\(\int \frac {1}{(3-x+2 x^2)^2 (2+3 x+5 x^2)} \, dx\) [77]

Optimal result

Integrand size = 25, antiderivative size = 94 \[ \int \frac {1}{\left (3-x+2 x^2\right )^2 \left (2+3 x+5 x^2\right )} \, dx=\frac {13-6 x}{506 \left (3-x+2 x^2\right )}+\frac {241 \arctan \left (\frac {1-4 x}{\sqrt {23}}\right )}{11132 \sqrt {23}}+\frac {69 \arctan \left (\frac {3+10 x}{\sqrt {31}}\right )}{484 \sqrt {31}}-\frac {13}{968} \log \left (3-x+2 x^2\right )+\frac {13}{968} \log \left (2+3 x+5 x^2\right ) \] Output:

(13-6*x)/(1012*x^2-506*x+1518)+241/256036*arctan(1/23*(1-4*x)*23^(1/2))*23 
^(1/2)+69/15004*arctan(1/31*(3+10*x)*31^(1/2))*31^(1/2)-13/968*ln(2*x^2-x+ 
3)+13/968*ln(5*x^2+3*x+2)
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (3-x+2 x^2\right )^2 \left (2+3 x+5 x^2\right )} \, dx=\frac {13-6 x}{506 \left (3-x+2 x^2\right )}-\frac {241 \arctan \left (\frac {-1+4 x}{\sqrt {23}}\right )}{11132 \sqrt {23}}+\frac {69 \arctan \left (\frac {3+10 x}{\sqrt {31}}\right )}{484 \sqrt {31}}-\frac {13}{968} \log \left (3-x+2 x^2\right )+\frac {13}{968} \log \left (2+3 x+5 x^2\right ) \] Input:

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

Output:

(13 - 6*x)/(506*(3 - x + 2*x^2)) - (241*ArcTan[(-1 + 4*x)/Sqrt[23]])/(1113 
2*Sqrt[23]) + (69*ArcTan[(3 + 10*x)/Sqrt[31]])/(484*Sqrt[31]) - (13*Log[3 
- x + 2*x^2])/968 + (13*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.

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

\(\Big \downarrow \) 1305

\(\displaystyle \frac {13-6 x}{506 \left (2 x^2-x+3\right )}-\frac {\int -\frac {11 \left (-30 x^2+97 x+172\right )}{\left (2 x^2-x+3\right ) \left (5 x^2+3 x+2\right )}dx}{5566}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{506} \int \frac {-30 x^2+97 x+172}{\left (2 x^2-x+3\right ) \left (5 x^2+3 x+2\right )}dx+\frac {13-6 x}{506 \left (2 x^2-x+3\right )}\)

