\(\int \frac {1}{(3-x+2 x^2) (2+3 x+5 x^2)^2} \, dx\) [71]

Optimal result

Integrand size = 25, antiderivative size = 94 \[ \int \frac {1}{\left (3-x+2 x^2\right ) \left (2+3 x+5 x^2\right )^2} \, dx=\frac {4+65 x}{682 \left (2+3 x+5 x^2\right )}+\frac {7 \arctan \left (\frac {1-4 x}{\sqrt {23}}\right )}{484 \sqrt {23}}+\frac {2891 \arctan \left (\frac {3+10 x}{\sqrt {31}}\right )}{15004 \sqrt {31}}+\frac {3}{968} \log \left (3-x+2 x^2\right )-\frac {3}{968} \log \left (2+3 x+5 x^2\right ) \] Output:

(4+65*x)/(3410*x^2+2046*x+1364)+7/11132*arctan(1/23*(1-4*x)*23^(1/2))*23^( 
1/2)+2891/465124*arctan(1/31*(3+10*x)*31^(1/2))*31^(1/2)+3/968*ln(2*x^2-x+ 
3)-3/968*ln(5*x^2+3*x+2)
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (3-x+2 x^2\right ) \left (2+3 x+5 x^2\right )^2} \, dx=\frac {4+65 x}{682 \left (2+3 x+5 x^2\right )}-\frac {7 \arctan \left (\frac {-1+4 x}{\sqrt {23}}\right )}{484 \sqrt {23}}+\frac {2891 \arctan \left (\frac {3+10 x}{\sqrt {31}}\right )}{15004 \sqrt {31}}+\frac {3}{968} \log \left (3-x+2 x^2\right )-\frac {3}{968} \log \left (2+3 x+5 x^2\right ) \] Input:

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

Output:

(4 + 65*x)/(682*(2 + 3*x + 5*x^2)) - (7*ArcTan[(-1 + 4*x)/Sqrt[23]])/(484* 
Sqrt[23]) + (2891*ArcTan[(3 + 10*x)/Sqrt[31]])/(15004*Sqrt[31]) + (3*Log[3 
 - x + 2*x^2])/968 - (3*Log[2 + 3*x + 5*x^2])/968
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {1305, 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.

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

\(\Big \downarrow \) 1305

\(\displaystyle \frac {65 x+4}{682 \left (5 x^2+3 x+2\right )}-\frac {\int -\frac {11 \left (130 x^2-127 x+164\right )}{\left (2 x^2-x+3\right ) \left (5 x^2+3 x+2\right )}dx}{7502}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{682} \int \frac {130 x^2-127 x+164}{\left (2 x^2-x+3\right ) \left (5 x^2+3 x+2\right )}dx+\frac {65 x+4}{682 \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 2141

