\(\int \frac {2-5 x}{x^{5/2} (2+5 x+3 x^2)^{3/2}} \, dx\) [215]

Optimal result

Integrand size = 25, antiderivative size = 199 \[ \int \frac {2-5 x}{x^{5/2} \left (2+5 x+3 x^2\right )^{3/2}} \, dx=-\frac {2}{3 x^{3/2} \sqrt {2+5 x+3 x^2}}+\frac {35}{3 \sqrt {x} \sqrt {2+5 x+3 x^2}}-\frac {170 \sqrt {x} (2+3 x)}{3 \sqrt {2+5 x+3 x^2}}+\frac {5 \sqrt {x} (101+102 x)}{3 \sqrt {2+5 x+3 x^2}}+\frac {170 \sqrt {2} \sqrt {2+5 x+3 x^2} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{3 \sqrt {1+x} \sqrt {2+3 x}}-\frac {115 \sqrt {1+x} \sqrt {2+3 x} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {2+5 x+3 x^2}} \] Output:

-2/3/x^(3/2)/(3*x^2+5*x+2)^(1/2)+35/3/x^(1/2)/(3*x^2+5*x+2)^(1/2)-170/3*x^ 
(1/2)*(2+3*x)/(3*x^2+5*x+2)^(1/2)+5/3*x^(1/2)*(101+102*x)/(3*x^2+5*x+2)^(1 
/2)+170/3*2^(1/2)*(3*x^2+5*x+2)^(1/2)*EllipticE(x^(1/2)/(1+x)^(1/2),1/2*I* 
2^(1/2))/(1+x)^(1/2)/(2+3*x)^(1/2)-115/2*2^(1/2)*(1+x)^(1/2)*(2+3*x)^(1/2) 
*InverseJacobiAM(arctan(x^(1/2)),1/2*I*2^(1/2))/(3*x^2+5*x+2)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 21.21 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.73 \[ \int \frac {2-5 x}{x^{5/2} \left (2+5 x+3 x^2\right )^{3/2}} \, dx=\frac {-4-610 x-690 x^2-340 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{5/2} E\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )-5 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{5/2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right ),\frac {3}{2}\right )}{6 x^{3/2} \sqrt {2+5 x+3 x^2}} \] Input:

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

Output:

(-4 - 610*x - 690*x^2 - (340*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^( 
5/2)*EllipticE[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2] - (5*I)*Sqrt[2]*Sqrt[1 + 
 x^(-1)]*Sqrt[3 + 2/x]*x^(5/2)*EllipticF[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2 
])/(6*x^(3/2)*Sqrt[2 + 5*x + 3*x^2])
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {1235, 27, 1237, 27, 1237, 27, 1240, 1503, 1413, 1456}

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.

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

\(\Big \downarrow \) 1235

\(\displaystyle \frac {2 (45 x+38)}{x^{3/2} \sqrt {3 x^2+5 x+2}}-\int -\frac {5 (27 x+23)}{x^{5/2} \sqrt {3 x^2+5 x+2}}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 5 \int \frac {27 x+23}{x^{5/2} \sqrt {3 x^2+5 x+2}}dx+\frac {2 (45 x+38)}{x^{3/2} \sqrt {3 x^2+5 x+2}}\)

\(\Big \downarrow \) 1237

\(\displaystyle 5 \left (-\frac {1}{3} \int \frac {69 x+68}{2 x^{3/2} \sqrt {3 x^2+5 x+2}}dx-\frac {23 \sqrt {3 x^2+5 x+2}}{3 x^{3/2}}\right )+\frac {2 (45 x+38)}{x^{3/2} \sqrt {3 x^2+5 x+2}}\)

\(\Big \downarrow \) 27

\(\displaystyle 5 \left (-\frac {1}{6} \int \frac {69 x+68}{x^{3/2} \sqrt {3 x^2+5 x+2}}dx-\frac {23 \sqrt {3 x^2+5 x+2}}{3 x^{3/2}}\right )+\frac {2 (45 x+38)}{x^{3/2} \sqrt {3 x^2+5 x+2}}\)

\(\Big \downarrow \) 1237

\(\displaystyle 5 \left (\frac {1}{6} \left (\int -\frac {3 (34 x+23)}{\sqrt {x} \sqrt {3 x^2+5 x+2}}dx+\frac {68 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}\right )-\frac {23 \sqrt {3 x^2+5 x+2}}{3 x^{3/2}}\right )+\frac {2 (45 x+38)}{x^{3/2} \sqrt {3 x^2+5 x+2}}\)

