\(\int \frac {1}{(3-x+2 x^2)^2 (2+3 x+5 x^2)^2} \, dx\) [78]

Optimal result

Integrand size = 25, antiderivative size = 127 \[ \int \frac {1}{\left (3-x+2 x^2\right )^2 \left (2+3 x+5 x^2\right )^2} \, dx=-\frac {25 (117-137 x)}{172546 \left (2+3 x+5 x^2\right )}+\frac {13-6 x}{506 \left (3-x+2 x^2\right ) \left (2+3 x+5 x^2\right )}+\frac {2769 \arctan \left (\frac {1-4 x}{\sqrt {23}}\right )}{122452 \sqrt {23}}+\frac {12643 \arctan \left (\frac {3+10 x}{\sqrt {31}}\right )}{165044 \sqrt {31}}+\frac {19 \log \left (3-x+2 x^2\right )}{10648}-\frac {19 \log \left (2+3 x+5 x^2\right )}{10648} \] Output:

(-2925+3425*x)/(862730*x^2+517638*x+345092)+1/506*(13-6*x)/(2*x^2-x+3)/(5* 
x^2+3*x+2)+2769/2816396*arctan(1/23*(1-4*x)*23^(1/2))*23^(1/2)+12643/51163 
64*arctan(1/31*(3+10*x)*31^(1/2))*31^(1/2)+19/10648*ln(2*x^2-x+3)-19/10648 
*ln(5*x^2+3*x+2)
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.83 \[ \int \frac {1}{\left (3-x+2 x^2\right )^2 \left (2+3 x+5 x^2\right )^2} \, dx=\frac {\frac {31372 \left (-4342+11154 x-9275 x^2+6850 x^3\right )}{6+7 x+16 x^2+x^3+10 x^4}-5322018 \sqrt {23} \arctan \left (\frac {-1+4 x}{\sqrt {23}}\right )[phone] \sqrt {31} \arctan \left (\frac {3+10 x}{\sqrt {31}}\right )+9659011 \log \left (3-x+2 x^2\right )-9659011 \log \left (2+3 x+5 x^2\right )}{5413113112} \] Input:

Integrate[1/((3 - x + 2*x^2)^2*(2 + 3*x + 5*x^2)^2),x]
 

Output:

((31372*(-4342 + 11154*x - 9275*x^2 + 6850*x^3))/(6 + 7*x + 16*x^2 + x^3 + 
 10*x^4) - 5322018*Sqrt[23]*ArcTan[(-1 + 4*x)/Sqrt[23]] + 13376294*Sqrt[31 
]*ArcTan[(3 + 10*x)/Sqrt[31]] + 9659011*Log[3 - x + 2*x^2] - 9659011*Log[2 
 + 3*x + 5*x^2])/5413113112
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.13, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {1305, 27, 2135, 27, 2141, 27, 1142, 25, 1083, 217, 1103}

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.

\(\displaystyle \int \frac {1}{\left (2 x^2-x+3\right )^2 \left (5 x^2+3 x+2\right )^2} \, dx\)

\(\Big \downarrow \) 1305

\(\displaystyle \frac {13-6 x}{506 \left (2 x^2-x+3\right ) \left (5 x^2+3 x+2\right )}-\frac {\int -\frac {11 \left (-90 x^2+209 x+211\right )}{\left (2 x^2-x+3\right ) \left (5 x^2+3 x+2\right )^2}dx}{5566}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{506} \int \frac {-90 x^2+209 x+211}{\left (2 x^2-x+3\right ) \left (5 x^2+3 x+2\right )^2}dx+\frac {13-6 x}{506 \left (2 x^2-x+3\right ) \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 2135