\(\displaystyle \frac {1}{682} \left (\frac {1}{242} \int -\frac {341 (5-6 x)}{2 x^2-x+3}dx+\frac {1}{242} \int \frac {11 (1306-465 x)}{5 x^2+3 x+2}dx\right )+\frac {65 x+4}{682 \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{682} \left (\frac {1}{22} \int \frac {1306-465 x}{5 x^2+3 x+2}dx-\frac {31}{22} \int \frac {5-6 x}{2 x^2-x+3}dx\right )+\frac {65 x+4}{682 \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {1}{682} \left (\frac {1}{22} \left (\frac {2891}{2} \int \frac {1}{5 x^2+3 x+2}dx-\frac {93}{2} \int \frac {10 x+3}{5 x^2+3 x+2}dx\right )-\frac {31}{22} \left (\frac {7}{2} \int \frac {1}{2 x^2-x+3}dx-\frac {3}{2} \int -\frac {1-4 x}{2 x^2-x+3}dx\right )\right )+\frac {65 x+4}{682 \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{682} \left (\frac {1}{22} \left (\frac {2891}{2} \int \frac {1}{5 x^2+3 x+2}dx-\frac {93}{2} \int \frac {10 x+3}{5 x^2+3 x+2}dx\right )-\frac {31}{22} \left (\frac {7}{2} \int \frac {1}{2 x^2-x+3}dx+\frac {3}{2} \int \frac {1-4 x}{2 x^2-x+3}dx\right )\right )+\frac {65 x+4}{682 \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {1}{682} \left (\frac {1}{22} \left (-\frac {93}{2} \int \frac {10 x+3}{5 x^2+3 x+2}dx-2891 \int \frac {1}{-(10 x+3)^2-31}d(10 x+3)\right )-\frac {31}{22} \left (\frac {3}{2} \int \frac {1-4 x}{2 x^2-x+3}dx-7 \int \frac {1}{-(4 x-1)^2-23}d(4 x-1)\right )\right )+\frac {65 x+4}{682 \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{682} \left (\frac {1}{22} \left (\frac {2891 \arctan \left (\frac {10 x+3}{\sqrt {31}}\right )}{\sqrt {31}}-\frac {93}{2} \int \frac {10 x+3}{5 x^2+3 x+2}dx\right )-\frac {31}{22} \left (\frac {3}{2} \int \frac {1-4 x}{2 x^2-x+3}dx+\frac {7 \arctan \left (\frac {4 x-1}{\sqrt {23}}\right )}{\sqrt {23}}\right )\right )+\frac {65 x+4}{682 \left (5 x^2+3 x+2\right )}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {1}{682} \left (\frac {1}{22} \left (\frac {2891 \arctan \left (\frac {10 x+3}{\sqrt {31}}\right )}{\sqrt {31}}-\frac {93}{2} \log \left (5 x^2+3 x+2\right )\right )-\frac {31}{22} \left (\frac {7 \arctan \left (\frac {4 x-1}{\sqrt {23}}\right )}{\sqrt {23}}-\frac {3}{2} \log \left (2 x^2-x+3\right )\right )\right )+\frac {65 x+4}{682 \left (5 x^2+3 x+2\right )}\)

Input:

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

Output:

(4 + 65*x)/(682*(2 + 3*x + 5*x^2)) + ((-31*((7*ArcTan[(-1 + 4*x)/Sqrt[23]] 
)/Sqrt[23] - (3*Log[3 - x + 2*x^2])/2))/22 + ((2891*ArcTan[(3 + 10*x)/Sqrt 
[31]])/Sqrt[31] - (93*Log[2 + 3*x + 5*x^2])/2)/22)/682
 

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 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 = 2.90 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.81

method	result	size
risch	\(\frac {\frac {13 x}{682}+\frac {2}{1705}}{x^{2}+\frac {3}{5} x +\frac {2}{5}}+\frac {3 \ln \left (16 x^{2}-8 x +24\right )}{968}-\frac {7 \sqrt {23}\, \arctan \left (\frac {\left (4 x -1\right ) \sqrt {23}}{23}\right )}{11132}-\frac {3 \ln \left (100 x^{2}+60 x +40\right )}{968}+\frac {2891 \arctan \left (\frac {\left (10 x +3\right ) \sqrt {31}}{31}\right ) \sqrt {31}}{465124}\)	\(76\)
default	\(\frac {3 \ln \left (2 x^{2}-x +3\right )}{968}-\frac {7 \sqrt {23}\, \arctan \left (\frac {\left (4 x -1\right ) \sqrt {23}}{23}\right )}{11132}-\frac {-\frac {286 x}{31}-\frac {88}{155}}{484 \left (x^{2}+\frac {3}{5} x +\frac {2}{5}\right )}-\frac {3 \ln \left (5 x^{2}+3 x +2\right )}{968}+\frac {2891 \arctan \left (\frac {\left (10 x +3\right ) \sqrt {31}}{31}\right ) \sqrt {31}}{465124}\)	\(77\)

Input:

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

Output:

(13/682*x+2/1705)/(x^2+3/5*x+2/5)+3/968*ln(16*x^2-8*x+24)-7/11132*23^(1/2) 
*arctan(1/23*(4*x-1)*23^(1/2))-3/968*ln(100*x^2+60*x+40)+2891/465124*arcta 
n(1/31*(10*x+3)*31^(1/2))*31^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.24 \[ \int \frac {1}{\left (3-x+2 x^2\right ) \left (2+3 x+5 x^2\right )^2} \, dx=\frac {132986 \, \sqrt {31} {\left (5 \, x^{2} + 3 \, x + 2\right )} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) - 13454 \, \sqrt {23} {\left (5 \, x^{2} + 3 \, x + 2\right )} \arctan \left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) - 66309 \, {\left (5 \, x^{2} + 3 \, x + 2\right )} \log \left (5 \, x^{2} + 3 \, x + 2\right ) + 66309 \, {\left (5 \, x^{2} + 3 \, x + 2\right )} \log \left (2 \, x^{2} - x + 3\right ) + 2039180 \, x + 125488}{21395704 \, {\left (5 \, x^{2} + 3 \, x + 2\right )}} \] Input:

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

Output:

1/21395704*(132986*sqrt(31)*(5*x^2 + 3*x + 2)*arctan(1/31*sqrt(31)*(10*x + 
 3)) - 13454*sqrt(23)*(5*x^2 + 3*x + 2)*arctan(1/23*sqrt(23)*(4*x - 1)) - 
66309*(5*x^2 + 3*x + 2)*log(5*x^2 + 3*x + 2) + 66309*(5*x^2 + 3*x + 2)*log 
(2*x^2 - x + 3) + 2039180*x + 125488)/(5*x^2 + 3*x + 2)
 

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.09 \[ \int \frac {1}{\left (3-x+2 x^2\right ) \left (2+3 x+5 x^2\right )^2} \, dx=\frac {65 x + 4}{3410 x^{2} + 2046 x + 1364} + \frac {3 \log {\left (x^{2} - \frac {x}{2} + \frac {3}{2} \right )}}{968} - \frac {3 \log {\left (x^{2} + \frac {3 x}{5} + \frac {2}{5} \right )}}{968} - \frac {7 \sqrt {23} \operatorname {atan}{\left (\frac {4 \sqrt {23} x}{23} - \frac {\sqrt {23}}{23} \right )}}{11132} + \frac {2891 \sqrt {31} \operatorname {atan}{\left (\frac {10 \sqrt {31} x}{31} + \frac {3 \sqrt {31}}{31} \right )}}{465124} \] Input:

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

Output:

(65*x + 4)/(3410*x**2 + 2046*x + 1364) + 3*log(x**2 - x/2 + 3/2)/968 - 3*l 
og(x**2 + 3*x/5 + 2/5)/968 - 7*sqrt(23)*atan(4*sqrt(23)*x/23 - sqrt(23)/23 
)/11132 + 2891*sqrt(31)*atan(10*sqrt(31)*x/31 + 3*sqrt(31)/31)/465124
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.83 \[ \int \frac {1}{\left (3-x+2 x^2\right ) \left (2+3 x+5 x^2\right )^2} \, dx=\frac {2891}{465124} \, \sqrt {31} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) - \frac {7}{11132} \, \sqrt {23} \arctan \left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) + \frac {65 \, x + 4}{682 \, {\left (5 \, x^{2} + 3 \, x + 2\right )}} - \frac {3}{968} \, \log \left (5 \, x^{2} + 3 \, x + 2\right ) + \frac {3}{968} \, \log \left (2 \, x^{2} - x + 3\right ) \] Input:

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

Output:

2891/465124*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) - 7/11132*sqrt(23)*a 
rctan(1/23*sqrt(23)*(4*x - 1)) + 1/682*(65*x + 4)/(5*x^2 + 3*x + 2) - 3/96 
8*log(5*x^2 + 3*x + 2) + 3/968*log(2*x^2 - x + 3)
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.83 \[ \int \frac {1}{\left (3-x+2 x^2\right ) \left (2+3 x+5 x^2\right )^2} \, dx=\frac {2891}{465124} \, \sqrt {31} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) - \frac {7}{11132} \, \sqrt {23} \arctan \left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) + \frac {65 \, x + 4}{682 \, {\left (5 \, x^{2} + 3 \, x + 2\right )}} - \frac {3}{968} \, \log \left (5 \, x^{2} + 3 \, x + 2\right ) + \frac {3}{968} \, \log \left (2 \, x^{2} - x + 3\right ) \] Input:

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

Output:

2891/465124*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) - 7/11132*sqrt(23)*a 
rctan(1/23*sqrt(23)*(4*x - 1)) + 1/682*(65*x + 4)/(5*x^2 + 3*x + 2) - 3/96 
8*log(5*x^2 + 3*x + 2) + 3/968*log(2*x^2 - x + 3)
 

Mupad [B] (verification not implemented)

