\(\int \frac {1+3 x+4 x^2}{(1+2 x)^3 (2+3 x^2)^{3/2}} \, dx\) [129]

Optimal result

Integrand size = 29, antiderivative size = 97 \[ \int \frac {1+3 x+4 x^2}{(1+2 x)^3 \left (2+3 x^2\right )^{3/2}} \, dx=\frac {358+351 x}{2662 \sqrt {2+3 x^2}}-\frac {2 \sqrt {2+3 x^2}}{121 (1+2 x)^2}+\frac {2 \sqrt {2+3 x^2}}{1331 (1+2 x)}-\frac {322 \text {arctanh}\left (\frac {4-3 x}{\sqrt {11} \sqrt {2+3 x^2}}\right )}{1331 \sqrt {11}} \] Output:

1/2662*(358+351*x)/(3*x^2+2)^(1/2)-2/121*(3*x^2+2)^(1/2)/(1+2*x)^2+2*(3*x^ 
2+2)^(1/2)/(1331+2662*x)-322/14641*arctanh(1/11*(4-3*x)*11^(1/2)/(3*x^2+2) 
^(1/2))*11^(1/2)
 

Mathematica [A] (verified)

Time = 0.88 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.84 \[ \int \frac {1+3 x+4 x^2}{(1+2 x)^3 \left (2+3 x^2\right )^{3/2}} \, dx=\frac {\frac {11 \left (278+1799 x+2716 x^2+1428 x^3\right )}{(1+2 x)^2 \sqrt {2+3 x^2}}+1288 \sqrt {11} \text {arctanh}\left (\frac {\sqrt {3}+2 \sqrt {3} x-2 \sqrt {2+3 x^2}}{\sqrt {11}}\right )}{29282} \] Input:

Integrate[(1 + 3*x + 4*x^2)/((1 + 2*x)^3*(2 + 3*x^2)^(3/2)),x]
 

Output:

((11*(278 + 1799*x + 2716*x^2 + 1428*x^3))/((1 + 2*x)^2*Sqrt[2 + 3*x^2]) + 
 1288*Sqrt[11]*ArcTanh[(Sqrt[3] + 2*Sqrt[3]*x - 2*Sqrt[2 + 3*x^2])/Sqrt[11 
]])/29282
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2178, 27, 2182, 27, 679, 488, 219}

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.

\(\displaystyle \int \frac {4 x^2+3 x+1}{(2 x+1)^3 \left (3 x^2+2\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 2178

\(\displaystyle \frac {351 x+358}{2662 \sqrt {3 x^2+2}}-\frac {1}{6} \int -\frac {12 \left (716 x^2+606 x+245\right )}{1331 (2 x+1)^3 \sqrt {3 x^2+2}}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \int \frac {716 x^2+606 x+245}{(2 x+1)^3 \sqrt {3 x^2+2}}dx}{1331}+\frac {351 x+358}{2662 \sqrt {3 x^2+2}}\)

\(\Big \downarrow \) 2182

\(\displaystyle \frac {2 \left (-\frac {1}{22} \int -\frac {22 (325 x+157)}{(2 x+1)^2 \sqrt {3 x^2+2}}dx-\frac {11 \sqrt {3 x^2+2}}{(2 x+1)^2}\right )}{1331}+\frac {351 x+358}{2662 \sqrt {3 x^2+2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \left (\int \frac {325 x+157}{(2 x+1)^2 \sqrt {3 x^2+2}}dx-\frac {11 \sqrt {3 x^2+2}}{(2 x+1)^2}\right )}{1331}+\frac {351 x+358}{2662 \sqrt {3 x^2+2}}\)

\(\Big \downarrow \) 679

\(\displaystyle \frac {2 \left (161 \int \frac {1}{(2 x+1) \sqrt {3 x^2+2}}dx+\frac {\sqrt {3 x^2+2}}{2 x+1}-\frac {11 \sqrt {3 x^2+2}}{(2 x+1)^2}\right )}{1331}+\frac {351 x+358}{2662 \sqrt {3 x^2+2}}\)

\(\Big \downarrow \) 488

\(\displaystyle \frac {2 \left (-161 \int \frac {1}{11-\frac {(4-3 x)^2}{3 x^2+2}}d\frac {4-3 x}{\sqrt {3 x^2+2}}+\frac {\sqrt {3 x^2+2}}{2 x+1}-\frac {11 \sqrt {3 x^2+2}}{(2 x+1)^2}\right )}{1331}+\frac {351 x+358}{2662 \sqrt {3 x^2+2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {2 \left (-\frac {161 \text {arctanh}\left (\frac {4-3 x}{\sqrt {11} \sqrt {3 x^2+2}}\right )}{\sqrt {11}}+\frac {\sqrt {3 x^2+2}}{2 x+1}-\frac {11 \sqrt {3 x^2+2}}{(2 x+1)^2}\right )}{1331}+\frac {351 x+358}{2662 \sqrt {3 x^2+2}}\)

Input:

Int[(1 + 3*x + 4*x^2)/((1 + 2*x)^3*(2 + 3*x^2)^(3/2)),x]
 

Output:

(358 + 351*x)/(2662*Sqrt[2 + 3*x^2]) + (2*((-11*Sqrt[2 + 3*x^2])/(1 + 2*x) 
^2 + Sqrt[2 + 3*x^2]/(1 + 2*x) - (161*ArcTanh[(4 - 3*x)/(Sqrt[11]*Sqrt[2 + 
 3*x^2])])/Sqrt[11]))/1331
 