\(\displaystyle \frac {1}{506} \left (\frac {\int \frac {22 \left (6850 x^2-19235 x+12538\right )}{\left (2 x^2-x+3\right ) \left (5 x^2+3 x+2\right )}dx}{7502}-\frac {25 (117-137 x)}{341 \left (5 x^2+3 x+2\right )}\right )+\frac {13-6 x}{506 \left (2 x^2-x+3\right ) \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{506} \left (\frac {1}{341} \int \frac {6850 x^2-19235 x+12538}{\left (2 x^2-x+3\right ) \left (5 x^2+3 x+2\right )}dx-\frac {25 (117-137 x)}{341 \left (5 x^2+3 x+2\right )}\right )+\frac {13-6 x}{506 \left (2 x^2-x+3\right ) \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 2141

\(\displaystyle \frac {1}{506} \left (\frac {1}{341} \left (\frac {1}{242} \int -\frac {341 (1603-874 x)}{2 x^2-x+3}dx+\frac {1}{242} \int \frac {253 (5438-2945 x)}{5 x^2+3 x+2}dx\right )-\frac {25 (117-137 x)}{341 \left (5 x^2+3 x+2\right )}\right )+\frac {13-6 x}{506 \left (2 x^2-x+3\right ) \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{506} \left (\frac {1}{341} \left (\frac {23}{22} \int \frac {5438-2945 x}{5 x^2+3 x+2}dx-\frac {31}{22} \int \frac {1603-874 x}{2 x^2-x+3}dx\right )-\frac {25 (117-137 x)}{341 \left (5 x^2+3 x+2\right )}\right )+\frac {13-6 x}{506 \left (2 x^2-x+3\right ) \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {1}{506} \left (\frac {1}{341} \left (\frac {23}{22} \left (\frac {12643}{2} \int \frac {1}{5 x^2+3 x+2}dx-\frac {589}{2} \int \frac {10 x+3}{5 x^2+3 x+2}dx\right )-\frac {31}{22} \left (\frac {2769}{2} \int \frac {1}{2 x^2-x+3}dx-\frac {437}{2} \int -\frac {1-4 x}{2 x^2-x+3}dx\right )\right )-\frac {25 (117-137 x)}{341 \left (5 x^2+3 x+2\right )}\right )+\frac {13-6 x}{506 \left (2 x^2-x+3\right ) \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{506} \left (\frac {1}{341} \left (\frac {23}{22} \left (\frac {12643}{2} \int \frac {1}{5 x^2+3 x+2}dx-\frac {589}{2} \int \frac {10 x+3}{5 x^2+3 x+2}dx\right )-\frac {31}{22} \left (\frac {2769}{2} \int \frac {1}{2 x^2-x+3}dx+\frac {437}{2} \int \frac {1-4 x}{2 x^2-x+3}dx\right )\right )-\frac {25 (117-137 x)}{341 \left (5 x^2+3 x+2\right )}\right )+\frac {13-6 x}{506 \left (2 x^2-x+3\right ) \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {1}{506} \left (\frac {1}{341} \left (\frac {23}{22} \left (-\frac {589}{2} \int \frac {10 x+3}{5 x^2+3 x+2}dx-12643 \int \frac {1}{-(10 x+3)^2-31}d(10 x+3)\right )-\frac {31}{22} \left (\frac {437}{2} \int \frac {1-4 x}{2 x^2-x+3}dx-2769 \int \frac {1}{-(4 x-1)^2-23}d(4 x-1)\right )\right )-\frac {25 (117-137 x)}{341 \left (5 x^2+3 x+2\right )}\right )+\frac {13-6 x}{506 \left (2 x^2-x+3\right ) \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{506} \left (\frac {1}{341} \left (\frac {23}{22} \left (\frac {12643 \arctan \left (\frac {10 x+3}{\sqrt {31}}\right )}{\sqrt {31}}-\frac {589}{2} \int \frac {10 x+3}{5 x^2+3 x+2}dx\right )-\frac {31}{22} \left (\frac {437}{2} \int \frac {1-4 x}{2 x^2-x+3}dx+\frac {2769 \arctan \left (\frac {4 x-1}{\sqrt {23}}\right )}{\sqrt {23}}\right )\right )-\frac {25 (117-137 x)}{341 \left (5 x^2+3 x+2\right )}\right )+\frac {13-6 x}{506 \left (2 x^2-x+3\right ) \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {1}{506} \left (\frac {1}{341} \left (\frac {23}{22} \left (\frac {12643 \arctan \left (\frac {10 x+3}{\sqrt {31}}\right )}{\sqrt {31}}-\frac {589}{2} \log \left (5 x^2+3 x+2\right )\right )-\frac {31}{22} \left (\frac {2769 \arctan \left (\frac {4 x-1}{\sqrt {23}}\right )}{\sqrt {23}}-\frac {437}{2} \log \left (2 x^2-x+3\right )\right )\right )-\frac {25 (117-137 x)}{341 \left (5 x^2+3 x+2\right )}\right )+\frac {13-6 x}{506 \left (2 x^2-x+3\right ) \left (5 x^2+3 x+2\right )}\)

