\(\int \frac {(5-x) (3+2 x)^{5/2}}{(2+5 x+3 x^2)^2} \, dx\) [836]

Optimal result

Integrand size = 27, antiderivative size = 83 \[ \int \frac {(5-x) (3+2 x)^{5/2}}{\left (2+5 x+3 x^2\right )^2} \, dx=-\frac {8}{9} \sqrt {3+2 x}-\frac {\sqrt {3+2 x} (533+587 x)}{9 \left (2+5 x+3 x^2\right )}-130 \text {arctanh}\left (\sqrt {3+2 x}\right )+100 \sqrt {\frac {5}{3}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right ) \] Output:

-8/9*(3+2*x)^(1/2)-(3+2*x)^(1/2)*(533+587*x)/(27*x^2+45*x+18)-130*arctanh( 
(3+2*x)^(1/2))+100/3*15^(1/2)*arctanh(1/5*15^(1/2)*(3+2*x)^(1/2))
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.89 \[ \int \frac {(5-x) (3+2 x)^{5/2}}{\left (2+5 x+3 x^2\right )^2} \, dx=-\frac {2}{3} \left (\frac {\sqrt {3+2 x} \left (183+209 x+8 x^2\right )}{4+10 x+6 x^2}+195 \text {arctanh}\left (\sqrt {3+2 x}\right )-50 \sqrt {15} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right )\right ) \] Input:

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

Output:

(-2*((Sqrt[3 + 2*x]*(183 + 209*x + 8*x^2))/(4 + 10*x + 6*x^2) + 195*ArcTan 
h[Sqrt[3 + 2*x]] - 50*Sqrt[15]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]]))/3
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {1233, 27, 1196, 27, 1197, 1480, 220}

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.

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

\(\Big \downarrow \) 1233

\(\displaystyle \frac {1}{3} \int -\frac {15 (4-9 x) \sqrt {2 x+3}}{3 x^2+5 x+2}dx-\frac {(2 x+3)^{3/2} (139 x+121)}{3 \left (3 x^2+5 x+2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle -5 \int \frac {(4-9 x) \sqrt {2 x+3}}{3 x^2+5 x+2}dx-\frac {(139 x+121) (2 x+3)^{3/2}}{3 \left (3 x^2+5 x+2\right )}\)

\(\Big \downarrow \) 1196

\(\displaystyle -5 \left (\frac {1}{3} \int \frac {3 (11 x+24)}{\sqrt {2 x+3} \left (3 x^2+5 x+2\right )}dx-6 \sqrt {2 x+3}\right )-\frac {(139 x+121) (2 x+3)^{3/2}}{3 \left (3 x^2+5 x+2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle -5 \left (\int \frac {11 x+24}{\sqrt {2 x+3} \left (3 x^2+5 x+2\right )}dx-6 \sqrt {2 x+3}\right )-\frac {(139 x+121) (2 x+3)^{3/2}}{3 \left (3 x^2+5 x+2\right )}\)

\(\Big \downarrow \) 1197

\(\displaystyle -5 \left (2 \int \frac {11 (2 x+3)+15}{3 (2 x+3)^2-8 (2 x+3)+5}d\sqrt {2 x+3}-6 \sqrt {2 x+3}\right )-\frac {(139 x+121) (2 x+3)^{3/2}}{3 \left (3 x^2+5 x+2\right )}\)

\(\Big \downarrow \) 1480

\(\displaystyle -5 \left (2 \left (50 \int \frac {1}{3 (2 x+3)-5}d\sqrt {2 x+3}-39 \int \frac {1}{3 (2 x+3)-3}d\sqrt {2 x+3}\right )-6 \sqrt {2 x+3}\right )-\frac {(139 x+121) (2 x+3)^{3/2}}{3 \left (3 x^2+5 x+2\right )}\)

\(\Big \downarrow \) 220

\(\displaystyle -5 \left (2 \left (13 \text {arctanh}\left (\sqrt {2 x+3}\right )-10 \sqrt {\frac {5}{3}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {2 x+3}\right )\right )-6 \sqrt {2 x+3}\right )-\frac {(139 x+121) (2 x+3)^{3/2}}{3 \left (3 x^2+5 x+2\right )}\)

Input:

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

Output:

-1/3*((3 + 2*x)^(3/2)*(121 + 139*x))/(2 + 5*x + 3*x^2) - 5*(-6*Sqrt[3 + 2* 
x] + 2*(13*ArcTanh[Sqrt[3 + 2*x]] - 10*Sqrt[5/3]*ArcTanh[Sqrt[3/5]*Sqrt[3 
+ 2*x]]))
 

