3.2.90 \(\int \frac {e^{-\frac {2500+e^6 x^2-500 x^3+25 x^6+e^3 (100 x-10 x^4)}{25 x^3}} (15000+2 e^6 x^2-150 x^6+e^3 (400 x+20 x^4))}{25 x^4} \, dx\)

Optimal. Leaf size=29 \[ 2 e^{-\frac {\left (\frac {e^3}{5}+\frac {10}{x}-x^2\right )^2}{x}} \]

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Rubi [A]  time = 0.80, antiderivative size = 25, normalized size of antiderivative = 0.86, number of steps used = 4, number of rules used = 3, integrand size = 76, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {12, 6688, 6706} \begin {gather*} 2 e^{-\frac {\left (-5 x^3+e^3 x+50\right )^2}{25 x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(15000 + 2*E^6*x^2 - 150*x^6 + E^3*(400*x + 20*x^4))/(25*E^((2500 + E^6*x^2 - 500*x^3 + 25*x^6 + E^3*(100*
x - 10*x^4))/(25*x^3))*x^4),x]

[Out]

2/E^((50 + E^3*x - 5*x^3)^2/(25*x^3))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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} &=\frac {1}{25} \int \frac {\exp \left (-\frac {2500+e^6 x^2-500 x^3+25 x^6+e^3 \left (100 x-10 x^4\right )}{25 x^3}\right ) \left (15000+2 e^6 x^2-150 x^6+e^3 \left (400 x+20 x^4\right )\right )}{x^4} \, dx\\ &=\frac {1}{25} \int \frac {2 e^{-\frac {\left (50+e^3 x-5 x^3\right )^2}{25 x^3}} \left (e^6 x^2+10 e^3 x \left (20+x^3\right )-75 \left (-100+x^6\right )\right )}{x^4} \, dx\\ &=\frac {2}{25} \int \frac {e^{-\frac {\left (50+e^3 x-5 x^3\right )^2}{25 x^3}} \left (e^6 x^2+10 e^3 x \left (20+x^3\right )-75 \left (-100+x^6\right )\right )}{x^4} \, dx\\ &=2 e^{-\frac {\left (50+e^3 x-5 x^3\right )^2}{25 x^3}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 25, normalized size = 0.86 \begin {gather*} 2 e^{-\frac {\left (50+e^3 x-5 x^3\right )^2}{25 x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(15000 + 2*E^6*x^2 - 150*x^6 + E^3*(400*x + 20*x^4))/(25*E^((2500 + E^6*x^2 - 500*x^3 + 25*x^6 + E^3
*(100*x - 10*x^4))/(25*x^3))*x^4),x]

[Out]

2/E^((50 + E^3*x - 5*x^3)^2/(25*x^3))

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fricas [A]  time = 0.88, size = 37, normalized size = 1.28 \begin {gather*} 2 \, e^{\left (-\frac {25 \, x^{6} - 500 \, x^{3} + x^{2} e^{6} - 10 \, {\left (x^{4} - 10 \, x\right )} e^{3} + 2500}{25 \, x^{3}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(2*x^2*exp(3)^2+(20*x^4+400*x)*exp(3)-150*x^6+15000)/x^4/exp(1/25*(x^2*exp(3)^2+(-10*x^4+100*x)
*exp(3)+25*x^6-500*x^3+2500)/x^3),x, algorithm="fricas")

[Out]

2*e^(-1/25*(25*x^6 - 500*x^3 + x^2*e^6 - 10*(x^4 - 10*x)*e^3 + 2500)/x^3)

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giac [A]  time = 0.30, size = 34, normalized size = 1.17 \begin {gather*} 2 \, e^{\left (-x^{3} + \frac {2}{5} \, x e^{3} - \frac {e^{6}}{25 \, x} - \frac {4 \, e^{3}}{x^{2}} - \frac {100}{x^{3}} + 20\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(2*x^2*exp(3)^2+(20*x^4+400*x)*exp(3)-150*x^6+15000)/x^4/exp(1/25*(x^2*exp(3)^2+(-10*x^4+100*x)
*exp(3)+25*x^6-500*x^3+2500)/x^3),x, algorithm="giac")

