Integrand size = 25, antiderivative size = 85 \[ \int \frac {\left (3-x+2 x^2\right )^2}{\left (2+3 x+5 x^2\right )^4} \, dx=\frac {121 (61+69 x)}{11625 \left (2+3 x+5 x^2\right )^3}+\frac {11 (4579+12060 x)}{120125 \left (2+3 x+5 x^2\right )^2}+\frac {16688 (3+10 x)}{148955 \left (2+3 x+5 x^2\right )}+\frac {66752 \arctan \left (\frac {3+10 x}{\sqrt {31}}\right )}{29791 \sqrt {31}} \]
121/11625*(61+69*x)/(5*x^2+3*x+2)^3+11/120125*(4579+12060*x)/(5*x^2+3*x+2) ^2+16688/148955*(3+10*x)/(5*x^2+3*x+2)+66752/923521*arctan(1/31*(3+10*x)*3 1^(1/2))*31^(1/2)
Time = 0.03 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.74 \[ \int \frac {\left (3-x+2 x^2\right )^2}{\left (2+3 x+5 x^2\right )^4} \, dx=\frac {1259239+5674908 x+12780597 x^2+21491796 x^3+18774000 x^4+12516000 x^5}{446865 \left (2+3 x+5 x^2\right )^3}+\frac {66752 \arctan \left (\frac {3+10 x}{\sqrt {31}}\right )}{29791 \sqrt {31}} \]
(1259239 + 5674908*x + 12780597*x^2 + 21491796*x^3 + 18774000*x^4 + 125160 00*x^5)/(446865*(2 + 3*x + 5*x^2)^3) + (66752*ArcTan[(3 + 10*x)/Sqrt[31]]) /(29791*Sqrt[31])
Time = 0.28 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {2191, 27, 2191, 27, 1086, 1083, 217}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (2 x^2-x+3\right )^2}{\left (5 x^2+3 x+2\right )^4} \, dx\) |
\(\Big \downarrow \) 2191 |
\(\displaystyle \frac {1}{93} \int \frac {6 \left (1550 x^2-2480 x+12863\right )}{125 \left (5 x^2+3 x+2\right )^3}dx+\frac {121 (69 x+61)}{11625 \left (5 x^2+3 x+2\right )^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 \int \frac {1550 x^2-2480 x+12863}{\left (5 x^2+3 x+2\right )^3}dx}{3875}+\frac {121 (69 x+61)}{11625 \left (5 x^2+3 x+2\right )^3}\) |
\(\Big \downarrow \) 2191 |
\(\displaystyle \frac {2 \left (\frac {1}{62} \int \frac {417200}{\left (5 x^2+3 x+2\right )^2}dx+\frac {11 (12060 x+4579)}{62 \left (5 x^2+3 x+2\right )^2}\right )}{3875}+\frac {121 (69 x+61)}{11625 \left (5 x^2+3 x+2\right )^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 \left (\frac {208600}{31} \int \frac {1}{\left (5 x^2+3 x+2\right )^2}dx+\frac {11 (12060 x+4579)}{62 \left (5 x^2+3 x+2\right )^2}\right )}{3875}+\frac {121 (69 x+61)}{11625 \left (5 x^2+3 x+2\right )^3}\) |
\(\Big \downarrow \) 1086 |
\(\displaystyle \frac {2 \left (\frac {208600}{31} \left (\frac {10}{31} \int \frac {1}{5 x^2+3 x+2}dx+\frac {10 x+3}{31 \left (5 x^2+3 x+2\right )}\right )+\frac {11 (12060 x+4579)}{62 \left (5 x^2+3 x+2\right )^2}\right )}{3875}+\frac {121 (69 x+61)}{11625 \left (5 x^2+3 x+2\right )^3}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {2 \left (\frac {208600}{31} \left (\frac {10 x+3}{31 \left (5 x^2+3 x+2\right )}-\frac {20}{31} \int \frac {1}{-(10 x+3)^2-31}d(10 x+3)\right )+\frac {11 (12060 x+4579)}{62 \left (5 x^2+3 x+2\right )^2}\right )}{3875}+\frac {121 (69 x+61)}{11625 \left (5 x^2+3 x+2\right )^3}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {2 \left (\frac {208600}{31} \left (\frac {20 \arctan \left (\frac {10 x+3}{\sqrt {31}}\right )}{31 \sqrt {31}}+\frac {10 x+3}{31 \left (5 x^2+3 x+2\right )}\right )+\frac {11 (12060 x+4579)}{62 \left (5 x^2+3 x+2\right )^2}\right )}{3875}+\frac {121 (69 x+61)}{11625 \left (5 x^2+3 x+2\right )^3}\) |