Input:

Int[1/((3 - x + 2*x^2)^2*(2 + 3*x + 5*x^2)^2),x]
 

Output:

(13 - 6*x)/(506*(3 - x + 2*x^2)*(2 + 3*x + 5*x^2)) + ((-25*(117 - 137*x))/ 
(341*(2 + 3*x + 5*x^2)) + ((-31*((2769*ArcTan[(-1 + 4*x)/Sqrt[23]])/Sqrt[2 
3] - (437*Log[3 - x + 2*x^2])/2))/22 + (23*((12643*ArcTan[(3 + 10*x)/Sqrt[ 
31]])/Sqrt[31] - (589*Log[2 + 3*x + 5*x^2])/2))/22)/341)/506
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

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])
 

rule 1083

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]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1142

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

rule 1305

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_.) + (e_.)*(x_) + (f_.)*(x 
_)^2)^(q_), x_Symbol] :> Simp[(2*a*c^2*e - b^2*c*e + b^3*f + b*c*(c*d - 3*a 
*f) + c*(2*c^2*d + b^2*f - c*(b*e + 2*a*f))*x)*(a + b*x + c*x^2)^(p + 1)*(( 
d + e*x + f*x^2)^(q + 1)/((b^2 - 4*a*c)*((c*d - a*f)^2 - (b*d - a*e)*(c*e - 
 b*f))*(p + 1))), x] - Simp[1/((b^2 - 4*a*c)*((c*d - a*f)^2 - (b*d - a*e)*( 
c*e - b*f))*(p + 1))   Int[(a + b*x + c*x^2)^(p + 1)*(d + e*x + f*x^2)^q*Si 
mp[2*c*((c*d - a*f)^2 - (b*d - a*e)*(c*e - b*f))*(p + 1) - (2*c^2*d + b^2*f 
 - c*(b*e + 2*a*f))*(a*f*(p + 1) - c*d*(p + 2)) - e*(b^2*c*e - 2*a*c^2*e - 
b^3*f - b*c*(c*d - 3*a*f))*(p + q + 2) + (2*f*(2*a*c^2*e - b^2*c*e + b^3*f 
+ b*c*(c*d - 3*a*f))*(p + q + 2) - (2*c^2*d + b^2*f - c*(b*e + 2*a*f))*(b*f 
*(p + 1) - c*e*(2*p + q + 4)))*x + c*f*(2*c^2*d + b^2*f - c*(b*e + 2*a*f))* 
(2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, q}, x] && NeQ[b 
^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0] && LtQ[p, -1] && NeQ[(c*d - a*f)^2 - 
(b*d - a*e)*(c*e - b*f), 0] &&  !( !IntegerQ[p] && ILtQ[q, -1]) &&  !IGtQ[q 
, 0]
 