\(\Big \downarrow \) 27

\(\displaystyle 5 \left (\frac {1}{6} \left (\frac {68 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}-3 \int \frac {34 x+23}{\sqrt {x} \sqrt {3 x^2+5 x+2}}dx\right )-\frac {23 \sqrt {3 x^2+5 x+2}}{3 x^{3/2}}\right )+\frac {2 (45 x+38)}{x^{3/2} \sqrt {3 x^2+5 x+2}}\)

\(\Big \downarrow \) 1240

\(\displaystyle 5 \left (\frac {1}{6} \left (\frac {68 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}-6 \int \frac {34 x+23}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}\right )-\frac {23 \sqrt {3 x^2+5 x+2}}{3 x^{3/2}}\right )+\frac {2 (45 x+38)}{x^{3/2} \sqrt {3 x^2+5 x+2}}\)

\(\Big \downarrow \) 1503

\(\displaystyle 5 \left (\frac {1}{6} \left (\frac {68 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}-6 \left (23 \int \frac {1}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+34 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}\right )\right )-\frac {23 \sqrt {3 x^2+5 x+2}}{3 x^{3/2}}\right )+\frac {2 (45 x+38)}{x^{3/2} \sqrt {3 x^2+5 x+2}}\)

\(\Big \downarrow \) 1413

\(\displaystyle 5 \left (\frac {1}{6} \left (\frac {68 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}-6 \left (34 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+\frac {23 (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^2+5 x+2}}\right )\right )-\frac {23 \sqrt {3 x^2+5 x+2}}{3 x^{3/2}}\right )+\frac {2 (45 x+38)}{x^{3/2} \sqrt {3 x^2+5 x+2}}\)

\(\Big \downarrow \) 1456

\(\displaystyle 5 \left (\frac {1}{6} \left (\frac {68 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}-6 \left (\frac {23 (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^2+5 x+2}}+34 \left (\frac {\sqrt {x} (3 x+2)}{3 \sqrt {3 x^2+5 x+2}}-\frac {\sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{3 \sqrt {3 x^2+5 x+2}}\right )\right )\right )-\frac {23 \sqrt {3 x^2+5 x+2}}{3 x^{3/2}}\right )+\frac {2 (45 x+38)}{x^{3/2} \sqrt {3 x^2+5 x+2}}\)

Input:

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

Output:

(2*(38 + 45*x))/(x^(3/2)*Sqrt[2 + 5*x + 3*x^2]) + 5*((-23*Sqrt[2 + 5*x + 3 
*x^2])/(3*x^(3/2)) + ((68*Sqrt[2 + 5*x + 3*x^2])/Sqrt[x] - 6*(34*((Sqrt[x] 
*(2 + 3*x))/(3*Sqrt[2 + 5*x + 3*x^2]) - (Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 
 + x)]*EllipticE[ArcTan[Sqrt[x]], -1/2])/(3*Sqrt[2 + 5*x + 3*x^2])) + (23* 
(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[ArcTan[Sqrt[x]], -1/2])/(Sqrt[2] 
*Sqrt[2 + 5*x + 3*x^2])))/6)
 

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 1235

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

rule 1237

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

rule 1240

Int[((f_) + (g_.)*(x_))/(Sqrt[x_]*Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2]), 
x_Symbol] :> Simp[2   Subst[Int[(f + g*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x, 
 Sqrt[x]], x] /; FreeQ[{a, b, c, f, g}, x]
 

rule 1413

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

rule 1456

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[b^2 - 4*a*c, 2]}, Simp[x*((b - q + 2*c*x^2)/(2*c*Sqrt[a + b*x^2 + c*x^4 
])), x] - Simp[Rt[(b - q)/(2*a), 2]*(2*a + (b - q)*x^2)*(Sqrt[(2*a + (b + q 
)*x^2)/(2*a + (b - q)*x^2)]/(2*c*Sqrt[a + b*x^2 + c*x^4]))*EllipticE[ArcTan 
[Rt[(b - q)/(2*a), 2]*x], -2*(q/(b - q))], x] /; PosQ[(b - q)/a]] /; FreeQ[ 
{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]
 

rule 1503

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

Maple [A] (verified)

