\(\int \frac {110+5 e^{20}+210 x-81 x^2+8 x^3}{100-40 x+4 x^2} \, dx\) [663]

Optimal result

Integrand size = 33, antiderivative size = 22 \[ \int \frac {110+5 e^{20}+210 x-81 x^2+8 x^3}{100-40 x+4 x^2} \, dx=x^2+\frac {x \left (22+e^{20}+x\right )}{4 (5-x)} \]

[Out]

x^2+1/4*(22+x+exp(5)^4)*x/(5-x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {27, 12, 1864} \[ \int \frac {110+5 e^{20}+210 x-81 x^2+8 x^3}{100-40 x+4 x^2} \, dx=x^2-\frac {x}{4}+\frac {5 \left (27+e^{20}\right )}{4 (5-x)} \]

[In]

Int[(110 + 5*E^20 + 210*x - 81*x^2 + 8*x^3)/(100 - 40*x + 4*x^2),x]

[Out]

(5*(27 + E^20))/(4*(5 - x)) - x/4 + 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 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 1864

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps \begin{align*} \text {integral}& = \int \frac {110+5 e^{20}+210 x-81 x^2+8 x^3}{4 (-5+x)^2} \, dx \\ & = \frac {1}{4} \int \frac {110+5 e^{20}+210 x-81 x^2+8 x^3}{(-5+x)^2} \, dx \\ & = \frac {1}{4} \int \left (-1+\frac {5 \left (27+e^{20}\right )}{(-5+x)^2}+8 x\right ) \, dx \\ & = \frac {5 \left (27+e^{20}\right )}{4 (5-x)}-\frac {x}{4}+x^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {110+5 e^{20}+210 x-81 x^2+8 x^3}{100-40 x+4 x^2} \, dx=\frac {1}{4} \left (-\frac {5 \left (27+e^{20}\right )}{-5+x}+39 (-5+x)+4 (-5+x)^2\right ) \]

[In]

Integrate[(110 + 5*E^20 + 210*x - 81*x^2 + 8*x^3)/(100 - 40*x + 4*x^2),x]

[Out]

((-5*(27 + E^20))/(-5 + x) + 39*(-5 + x) + 4*(-5 + x)^2)/4

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95

method	result	size
default	\(x^{2}-\frac {x}{4}-\frac {135+5 \,{\mathrm e}^{20}}{4 \left (-5+x \right )}\)	\(21\)
norman	\(\frac {x^{3}-\frac {21 x^{2}}{4}-\frac {55}{2}-\frac {5 \,{\mathrm e}^{20}}{4}}{-5+x}\)	\(23\)
risch	\(x^{2}-\frac {x}{4}-\frac {5 \,{\mathrm e}^{20}}{4 \left (-5+x \right )}-\frac {135}{4 \left (-5+x \right )}\)	\(24\)
gosper	\(-\frac {5 \,{\mathrm e}^{20}-4 x^{3}+21 x^{2}+110}{4 \left (-5+x \right )}\)	\(26\)
parallelrisch	\(-\frac {5 \,{\mathrm e}^{20}-4 x^{3}+21 x^{2}+110}{4 \left (-5+x \right )}\)	\(26\)
meijerg	\(\frac {58 x}{5 \left (1-\frac {x}{5}\right )}+\frac {x \,{\mathrm e}^{20}}{-4 x +20}+\frac {5 x \left (-\frac {2}{25} x^{2}-\frac {6}{5} x +12\right )}{2 \left (1-\frac {x}{5}\right )}-\frac {27 x \left (-\frac {3 x}{5}+6\right )}{4 \left (1-\frac {x}{5}\right )}\)	\(59\)

[In]

int((5*exp(5)^4+8*x^3-81*x^2+210*x+110)/(4*x^2-40*x+100),x,method=_RETURNVERBOSE)

[Out]

x^2-1/4*x-1/4*(135+5*exp(20))/(-5+x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {110+5 e^{20}+210 x-81 x^2+8 x^3}{100-40 x+4 x^2} \, dx=\frac {4 \, x^{3} - 21 \, x^{2} + 5 \, x - 5 \, e^{20} - 135}{4 \, {\left (x - 5\right )}} \]

[In]

integrate((5*exp(5)^4+8*x^3-81*x^2+210*x+110)/(4*x^2-40*x+100),x, algorithm="fricas")

[Out]

1/4*(4*x^3 - 21*x^2 + 5*x - 5*e^20 - 135)/(x - 5)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {110+5 e^{20}+210 x-81 x^2+8 x^3}{100-40 x+4 x^2} \, dx=x^{2} - \frac {x}{4} + \frac {- 5 e^{20} - 135}{4 x - 20} \]

[In]

integrate((5*exp(5)**4+8*x**3-81*x**2+210*x+110)/(4*x**2-40*x+100),x)

[Out]

x**2 - x/4 + (-5*exp(20) - 135)/(4*x - 20)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {110+5 e^{20}+210 x-81 x^2+8 x^3}{100-40 x+4 x^2} \, dx=x^{2} - \frac {1}{4} \, x - \frac {5 \, {\left (e^{20} + 27\right )}}{4 \, {\left (x - 5\right )}} \]

[In]

integrate((5*exp(5)^4+8*x^3-81*x^2+210*x+110)/(4*x^2-40*x+100),x, algorithm="maxima")

[Out]

x^2 - 1/4*x - 5/4*(e^20 + 27)/(x - 5)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {110+5 e^{20}+210 x-81 x^2+8 x^3}{100-40 x+4 x^2} \, dx=x^{2} - \frac {1}{4} \, x - \frac {5 \, {\left (e^{20} + 27\right )}}{4 \, {\left (x - 5\right )}} \]

[In]

integrate((5*exp(5)^4+8*x^3-81*x^2+210*x+110)/(4*x^2-40*x+100),x, algorithm="giac")

[Out]

x^2 - 1/4*x - 5/4*(e^20 + 27)/(x - 5)

Mupad [B] (verification not implemented)

Time = 8.36 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {110+5 e^{20}+210 x-81 x^2+8 x^3}{100-40 x+4 x^2} \, dx=x^2-\frac {5\,{\mathrm {e}}^{20}+135}{4\,x-20}-\frac {x}{4} \]

[In]

int((210*x + 5*exp(20) - 81*x^2 + 8*x^3 + 110)/(4*x^2 - 40*x + 100),x)

[Out]

x^2 - (5*exp(20) + 135)/(4*x - 20) - x/4