rule 2135

Int[(Px_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_) + (e_.)*(x_) + (f_. 
)*(x_)^2)^(q_), x_Symbol] :> With[{A = Coeff[Px, x, 0], B = Coeff[Px, x, 1] 
, C = Coeff[Px, x, 2]}, Simp[(a + b*x + c*x^2)^(p + 1)*((d + e*x + f*x^2)^( 
q + 1)/((b^2 - 4*a*c)*((c*d - a*f)^2 - (b*d - a*e)*(c*e - b*f))*(p + 1)))*( 
(A*c - a*C)*(2*a*c*e - b*(c*d + a*f)) + (A*b - a*B)*(2*c^2*d + b^2*f - c*(b 
*e + 2*a*f)) + c*(A*(2*c^2*d + b^2*f - c*(b*e + 2*a*f)) - B*(b*c*d - 2*a*c* 
e + a*b*f) + C*(b^2*d - a*b*e - 2*a*(c*d - a*f)))*x), x] + Simp[1/((b^2 - 4 
*a*c)*((c*d - a*f)^2 - (b*d - a*e)*(c*e - b*f))*(p + 1))   Int[(a + b*x + c 
*x^2)^(p + 1)*(d + e*x + f*x^2)^q*Simp[(b*B - 2*A*c - 2*a*C)*((c*d - a*f)^2 
 - (b*d - a*e)*(c*e - b*f))*(p + 1) + (b^2*(C*d + A*f) - b*(B*c*d + A*c*e + 
 a*C*e + a*B*f) + 2*(A*c*(c*d - a*f) - a*(c*C*d - B*c*e - a*C*f)))*(a*f*(p 
+ 1) - c*d*(p + 2)) - e*((A*c - a*C)*(2*a*c*e - b*(c*d + a*f)) + (A*b - a*B 
)*(2*c^2*d + b^2*f - c*(b*e + 2*a*f)))*(p + q + 2) - (2*f*((A*c - a*C)*(2*a 
*c*e - b*(c*d + a*f)) + (A*b - a*B)*(2*c^2*d + b^2*f - c*(b*e + 2*a*f)))*(p 
 + q + 2) - (b^2*(C*d + A*f) - b*(B*c*d + A*c*e + a*C*e + a*B*f) + 2*(A*c*( 
c*d - a*f) - a*(c*C*d - B*c*e - a*C*f)))*(b*f*(p + 1) - c*e*(2*p + q + 4))) 
*x - c*f*(b^2*(C*d + A*f) - b*(B*c*d + A*c*e + a*C*e + a*B*f) + 2*(A*c*(c*d 
 - a*f) - a*(c*C*d - B*c*e - a*C*f)))*(2*p + 2*q + 5)*x^2, x], x], x]] /; F 
reeQ[{a, b, c, d, e, f, q}, x] && PolyQ[Px, x, 2] && LtQ[p, -1] && NeQ[(c*d 
 - a*f)^2 - (b*d - a*e)*(c*e - b*f), 0] &&  !( !IntegerQ[p] && ILtQ[q, -1]) 
 &&  !IGtQ[q, 0]
 

rule 2141

Int[(Px_)/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_) + (e_.)*(x_) + (f_.)*(x 
_)^2)), x_Symbol] :> With[{A = Coeff[Px, x, 0], B = Coeff[Px, x, 1], C = Co 
eff[Px, x, 2], q = c^2*d^2 - b*c*d*e + a*c*e^2 + b^2*d*f - 2*a*c*d*f - a*b* 
e*f + a^2*f^2}, Simp[1/q   Int[(A*c^2*d - a*c*C*d - A*b*c*e + a*B*c*e + A*b 
^2*f - a*b*B*f - a*A*c*f + a^2*C*f + c*(B*c*d - b*C*d - A*c*e + a*C*e + A*b 
*f - a*B*f)*x)/(a + b*x + c*x^2), x], x] + Simp[1/q   Int[(c*C*d^2 - B*c*d* 
e + A*c*e^2 + b*B*d*f - A*c*d*f - a*C*d*f - A*b*e*f + a*A*f^2 - f*(B*c*d - 
b*C*d - A*c*e + a*C*e + A*b*f - a*B*f)*x)/(d + e*x + f*x^2), x], x] /; NeQ[ 
q, 0]] /; FreeQ[{a, b, c, d, e, f}, x] && PolyQ[Px, x, 2]
 