(121*(61 + 69*x))/(11625*(2 + 3*x + 5*x^2)^3) + (2*((11*(4579 + 12060*x))/ (62*(2 + 3*x + 5*x^2)^2) + (208600*((3 + 10*x)/(31*(2 + 3*x + 5*x^2)) + (2 0*ArcTan[(3 + 10*x)/Sqrt[31]])/(31*Sqrt[31])))/31))/3875
3.1.29.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] - Simp[2*c*((2*p + 3)/((p + 1)*(b^2 - 4*a*c))) Int[(a + b*x + c*x^2)^(p + 1), x], x] /; Fre eQ[{a, b, c}, x] && ILtQ[p, -1]
Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*x^2, x], f = Coeff[PolynomialRemainder[P q, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 1]}, Simp[(b*f - 2*a*g + (2*c*f - b*g)*x)*((a + b*x + c*x^2)^ (p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)) Int [(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (2*p + 3)* (2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^ 2 - 4*a*c, 0] && LtQ[p, -1]
Time = 0.72 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.67
method | result | size |
default | \(\frac {\frac {834400}{29791} x^{5}+\frac {1251600}{29791} x^{4}+\frac {7163932}{148955} x^{3}+\frac {4260199}{148955} x^{2}+\frac {1891636}{148955} x +\frac {1259239}{446865}}{\left (5 x^{2}+3 x +2\right )^{3}}+\frac {66752 \arctan \left (\frac {\left (10 x +3\right ) \sqrt {31}}{31}\right ) \sqrt {31}}{923521}\) | \(57\) |
risch | \(\frac {\frac {834400}{29791} x^{5}+\frac {1251600}{29791} x^{4}+\frac {7163932}{148955} x^{3}+\frac {4260199}{148955} x^{2}+\frac {1891636}{148955} x +\frac {1259239}{446865}}{\left (5 x^{2}+3 x +2\right )^{3}}+\frac {66752 \arctan \left (\frac {\left (10 x +3\right ) \sqrt {31}}{31}\right ) \sqrt {31}}{923521}\) | \(57\) |
125*(33376/148955*x^5+50064/148955*x^4+7163932/18619375*x^3+4260199/186193 75*x^2+1891636/18619375*x+1259239/55858125)/(5*x^2+3*x+2)^3+66752/923521*a rctan(1/31*(10*x+3)*31^(1/2))*31^(1/2)
Time = 0.41 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.24 \[ \int \frac {\left (3-x+2 x^2\right )^2}{\left (2+3 x+5 x^2\right )^4} \, dx=\frac {387996000 \, x^{5} + 581994000 \, x^{4} + 666245676 \, x^{3} + 1001280 \, \sqrt {31} {\left (125 \, x^{6} + 225 \, x^{5} + 285 \, x^{4} + 207 \, x^{3} + 114 \, x^{2} + 36 \, x + 8\right )} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) + 396198507 \, x^{2} + 175922148 \, x + 39036409}{13852815 \, {\left (125 \, x^{6} + 225 \, x^{5} + 285 \, x^{4} + 207 \, x^{3} + 114 \, x^{2} + 36 \, x + 8\right )}} \]