\(\Big \downarrow \) 2141

\(\displaystyle \frac {1}{506} \left (\frac {1}{242} \int \frac {11 (29-598 x)}{2 x^2-x+3}dx+\frac {1}{242} \int \frac {253 (65 x+54)}{5 x^2+3 x+2}dx\right )+\frac {13-6 x}{506 \left (2 x^2-x+3\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{506} \left (\frac {1}{22} \int \frac {29-598 x}{2 x^2-x+3}dx+\frac {23}{22} \int \frac {65 x+54}{5 x^2+3 x+2}dx\right )+\frac {13-6 x}{506 \left (2 x^2-x+3\right )}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {1}{506} \left (\frac {1}{22} \left (-\frac {241}{2} \int \frac {1}{2 x^2-x+3}dx-\frac {299}{2} \int -\frac {1-4 x}{2 x^2-x+3}dx\right )+\frac {23}{22} \left (\frac {69}{2} \int \frac {1}{5 x^2+3 x+2}dx+\frac {13}{2} \int \frac {10 x+3}{5 x^2+3 x+2}dx\right )\right )+\frac {13-6 x}{506 \left (2 x^2-x+3\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{506} \left (\frac {1}{22} \left (\frac {299}{2} \int \frac {1-4 x}{2 x^2-x+3}dx-\frac {241}{2} \int \frac {1}{2 x^2-x+3}dx\right )+\frac {23}{22} \left (\frac {69}{2} \int \frac {1}{5 x^2+3 x+2}dx+\frac {13}{2} \int \frac {10 x+3}{5 x^2+3 x+2}dx\right )\right )+\frac {13-6 x}{506 \left (2 x^2-x+3\right )}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {1}{506} \left (\frac {1}{22} \left (\frac {299}{2} \int \frac {1-4 x}{2 x^2-x+3}dx+241 \int \frac {1}{-(4 x-1)^2-23}d(4 x-1)\right )+\frac {23}{22} \left (\frac {13}{2} \int \frac {10 x+3}{5 x^2+3 x+2}dx-69 \int \frac {1}{-(10 x+3)^2-31}d(10 x+3)\right )\right )+\frac {13-6 x}{506 \left (2 x^2-x+3\right )}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{506} \left (\frac {1}{22} \left (\frac {299}{2} \int \frac {1-4 x}{2 x^2-x+3}dx-\frac {241 \arctan \left (\frac {4 x-1}{\sqrt {23}}\right )}{\sqrt {23}}\right )+\frac {23}{22} \left (\frac {13}{2} \int \frac {10 x+3}{5 x^2+3 x+2}dx+\frac {69 \arctan \left (\frac {10 x+3}{\sqrt {31}}\right )}{\sqrt {31}}\right )\right )+\frac {13-6 x}{506 \left (2 x^2-x+3\right )}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {1}{506} \left (\frac {1}{22} \left (-\frac {241 \arctan \left (\frac {4 x-1}{\sqrt {23}}\right )}{\sqrt {23}}-\frac {299}{2} \log \left (2 x^2-x+3\right )\right )+\frac {23}{22} \left (\frac {69 \arctan \left (\frac {10 x+3}{\sqrt {31}}\right )}{\sqrt {31}}+\frac {13}{2} \log \left (5 x^2+3 x+2\right )\right )\right )+\frac {13-6 x}{506 \left (2 x^2-x+3\right )}\)

Input:

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

Output:

(13 - 6*x)/(506*(3 - x + 2*x^2)) + (((-241*ArcTan[(-1 + 4*x)/Sqrt[23]])/Sq 
rt[23] - (299*Log[3 - x + 2*x^2])/2)/22 + (23*((69*ArcTan[(3 + 10*x)/Sqrt[ 
31]])/Sqrt[31] + (13*Log[2 + 3*x + 5*x^2])/2))/22)/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 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 = 2.74 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.81

method	result	size
risch	\(\frac {-\frac {3 x}{506}+\frac {13}{1012}}{x^{2}-\frac {1}{2} x +\frac {3}{2}}+\frac {13 \ln \left (100 x^{2}+60 x +40\right )}{968}+\frac {69 \arctan \left (\frac {\left (10 x +3\right ) \sqrt {31}}{31}\right ) \sqrt {31}}{15004}-\frac {13 \ln \left (16 x^{2}-8 x +24\right )}{968}-\frac {241 \sqrt {23}\, \arctan \left (\frac {\left (4 x -1\right ) \sqrt {23}}{23}\right )}{256036}\)	\(76\)
default	\(-\frac {\frac {66 x}{23}-\frac {143}{23}}{484 \left (x^{2}-\frac {1}{2} x +\frac {3}{2}\right )}-\frac {13 \ln \left (2 x^{2}-x +3\right )}{968}-\frac {241 \sqrt {23}\, \arctan \left (\frac {\left (4 x -1\right ) \sqrt {23}}{23}\right )}{256036}+\frac {13 \ln \left (5 x^{2}+3 x +2\right )}{968}+\frac {69 \arctan \left (\frac {\left (10 x +3\right ) \sqrt {31}}{31}\right ) \sqrt {31}}{15004}\)	\(77\)

Input:

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

Output:

(-3/506*x+13/1012)/(x^2-1/2*x+3/2)+13/968*ln(100*x^2+60*x+40)+69/15004*arc 
tan(1/31*(10*x+3)*31^(1/2))*31^(1/2)-13/968*ln(16*x^2-8*x+24)-241/256036*2 
3^(1/2)*arctan(1/23*(4*x-1)*23^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.24 \[ \int \frac {1}{\left (3-x+2 x^2\right )^2 \left (2+3 x+5 x^2\right )} \, dx=\frac {73002 \, \sqrt {31} {\left (2 \, x^{2} - x + 3\right )} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) - 14942 \, \sqrt {23} {\left (2 \, x^{2} - x + 3\right )} \arctan \left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) + 213187 \, {\left (2 \, x^{2} - x + 3\right )} \log \left (5 \, x^{2} + 3 \, x + 2\right ) - 213187 \, {\left (2 \, x^{2} - x + 3\right )} \log \left (2 \, x^{2} - x + 3\right ) - 188232 \, x + 407836}{15874232 \, {\left (2 \, x^{2} - x + 3\right )}} \] Input:

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

Output:

1/15874232*(73002*sqrt(31)*(2*x^2 - x + 3)*arctan(1/31*sqrt(31)*(10*x + 3) 
) - 14942*sqrt(23)*(2*x^2 - x + 3)*arctan(1/23*sqrt(23)*(4*x - 1)) + 21318 
7*(2*x^2 - x + 3)*log(5*x^2 + 3*x + 2) - 213187*(2*x^2 - x + 3)*log(2*x^2 
- x + 3) - 188232*x + 407836)/(2*x^2 - x + 3)
 

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.09 \[ \int \frac {1}{\left (3-x+2 x^2\right )^2 \left (2+3 x+5 x^2\right )} \, dx=\frac {13 - 6 x}{1012 x^{2} - 506 x + 1518} - \frac {13 \log {\left (x^{2} - \frac {x}{2} + \frac {3}{2} \right )}}{968} + \frac {13 \log {\left (x^{2} + \frac {3 x}{5} + \frac {2}{5} \right )}}{968} - \frac {241 \sqrt {23} \operatorname {atan}{\left (\frac {4 \sqrt {23} x}{23} - \frac {\sqrt {23}}{23} \right )}}{256036} + \frac {69 \sqrt {31} \operatorname {atan}{\left (\frac {10 \sqrt {31} x}{31} + \frac {3 \sqrt {31}}{31} \right )}}{15004} \] Input:

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

Output:

(13 - 6*x)/(1012*x**2 - 506*x + 1518) - 13*log(x**2 - x/2 + 3/2)/968 + 13* 
log(x**2 + 3*x/5 + 2/5)/968 - 241*sqrt(23)*atan(4*sqrt(23)*x/23 - sqrt(23) 
/23)/256036 + 69*sqrt(31)*atan(10*sqrt(31)*x/31 + 3*sqrt(31)/31)/15004
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.83 \[ \int \frac {1}{\left (3-x+2 x^2\right )^2 \left (2+3 x+5 x^2\right )} \, dx=\frac {69}{15004} \, \sqrt {31} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) - \frac {241}{256036} \, \sqrt {23} \arctan \left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) - \frac {6 \, x - 13}{506 \, {\left (2 \, x^{2} - x + 3\right )}} + \frac {13}{968} \, \log \left (5 \, x^{2} + 3 \, x + 2\right ) - \frac {13}{968} \, \log \left (2 \, x^{2} - x + 3\right ) \] Input:

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

Output:

69/15004*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) - 241/256036*sqrt(23)*a 
rctan(1/23*sqrt(23)*(4*x - 1)) - 1/506*(6*x - 13)/(2*x^2 - x + 3) + 13/968 
*log(5*x^2 + 3*x + 2) - 13/968*log(2*x^2 - x + 3)
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.83 \[ \int \frac {1}{\left (3-x+2 x^2\right )^2 \left (2+3 x+5 x^2\right )} \, dx=\frac {69}{15004} \, \sqrt {31} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) - \frac {241}{256036} \, \sqrt {23} \arctan \left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) - \frac {6 \, x - 13}{506 \, {\left (2 \, x^{2} - x + 3\right )}} + \frac {13}{968} \, \log \left (5 \, x^{2} + 3 \, x + 2\right ) - \frac {13}{968} \, \log \left (2 \, x^{2} - x + 3\right ) \] Input:

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

Output:

69/15004*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) - 241/256036*sqrt(23)*a 
rctan(1/23*sqrt(23)*(4*x - 1)) - 1/506*(6*x - 13)/(2*x^2 - x + 3) + 13/968 
*log(5*x^2 + 3*x + 2) - 13/968*log(2*x^2 - x + 3)
 