Maple [A] (verified)

Time = 3.08 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.74

method	result	size
default	\(\frac {-\frac {77 x}{23}-\frac {341}{46}}{5324 x^{2}-2662 x +7986}+\frac {19 \ln \left (2 x^{2}-x +3\right )}{10648}-\frac {2769 \sqrt {23}\, \arctan \left (\frac {\left (4 x -1\right ) \sqrt {23}}{23}\right )}{2816396}-\frac {-\frac {759 x}{31}+\frac {1078}{155}}{5324 \left (x^{2}+\frac {3}{5} x +\frac {2}{5}\right )}-\frac {19 \ln \left (5 x^{2}+3 x +2\right )}{10648}+\frac {12643 \arctan \left (\frac {\left (10 x +3\right ) \sqrt {31}}{31}\right ) \sqrt {31}}{5116364}\)	\(94\)
risch	\(\frac {\frac {3425}{86273} x^{3}-\frac {9275}{172546} x^{2}+\frac {507}{7843} x -\frac {2171}{86273}}{\left (2 x^{2}-x +3\right ) \left (5 x^{2}+3 x +2\right )}+\frac {19 \ln \left (16 x^{2}-8 x +24\right )}{10648}-\frac {2769 \sqrt {23}\, \arctan \left (\frac {\left (4 x -1\right ) \sqrt {23}}{23}\right )}{2816396}-\frac {19 \ln \left (100 x^{2}+60 x +40\right )}{10648}+\frac {12643 \arctan \left (\frac {\left (10 x +3\right ) \sqrt {31}}{31}\right ) \sqrt {31}}{5116364}\)	\(101\)

Input:

int(1/(2*x^2-x+3)^2/(5*x^2+3*x+2)^2,x,method=_RETURNVERBOSE)
 

Output:

1/5324*(-77/23*x-341/46)/(x^2-1/2*x+3/2)+19/10648*ln(2*x^2-x+3)-2769/28163 
96*23^(1/2)*arctan(1/23*(4*x-1)*23^(1/2))-1/5324*(-759/31*x+1078/155)/(x^2 
+3/5*x+2/5)-19/10648*ln(5*x^2+3*x+2)+12643/5116364*arctan(1/31*(10*x+3)*31 
^(1/2))*31^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.31 \[ \int \frac {1}{\left (3-x+2 x^2\right )^2 \left (2+3 x+5 x^2\right )^2} \, dx=\frac {214898200 \, x^{3} + 13376294 \, \sqrt {31} {\left (10 \, x^{4} + x^{3} + 16 \, x^{2} + 7 \, x + 6\right )} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) - 5322018 \, \sqrt {23} {\left (10 \, x^{4} + x^{3} + 16 \, x^{2} + 7 \, x + 6\right )} \arctan \left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) - 290975300 \, x^{2} - 9659011 \, {\left (10 \, x^{4} + x^{3} + 16 \, x^{2} + 7 \, x + 6\right )} \log \left (5 \, x^{2} + 3 \, x + 2\right ) + 9659011 \, {\left (10 \, x^{4} + x^{3} + 16 \, x^{2} + 7 \, x + 6\right )} \log \left (2 \, x^{2} - x + 3\right ) + 349923288 \, x - 136217224}{5413113112 \, {\left (10 \, x^{4} + x^{3} + 16 \, x^{2} + 7 \, x + 6\right )}} \] Input:

integrate(1/(2*x^2-x+3)^2/(5*x^2+3*x+2)^2,x, algorithm="fricas")
 

Output:

