\(\int 3 e^{5 x-125 x^2-50 x^3-5 x^4} (15-750 x-450 x^2-60 x^3) \, dx\) [6597]

Optimal result

Integrand size = 38, antiderivative size = 23 \[ \int 3 e^{5 x-125 x^2-50 x^3-5 x^4} \left (15-750 x-450 x^2-60 x^3\right ) \, dx=3 \left (-4+3 e^{5 \left (x-(x+x (4+x))^2\right )}\right ) \]

[Out]

3*exp(5*x-5*((4+x)*x+x)^2+ln(3))-12

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {12, 6838} \[ \int 3 e^{5 x-125 x^2-50 x^3-5 x^4} \left (15-750 x-450 x^2-60 x^3\right ) \, dx=9 e^{-5 x^4-50 x^3-125 x^2+5 x} \]

[In]

Int[3*E^(5*x - 125*x^2 - 50*x^3 - 5*x^4)*(15 - 750*x - 450*x^2 - 60*x^3),x]

[Out]

9*E^(5*x - 125*x^2 - 50*x^3 - 5*x^4)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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}& = 3 \int e^{5 x-125 x^2-50 x^3-5 x^4} \left (15-750 x-450 x^2-60 x^3\right ) \, dx \\ & = 9 e^{5 x-125 x^2-50 x^3-5 x^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int 3 e^{5 x-125 x^2-50 x^3-5 x^4} \left (15-750 x-450 x^2-60 x^3\right ) \, dx=9 e^{-5 x \left (-1+25 x+10 x^2+x^3\right )} \]

[In]

Integrate[3*E^(5*x - 125*x^2 - 50*x^3 - 5*x^4)*(15 - 750*x - 450*x^2 - 60*x^3),x]

[Out]

9/E^(5*x*(-1 + 25*x + 10*x^2 + x^3))

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87

method	result	size
risch	\(9 \,{\mathrm e}^{-5 x \left (x^{3}+10 x^{2}+25 x -1\right )}\)	\(20\)
gosper	\(3 \,{\mathrm e}^{\ln \left (3\right )-5 x^{4}-50 x^{3}-125 x^{2}+5 x}\)	\(25\)
norman	\(3 \,{\mathrm e}^{\ln \left (3\right )-5 x^{4}-50 x^{3}-125 x^{2}+5 x}\)	\(25\)
parallelrisch	\(3 \,{\mathrm e}^{\ln \left (3\right )-5 x^{4}-50 x^{3}-125 x^{2}+5 x}\)	\(25\)

[In]

int((-60*x^3-450*x^2-750*x+15)*exp(ln(3)-5*x^4-50*x^3-125*x^2+5*x),x,method=_RETURNVERBOSE)

[Out]

9*exp(-5*x*(x^3+10*x^2+25*x-1))

Fricas [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int 3 e^{5 x-125 x^2-50 x^3-5 x^4} \left (15-750 x-450 x^2-60 x^3\right ) \, dx=3 \, e^{\left (-5 \, x^{4} - 50 \, x^{3} - 125 \, x^{2} + 5 \, x + \log \left (3\right )\right )} \]

[In]

integrate((-60*x^3-450*x^2-750*x+15)*exp(log(3)-5*x^4-50*x^3-125*x^2+5*x),x, algorithm="fricas")

[Out]

3*e^(-5*x^4 - 50*x^3 - 125*x^2 + 5*x + log(3))

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int 3 e^{5 x-125 x^2-50 x^3-5 x^4} \left (15-750 x-450 x^2-60 x^3\right ) \, dx=9 e^{- 5 x^{4} - 50 x^{3} - 125 x^{2} + 5 x} \]

[In]

integrate((-60*x**3-450*x**2-750*x+15)*exp(ln(3)-5*x**4-50*x**3-125*x**2+5*x),x)

[Out]

9*exp(-5*x**4 - 50*x**3 - 125*x**2 + 5*x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int 3 e^{5 x-125 x^2-50 x^3-5 x^4} \left (15-750 x-450 x^2-60 x^3\right ) \, dx=9 \, e^{\left (-5 \, x^{4} - 50 \, x^{3} - 125 \, x^{2} + 5 \, x\right )} \]

[In]

integrate((-60*x^3-450*x^2-750*x+15)*exp(log(3)-5*x^4-50*x^3-125*x^2+5*x),x, algorithm="maxima")

[Out]

9*e^(-5*x^4 - 50*x^3 - 125*x^2 + 5*x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int 3 e^{5 x-125 x^2-50 x^3-5 x^4} \left (15-750 x-450 x^2-60 x^3\right ) \, dx=3 \, e^{\left (-5 \, x^{4} - 50 \, x^{3} - 125 \, x^{2} + 5 \, x + \log \left (3\right )\right )} \]

[In]

integrate((-60*x^3-450*x^2-750*x+15)*exp(log(3)-5*x^4-50*x^3-125*x^2+5*x),x, algorithm="giac")

[Out]

3*e^(-5*x^4 - 50*x^3 - 125*x^2 + 5*x + log(3))

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int 3 e^{5 x-125 x^2-50 x^3-5 x^4} \left (15-750 x-450 x^2-60 x^3\right ) \, dx=9\,{\mathrm {e}}^{5\,x}\,{\mathrm {e}}^{-5\,x^4}\,{\mathrm {e}}^{-50\,x^3}\,{\mathrm {e}}^{-125\,x^2} \]

[In]

int(-exp(5*x + log(3) - 125*x^2 - 50*x^3 - 5*x^4)*(750*x + 450*x^2 + 60*x^3 - 15),x)

[Out]

9*exp(5*x)*exp(-5*x^4)*exp(-50*x^3)*exp(-125*x^2)