Integrand size = 68, antiderivative size = 24 \[ \int \frac {e^{4 x-5 x^2-x^3+x \log (5)-x \log (3-x)} \left (-12+33 x-x^2-3 x^3+(-3+x) \log (5)+(3-x) \log (3-x)\right )}{-3+x} \, dx=e^{x \left (4+\log (5)-x \left (5+x+\frac {\log (3-x)}{x}\right )\right )} \] Output:
exp((4-x*(ln(3-x)/x+x+5)+ln(5))*x)
Time = 1.99 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{4 x-5 x^2-x^3+x \log (5)-x \log (3-x)} \left (-12+33 x-x^2-3 x^3+(-3+x) \log (5)+(3-x) \log (3-x)\right )}{-3+x} \, dx=5^x e^{-x \left (-4+5 x+x^2\right )} (3-x)^{-x} \] Input:
Integrate[(E^(4*x - 5*x^2 - x^3 + x*Log[5] - x*Log[3 - x])*(-12 + 33*x - x ^2 - 3*x^3 + (-3 + x)*Log[5] + (3 - x)*Log[3 - x]))/(-3 + x),x]
Output:
5^x/(E^(x*(-4 + 5*x + x^2))*(3 - x)^x)
Time = 0.91 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.015, Rules used = {7257}
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.
\(\displaystyle \int \frac {e^{-x^3-5 x^2+4 x-x \log (3-x)+x \log (5)} \left (-3 x^3-x^2+33 x+(3-x) \log (3-x)+(x-3) \log (5)-12\right )}{x-3} \, dx\) |
\(\Big \downarrow \) 7257 |
\(\displaystyle 5^x e^{-x^3-5 x^2+4 x} (3-x)^{-x}\) |
Input:
Int[(E^(4*x - 5*x^2 - x^3 + x*Log[5] - x*Log[3 - x])*(-12 + 33*x - x^2 - 3 *x^3 + (-3 + x)*Log[5] + (3 - x)*Log[3 - x]))/(-3 + x),x]
Output:
(5^x*E^(4*x - 5*x^2 - x^3))/(3 - x)^x
Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Sim p[q*(F^v/Log[F]), x] /; !FalseQ[q]] /; FreeQ[F, x]
Time = 0.56 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96
method | result | size |
parallelrisch | \({\mathrm e}^{-x \left (x^{2}+\ln \left (-x +3\right )-\ln \left (5\right )+5 x -4\right )}\) | \(23\) |
risch | \(\left (-x +3\right )^{-x} 5^{x} {\mathrm e}^{-x \left (x^{2}+5 x -4\right )}\) | \(26\) |
norman | \({\mathrm e}^{-x \ln \left (-x +3\right )+x \ln \left (5\right )-x^{3}-5 x^{2}+4 x}\) | \(29\) |
Input:
int(((-x+3)*ln(-x+3)+(-3+x)*ln(5)-3*x^3-x^2+33*x-12)*exp(-x*ln(-x+3)+x*ln( 5)-x^3-5*x^2+4*x)/(-3+x),x,method=_RETURNVERBOSE)
Output:
exp(-x*(x^2+ln(-x+3)-ln(5)+5*x-4))
Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {e^{4 x-5 x^2-x^3+x \log (5)-x \log (3-x)} \left (-12+33 x-x^2-3 x^3+(-3+x) \log (5)+(3-x) \log (3-x)\right )}{-3+x} \, dx=e^{\left (-x^{3} - 5 \, x^{2} + x \log \left (5\right ) - x \log \left (-x + 3\right ) + 4 \, x\right )} \] Input:
integrate(((3-x)*log(3-x)+(-3+x)*log(5)-3*x^3-x^2+33*x-12)*exp(-x*log(3-x) +x*log(5)-x^3-5*x^2+4*x)/(-3+x),x, algorithm="fricas")
Output:
e^(-x^3 - 5*x^2 + x*log(5) - x*log(-x + 3) + 4*x)
Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{4 x-5 x^2-x^3+x \log (5)-x \log (3-x)} \left (-12+33 x-x^2-3 x^3+(-3+x) \log (5)+(3-x) \log (3-x)\right )}{-3+x} \, dx=e^{- x^{3} - 5 x^{2} - x \log {\left (3 - x \right )} + x \log {\left (5 \right )} + 4 x} \] Input:
integrate(((3-x)*ln(3-x)+(-3+x)*ln(5)-3*x**3-x**2+33*x-12)*exp(-x*ln(3-x)+ x*ln(5)-x**3-5*x**2+4*x)/(-3+x),x)
Output:
exp(-x**3 - 5*x**2 - x*log(3 - x) + x*log(5) + 4*x)
Time = 0.19 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {e^{4 x-5 x^2-x^3+x \log (5)-x \log (3-x)} \left (-12+33 x-x^2-3 x^3+(-3+x) \log (5)+(3-x) \log (3-x)\right )}{-3+x} \, dx=e^{\left (-x^{3} - 5 \, x^{2} + x \log \left (5\right ) - x \log \left (-x + 3\right ) + 4 \, x\right )} \] Input:
integrate(((3-x)*log(3-x)+(-3+x)*log(5)-3*x^3-x^2+33*x-12)*exp(-x*log(3-x) +x*log(5)-x^3-5*x^2+4*x)/(-3+x),x, algorithm="maxima")
Output:
e^(-x^3 - 5*x^2 + x*log(5) - x*log(-x + 3) + 4*x)