1/5413113112*(214898200*x^3 + 13376294*sqrt(31)*(10*x^4 + x^3 + 16*x^2 + 7 
*x + 6)*arctan(1/31*sqrt(31)*(10*x + 3)) - 5322018*sqrt(23)*(10*x^4 + x^3 
+ 16*x^2 + 7*x + 6)*arctan(1/23*sqrt(23)*(4*x - 1)) - 290975300*x^2 - 9659 
011*(10*x^4 + x^3 + 16*x^2 + 7*x + 6)*log(5*x^2 + 3*x + 2) + 9659011*(10*x 
^4 + x^3 + 16*x^2 + 7*x + 6)*log(2*x^2 - x + 3) + 349923288*x - 136217224) 
/(10*x^4 + x^3 + 16*x^2 + 7*x + 6)
 

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\left (3-x+2 x^2\right )^2 \left (2+3 x+5 x^2\right )^2} \, dx=\frac {6850 x^{3} - 9275 x^{2} + 11154 x - 4342}{1725460 x^{4} + 172546 x^{3} + 2760736 x^{2} + 1207822 x + 1035276} + \frac {19 \log {\left (x^{2} - \frac {x}{2} + \frac {3}{2} \right )}}{10648} - \frac {19 \log {\left (x^{2} + \frac {3 x}{5} + \frac {2}{5} \right )}}{10648} - \frac {2769 \sqrt {23} \operatorname {atan}{\left (\frac {4 \sqrt {23} x}{23} - \frac {\sqrt {23}}{23} \right )}}{2816396} + \frac {12643 \sqrt {31} \operatorname {atan}{\left (\frac {10 \sqrt {31} x}{31} + \frac {3 \sqrt {31}}{31} \right )}}{5116364} \] Input:

integrate(1/(2*x**2-x+3)**2/(5*x**2+3*x+2)**2,x)
 

Output:

(6850*x**3 - 9275*x**2 + 11154*x - 4342)/(1725460*x**4 + 172546*x**3 + 276 
0736*x**2 + 1207822*x + 1035276) + 19*log(x**2 - x/2 + 3/2)/10648 - 19*log 
(x**2 + 3*x/5 + 2/5)/10648 - 2769*sqrt(23)*atan(4*sqrt(23)*x/23 - sqrt(23) 
/23)/2816396 + 12643*sqrt(31)*atan(10*sqrt(31)*x/31 + 3*sqrt(31)/31)/51163 
64
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\left (3-x+2 x^2\right )^2 \left (2+3 x+5 x^2\right )^2} \, dx=\frac {12643}{5116364} \, \sqrt {31} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) - \frac {2769}{2816396} \, \sqrt {23} \arctan \left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) + \frac {6850 \, x^{3} - 9275 \, x^{2} + 11154 \, x - 4342}{172546 \, {\left (10 \, x^{4} + x^{3} + 16 \, x^{2} + 7 \, x + 6\right )}} - \frac {19}{10648} \, \log \left (5 \, x^{2} + 3 \, x + 2\right ) + \frac {19}{10648} \, \log \left (2 \, x^{2} - x + 3\right ) \] Input:

integrate(1/(2*x^2-x+3)^2/(5*x^2+3*x+2)^2,x, algorithm="maxima")
 

Output:

12643/5116364*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) - 2769/2816396*sqr 
t(23)*arctan(1/23*sqrt(23)*(4*x - 1)) + 1/172546*(6850*x^3 - 9275*x^2 + 11 
154*x - 4342)/(10*x^4 + x^3 + 16*x^2 + 7*x + 6) - 19/10648*log(5*x^2 + 3*x 
 + 2) + 19/10648*log(2*x^2 - x + 3)
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\left (3-x+2 x^2\right )^2 \left (2+3 x+5 x^2\right )^2} \, dx=\frac {12643}{5116364} \, \sqrt {31} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) - \frac {2769}{2816396} \, \sqrt {23} \arctan \left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) + \frac {6850 \, x^{3} - 9275 \, x^{2} + 11154 \, x - 4342}{172546 \, {\left (10 \, x^{4} + x^{3} + 16 \, x^{2} + 7 \, x + 6\right )}} - \frac {19}{10648} \, \log \left (5 \, x^{2} + 3 \, x + 2\right ) + \frac {19}{10648} \, \log \left (2 \, x^{2} - x + 3\right ) \] Input:

integrate(1/(2*x^2-x+3)^2/(5*x^2+3*x+2)^2,x, algorithm="giac")
 

