3.72.71 \(\int \frac {e^2 (-490-49 x)}{15 x^3} \, dx\)

Optimal. Leaf size=13 \[ \frac {49 e^2 (5+x)}{15 x^2} \]

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Rubi [A]  time = 0.00, antiderivative size = 15, normalized size of antiderivative = 1.15, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {12, 37} \begin {gather*} \frac {49 e^2 (x+10)^2}{300 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^2*(-490 - 49*x))/(15*x^3),x]

[Out]

(49*E^2*(10 + x)^2)/(300*x^2)

Rule 12

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

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{15} e^2 \int \frac {-490-49 x}{x^3} \, dx\\ &=\frac {49 e^2 (10+x)^2}{300 x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 18, normalized size = 1.38 \begin {gather*} -\frac {49}{15} e^2 \left (-\frac {5}{x^2}-\frac {1}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^2*(-490 - 49*x))/(15*x^3),x]

[Out]

(-49*E^2*(-5/x^2 - x^(-1)))/15

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fricas [A]  time = 0.61, size = 10, normalized size = 0.77 \begin {gather*} \frac {49 \, {\left (x + 5\right )} e^{2}}{15 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*(-49*x-490)*exp(1)^2/x^3,x, algorithm="fricas")

[Out]

49/15*(x + 5)*e^2/x^2

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giac [A]  time = 0.12, size = 10, normalized size = 0.77 \begin {gather*} \frac {49 \, {\left (x + 5\right )} e^{2}}{15 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*(-49*x-490)*exp(1)^2/x^3,x, algorithm="giac")

[Out]

49/15*(x + 5)*e^2/x^2

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maple [A]  time = 0.02, size = 13, normalized size = 1.00

method	result	size
gosper	\(\frac {49 \,{\mathrm e}^{2} \left (5+x \right )}{15 x^{2}}\)	\(13\)
risch	\(\frac {{\mathrm e}^{2} \left (245+49 x \right )}{15 x^{2}}\)	\(13\)
default	\(\frac {49 \,{\mathrm e}^{2} \left (\frac {5}{x^{2}}+\frac {1}{x}\right )}{15}\)	\(16\)
norman	\(\frac {\frac {49 \,{\mathrm e}^{2}}{3}+\frac {49 \,{\mathrm e}^{2} x}{15}}{x^{2}}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/15*(-49*x-490)*exp(1)^2/x^3,x,method=_RETURNVERBOSE)

[Out]

49/15*exp(1)^2*(5+x)/x^2

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maxima [A]  time = 0.35, size = 10, normalized size = 0.77 \begin {gather*} \frac {49 \, {\left (x + 5\right )} e^{2}}{15 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*(-49*x-490)*exp(1)^2/x^3,x, algorithm="maxima")

[Out]

49/15*(x + 5)*e^2/x^2

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mupad [B]  time = 0.03, size = 15, normalized size = 1.15 \begin {gather*} \frac {245\,{\mathrm {e}}^2+49\,x\,{\mathrm {e}}^2}{15\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2)*(49*x + 490))/(15*x^3),x)

[Out]

(245*exp(2) + 49*x*exp(2))/(15*x^2)

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sympy [A]  time = 0.11, size = 19, normalized size = 1.46 \begin {gather*} - \frac {- 49 x e^{2} - 245 e^{2}}{15 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*(-49*x-490)*exp(1)**2/x**3,x)

[Out]

-(-49*x*exp(2) - 245*exp(2))/(15*x**2)

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