[Out]

2*e^(-x^3 + 2/5*x*e^3 - 1/25*e^6/x - 4*e^3/x^2 - 100/x^3 + 20)

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maple [A]  time = 0.11, size = 40, normalized size = 1.38

method	result	size
risch	\(2 \,{\mathrm e}^{\frac {-25 x^{6}+10 x^{4} {\mathrm e}^{3}-x^{2} {\mathrm e}^{6}+500 x^{3}-100 x \,{\mathrm e}^{3}-2500}{25 x^{3}}}\)	\(40\)
gosper	\(2 \,{\mathrm e}^{-\frac {25 x^{6}-10 x^{4} {\mathrm e}^{3}+x^{2} {\mathrm e}^{6}-500 x^{3}+100 x \,{\mathrm e}^{3}+2500}{25 x^{3}}}\)	\(43\)
norman	\(2 \,{\mathrm e}^{-\frac {x^{2} {\mathrm e}^{6}+\left (-10 x^{4}+100 x \right ) {\mathrm e}^{3}+25 x^{6}-500 x^{3}+2500}{25 x^{3}}}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/25*(2*x^2*exp(3)^2+(20*x^4+400*x)*exp(3)-150*x^6+15000)/x^4/exp(1/25*(x^2*exp(3)^2+(-10*x^4+100*x)*exp(3
)+25*x^6-500*x^3+2500)/x^3),x,method=_RETURNVERBOSE)

[Out]

2*exp(1/25*(-25*x^6+10*x^4*exp(3)-x^2*exp(6)+500*x^3-100*x*exp(3)-2500)/x^3)

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maxima [A]  time = 0.69, size = 34, normalized size = 1.17 \begin {gather*} 2 \, e^{\left (-x^{3} + \frac {2}{5} \, x e^{3} - \frac {e^{6}}{25 \, x} - \frac {4 \, e^{3}}{x^{2}} - \frac {100}{x^{3}} + 20\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(2*x^2*exp(3)^2+(20*x^4+400*x)*exp(3)-150*x^6+15000)/x^4/exp(1/25*(x^2*exp(3)^2+(-10*x^4+100*x)
*exp(3)+25*x^6-500*x^3+2500)/x^3),x, algorithm="maxima")

[Out]

2*e^(-x^3 + 2/5*x*e^3 - 1/25*e^6/x - 4*e^3/x^2 - 100/x^3 + 20)

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mupad [B]  time = 0.45, size = 38, normalized size = 1.31 \begin {gather*} 2\,{\mathrm {e}}^{-\frac {4\,{\mathrm {e}}^3}{x^2}}\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^6}{25\,x}}\,{\mathrm {e}}^{20}\,{\mathrm {e}}^{-x^3}\,{\mathrm {e}}^{-\frac {100}{x^3}}\,{\mathrm {e}}^{\frac {2\,x\,{\mathrm {e}}^3}{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-((exp(3)*(100*x - 10*x^4))/25 + (x^2*exp(6))/25 - 20*x^3 + x^6 + 100)/x^3)*((exp(3)*(400*x + 20*x^4)
)/25 + (2*x^2*exp(6))/25 - 6*x^6 + 600))/x^4,x)

[Out]

2*exp(-(4*exp(3))/x^2)*exp(-exp(6)/(25*x))*exp(20)*exp(-x^3)*exp(-100/x^3)*exp((2*x*exp(3))/5)

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sympy [B]  time = 0.31, size = 37, normalized size = 1.28 \begin {gather*} 2 e^{- \frac {x^{6} - 20 x^{3} + \frac {x^{2} e^{6}}{25} + \frac {\left (- 10 x^{4} + 100 x\right ) e^{3}}{25} + 100}{x^{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(2*x**2*exp(3)**2+(20*x**4+400*x)*exp(3)-150*x**6+15000)/x**4/exp(1/25*(x**2*exp(3)**2+(-10*x**
4+100*x)*exp(3)+25*x**6-500*x**3+2500)/x**3),x)

[Out]

2*exp(-(x**6 - 20*x**3 + x**2*exp(6)/25 + (-10*x**4 + 100*x)*exp(3)/25 + 100)/x**3)

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