Output:

12643/5116364*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) - 2769/2816396*sqr 
t(23)*arctan(1/23*sqrt(23)*(4*x - 1)) + 1/172546*(6850*x^3 - 9275*x^2 + 11 
154*x - 4342)/(10*x^4 + x^3 + 16*x^2 + 7*x + 6) - 19/10648*log(5*x^2 + 3*x 
 + 2) + 19/10648*log(2*x^2 - x + 3)
 

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\left (3-x+2 x^2\right )^2 \left (2+3 x+5 x^2\right )^2} \, dx=\ln \left (x-\frac {1}{4}-\frac {\sqrt {23}\,1{}\mathrm {i}}{4}\right )\,\left (\frac {19}{10648}+\frac {\sqrt {23}\,2769{}\mathrm {i}}{5632792}\right )-\ln \left (x-\frac {1}{4}+\frac {\sqrt {23}\,1{}\mathrm {i}}{4}\right )\,\left (-\frac {19}{10648}+\frac {\sqrt {23}\,2769{}\mathrm {i}}{5632792}\right )-\ln \left (x+\frac {3}{10}-\frac {\sqrt {31}\,1{}\mathrm {i}}{10}\right )\,\left (\frac {19}{10648}+\frac {\sqrt {31}\,12643{}\mathrm {i}}{10232728}\right )+\ln \left (x+\frac {3}{10}+\frac {\sqrt {31}\,1{}\mathrm {i}}{10}\right )\,\left (-\frac {19}{10648}+\frac {\sqrt {31}\,12643{}\mathrm {i}}{10232728}\right )+\frac {\frac {685\,x^3}{172546}-\frac {1855\,x^2}{345092}+\frac {507\,x}{78430}-\frac {2171}{862730}}{x^4+\frac {x^3}{10}+\frac {8\,x^2}{5}+\frac {7\,x}{10}+\frac {3}{5}} \] Input:

int(1/((2*x^2 - x + 3)^2*(3*x + 5*x^2 + 2)^2),x)
 

Output:

log(x - (23^(1/2)*1i)/4 - 1/4)*((23^(1/2)*2769i)/5632792 + 19/10648) - log 
(x + (23^(1/2)*1i)/4 - 1/4)*((23^(1/2)*2769i)/5632792 - 19/10648) - log(x 
- (31^(1/2)*1i)/10 + 3/10)*((31^(1/2)*12643i)/10232728 + 19/10648) + log(x 
 + (31^(1/2)*1i)/10 + 3/10)*((31^(1/2)*12643i)/10232728 - 19/10648) + ((50 
7*x)/78430 - (1855*x^2)/345092 + (685*x^3)/172546 - 2171/862730)/((7*x)/10 
 + (8*x^2)/5 + x^3/10 + x^4 + 3/5)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 358, normalized size of antiderivative = 2.82 \[ \int \frac {1}{\left (3-x+2 x^2\right )^2 \left (2+3 x+5 x^2\right )^2} \, dx=\frac {133762940 \sqrt {31}\, \mathit {atan} \left (\frac {10 x +3}{\sqrt {31}}\right ) x^{4}[phone] \sqrt {31}\, \mathit {atan} \left (\frac {10 x +3}{\sqrt {31}}\right ) x^{3}[phone] \sqrt {31}\, \mathit {atan} \left (\frac {10 x +3}{\sqrt {31}}\right ) x^{2}[phone] \sqrt {31}\, \mathit {atan} \left (\frac {10 x +3}{\sqrt {31}}\right ) x [phone] \sqrt {31}\, \mathit {atan} \left (\frac {10 x +3}{\sqrt {31}}\right )-53220180 \sqrt {23}\, \mathit {atan} \left (\frac {4 x -1}{\sqrt {23}}\right ) x^{4}-5322018 \sqrt {23}\, \mathit {atan} \left (\frac {4 x -1}{\sqrt {23}}\right ) x^{3}-85152288 \sqrt {23}\, \mathit {atan} \left (\frac {4 x -1}{\sqrt {23}}\right ) x^{2}-37254126 \sqrt {23}\, \mathit {atan} \left (\frac {4 x -1}{\sqrt {23}}\right ) x -31932108 \sqrt {23}\, \mathit {atan} \left (\frac {4 x -1}{\sqrt {23}}\right )-96590110 \,\mathrm {log}\left (5 x^{2}+3 x +2\right ) x^{4}-9659011 \,\mathrm {log}\left (5 x^{2}+3 x +2\right ) x^{3}-154544176 \,\mathrm {log}\left (5 x^{2}+3 x +2\right ) x^{2}-67613077 \,\mathrm {log}\left (5 x^{2}+3 x +2\right ) x -57954066 \,\mathrm {log}\left (5 x^{2}+3 x +2\right )[phone] \,\mathrm {log}\left (2 x^{2}-x +3\right ) x^{4}+9659011 \,\mathrm {log}\left (2 x^{2}-x +3\right ) x^{3}[phone] \,\mathrm {log}\left (2 x^{2}-x +3\right ) x^{2}[phone] \,\mathrm {log}\left (2 x^{2}-x +3\right ) x [phone] \,\mathrm {log}\left (2 x^{2}-x +3\right )-2148982000 x^{4}-3729346500 x^{2}-1154364112 x -1425606424}{54131131120 x^{4}[phone] x^{3}[phone] x^{2}[phone] x [phone]} \] Input:

int(1/(2*x^2-x+3)^2/(5*x^2+3*x+2)^2,x)
 

Output:

(133762940*sqrt(31)*atan((10*x + 3)/sqrt(31))*x**4 + 13376294*sqrt(31)*ata 
n((10*x + 3)/sqrt(31))*x**3 + 214020704*sqrt(31)*atan((10*x + 3)/sqrt(31)) 
*x**2 + 93634058*sqrt(31)*atan((10*x + 3)/sqrt(31))*x + 80257764*sqrt(31)* 
atan((10*x + 3)/sqrt(31)) - 53220180*sqrt(23)*atan((4*x - 1)/sqrt(23))*x** 
4 - 5322018*sqrt(23)*atan((4*x - 1)/sqrt(23))*x**3 - 85152288*sqrt(23)*ata 
n((4*x - 1)/sqrt(23))*x**2 - 37254126*sqrt(23)*atan((4*x - 1)/sqrt(23))*x 
- 31932108*sqrt(23)*atan((4*x - 1)/sqrt(23)) - 96590110*log(5*x**2 + 3*x + 
 2)*x**4 - 9659011*log(5*x**2 + 3*x + 2)*x**3 - 154544176*log(5*x**2 + 3*x 
 + 2)*x**2 - 67613077*log(5*x**2 + 3*x + 2)*x - 57954066*log(5*x**2 + 3*x 
+ 2) + 96590110*log(2*x**2 - x + 3)*x**4 + 9659011*log(2*x**2 - x + 3)*x** 
3 + 154544176*log(2*x**2 - x + 3)*x**2 + 67613077*log(2*x**2 - x + 3)*x + 
57954066*log(2*x**2 - x + 3) - 2148982000*x**4 - 3729346500*x**2 - 1154364 
112*x - 1425606424)/(5413113112*(10*x**4 + x**3 + 16*x**2 + 7*x + 6))