\(\int e^{x-25 x^3+10 x^4-x^5} (1+x-75 x^3+40 x^4-5 x^5) \, dx\) [8631]

Optimal result

Integrand size = 38, antiderivative size = 18 \[ \int e^{x-25 x^3+10 x^4-x^5} \left (1+x-75 x^3+40 x^4-5 x^5\right ) \, dx=e^{x-(5-x)^2 x^3} x \]

[Out]

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(56\) vs. \(2(18)=36\).

Time = 0.07 (sec) , antiderivative size = 56, normalized size of antiderivative = 3.11, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {2326} \[ \int e^{x-25 x^3+10 x^4-x^5} \left (1+x-75 x^3+40 x^4-5 x^5\right ) \, dx=\frac {e^{-x^5+10 x^4-25 x^3+x} \left (-5 x^5+40 x^4-75 x^3+x\right )}{-5 x^4+40 x^3-75 x^2+1} \]

[In]

Int[E^(x - 25*x^3 + 10*x^4 - x^5)*(1 + x - 75*x^3 + 40*x^4 - 5*x^5),x]

[Out]

(E^(x - 25*x^3 + 10*x^4 - x^5)*(x - 75*x^3 + 40*x^4 - 5*x^5))/(1 - 75*x^2 + 40*x^3 - 5*x^4)

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \frac {e^{x-25 x^3+10 x^4-x^5} \left (x-75 x^3+40 x^4-5 x^5\right )}{1-75 x^2+40 x^3-5 x^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.77 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.17 \[ \int e^{x-25 x^3+10 x^4-x^5} \left (1+x-75 x^3+40 x^4-5 x^5\right ) \, dx=e^{x-25 x^3+10 x^4-x^5} x \]

[In]

Integrate[E^(x - 25*x^3 + 10*x^4 - x^5)*(1 + x - 75*x^3 + 40*x^4 - 5*x^5),x]

[Out]

E^(x - 25*x^3 + 10*x^4 - x^5)*x

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.17

method	result	size
gosper	\({\mathrm e}^{\ln \left (x \right )-x^{5}+10 x^{4}-25 x^{3}+x}\)	\(21\)
default	\({\mathrm e}^{\ln \left (x \right )-x^{5}+10 x^{4}-25 x^{3}+x}\)	\(21\)
norman	\({\mathrm e}^{\ln \left (x \right )-x^{5}+10 x^{4}-25 x^{3}+x}\)	\(21\)
parallelrisch	\({\mathrm e}^{\ln \left (x \right )-x^{5}+10 x^{4}-25 x^{3}+x}\)	\(21\)
risch	\(x \,{\mathrm e}^{-x \left (x^{2}-5 x +1\right ) \left (x^{2}-5 x -1\right )}\)	\(23\)

[In]

int((-5*x^5+40*x^4-75*x^3+x+1)*exp(ln(x)-x^5+10*x^4-25*x^3+x)/x,x,method=_RETURNVERBOSE)

[Out]

exp(ln(x)-x^5+10*x^4-25*x^3+x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int e^{x-25 x^3+10 x^4-x^5} \left (1+x-75 x^3+40 x^4-5 x^5\right ) \, dx=e^{\left (-x^{5} + 10 \, x^{4} - 25 \, x^{3} + x + \log \left (x\right )\right )} \]

[In]

integrate((-5*x^5+40*x^4-75*x^3+x+1)*exp(log(x)-x^5+10*x^4-25*x^3+x)/x,x, algorithm="fricas")

[Out]

e^(-x^5 + 10*x^4 - 25*x^3 + x + log(x))

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int e^{x-25 x^3+10 x^4-x^5} \left (1+x-75 x^3+40 x^4-5 x^5\right ) \, dx=x e^{- x^{5} + 10 x^{4} - 25 x^{3} + x} \]

[In]

integrate((-5*x**5+40*x**4-75*x**3+x+1)*exp(ln(x)-x**5+10*x**4-25*x**3+x)/x,x)

[Out]

x*exp(-x**5 + 10*x**4 - 25*x**3 + x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int e^{x-25 x^3+10 x^4-x^5} \left (1+x-75 x^3+40 x^4-5 x^5\right ) \, dx=x e^{\left (-x^{5} + 10 \, x^{4} - 25 \, x^{3} + x\right )} \]

[In]

integrate((-5*x^5+40*x^4-75*x^3+x+1)*exp(log(x)-x^5+10*x^4-25*x^3+x)/x,x, algorithm="maxima")

[Out]

x*e^(-x^5 + 10*x^4 - 25*x^3 + x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int e^{x-25 x^3+10 x^4-x^5} \left (1+x-75 x^3+40 x^4-5 x^5\right ) \, dx=e^{\left (-x^{5} + 10 \, x^{4} - 25 \, x^{3} + x + \log \left (x\right )\right )} \]

[In]

integrate((-5*x^5+40*x^4-75*x^3+x+1)*exp(log(x)-x^5+10*x^4-25*x^3+x)/x,x, algorithm="giac")

[Out]

e^(-x^5 + 10*x^4 - 25*x^3 + x + log(x))

Mupad [B] (verification not implemented)

Time = 13.13 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \[ \int e^{x-25 x^3+10 x^4-x^5} \left (1+x-75 x^3+40 x^4-5 x^5\right ) \, dx=x\,{\mathrm {e}}^{-x^5}\,{\mathrm {e}}^{10\,x^4}\,{\mathrm {e}}^{-25\,x^3}\,{\mathrm {e}}^x \]

[In]

int((exp(x + log(x) - 25*x^3 + 10*x^4 - x^5)*(x - 75*x^3 + 40*x^4 - 5*x^5 + 1))/x,x)

[Out]

x*exp(-x^5)*exp(10*x^4)*exp(-25*x^3)*exp(x)