Time = 0.99 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.58

method	result	size
default	\(\frac {165 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, \operatorname {EllipticF}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right ) x -170 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, \operatorname {EllipticE}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right ) x +3060 x^{3}+3030 x^{2}+210 x -12}{18 x^{\frac {3}{2}} \sqrt {3 x^{2}+5 x +2}}\)	\(115\)
elliptic	\(\frac {\sqrt {\left (3 x^{2}+5 x +2\right ) x}\, \left (-\frac {\sqrt {3 x^{3}+5 x^{2}+2 x}}{3 x^{2}}+\frac {20 x^{2}+\frac {100}{3} x +\frac {40}{3}}{\sqrt {\left (3 x^{2}+5 x +2\right ) x}}-\frac {2 x \left (-\frac {68}{3}-25 x \right ) \sqrt {3}}{\sqrt {x \left (x^{2}+\frac {5}{3} x +\frac {2}{3}\right )}}-\frac {115 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \operatorname {EllipticF}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{6 \sqrt {3 x^{3}+5 x^{2}+2 x}}-\frac {85 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \left (\frac {\operatorname {EllipticE}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3}-\operatorname {EllipticF}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right )}{3 \sqrt {3 x^{3}+5 x^{2}+2 x}}\right )}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\)	\(229\)

Input:

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

Output:

1/18*(165*(6*x+4)^(1/2)*(3+3*x)^(1/2)*6^(1/2)*(-x)^(1/2)*EllipticF(1/2*(6* 
x+4)^(1/2),I*2^(1/2))*x-170*(6*x+4)^(1/2)*(3+3*x)^(1/2)*6^(1/2)*(-x)^(1/2) 
*EllipticE(1/2*(6*x+4)^(1/2),I*2^(1/2))*x+3060*x^3+3030*x^2+210*x-12)/x^(3 
/2)/(3*x^2+5*x+2)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.55 \[ \int \frac {2-5 x}{x^{5/2} \left (2+5 x+3 x^2\right )^{3/2}} \, dx=-\frac {185 \, \sqrt {3} {\left (3 \, x^{4} + 5 \, x^{3} + 2 \, x^{2}\right )} {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right ) - 1530 \, \sqrt {3} {\left (3 \, x^{4} + 5 \, x^{3} + 2 \, x^{2}\right )} {\rm weierstrassZeta}\left (\frac {28}{27}, \frac {80}{729}, {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right )\right ) - 9 \, {\left (510 \, x^{3} + 505 \, x^{2} + 35 \, x - 2\right )} \sqrt {3 \, x^{2} + 5 \, x + 2} \sqrt {x}}{27 \, {\left (3 \, x^{4} + 5 \, x^{3} + 2 \, x^{2}\right )}} \] Input:

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

Output:

-1/27*(185*sqrt(3)*(3*x^4 + 5*x^3 + 2*x^2)*weierstrassPInverse(28/27, 80/7 
29, x + 5/9) - 1530*sqrt(3)*(3*x^4 + 5*x^3 + 2*x^2)*weierstrassZeta(28/27, 
 80/729, weierstrassPInverse(28/27, 80/729, x + 5/9)) - 9*(510*x^3 + 505*x 
^2 + 35*x - 2)*sqrt(3*x^2 + 5*x + 2)*sqrt(x))/(3*x^4 + 5*x^3 + 2*x^2)
 

Sympy [F]

\[ \int \frac {2-5 x}{x^{5/2} \left (2+5 x+3 x^2\right )^{3/2}} \, dx=- \int \frac {5}{3 x^{\frac {7}{2}} \sqrt {3 x^{2} + 5 x + 2} + 5 x^{\frac {5}{2}} \sqrt {3 x^{2} + 5 x + 2} + 2 x^{\frac {3}{2}} \sqrt {3 x^{2} + 5 x + 2}}\, dx - \int \left (- \frac {2}{3 x^{\frac {9}{2}} \sqrt {3 x^{2} + 5 x + 2} + 5 x^{\frac {7}{2}} \sqrt {3 x^{2} + 5 x + 2} + 2 x^{\frac {5}{2}} \sqrt {3 x^{2} + 5 x + 2}}\right )\, dx \] Input:

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

Output:

-Integral(5/(3*x**(7/2)*sqrt(3*x**2 + 5*x + 2) + 5*x**(5/2)*sqrt(3*x**2 + 
5*x + 2) + 2*x**(3/2)*sqrt(3*x**2 + 5*x + 2)), x) - Integral(-2/(3*x**(9/2 
)*sqrt(3*x**2 + 5*x + 2) + 5*x**(7/2)*sqrt(3*x**2 + 5*x + 2) + 2*x**(5/2)* 
sqrt(3*x**2 + 5*x + 2)), x)
 

