\(\int \frac {-2112 x+3504 x^2-1756 x^3+3 x^4}{-64+288 x-432 x^2+216 x^3} \, dx\) [8489]

Optimal result

Integrand size = 37, antiderivative size = 26 \[ \int \frac {-2112 x+3504 x^2-1756 x^3+3 x^4}{-64+288 x-432 x^2+216 x^3} \, dx=-3+e^2-8 x+\frac {(4-x)^4}{16 (-2+3 x)^2} \]

[Out]

1/16*(-x+4)^4/(-2+3*x)^2-8*x+exp(2)-3

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {2099} \[ \int \frac {-2112 x+3504 x^2-1756 x^3+3 x^4}{-64+288 x-432 x^2+216 x^3} \, dx=\frac {x^2}{144}-\frac {875 x}{108}+\frac {250}{81 (2-3 x)}+\frac {625}{81 (2-3 x)^2} \]

[In]

Int[(-2112*x + 3504*x^2 - 1756*x^3 + 3*x^4)/(-64 + 288*x - 432*x^2 + 216*x^3),x]

[Out]

625/(81*(2 - 3*x)^2) + 250/(81*(2 - 3*x)) - (875*x)/108 + x^2/144

Rule 2099

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {875}{108}+\frac {x}{72}-\frac {1250}{27 (-2+3 x)^3}+\frac {250}{27 (-2+3 x)^2}\right ) \, dx \\ & = \frac {625}{81 (2-3 x)^2}+\frac {250}{81 (2-3 x)}-\frac {875 x}{108}+\frac {x^2}{144} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {-2112 x+3504 x^2-1756 x^3+3 x^4}{-64+288 x-432 x^2+216 x^3} \, dx=\frac {15328-45984 x+63000 x^2-31536 x^3+27 x^4}{432 (2-3 x)^2} \]

[In]

Integrate[(-2112*x + 3504*x^2 - 1756*x^3 + 3*x^4)/(-64 + 288*x - 432*x^2 + 216*x^3),x]

[Out]

(15328 - 45984*x + 63000*x^2 - 31536*x^3 + 27*x^4)/(432*(2 - 3*x)^2)

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96

method	result	size
norman	\(\frac {66 x^{2}-73 x^{3}+\frac {1}{16} x^{4}}{\left (-2+3 x \right )^{2}}\)	\(25\)
gosper	\(\frac {x^{2} \left (x^{2}-1168 x +1056\right )}{144 x^{2}-192 x +64}\)	\(26\)
risch	\(\frac {x^{2}}{144}-\frac {875 x}{108}+\frac {-\frac {250 x}{243}+\frac {125}{81}}{x^{2}-\frac {4}{3} x +\frac {4}{9}}\)	\(26\)
default	\(\frac {x^{2}}{144}-\frac {875 x}{108}+\frac {625}{81 \left (-2+3 x \right )^{2}}-\frac {250}{81 \left (-2+3 x \right )}\)	\(28\)
parallelrisch	\(\frac {2 x^{4}-2336 x^{3}+2112 x^{2}}{288 x^{2}-384 x +128}\)	\(31\)

[In]

int((3*x^4-1756*x^3+3504*x^2-2112*x)/(216*x^3-432*x^2+288*x-64),x,method=_RETURNVERBOSE)

[Out]

(66*x^2-73*x^3+1/16*x^4)/(-2+3*x)^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {-2112 x+3504 x^2-1756 x^3+3 x^4}{-64+288 x-432 x^2+216 x^3} \, dx=\frac {9 \, x^{4} - 10512 \, x^{3} + 14004 \, x^{2} - 6000 \, x + 2000}{144 \, {\left (9 \, x^{2} - 12 \, x + 4\right )}} \]

[In]

integrate((3*x^4-1756*x^3+3504*x^2-2112*x)/(216*x^3-432*x^2+288*x-64),x, algorithm="fricas")

[Out]

1/144*(9*x^4 - 10512*x^3 + 14004*x^2 - 6000*x + 2000)/(9*x^2 - 12*x + 4)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {-2112 x+3504 x^2-1756 x^3+3 x^4}{-64+288 x-432 x^2+216 x^3} \, dx=\frac {x^{2}}{144} - \frac {875 x}{108} + \frac {375 - 250 x}{243 x^{2} - 324 x + 108} \]

[In]

integrate((3*x**4-1756*x**3+3504*x**2-2112*x)/(216*x**3-432*x**2+288*x-64),x)

[Out]

x**2/144 - 875*x/108 + (375 - 250*x)/(243*x**2 - 324*x + 108)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {-2112 x+3504 x^2-1756 x^3+3 x^4}{-64+288 x-432 x^2+216 x^3} \, dx=\frac {1}{144} \, x^{2} - \frac {875}{108} \, x - \frac {125 \, {\left (2 \, x - 3\right )}}{27 \, {\left (9 \, x^{2} - 12 \, x + 4\right )}} \]

[In]

integrate((3*x^4-1756*x^3+3504*x^2-2112*x)/(216*x^3-432*x^2+288*x-64),x, algorithm="maxima")

[Out]

1/144*x^2 - 875/108*x - 125/27*(2*x - 3)/(9*x^2 - 12*x + 4)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {-2112 x+3504 x^2-1756 x^3+3 x^4}{-64+288 x-432 x^2+216 x^3} \, dx=\frac {1}{144} \, x^{2} - \frac {875}{108} \, x - \frac {125 \, {\left (2 \, x - 3\right )}}{27 \, {\left (3 \, x - 2\right )}^{2}} \]

[In]

integrate((3*x^4-1756*x^3+3504*x^2-2112*x)/(216*x^3-432*x^2+288*x-64),x, algorithm="giac")

[Out]

1/144*x^2 - 875/108*x - 125/27*(2*x - 3)/(3*x - 2)^2

Mupad [B] (verification not implemented)

Time = 12.41 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {-2112 x+3504 x^2-1756 x^3+3 x^4}{-64+288 x-432 x^2+216 x^3} \, dx=\frac {x^2}{144}-\frac {\frac {250\,x}{27}-\frac {125}{9}}{{\left (3\,x-2\right )}^2}-\frac {875\,x}{108} \]

[In]

int(-(2112*x - 3504*x^2 + 1756*x^3 - 3*x^4)/(288*x - 432*x^2 + 216*x^3 - 64),x)

[Out]

x^2/144 - ((250*x)/27 - 125/9)/(3*x - 2)^2 - (875*x)/108