\(\int \frac {-2592 x^2+144 x^3+576 x^5+e^4 (-18+4 x^3)+e^2 (432 x-96 x^4)}{3 e^4 x^3-72 e^2 x^4+432 x^5} \, dx\) [7282]

Optimal result

Integrand size = 65, antiderivative size = 22 \[ \int \frac {-2592 x^2+144 x^3+576 x^5+e^4 \left (-18+4 x^3\right )+e^2 \left (432 x-96 x^4\right )}{3 e^4 x^3-72 e^2 x^4+432 x^5} \, dx=\frac {4}{e^2-12 x}+\frac {3}{x^2}+\frac {4 x}{3} \]

[Out]

4/3*x+4/(-12*x+exp(2))+3/x^2

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {1608, 27, 12, 1634} \[ \int \frac {-2592 x^2+144 x^3+576 x^5+e^4 \left (-18+4 x^3\right )+e^2 \left (432 x-96 x^4\right )}{3 e^4 x^3-72 e^2 x^4+432 x^5} \, dx=\frac {3}{x^2}+\frac {4 x}{3}+\frac {4}{e^2-12 x} \]

[In]

Int[(-2592*x^2 + 144*x^3 + 576*x^5 + E^4*(-18 + 4*x^3) + E^2*(432*x - 96*x^4))/(3*E^4*x^3 - 72*E^2*x^4 + 432*x
^5),x]

[Out]

4/(E^2 - 12*x) + 3/x^2 + (4*x)/3

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 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]

Rule 1634

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-2592 x^2+144 x^3+576 x^5+e^4 \left (-18+4 x^3\right )+e^2 \left (432 x-96 x^4\right )}{x^3 \left (3 e^4-72 e^2 x+432 x^2\right )} \, dx \\ & = \int \frac {-2592 x^2+144 x^3+576 x^5+e^4 \left (-18+4 x^3\right )+e^2 \left (432 x-96 x^4\right )}{3 \left (e^2-12 x\right )^2 x^3} \, dx \\ & = \frac {1}{3} \int \frac {-2592 x^2+144 x^3+576 x^5+e^4 \left (-18+4 x^3\right )+e^2 \left (432 x-96 x^4\right )}{\left (e^2-12 x\right )^2 x^3} \, dx \\ & = \frac {1}{3} \int \left (4+\frac {144}{\left (e^2-12 x\right )^2}-\frac {18}{x^3}\right ) \, dx \\ & = \frac {4}{e^2-12 x}+\frac {3}{x^2}+\frac {4 x}{3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {-2592 x^2+144 x^3+576 x^5+e^4 \left (-18+4 x^3\right )+e^2 \left (432 x-96 x^4\right )}{3 e^4 x^3-72 e^2 x^4+432 x^5} \, dx=\frac {2}{3} \left (\frac {9}{2 x^2}+2 x-\frac {6}{-e^2+12 x}\right ) \]

[In]

Integrate[(-2592*x^2 + 144*x^3 + 576*x^5 + E^4*(-18 + 4*x^3) + E^2*(432*x - 96*x^4))/(3*E^4*x^3 - 72*E^2*x^4 +
432*x^5),x]

[Out]

(2*(9/(2*x^2) + 2*x - 6/(-E^2 + 12*x)))/3

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36

method	result	size
risch	\(\frac {4 x}{3}+\frac {4 x^{2}+3 \,{\mathrm e}^{2}-36 x}{x^{2} \left (-12 x +{\mathrm e}^{2}\right )}\)	\(30\)
norman	\(\frac {\left (\frac {{\mathrm e}^{4}}{9}+4\right ) x^{2}-36 x -16 x^{4}+3 \,{\mathrm e}^{2}}{x^{2} \left (-12 x +{\mathrm e}^{2}\right )}\)	\(38\)
gosper	\(\frac {x^{2} {\mathrm e}^{4}-144 x^{4}+36 x^{2}+27 \,{\mathrm e}^{2}-324 x}{9 x^{2} \left (-12 x +{\mathrm e}^{2}\right )}\)	\(40\)
parallelrisch	\(\frac {4 x^{2} {\mathrm e}^{4}-576 x^{4}+144 x^{2}+108 \,{\mathrm e}^{2}-1296 x}{36 x^{2} \left (-12 x +{\mathrm e}^{2}\right )}\)	\(41\)

