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

Optimal. Leaf size=23 \[ 3 \left (-4+3 e^{5 \left (x-(x+x (4+x))^2\right )}\right ) \]

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Rubi [A]  time = 0.08, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {12, 6706} \begin {gather*} 9 e^{-5 x^4-50 x^3-125 x^2+5 x} \end {gather*}

Antiderivative was successfully verified.

[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 6706

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 20, normalized size = 0.87 \begin {gather*} 9 e^{-5 x \left (-1+25 x+10 x^2+x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[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))

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fricas [A]  time = 0.49, size = 24, normalized size = 1.04 \begin {gather*} 3 \, e^{\left (-5 \, x^{4} - 50 \, x^{3} - 125 \, x^{2} + 5 \, x + \log \relax (3)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 0.15, size = 24, normalized size = 1.04 \begin {gather*} 3 \, e^{\left (-5 \, x^{4} - 50 \, x^{3} - 125 \, x^{2} + 5 \, x + \log \relax (3)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [A]  time = 0.05, size = 20, normalized size = 0.87

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

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maxima [A]  time = 0.38, size = 22, normalized size = 0.96 \begin {gather*} 9 \, e^{\left (-5 \, x^{4} - 50 \, x^{3} - 125 \, x^{2} + 5 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.07, size = 24, normalized size = 1.04 \begin {gather*} 9\,{\mathrm {e}}^{5\,x}\,{\mathrm {e}}^{-5\,x^4}\,{\mathrm {e}}^{-50\,x^3}\,{\mathrm {e}}^{-125\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.12, size = 20, normalized size = 0.87 \begin {gather*} 9 e^{- 5 x^{4} - 50 x^{3} - 125 x^{2} + 5 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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