Defintions of rubi rules used

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 220

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 
1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && 
 (LtQ[a, 0] || GtQ[b, 0])
 

rule 1196

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + 
(c_.)*(x_)^2), x_Symbol] :> Simp[g*((d + e*x)^m/(c*m)), x] + Simp[1/c   Int 
[(d + e*x)^(m - 1)*(Simp[c*d*f - a*e*g + (g*c*d - b*e*g + c*e*f)*x, x]/(a + 
 b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && FractionQ[m] & 
& GtQ[m, 0]
 

rule 1197

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)), x_Symbol] :> Simp[2   Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - 
b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; Fr 
eeQ[{a, b, c, d, e, f, g}, x]
 

rule 1233

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(d + e*x)^(m - 1))*(a + b*x + c*x^2) 
^(p + 1)*((2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g - c 
*(b*e*f + b*d*g + 2*a*e*g))*x)/(c*(p + 1)*(b^2 - 4*a*c))), x] - Simp[1/(c*( 
p + 1)*(b^2 - 4*a*c))   Int[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Sim 
p[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a*e*(e*f 
*(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*( 
m + p + 1) + 2*c^2*d*f*(m + 2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2* 
p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && LtQ[p, -1] && 
GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, b, c, d, e, f, g]) | 
|  !ILtQ[m + 2*p + 3, 0])
 

rule 1480

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : 
> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q))   Int[1/( 
b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q))   Int[1/(b/2 
+ q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] 
 && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
 

Maple [A] (verified)

Time = 1.63 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.90

method	result	size
risch	\(-\frac {\left (8 x^{2}+209 x +183\right ) \sqrt {2 x +3}}{3 \left (3 x^{2}+5 x +2\right )}-65 \ln \left (\sqrt {2 x +3}+1\right )+65 \ln \left (\sqrt {2 x +3}-1\right )+\frac {100 \sqrt {15}\, \operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {2 x +3}}{5}\right )}{3}\)	\(75\)
derivativedivides	\(-\frac {8 \sqrt {2 x +3}}{9}-\frac {850 \sqrt {2 x +3}}{27 \left (2 x +\frac {4}{3}\right )}+\frac {100 \sqrt {15}\, \operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {2 x +3}}{5}\right )}{3}-\frac {6}{\sqrt {2 x +3}+1}-65 \ln \left (\sqrt {2 x +3}+1\right )-\frac {6}{\sqrt {2 x +3}-1}+65 \ln \left (\sqrt {2 x +3}-1\right )\)	\(95\)
default	\(-\frac {8 \sqrt {2 x +3}}{9}-\frac {850 \sqrt {2 x +3}}{27 \left (2 x +\frac {4}{3}\right )}+\frac {100 \sqrt {15}\, \operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {2 x +3}}{5}\right )}{3}-\frac {6}{\sqrt {2 x +3}+1}-65 \ln \left (\sqrt {2 x +3}+1\right )-\frac {6}{\sqrt {2 x +3}-1}+65 \ln \left (\sqrt {2 x +3}-1\right )\)	\(95\)
trager	\(-\frac {\left (8 x^{2}+209 x +183\right ) \sqrt {2 x +3}}{3 \left (3 x^{2}+5 x +2\right )}-65 \ln \left (\frac {\sqrt {2 x +3}+2+x}{x +1}\right )+\frac {50 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right ) x +15 \sqrt {2 x +3}+7 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right )}{3 x +2}\right )}{3}\)	\(96\)
pseudoelliptic	\(\frac {300 \sqrt {15}\, \left (x +1\right ) \left (x +\frac {2}{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {2 x +3}}{5}\right )+\left (585 x^{2}+975 x +390\right ) \ln \left (\sqrt {2 x +3}-1\right )+\left (-585 x^{2}-975 x -390\right ) \ln \left (\sqrt {2 x +3}+1\right )-8 \sqrt {2 x +3}\, \left (x^{2}+\frac {209}{8} x +\frac {183}{8}\right )}{9 x^{2}+15 x +6}\)	\(98\)

Input:

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

Output:

-1/3*(8*x^2+209*x+183)/(3*x^2+5*x+2)*(2*x+3)^(1/2)-65*ln((2*x+3)^(1/2)+1)+ 
65*ln((2*x+3)^(1/2)-1)+100/3*15^(1/2)*arctanh(1/5*15^(1/2)*(2*x+3)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.43 \[ \int \frac {(5-x) (3+2 x)^{5/2}}{\left (2+5 x+3 x^2\right )^2} \, dx=\frac {150 \, \sqrt {\frac {5}{3}} {\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (\frac {3 \, \sqrt {\frac {5}{3}} \sqrt {2 \, x + 3} + 3 \, x + 7}{3 \, x + 2}\right ) - 195 \, {\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (\sqrt {2 \, x + 3} + 1\right ) + 195 \, {\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (\sqrt {2 \, x + 3} - 1\right ) - {\left (8 \, x^{2} + 209 \, x + 183\right )} \sqrt {2 \, x + 3}}{3 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}} \] Input:

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

Output:

1/3*(150*sqrt(5/3)*(3*x^2 + 5*x + 2)*log((3*sqrt(5/3)*sqrt(2*x + 3) + 3*x 
+ 7)/(3*x + 2)) - 195*(3*x^2 + 5*x + 2)*log(sqrt(2*x + 3) + 1) + 195*(3*x^ 
2 + 5*x + 2)*log(sqrt(2*x + 3) - 1) - (8*x^2 + 209*x + 183)*sqrt(2*x + 3)) 
/(3*x^2 + 5*x + 2)
 

Sympy [A] (verification not implemented)

Time = 40.01 (sec) , antiderivative size = 224, normalized size of antiderivative = 2.70 \[ \int \frac {(5-x) (3+2 x)^{5/2}}{\left (2+5 x+3 x^2\right )^2} \, dx=- \frac {8 \sqrt {2 x + 3}}{9} - \frac {365 \sqrt {15} \left (\log {\left (\sqrt {2 x + 3} - \frac {\sqrt {15}}{3} \right )} - \log {\left (\sqrt {2 x + 3} + \frac {\sqrt {15}}{3} \right )}\right )}{27} + \frac {8500 \left (\begin {cases} \frac {\sqrt {15} \left (- \frac {\log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1\right )}\right )}{75} & \text {for}\: \sqrt {2 x + 3} > - \frac {\sqrt {15}}{3} \wedge \sqrt {2 x + 3} < \frac {\sqrt {15}}{3} \end {cases}\right )}{9} + 65 \log {\left (\sqrt {2 x + 3} - 1 \right )} - 65 \log {\left (\sqrt {2 x + 3} + 1 \right )} - \frac {6}{\sqrt {2 x + 3} + 1} - \frac {6}{\sqrt {2 x + 3} - 1} \] Input:

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

Output:

-8*sqrt(2*x + 3)/9 - 365*sqrt(15)*(log(sqrt(2*x + 3) - sqrt(15)/3) - log(s 
qrt(2*x + 3) + sqrt(15)/3))/27 + 8500*Piecewise((sqrt(15)*(-log(sqrt(15)*s 
qrt(2*x + 3)/5 - 1)/4 + log(sqrt(15)*sqrt(2*x + 3)/5 + 1)/4 - 1/(4*(sqrt(1 
5)*sqrt(2*x + 3)/5 + 1)) - 1/(4*(sqrt(15)*sqrt(2*x + 3)/5 - 1)))/75, (sqrt 
(2*x + 3) > -sqrt(15)/3) & (sqrt(2*x + 3) < sqrt(15)/3)))/9 + 65*log(sqrt( 
2*x + 3) - 1) - 65*log(sqrt(2*x + 3) + 1) - 6/(sqrt(2*x + 3) + 1) - 6/(sqr 
t(2*x + 3) - 1)
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.29 \[ \int \frac {(5-x) (3+2 x)^{5/2}}{\left (2+5 x+3 x^2\right )^2} \, dx=-\frac {50}{3} \, \sqrt {15} \log \left (-\frac {\sqrt {15} - 3 \, \sqrt {2 \, x + 3}}{\sqrt {15} + 3 \, \sqrt {2 \, x + 3}}\right ) - \frac {8}{9} \, \sqrt {2 \, x + 3} - \frac {2 \, {\left (587 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}} - 695 \, \sqrt {2 \, x + 3}\right )}}{9 \, {\left (3 \, {\left (2 \, x + 3\right )}^{2} - 16 \, x - 19\right )}} - 65 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) + 65 \, \log \left (\sqrt {2 \, x + 3} - 1\right ) \] Input:

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

Output:

-50/3*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 
3))) - 8/9*sqrt(2*x + 3) - 2/9*(587*(2*x + 3)^(3/2) - 695*sqrt(2*x + 3))/( 
3*(2*x + 3)^2 - 16*x - 19) - 65*log(sqrt(2*x + 3) + 1) + 65*log(sqrt(2*x + 
 3) - 1)
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.34 \[ \int \frac {(5-x) (3+2 x)^{5/2}}{\left (2+5 x+3 x^2\right )^2} \, dx=-\frac {50}{3} \, \sqrt {15} \log \left (\frac {{\left | -2 \, \sqrt {15} + 6 \, \sqrt {2 \, x + 3} \right |}}{2 \, {\left (\sqrt {15} + 3 \, \sqrt {2 \, x + 3}\right )}}\right ) - \frac {8}{9} \, \sqrt {2 \, x + 3} - \frac {2 \, {\left (587 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}} - 695 \, \sqrt {2 \, x + 3}\right )}}{9 \, {\left (3 \, {\left (2 \, x + 3\right )}^{2} - 16 \, x - 19\right )}} - 65 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) + 65 \, \log \left ({\left | \sqrt {2 \, x + 3} - 1 \right |}\right ) \] Input:

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

Output:

-50/3*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3*sq 
rt(2*x + 3))) - 8/9*sqrt(2*x + 3) - 2/9*(587*(2*x + 3)^(3/2) - 695*sqrt(2* 
x + 3))/(3*(2*x + 3)^2 - 16*x - 19) - 65*log(sqrt(2*x + 3) + 1) + 65*log(a 
bs(sqrt(2*x + 3) - 1))
 

Mupad [B] (verification not implemented)

Time = 11.63 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.98 \[ \int \frac {(5-x) (3+2 x)^{5/2}}{\left (2+5 x+3 x^2\right )^2} \, dx=-\frac {\frac {1390\,\sqrt {2\,x+3}}{27}-\frac {1174\,{\left (2\,x+3\right )}^{3/2}}{27}}{\frac {16\,x}{3}-{\left (2\,x+3\right )}^2+\frac {19}{3}}-\frac {8\,\sqrt {2\,x+3}}{9}+\mathrm {atan}\left (\sqrt {2\,x+3}\,1{}\mathrm {i}\right )\,130{}\mathrm {i}-\frac {\sqrt {15}\,\mathrm {atan}\left (\frac {\sqrt {15}\,\sqrt {2\,x+3}\,1{}\mathrm {i}}{5}\right )\,100{}\mathrm {i}}{3} \] Input:

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

Output:

atan((2*x + 3)^(1/2)*1i)*130i - ((1390*(2*x + 3)^(1/2))/27 - (1174*(2*x + 
3)^(3/2))/27)/((16*x)/3 - (2*x + 3)^2 + 19/3) - (15^(1/2)*atan((15^(1/2)*( 
2*x + 3)^(1/2)*1i)/5)*100i)/3 - (8*(2*x + 3)^(1/2))/9
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 226, normalized size of antiderivative = 2.72 \[ \int \frac {(5-x) (3+2 x)^{5/2}}{\left (2+5 x+3 x^2\right )^2} \, dx=\frac {-8 \sqrt {2 x +3}\, x^{2}-209 \sqrt {2 x +3}\, x -183 \sqrt {2 x +3}-150 \sqrt {15}\, \mathrm {log}\left (3 \sqrt {2 x +3}-\sqrt {15}\right ) x^{2}-250 \sqrt {15}\, \mathrm {log}\left (3 \sqrt {2 x +3}-\sqrt {15}\right ) x -100 \sqrt {15}\, \mathrm {log}\left (3 \sqrt {2 x +3}-\sqrt {15}\right )+150 \sqrt {15}\, \mathrm {log}\left (3 \sqrt {2 x +3}+\sqrt {15}\right ) x^{2}+250 \sqrt {15}\, \mathrm {log}\left (3 \sqrt {2 x +3}+\sqrt {15}\right ) x +100 \sqrt {15}\, \mathrm {log}\left (3 \sqrt {2 x +3}+\sqrt {15}\right )+585 \,\mathrm {log}\left (\sqrt {2 x +3}-1\right ) x^{2}+975 \,\mathrm {log}\left (\sqrt {2 x +3}-1\right ) x +390 \,\mathrm {log}\left (\sqrt {2 x +3}-1\right )-585 \,\mathrm {log}\left (\sqrt {2 x +3}+1\right ) x^{2}-975 \,\mathrm {log}\left (\sqrt {2 x +3}+1\right ) x -390 \,\mathrm {log}\left (\sqrt {2 x +3}+1\right )}{9 x^{2}+15 x +6} \] Input:

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

Output:

( - 8*sqrt(2*x + 3)*x**2 - 209*sqrt(2*x + 3)*x - 183*sqrt(2*x + 3) - 150*s 
qrt(15)*log(3*sqrt(2*x + 3) - sqrt(15))*x**2 - 250*sqrt(15)*log(3*sqrt(2*x 
 + 3) - sqrt(15))*x - 100*sqrt(15)*log(3*sqrt(2*x + 3) - sqrt(15)) + 150*s 
qrt(15)*log(3*sqrt(2*x + 3) + sqrt(15))*x**2 + 250*sqrt(15)*log(3*sqrt(2*x 
 + 3) + sqrt(15))*x + 100*sqrt(15)*log(3*sqrt(2*x + 3) + sqrt(15)) + 585*l 
og(sqrt(2*x + 3) - 1)*x**2 + 975*log(sqrt(2*x + 3) - 1)*x + 390*log(sqrt(2 
*x + 3) - 1) - 585*log(sqrt(2*x + 3) + 1)*x**2 - 975*log(sqrt(2*x + 3) + 1 
)*x - 390*log(sqrt(2*x + 3) + 1))/(3*(3*x**2 + 5*x + 2))