3.65.21 \(\int \frac {25 x}{4 e^6} \, dx\)

Optimal. Leaf size=10 \[ \frac {25 x^2}{8 e^6} \]

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Rubi [A]  time = 0.00, antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {12, 30} \begin {gather*} \frac {25 x^2}{8 e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(25*x)/(4*E^6),x]

[Out]

(25*x^2)/(8*E^6)

Rule 12

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {25 \int x \, dx}{4 e^6}\\ &=\frac {25 x^2}{8 e^6}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 10, normalized size = 1.00 \begin {gather*} \frac {25 x^2}{8 e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25*x)/(4*E^6),x]

[Out]

(25*x^2)/(8*E^6)

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fricas [A]  time = 0.59, size = 7, normalized size = 0.70 \begin {gather*} \frac {25}{8} \, x^{2} e^{\left (-6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(25/4*x*exp(x)^2/exp(3+x)^2,x, algorithm="fricas")

[Out]

25/8*x^2*e^(-6)

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giac [A]  time = 0.21, size = 7, normalized size = 0.70 \begin {gather*} \frac {25}{8} \, x^{2} e^{\left (-6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(25/4*x*exp(x)^2/exp(3+x)^2,x, algorithm="giac")

[Out]

25/8*x^2*e^(-6)

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maple [A]  time = 0.06, size = 8, normalized size = 0.80

method	result	size
risch	\(\frac {25 x^{2} {\mathrm e}^{-6}}{8}\)	\(8\)
default	\(\frac {25 x^{2} {\mathrm e}^{-6}}{8}\)	\(10\)
norman	\(\frac {25 x^{2} {\mathrm e}^{-6}}{8}\)	\(10\)
gosper	\(\frac {25 \,{\mathrm e}^{-2 x -6} x^{2} {\mathrm e}^{2 x}}{8}\)	\(16\)
meijerg	\(\frac {25 \,{\mathrm e}^{2 x -6+2 x \,{\mathrm e}^{-6}} {\mathrm e}^{-4 x} \left (1-\frac {\left (2-4 x \left (1-{\mathrm e}^{-6}\right )\right ) {\mathrm e}^{2 x \left (1-{\mathrm e}^{-6}\right )}}{2}\right )}{16 \left (1-{\mathrm e}^{-6}\right )^{2}}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(25/4*x*exp(x)^2/exp(3+x)^2,x,method=_RETURNVERBOSE)

[Out]

25/8*x^2*exp(-6)

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maxima [A]  time = 0.36, size = 7, normalized size = 0.70 \begin {gather*} \frac {25}{8} \, x^{2} e^{\left (-6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(25/4*x*exp(x)^2/exp(3+x)^2,x, algorithm="maxima")

[Out]

25/8*x^2*e^(-6)

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mupad [B]  time = 4.83, size = 7, normalized size = 0.70 \begin {gather*} \frac {25\,x^2\,{\mathrm {e}}^{-6}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((25*x*exp(2*x)*exp(- 2*x - 6))/4,x)

[Out]

(25*x^2*exp(-6))/8

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sympy [A]  time = 0.06, size = 8, normalized size = 0.80 \begin {gather*} \frac {25 x^{2}}{8 e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(25/4*x*exp(x)**2/exp(3+x)**2,x)

[Out]

25*x**2*exp(-6)/8

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