - 24)^2*e^24*log(1/3*x - 8) + 72*(x - 24)^2*e^24 + 72*(x - 24)^2*e^(-x + 24) - 48*(x - 24)*e^24*log (1/3*x - 8) + 1152*(x - 24)*e^24 + 1744*(x - 24)*e^(-x + 24) - 592*e^24*lo g(1/3*x - 8) + 14208*e^(-x + 24))/((x - 24)*e^24 + 24*e^24)
Time = 0.24 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.31 \[ \int \frac {e^{-x} \left (768 x^2-128 x^3+52 x^4-2 x^5+e^x \left (-32 x-98 x^3+4 x^4\right )+e^x \left (-768+32 x+48 x^2-2 x^3\right ) \log \left (\frac {1}{3} (-24+x)\right )\right )}{-24 x^2+x^3} \, dx={\mathrm {e}}^{-x}\,\left (2\,x^2+32\right )-\ln \left (\frac {x}{3}-8\right )\,\left (4\,x-\frac {2\,x^2-32}{x}\right )+2\,x^2 \]