Maxima [F]

\[ \int \frac {2-5 x}{x^{5/2} \left (2+5 x+3 x^2\right )^{3/2}} \, dx=\int { -\frac {5 \, x - 2}{{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}} x^{\frac {5}{2}}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {2-5 x}{x^{5/2} \left (2+5 x+3 x^2\right )^{3/2}} \, dx=\int { -\frac {5 \, x - 2}{{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}} x^{\frac {5}{2}}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {2-5 x}{x^{5/2} \left (2+5 x+3 x^2\right )^{3/2}} \, dx=\int -\frac {5\,x-2}{x^{5/2}\,{\left (3\,x^2+5\,x+2\right )}^{3/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {2-5 x}{x^{5/2} \left (2+5 x+3 x^2\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 30*sqrt(3*x**2 + 5*x + 2)*x**2 + 30*sqrt(3*x**2 + 5*x + 2)*x - 4*sqrt( 
3*x**2 + 5*x + 2) - 120*sqrt(x)*int(sqrt(3*x**2 + 5*x + 2)/(9*sqrt(x)*x**5 
 + 30*sqrt(x)*x**4 + 37*sqrt(x)*x**3 + 20*sqrt(x)*x**2 + 4*sqrt(x)*x),x)*x 
**3 - 200*sqrt(x)*int(sqrt(3*x**2 + 5*x + 2)/(9*sqrt(x)*x**5 + 30*sqrt(x)* 
x**4 + 37*sqrt(x)*x**3 + 20*sqrt(x)*x**2 + 4*sqrt(x)*x),x)*x**2 - 80*sqrt( 
x)*int(sqrt(3*x**2 + 5*x + 2)/(9*sqrt(x)*x**5 + 30*sqrt(x)*x**4 + 37*sqrt( 
x)*x**3 + 20*sqrt(x)*x**2 + 4*sqrt(x)*x),x)*x + 450*sqrt(x)*int(sqrt(3*x** 
2 + 5*x + 2)/(9*sqrt(x)*x**4 + 30*sqrt(x)*x**3 + 37*sqrt(x)*x**2 + 20*sqrt 
(x)*x + 4*sqrt(x)),x)*x**3 + 750*sqrt(x)*int(sqrt(3*x**2 + 5*x + 2)/(9*sqr 
t(x)*x**4 + 30*sqrt(x)*x**3 + 37*sqrt(x)*x**2 + 20*sqrt(x)*x + 4*sqrt(x)), 
x)*x**2 + 300*sqrt(x)*int(sqrt(3*x**2 + 5*x + 2)/(9*sqrt(x)*x**4 + 30*sqrt 
(x)*x**3 + 37*sqrt(x)*x**2 + 20*sqrt(x)*x + 4*sqrt(x)),x)*x - 135*sqrt(x)* 
int((sqrt(3*x**2 + 5*x + 2)*x**2)/(9*sqrt(x)*x**4 + 30*sqrt(x)*x**3 + 37*s 
qrt(x)*x**2 + 20*sqrt(x)*x + 4*sqrt(x)),x)*x**3 - 225*sqrt(x)*int((sqrt(3* 
x**2 + 5*x + 2)*x**2)/(9*sqrt(x)*x**4 + 30*sqrt(x)*x**3 + 37*sqrt(x)*x**2 
+ 20*sqrt(x)*x + 4*sqrt(x)),x)*x**2 - 90*sqrt(x)*int((sqrt(3*x**2 + 5*x + 
2)*x**2)/(9*sqrt(x)*x**4 + 30*sqrt(x)*x**3 + 37*sqrt(x)*x**2 + 20*sqrt(x)* 
x + 4*sqrt(x)),x)*x + 405*sqrt(x)*int((sqrt(3*x**2 + 5*x + 2)*x)/(9*sqrt(x 
)*x**4 + 30*sqrt(x)*x**3 + 37*sqrt(x)*x**2 + 20*sqrt(x)*x + 4*sqrt(x)),x)* 
x**3 + 675*sqrt(x)*int((sqrt(3*x**2 + 5*x + 2)*x)/(9*sqrt(x)*x**4 + 30*...