\(\int \frac {-45-9 e^2 x}{900 x^3+240 e^2 x^4+16 e^4 x^5} \, dx\) [4795]

Optimal result

Integrand size = 33, antiderivative size = 19 \[ \int \frac {-45-9 e^2 x}{900 x^3+240 e^2 x^4+16 e^4 x^5} \, dx=\frac {1}{4 x^2 \left (10+\frac {4 e^2 x}{3}\right )} \]

[Out]

1/4/x^2/(10+4/3*exp(2)*x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {1608, 27, 12, 75} \[ \int \frac {-45-9 e^2 x}{900 x^3+240 e^2 x^4+16 e^4 x^5} \, dx=\frac {3}{8 x^2 \left (2 e^2 x+15\right )} \]

[In]

Int[(-45 - 9*E^2*x)/(900*x^3 + 240*E^2*x^4 + 16*E^4*x^5),x]

[Out]

3/(8*x^2*(15 + 2*E^2*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 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 75

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

Rule 1608

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-45-9 e^2 x}{x^3 \left (900+240 e^2 x+16 e^4 x^2\right )} \, dx \\ & = \int \frac {-45-9 e^2 x}{4 x^3 \left (15+2 e^2 x\right )^2} \, dx \\ & = \frac {1}{4} \int \frac {-45-9 e^2 x}{x^3 \left (15+2 e^2 x\right )^2} \, dx \\ & = \frac {3}{8 x^2 \left (15+2 e^2 x\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-45-9 e^2 x}{900 x^3+240 e^2 x^4+16 e^4 x^5} \, dx=\frac {3}{8 x^2 \left (15+2 e^2 x\right )} \]

[In]

Integrate[(-45 - 9*E^2*x)/(900*x^3 + 240*E^2*x^4 + 16*E^4*x^5),x]

[Out]

3/(8*x^2*(15 + 2*E^2*x))

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79

method	result	size
gosper	\(\frac {3}{8 x^{2} \left (2 \,{\mathrm e}^{2} x +15\right )}\)	\(15\)
norman	\(\frac {3}{8 x^{2} \left (2 \,{\mathrm e}^{2} x +15\right )}\)	\(15\)
risch	\(\frac {3}{8 x^{2} \left (2 \,{\mathrm e}^{2} x +15\right )}\)	\(15\)
parallelrisch	\(\frac {3}{8 x^{2} \left (2 \,{\mathrm e}^{2} x +15\right )}\)	\(15\)

[In]

int((-9*exp(2)*x-45)/(16*x^5*exp(2)^2+240*x^4*exp(2)+900*x^3),x,method=_RETURNVERBOSE)

[Out]

3/8/x^2/(2*exp(2)*x+15)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-45-9 e^2 x}{900 x^3+240 e^2 x^4+16 e^4 x^5} \, dx=\frac {3}{8 \, {\left (2 \, x^{3} e^{2} + 15 \, x^{2}\right )}} \]

[In]

integrate((-9*exp(2)*x-45)/(16*x^5*exp(2)^2+240*x^4*exp(2)+900*x^3),x, algorithm="fricas")

[Out]

3/8/(2*x^3*e^2 + 15*x^2)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {-45-9 e^2 x}{900 x^3+240 e^2 x^4+16 e^4 x^5} \, dx=\frac {3}{16 x^{3} e^{2} + 120 x^{2}} \]

[In]

integrate((-9*exp(2)*x-45)/(16*x**5*exp(2)**2+240*x**4*exp(2)+900*x**3),x)

[Out]

3/(16*x**3*exp(2) + 120*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-45-9 e^2 x}{900 x^3+240 e^2 x^4+16 e^4 x^5} \, dx=\frac {3}{8 \, {\left (2 \, x^{3} e^{2} + 15 \, x^{2}\right )}} \]

[In]

integrate((-9*exp(2)*x-45)/(16*x^5*exp(2)^2+240*x^4*exp(2)+900*x^3),x, algorithm="maxima")

[Out]

3/8/(2*x^3*e^2 + 15*x^2)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37 \[ \int \frac {-45-9 e^2 x}{900 x^3+240 e^2 x^4+16 e^4 x^5} \, dx=\frac {e^{4}}{150 \, {\left (2 \, x e^{2} + 15\right )}} - \frac {2 \, x e^{2} - 15}{600 \, x^{2}} \]

[In]

integrate((-9*exp(2)*x-45)/(16*x^5*exp(2)^2+240*x^4*exp(2)+900*x^3),x, algorithm="giac")

[Out]

1/150*e^4/(2*x*e^2 + 15) - 1/600*(2*x*e^2 - 15)/x^2

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-45-9 e^2 x}{900 x^3+240 e^2 x^4+16 e^4 x^5} \, dx=\frac {3}{16\,{\mathrm {e}}^2\,x^3+120\,x^2} \]

[In]

int(-(9*x*exp(2) + 45)/(240*x^4*exp(2) + 16*x^5*exp(4) + 900*x^3),x)

[Out]

3/(16*x^3*exp(2) + 120*x^2)