3.104.16 \(\int \frac {1}{125} e^{\frac {7500-9 x^4}{2500}} (750 e^{\frac {-7500+9 x^4}{2500}}-9 x^3) \, dx\)

Optimal. Leaf size=18 \[ -2+x+5 \left (e^{3-\frac {9 x^4}{2500}}+x\right ) \]

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Rubi [A]  time = 0.08, antiderivative size = 17, normalized size of antiderivative = 0.94, number of steps used = 4, number of rules used = 3, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {12, 6688, 2209} \begin {gather*} 5 e^{3-\frac {9 x^4}{2500}}+6 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((7500 - 9*x^4)/2500)*(750*E^((-7500 + 9*x^4)/2500) - 9*x^3))/125,x]

[Out]

5*E^(3 - (9*x^4)/2500) + 6*x

Rule 12

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

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{125} \int e^{\frac {7500-9 x^4}{2500}} \left (750 e^{\frac {-7500+9 x^4}{2500}}-9 x^3\right ) \, dx\\ &=\frac {1}{125} \int \left (750-9 e^{3-\frac {9 x^4}{2500}} x^3\right ) \, dx\\ &=6 x-\frac {9}{125} \int e^{3-\frac {9 x^4}{2500}} x^3 \, dx\\ &=5 e^{3-\frac {9 x^4}{2500}}+6 x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 17, normalized size = 0.94 \begin {gather*} 5 e^{3-\frac {9 x^4}{2500}}+6 x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((7500 - 9*x^4)/2500)*(750*E^((-7500 + 9*x^4)/2500) - 9*x^3))/125,x]

[Out]

5*E^(3 - (9*x^4)/2500) + 6*x

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fricas [A]  time = 1.08, size = 22, normalized size = 1.22 \begin {gather*} {\left (6 \, x e^{\left (\frac {9}{2500} \, x^{4} - 3\right )} + 5\right )} e^{\left (-\frac {9}{2500} \, x^{4} + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/125*(750*exp(9/2500*x^4-3)-9*x^3)/exp(9/2500*x^4-3),x, algorithm="fricas")

[Out]

(6*x*e^(9/2500*x^4 - 3) + 5)*e^(-9/2500*x^4 + 3)

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giac [A]  time = 0.17, size = 14, normalized size = 0.78 \begin {gather*} 6 \, x + 5 \, e^{\left (-\frac {9}{2500} \, x^{4} + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/125*(750*exp(9/2500*x^4-3)-9*x^3)/exp(9/2500*x^4-3),x, algorithm="giac")

[Out]

6*x + 5*e^(-9/2500*x^4 + 3)

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maple [A]  time = 0.06, size = 15, normalized size = 0.83

method	result	size
risch	\(6 x +5 \,{\mathrm e}^{-\frac {9 x^{4}}{2500}+3}\)	\(15\)
default	\(6 x +5 \,{\mathrm e}^{-\frac {9 x^{4}}{2500}} {\mathrm e}^{3}\)	\(19\)
norman	\(\left (5+6 x \,{\mathrm e}^{\frac {9 x^{4}}{2500}-3}\right ) {\mathrm e}^{-\frac {9 x^{4}}{2500}+3}\)	\(25\)
meijerg	\(-\frac {5 \sqrt {3}\, \sqrt {2}\, \left (-1\right )^{\frac {3}{4}} {\mathrm e}^{-\frac {9 x^{4}}{2500}+\frac {9 x^{4} {\mathrm e}^{3}}{2500}} \left (\frac {\sqrt {2}\, x \left (-1\right )^{\frac {1}{4}} \left (-{\mathrm e}^{3}+1\right )^{\frac {1}{4}} \pi }{\left (-x^{4} \left (-{\mathrm e}^{3}+1\right )\right )^{\frac {1}{4}} \Gamma \left (\frac {3}{4}\right )}-\frac {\left (-1\right )^{\frac {1}{4}} x \left (-{\mathrm e}^{3}+1\right )^{\frac {1}{4}} \Gamma \left (\frac {1}{4}, -\frac {9 x^{4} \left (-{\mathrm e}^{3}+1\right )}{2500}\right )}{\left (-x^{4} \left (-{\mathrm e}^{3}+1\right )\right )^{\frac {1}{4}}}\right )}{2 \left (-{\mathrm e}^{3}+1\right )^{\frac {1}{4}}}-5 \,{\mathrm e}^{-\frac {9 x^{4}}{2500}+\frac {9 x^{4} {\mathrm e}^{3}}{2500}} \left (1-{\mathrm e}^{-\frac {9 x^{4} {\mathrm e}^{3}}{2500}}\right )\)	\(138\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/125*(750*exp(9/2500*x^4-3)-9*x^3)/exp(9/2500*x^4-3),x,method=_RETURNVERBOSE)

[Out]

6*x+5*exp(-9/2500*x^4+3)

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maxima [A]  time = 0.34, size = 14, normalized size = 0.78 \begin {gather*} 6 \, x + 5 \, e^{\left (-\frac {9}{2500} \, x^{4} + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/125*(750*exp(9/2500*x^4-3)-9*x^3)/exp(9/2500*x^4-3),x, algorithm="maxima")

[Out]

6*x + 5*e^(-9/2500*x^4 + 3)

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mupad [B]  time = 0.11, size = 14, normalized size = 0.78 \begin {gather*} 6\,x+5\,{\mathrm {e}}^{3-\frac {9\,x^4}{2500}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(3 - (9*x^4)/2500)*(6*exp((9*x^4)/2500 - 3) - (9*x^3)/125),x)

[Out]

6*x + 5*exp(3 - (9*x^4)/2500)

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sympy [A]  time = 0.11, size = 14, normalized size = 0.78 \begin {gather*} 6 x + 5 e^{3 - \frac {9 x^{4}}{2500}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/125*(750*exp(9/2500*x**4-3)-9*x**3)/exp(9/2500*x**4-3),x)

[Out]

6*x + 5*exp(3 - 9*x**4/2500)

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