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.02 \[ \int \frac {1}{\left (3-x+2 x^2\right )^2 \left (2+3 x+5 x^2\right )} \, dx=-\frac {\frac {3\,x}{506}-\frac {13}{1012}}{x^2-\frac {x}{2}+\frac {3}{2}}-\ln \left (x+\frac {3}{10}-\frac {\sqrt {31}\,1{}\mathrm {i}}{10}\right )\,\left (-\frac {13}{968}+\frac {\sqrt {31}\,69{}\mathrm {i}}{30008}\right )+\ln \left (x+\frac {3}{10}+\frac {\sqrt {31}\,1{}\mathrm {i}}{10}\right )\,\left (\frac {13}{968}+\frac {\sqrt {31}\,69{}\mathrm {i}}{30008}\right )+\ln \left (x-\frac {1}{4}-\frac {\sqrt {23}\,1{}\mathrm {i}}{4}\right )\,\left (-\frac {13}{968}+\frac {\sqrt {23}\,241{}\mathrm {i}}{512072}\right )-\ln \left (x-\frac {1}{4}+\frac {\sqrt {23}\,1{}\mathrm {i}}{4}\right )\,\left (\frac {13}{968}+\frac {\sqrt {23}\,241{}\mathrm {i}}{512072}\right ) \] Input:

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

Output:

log(x + (31^(1/2)*1i)/10 + 3/10)*((31^(1/2)*69i)/30008 + 13/968) - log(x - 
 (31^(1/2)*1i)/10 + 3/10)*((31^(1/2)*69i)/30008 - 13/968) - ((3*x)/506 - 1 
3/1012)/(x^2 - x/2 + 3/2) + log(x - (23^(1/2)*1i)/4 - 1/4)*((23^(1/2)*241i 
)/512072 - 13/968) - log(x + (23^(1/2)*1i)/4 - 1/4)*((23^(1/2)*241i)/51207 
2 + 13/968)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.17 \[ \int \frac {1}{\left (3-x+2 x^2\right )^2 \left (2+3 x+5 x^2\right )} \, dx=\frac {146004 \sqrt {31}\, \mathit {atan} \left (\frac {10 x +3}{\sqrt {31}}\right ) x^{2}-73002 \sqrt {31}\, \mathit {atan} \left (\frac {10 x +3}{\sqrt {31}}\right ) x +219006 \sqrt {31}\, \mathit {atan} \left (\frac {10 x +3}{\sqrt {31}}\right )-29884 \sqrt {23}\, \mathit {atan} \left (\frac {4 x -1}{\sqrt {23}}\right ) x^{2}+14942 \sqrt {23}\, \mathit {atan} \left (\frac {4 x -1}{\sqrt {23}}\right ) x -44826 \sqrt {23}\, \mathit {atan} \left (\frac {4 x -1}{\sqrt {23}}\right )+426374 \,\mathrm {log}\left (5 x^{2}+3 x +2\right ) x^{2}-213187 \,\mathrm {log}\left (5 x^{2}+3 x +2\right ) x +639561 \,\mathrm {log}\left (5 x^{2}+3 x +2\right )-426374 \,\mathrm {log}\left (2 x^{2}-x +3\right ) x^{2}+213187 \,\mathrm {log}\left (2 x^{2}-x +3\right ) x -639561 \,\mathrm {log}\left (2 x^{2}-x +3\right )-376464 x^{2}-156860}{31748464 x^{2}-15874232 x [phone]} \] Input:

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

Output:

(146004*sqrt(31)*atan((10*x + 3)/sqrt(31))*x**2 - 73002*sqrt(31)*atan((10* 
x + 3)/sqrt(31))*x + 219006*sqrt(31)*atan((10*x + 3)/sqrt(31)) - 29884*sqr 
t(23)*atan((4*x - 1)/sqrt(23))*x**2 + 14942*sqrt(23)*atan((4*x - 1)/sqrt(2 
3))*x - 44826*sqrt(23)*atan((4*x - 1)/sqrt(23)) + 426374*log(5*x**2 + 3*x 
+ 2)*x**2 - 213187*log(5*x**2 + 3*x + 2)*x + 639561*log(5*x**2 + 3*x + 2) 
- 426374*log(2*x**2 - x + 3)*x**2 + 213187*log(2*x**2 - x + 3)*x - 639561* 
log(2*x**2 - x + 3) - 376464*x**2 - 156860)/(15874232*(2*x**2 - x + 3))