Time = 16.64 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.01 \[ \int \frac {1}{\left (3-x+2 x^2\right ) \left (2+3 x+5 x^2\right )^2} \, dx=\frac {\frac {13\,x}{682}+\frac {2}{1705}}{x^2+\frac {3\,x}{5}+\frac {2}{5}}+\ln \left (x-\frac {1}{4}-\frac {\sqrt {23}\,1{}\mathrm {i}}{4}\right )\,\left (\frac {3}{968}+\frac {\sqrt {23}\,7{}\mathrm {i}}{22264}\right )-\ln \left (x-\frac {1}{4}+\frac {\sqrt {23}\,1{}\mathrm {i}}{4}\right )\,\left (-\frac {3}{968}+\frac {\sqrt {23}\,7{}\mathrm {i}}{22264}\right )-\ln \left (x+\frac {3}{10}-\frac {\sqrt {31}\,1{}\mathrm {i}}{10}\right )\,\left (\frac {3}{968}+\frac {\sqrt {31}\,2891{}\mathrm {i}}{930248}\right )+\ln \left (x+\frac {3}{10}+\frac {\sqrt {31}\,1{}\mathrm {i}}{10}\right )\,\left (-\frac {3}{968}+\frac {\sqrt {31}\,2891{}\mathrm {i}}{930248}\right ) \] Input:

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

Output:

((13*x)/682 + 2/1705)/((3*x)/5 + x^2 + 2/5) + log(x - (23^(1/2)*1i)/4 - 1/ 
4)*((23^(1/2)*7i)/22264 + 3/968) - log(x + (23^(1/2)*1i)/4 - 1/4)*((23^(1/ 
2)*7i)/22264 - 3/968) - log(x - (31^(1/2)*1i)/10 + 3/10)*((31^(1/2)*2891i) 
/930248 + 3/968) + log(x + (31^(1/2)*1i)/10 + 3/10)*((31^(1/2)*2891i)/9302 
48 - 3/968)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.17 \[ \int \frac {1}{\left (3-x+2 x^2\right ) \left (2+3 x+5 x^2\right )^2} \, dx=\frac {1994790 \sqrt {31}\, \mathit {atan} \left (\frac {10 x +3}{\sqrt {31}}\right ) x^{2}+1196874 \sqrt {31}\, \mathit {atan} \left (\frac {10 x +3}{\sqrt {31}}\right ) x +797916 \sqrt {31}\, \mathit {atan} \left (\frac {10 x +3}{\sqrt {31}}\right )-201810 \sqrt {23}\, \mathit {atan} \left (\frac {4 x -1}{\sqrt {23}}\right ) x^{2}-121086 \sqrt {23}\, \mathit {atan} \left (\frac {4 x -1}{\sqrt {23}}\right ) x -80724 \sqrt {23}\, \mathit {atan} \left (\frac {4 x -1}{\sqrt {23}}\right )-994635 \,\mathrm {log}\left (5 x^{2}+3 x +2\right ) x^{2}-596781 \,\mathrm {log}\left (5 x^{2}+3 x +2\right ) x -397854 \,\mathrm {log}\left (5 x^{2}+3 x +2\right )+994635 \,\mathrm {log}\left (2 x^{2}-x +3\right ) x^{2}+596781 \,\mathrm {log}\left (2 x^{2}-x +3\right ) x +397854 \,\mathrm {log}\left (2 x^{2}-x +3\right )-10195900 x^{2}-3701896}{320935560 x^{2}[phone] x [phone]} \] Input:

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

Output:

(1994790*sqrt(31)*atan((10*x + 3)/sqrt(31))*x**2 + 1196874*sqrt(31)*atan(( 
10*x + 3)/sqrt(31))*x + 797916*sqrt(31)*atan((10*x + 3)/sqrt(31)) - 201810 
*sqrt(23)*atan((4*x - 1)/sqrt(23))*x**2 - 121086*sqrt(23)*atan((4*x - 1)/s 
qrt(23))*x - 80724*sqrt(23)*atan((4*x - 1)/sqrt(23)) - 994635*log(5*x**2 + 
 3*x + 2)*x**2 - 596781*log(5*x**2 + 3*x + 2)*x - 397854*log(5*x**2 + 3*x 
+ 2) + 994635*log(2*x**2 - x + 3)*x**2 + 596781*log(2*x**2 - x + 3)*x + 39 
7854*log(2*x**2 - x + 3) - 10195900*x**2 - 3701896)/(64187112*(5*x**2 + 3* 
x + 2))