1/13852815*(387996000*x^5 + 581994000*x^4 + 666245676*x^3 + 1001280*sqrt(3 1)*(125*x^6 + 225*x^5 + 285*x^4 + 207*x^3 + 114*x^2 + 36*x + 8)*arctan(1/3 1*sqrt(31)*(10*x + 3)) + 396198507*x^2 + 175922148*x + 39036409)/(125*x^6 + 225*x^5 + 285*x^4 + 207*x^3 + 114*x^2 + 36*x + 8)
Time = 0.09 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.98 \[ \int \frac {\left (3-x+2 x^2\right )^2}{\left (2+3 x+5 x^2\right )^4} \, dx=\frac {12516000 x^{5} + 18774000 x^{4} + 21491796 x^{3} + 12780597 x^{2} + 5674908 x + 1259239}{55858125 x^{6} + 100544625 x^{5} + 127356525 x^{4} + 92501055 x^{3} + 50942610 x^{2} + 16087140 x + 3574920} + \frac {66752 \sqrt {31} \operatorname {atan}{\left (\frac {10 \sqrt {31} x}{31} + \frac {3 \sqrt {31}}{31} \right )}}{923521} \]
(12516000*x**5 + 18774000*x**4 + 21491796*x**3 + 12780597*x**2 + 5674908*x + 1259239)/(55858125*x**6 + 100544625*x**5 + 127356525*x**4 + 92501055*x* *3 + 50942610*x**2 + 16087140*x + 3574920) + 66752*sqrt(31)*atan(10*sqrt(3 1)*x/31 + 3*sqrt(31)/31)/923521
Time = 0.27 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.89 \[ \int \frac {\left (3-x+2 x^2\right )^2}{\left (2+3 x+5 x^2\right )^4} \, dx=\frac {66752}{923521} \, \sqrt {31} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) + \frac {12516000 \, x^{5} + 18774000 \, x^{4} + 21491796 \, x^{3} + 12780597 \, x^{2} + 5674908 \, x + 1259239}{446865 \, {\left (125 \, x^{6} + 225 \, x^{5} + 285 \, x^{4} + 207 \, x^{3} + 114 \, x^{2} + 36 \, x + 8\right )}} \]
66752/923521*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) + 1/446865*(1251600 0*x^5 + 18774000*x^4 + 21491796*x^3 + 12780597*x^2 + 5674908*x + 1259239)/ (125*x^6 + 225*x^5 + 285*x^4 + 207*x^3 + 114*x^2 + 36*x + 8)
Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.66 \[ \int \frac {\left (3-x+2 x^2\right )^2}{\left (2+3 x+5 x^2\right )^4} \, dx=\frac {66752}{923521} \, \sqrt {31} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) + \frac {12516000 \, x^{5} + 18774000 \, x^{4} + 21491796 \, x^{3} + 12780597 \, x^{2} + 5674908 \, x + 1259239}{446865 \, {\left (5 \, x^{2} + 3 \, x + 2\right )}^{3}} \]
66752/923521*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) + 1/446865*(1251600 0*x^5 + 18774000*x^4 + 21491796*x^3 + 12780597*x^2 + 5674908*x + 1259239)/ (5*x^2 + 3*x + 2)^3
Time = 12.44 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.88 \[ \int \frac {\left (3-x+2 x^2\right )^2}{\left (2+3 x+5 x^2\right )^4} \, dx=\frac {66752\,\sqrt {31}\,\mathrm {atan}\left (\frac {10\,\sqrt {31}\,x}{31}+\frac {3\,\sqrt {31}}{31}\right )}{923521}+\frac {\frac {33376\,x^5}{148955}+\frac {50064\,x^4}{148955}+\frac {7163932\,x^3}{18619375}+\frac {4260199\,x^2}{18619375}+\frac {1891636\,x}{18619375}+\frac {1259239}{55858125}}{x^6+\frac {9\,x^5}{5}+\frac {57\,x^4}{25}+\frac {207\,x^3}{125}+\frac {114\,x^2}{125}+\frac {36\,x}{125}+\frac {8}{125}} \]