\(\int \frac {e^2 (-490-49 x)}{15 x^3} \, dx\) [7171]

Optimal result

Integrand size = 15, antiderivative size = 13 \[ \int \frac {e^2 (-490-49 x)}{15 x^3} \, dx=\frac {49 e^2 (5+x)}{15 x^2} \]

[Out]

1/15*exp(1)^2/x^2*(245+49*x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {12, 37} \[ \int \frac {e^2 (-490-49 x)}{15 x^3} \, dx=\frac {49 e^2 (x+10)^2}{300 x^2} \]

[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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.38 \[ \int \frac {e^2 (-490-49 x)}{15 x^3} \, dx=-\frac {49}{15} e^2 \left (-\frac {5}{x^2}-\frac {1}{x}\right ) \]

[In]

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

[Out]

(-49*E^2*(-5/x^2 - x^(-1)))/15

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00

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

[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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int \frac {e^2 (-490-49 x)}{15 x^3} \, dx=\frac {49 \, {\left (x + 5\right )} e^{2}}{15 \, x^{2}} \]

[In]

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

[Out]

49/15*(x + 5)*e^2/x^2

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.46 \[ \int \frac {e^2 (-490-49 x)}{15 x^3} \, dx=- \frac {- 49 x e^{2} - 245 e^{2}}{15 x^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int \frac {e^2 (-490-49 x)}{15 x^3} \, dx=\frac {49 \, {\left (x + 5\right )} e^{2}}{15 \, x^{2}} \]

[In]

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

[Out]

49/15*(x + 5)*e^2/x^2

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int \frac {e^2 (-490-49 x)}{15 x^3} \, dx=\frac {49 \, {\left (x + 5\right )} e^{2}}{15 \, x^{2}} \]

[In]

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

[Out]

49/15*(x + 5)*e^2/x^2

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \frac {e^2 (-490-49 x)}{15 x^3} \, dx=\frac {245\,{\mathrm {e}}^2+49\,x\,{\mathrm {e}}^2}{15\,x^2} \]

[In]

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

[Out]

(245*exp(2) + 49*x*exp(2))/(15*x^2)