[In]

int(((4*x^3-18)*exp(2)^2+(-96*x^4+432*x)*exp(2)+576*x^5+144*x^3-2592*x^2)/(3*x^3*exp(2)^2-72*x^4*exp(2)+432*x^
5),x,method=_RETURNVERBOSE)

[Out]

4/3*x+(4*x^2+3*exp(2)-36*x)/x^2/(-12*x+exp(2))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.91 \[ \int \frac {-2592 x^2+144 x^3+576 x^5+e^4 \left (-18+4 x^3\right )+e^2 \left (432 x-96 x^4\right )}{3 e^4 x^3-72 e^2 x^4+432 x^5} \, dx=\frac {48 \, x^{4} - 12 \, x^{2} - {\left (4 \, x^{3} + 9\right )} e^{2} + 108 \, x}{3 \, {\left (12 \, x^{3} - x^{2} e^{2}\right )}} \]

[In]

integrate(((4*x^3-18)*exp(2)^2+(-96*x^4+432*x)*exp(2)+576*x^5+144*x^3-2592*x^2)/(3*x^3*exp(2)^2-72*x^4*exp(2)+
432*x^5),x, algorithm="fricas")

[Out]

1/3*(48*x^4 - 12*x^2 - (4*x^3 + 9)*e^2 + 108*x)/(12*x^3 - x^2*e^2)

Sympy [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {-2592 x^2+144 x^3+576 x^5+e^4 \left (-18+4 x^3\right )+e^2 \left (432 x-96 x^4\right )}{3 e^4 x^3-72 e^2 x^4+432 x^5} \, dx=\frac {4 x}{3} + \frac {- 4 x^{2} + 36 x - 3 e^{2}}{12 x^{3} - x^{2} e^{2}} \]

[In]

integrate(((4*x**3-18)*exp(2)**2+(-96*x**4+432*x)*exp(2)+576*x**5+144*x**3-2592*x**2)/(3*x**3*exp(2)**2-72*x**
4*exp(2)+432*x**5),x)

[Out]

4*x/3 + (-4*x**2 + 36*x - 3*exp(2))/(12*x**3 - x**2*exp(2))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.55 \[ \int \frac {-2592 x^2+144 x^3+576 x^5+e^4 \left (-18+4 x^3\right )+e^2 \left (432 x-96 x^4\right )}{3 e^4 x^3-72 e^2 x^4+432 x^5} \, dx=\frac {4}{3} \, x - \frac {4 \, x^{2} - 36 \, x + 3 \, e^{2}}{12 \, x^{3} - x^{2} e^{2}} \]

[In]

integrate(((4*x^3-18)*exp(2)^2+(-96*x^4+432*x)*exp(2)+576*x^5+144*x^3-2592*x^2)/(3*x^3*exp(2)^2-72*x^4*exp(2)+
432*x^5),x, algorithm="maxima")

[Out]

4/3*x - (4*x^2 - 36*x + 3*e^2)/(12*x^3 - x^2*e^2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {-2592 x^2+144 x^3+576 x^5+e^4 \left (-18+4 x^3\right )+e^2 \left (432 x-96 x^4\right )}{3 e^4 x^3-72 e^2 x^4+432 x^5} \, dx=\frac {4}{3} \, x - \frac {4}{12 \, x - e^{2}} + \frac {3}{x^{2}} \]

[In]

integrate(((4*x^3-18)*exp(2)^2+(-96*x^4+432*x)*exp(2)+576*x^5+144*x^3-2592*x^2)/(3*x^3*exp(2)^2-72*x^4*exp(2)+
432*x^5),x, algorithm="giac")

[Out]

4/3*x - 4/(12*x - e^2) + 3/x^2

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45 \[ \int \frac {-2592 x^2+144 x^3+576 x^5+e^4 \left (-18+4 x^3\right )+e^2 \left (432 x-96 x^4\right )}{3 e^4 x^3-72 e^2 x^4+432 x^5} \, dx=\frac {4\,x}{3}-\frac {4\,x^2-36\,x+3\,{\mathrm {e}}^2}{x^2\,\left (12\,x-{\mathrm {e}}^2\right )} \]

[In]

int((exp(2)*(432*x - 96*x^4) + exp(4)*(4*x^3 - 18) - 2592*x^2 + 144*x^3 + 576*x^5)/(3*x^3*exp(4) - 72*x^4*exp(
2) + 432*x^5),x)

[Out]

(4*x)/3 - (3*exp(2) - 36*x + 4*x^2)/(x^2*(12*x - exp(2)))