\(\int \frac {e^{4 x-5 x^2-x^3+x \log (5)-x \log (3-x)} (-12+33 x-x^2-3 x^3+(-3+x) \log (5)+(3-x) \log (3-x))}{-3+x} \, dx\) [2192]

Optimal result

Integrand size = 68, antiderivative size = 24 \[ \int \frac {e^{4 x-5 x^2-x^3+x \log (5)-x \log (3-x)} \left (-12+33 x-x^2-3 x^3+(-3+x) \log (5)+(3-x) \log (3-x)\right )}{-3+x} \, dx=e^{x \left (4+\log (5)-x \left (5+x+\frac {\log (3-x)}{x}\right )\right )} \]

[Out]

exp((4-x*(ln(-x+3)/x+x+5)+ln(5))*x)

Rubi [A] (verified)

Time = 0.34 (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 = {6838} \[ \int \frac {e^{4 x-5 x^2-x^3+x \log (5)-x \log (3-x)} \left (-12+33 x-x^2-3 x^3+(-3+x) \log (5)+(3-x) \log (3-x)\right )}{-3+x} \, dx=5^x e^{-x^3-5 x^2+4 x} (3-x)^{-x} \]

[In]

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]

[Out]

(5^x*E^(4*x - 5*x^2 - x^3))/(3 - x)^x

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = 5^x e^{4 x-5 x^2-x^3} (3-x)^{-x} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.92 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{4 x-5 x^2-x^3+x \log (5)-x \log (3-x)} \left (-12+33 x-x^2-3 x^3+(-3+x) \log (5)+(3-x) \log (3-x)\right )}{-3+x} \, dx=5^x e^{-x \left (-4+5 x+x^2\right )} (3-x)^{-x} \]

[In]

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]

[Out]

5^x/(E^(x*(-4 + 5*x + x^2))*(3 - x)^x)

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00

method	result	size
parallelrisch	\({\mathrm e}^{x \left (-x^{2}+\ln \left (5\right )-\ln \left (-x +3\right )-5 x +4\right )}\)	\(24\)
risch	\(\left (-x +3\right )^{-x} 5^{x} {\mathrm e}^{-x \left (x^{2}+5 x -4\right )}\)	\(26\)
norman	\({\mathrm e}^{-x \ln \left (-x +3\right )+x \ln \left (5\right )-x^{3}-5 x^{2}+4 x}\)	\(29\)

[In]

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=_R
ETURNVERBOSE)

[Out]

exp(x*(-x^2+ln(5)-ln(-x+3)-5*x+4))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {e^{4 x-5 x^2-x^3+x \log (5)-x \log (3-x)} \left (-12+33 x-x^2-3 x^3+(-3+x) \log (5)+(3-x) \log (3-x)\right )}{-3+x} \, dx=e^{\left (-x^{3} - 5 \, x^{2} + x \log \left (5\right ) - x \log \left (-x + 3\right ) + 4 \, x\right )} \]

[In]

integrate(((-x+3)*log(-x+3)+(-3+x)*log(5)-3*x^3-x^2+33*x-12)*exp(-x*log(-x+3)+x*log(5)-x^3-5*x^2+4*x)/(-3+x),x
, algorithm="fricas")

[Out]

e^(-x^3 - 5*x^2 + x*log(5) - x*log(-x + 3) + 4*x)

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{4 x-5 x^2-x^3+x \log (5)-x \log (3-x)} \left (-12+33 x-x^2-3 x^3+(-3+x) \log (5)+(3-x) \log (3-x)\right )}{-3+x} \, dx=e^{- x^{3} - 5 x^{2} - x \log {\left (3 - x \right )} + x \log {\left (5 \right )} + 4 x} \]

[In]

integrate(((-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
)

[Out]

exp(-x**3 - 5*x**2 - x*log(3 - x) + x*log(5) + 4*x)

Maxima [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {e^{4 x-5 x^2-x^3+x \log (5)-x \log (3-x)} \left (-12+33 x-x^2-3 x^3+(-3+x) \log (5)+(3-x) \log (3-x)\right )}{-3+x} \, dx=e^{\left (-x^{3} - 5 \, x^{2} + x \log \left (5\right ) - x \log \left (-x + 3\right ) + 4 \, x\right )} \]

[In]

integrate(((-x+3)*log(-x+3)+(-3+x)*log(5)-3*x^3-x^2+33*x-12)*exp(-x*log(-x+3)+x*log(5)-x^3-5*x^2+4*x)/(-3+x),x
, algorithm="maxima")

[Out]

e^(-x^3 - 5*x^2 + x*log(5) - x*log(-x + 3) + 4*x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {e^{4 x-5 x^2-x^3+x \log (5)-x \log (3-x)} \left (-12+33 x-x^2-3 x^3+(-3+x) \log (5)+(3-x) \log (3-x)\right )}{-3+x} \, dx=e^{\left (-x^{3} - 5 \, x^{2} + x \log \left (5\right ) - x \log \left (-x + 3\right ) + 4 \, x\right )} \]

[In]

integrate(((-x+3)*log(-x+3)+(-3+x)*log(5)-3*x^3-x^2+33*x-12)*exp(-x*log(-x+3)+x*log(5)-x^3-5*x^2+4*x)/(-3+x),x
, algorithm="giac")

[Out]

e^(-x^3 - 5*x^2 + x*log(5) - x*log(-x + 3) + 4*x)

Mupad [B] (verification not implemented)

Time = 9.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {e^{4 x-5 x^2-x^3+x \log (5)-x \log (3-x)} \left (-12+33 x-x^2-3 x^3+(-3+x) \log (5)+(3-x) \log (3-x)\right )}{-3+x} \, dx=\frac {5^x\,{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^{-x^3}\,{\mathrm {e}}^{-5\,x^2}}{{\left (3-x\right )}^x} \]

[In]

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)

[Out]

(5^x*exp(4*x)*exp(-x^3)*exp(-5*x^2))/(3 - x)^x