Time = 0.11 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {e^{4 x-5 x^2-x^3+x \log (5)-x \log (3-x)} \left (-12+33 x-x^2-3 x^3+(-3+x) \log (5)+(3-x) \log (3-x)\right )}{-3+x} \, dx=e^{\left (-x^{3} - 5 \, x^{2} + x \log \left (5\right ) - x \log \left (-x + 3\right ) + 4 \, x\right )} \] Input:
integrate(((3-x)*log(3-x)+(-3+x)*log(5)-3*x^3-x^2+33*x-12)*exp(-x*log(3-x) +x*log(5)-x^3-5*x^2+4*x)/(-3+x),x, algorithm="giac")
Output:
e^(-x^3 - 5*x^2 + x*log(5) - x*log(-x + 3) + 4*x)
Time = 1.93 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {e^{4 x-5 x^2-x^3+x \log (5)-x \log (3-x)} \left (-12+33 x-x^2-3 x^3+(-3+x) \log (5)+(3-x) \log (3-x)\right )}{-3+x} \, dx=\frac {5^x\,{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^{-x^3}\,{\mathrm {e}}^{-5\,x^2}}{{\left (3-x\right )}^x} \] Input:
int(-(exp(4*x + x*log(5) - x*log(3 - x) - 5*x^2 - x^3)*(x^2 - log(5)*(x - 3) - 33*x + 3*x^3 + log(3 - x)*(x - 3) + 12))/(x - 3),x)
Output:
(5^x*exp(4*x)*exp(-x^3)*exp(-5*x^2))/(3 - x)^x
\[ \int \frac {e^{4 x-5 x^2-x^3+x \log (5)-x \log (3-x)} \left (-12+33 x-x^2-3 x^3+(-3+x) \log (5)+(3-x) \log (3-x)\right )}{-3+x} \, dx=-3 \left (\int \frac {e^{4 x} 5^{x} x^{3}}{e^{x^{3}+5 x^{2}} \left (-x +3\right )^{x} x -3 e^{x^{3}+5 x^{2}} \left (-x +3\right )^{x}}d x \right )-\left (\int \frac {e^{4 x} 5^{x} x^{2}}{e^{x^{3}+5 x^{2}} \left (-x +3\right )^{x} x -3 e^{x^{3}+5 x^{2}} \left (-x +3\right )^{x}}d x \right )-\left (\int \frac {e^{4 x} 5^{x} \mathrm {log}\left (-x +3\right ) x}{e^{x^{3}+5 x^{2}} \left (-x +3\right )^{x} x -3 e^{x^{3}+5 x^{2}} \left (-x +3\right )^{x}}d x \right )+3 \left (\int \frac {e^{4 x} 5^{x} \mathrm {log}\left (-x +3\right )}{e^{x^{3}+5 x^{2}} \left (-x +3\right )^{x} x -3 e^{x^{3}+5 x^{2}} \left (-x +3\right )^{x}}d x \right )+\left (\int \frac {e^{4 x} 5^{x} x}{e^{x^{3}+5 x^{2}} \left (-x +3\right )^{x} x -3 e^{x^{3}+5 x^{2}} \left (-x +3\right )^{x}}d x \right ) \mathrm {log}\left (5\right )+33 \left (\int \frac {e^{4 x} 5^{x} x}{e^{x^{3}+5 x^{2}} \left (-x +3\right )^{x} x -3 e^{x^{3}+5 x^{2}} \left (-x +3\right )^{x}}d x \right )-3 \left (\int \frac {e^{4 x} 5^{x}}{e^{x^{3}+5 x^{2}} \left (-x +3\right )^{x} x -3 e^{x^{3}+5 x^{2}} \left (-x +3\right )^{x}}d x \right ) \mathrm {log}\left (5\right )-12 \left (\int \frac {e^{4 x} 5^{x}}{e^{x^{3}+5 x^{2}} \left (-x +3\right )^{x} x -3 e^{x^{3}+5 x^{2}} \left (-x +3\right )^{x}}d x \right ) \] Input:
int(((3-x)*log(3-x)+(-3+x)*log(5)-3*x^3-x^2+33*x-12)*exp(-x*log(3-x)+x*log (5)-x^3-5*x^2+4*x)/(-3+x),x)
Output:
- 3*int((e**(4*x)*5**x*x**3)/(e**(x**3 + 5*x**2)*( - x + 3)**x*x - 3*e**( x**3 + 5*x**2)*( - x + 3)**x),x) - int((e**(4*x)*5**x*x**2)/(e**(x**3 + 5* x**2)*( - x + 3)**x*x - 3*e**(x**3 + 5*x**2)*( - x + 3)**x),x) - int((e**( 4*x)*5**x*log( - x + 3)*x)/(e**(x**3 + 5*x**2)*( - x + 3)**x*x - 3*e**(x** 3 + 5*x**2)*( - x + 3)**x),x) + 3*int((e**(4*x)*5**x*log( - x + 3))/(e**(x **3 + 5*x**2)*( - x + 3)**x*x - 3*e**(x**3 + 5*x**2)*( - x + 3)**x),x) + i nt((e**(4*x)*5**x*x)/(e**(x**3 + 5*x**2)*( - x + 3)**x*x - 3*e**(x**3 + 5* x**2)*( - x + 3)**x),x)*log(5) + 33*int((e**(4*x)*5**x*x)/(e**(x**3 + 5*x* *2)*( - x + 3)**x*x - 3*e**(x**3 + 5*x**2)*( - x + 3)**x),x) - 3*int((e**( 4*x)*5**x)/(e**(x**3 + 5*x**2)*( - x + 3)**x*x - 3*e**(x**3 + 5*x**2)*( - x + 3)**x),x)*log(5) - 12*int((e**(4*x)*5**x)/(e**(x**3 + 5*x**2)*( - x + 3)**x*x - 3*e**(x**3 + 5*x**2)*( - x + 3)**x),x)