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 219

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

rule 488

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

rule 679

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1 
)/(2*(p + 1)*(c*d^2 + a*e^2))), x] + Simp[(c*d*f + a*e*g)/(c*d^2 + a*e^2) 
 Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, 
 p}, x] && EqQ[Simplify[m + 2*p + 3], 0]
 

rule 2178

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{Qx = PolynomialQuotient[(d + e*x)^m*Pq, a + b*x^2, x], R = Coeff[Po 
lynomialRemainder[(d + e*x)^m*Pq, a + b*x^2, x], x, 0], S = Coeff[Polynomia 
lRemainder[(d + e*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*S - b*R*x)*((a + 
b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1))   Int[(d + e*x 
)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*b*(p + 1)*Qx)/(d + e*x)^m + (b*R*( 
2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, b, d, e}, x] && PolyQ[Pq, x 
] && NeQ[b*d^2 + a*e^2, 0] && LtQ[p, -1] && ILtQ[m, 0]
 

rule 2182

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
 With[{Qx = PolynomialQuotient[Pq, d + e*x, x], R = PolynomialRemainder[Pq, 
 d + e*x, x]}, Simp[e*R*(d + e*x)^(m + 1)*((a + b*x^2)^(p + 1)/((m + 1)*(b* 
d^2 + a*e^2))), x] + Simp[1/((m + 1)*(b*d^2 + a*e^2))   Int[(d + e*x)^(m + 
1)*(a + b*x^2)^p*ExpandToSum[(m + 1)*(b*d^2 + a*e^2)*Qx + b*d*R*(m + 1) - b 
*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, d, e, p}, x] && PolyQ[Pq, 
 x] && NeQ[b*d^2 + a*e^2, 0] && LtQ[m, -1]
 

Maple [A] (verified)

Time = 0.95 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.67

method	result	size
risch	\(\frac {1428 x^{3}+2716 x^{2}+1799 x +278}{2662 \left (1+2 x \right )^{2} \sqrt {3 x^{2}+2}}-\frac {322 \sqrt {11}\, \operatorname {arctanh}\left (\frac {2 \left (4-3 x \right ) \sqrt {11}}{11 \sqrt {12 \left (\frac {1}{2}+x \right )^{2}+5-12 x}}\right )}{14641}\)	\(65\)
trager	\(\frac {1428 x^{3}+2716 x^{2}+1799 x +278}{2662 \left (1+2 x \right )^{2} \sqrt {3 x^{2}+2}}+\frac {322 \operatorname {RootOf}\left (\textit {\_Z}^{2}-11\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-11\right ) x +11 \sqrt {3 x^{2}+2}-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-11\right )}{1+2 x}\right )}{14641}\)	\(81\)
default	\(\frac {161}{1331 \sqrt {3 \left (\frac {1}{2}+x \right )^{2}+\frac {5}{4}-3 x}}+\frac {357 x}{2662 \sqrt {3 \left (\frac {1}{2}+x \right )^{2}+\frac {5}{4}-3 x}}-\frac {322 \sqrt {11}\, \operatorname {arctanh}\left (\frac {2 \left (4-3 x \right ) \sqrt {11}}{11 \sqrt {12 \left (\frac {1}{2}+x \right )^{2}+5-12 x}}\right )}{14641}-\frac {1}{88 \left (\frac {1}{2}+x \right )^{2} \sqrt {3 \left (\frac {1}{2}+x \right )^{2}+\frac {5}{4}-3 x}}+\frac {7}{484 \left (\frac {1}{2}+x \right ) \sqrt {3 \left (\frac {1}{2}+x \right )^{2}+\frac {5}{4}-3 x}}\)	\(107\)

Input:

int((4*x^2+3*x+1)/(1+2*x)^3/(3*x^2+2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/2662*(1428*x^3+2716*x^2+1799*x+278)/(1+2*x)^2/(3*x^2+2)^(1/2)-322/14641* 
11^(1/2)*arctanh(2/11*(4-3*x)*11^(1/2)/(12*(1/2+x)^2+5-12*x)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.23 \[ \int \frac {1+3 x+4 x^2}{(1+2 x)^3 \left (2+3 x^2\right )^{3/2}} \, dx=\frac {322 \, \sqrt {11} {\left (12 \, x^{4} + 12 \, x^{3} + 11 \, x^{2} + 8 \, x + 2\right )} \log \left (-\frac {\sqrt {11} \sqrt {3 \, x^{2} + 2} {\left (3 \, x - 4\right )} + 21 \, x^{2} - 12 \, x + 19}{4 \, x^{2} + 4 \, x + 1}\right ) + 11 \, {\left (1428 \, x^{3} + 2716 \, x^{2} + 1799 \, x + 278\right )} \sqrt {3 \, x^{2} + 2}}{29282 \, {\left (12 \, x^{4} + 12 \, x^{3} + 11 \, x^{2} + 8 \, x + 2\right )}} \] Input:

integrate((4*x^2+3*x+1)/(1+2*x)^3/(3*x^2+2)^(3/2),x, algorithm="fricas")
 

Output:

1/29282*(322*sqrt(11)*(12*x^4 + 12*x^3 + 11*x^2 + 8*x + 2)*log(-(sqrt(11)* 
sqrt(3*x^2 + 2)*(3*x - 4) + 21*x^2 - 12*x + 19)/(4*x^2 + 4*x + 1)) + 11*(1 
428*x^3 + 2716*x^2 + 1799*x + 278)*sqrt(3*x^2 + 2))/(12*x^4 + 12*x^3 + 11* 
x^2 + 8*x + 2)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1+3 x+4 x^2}{(1+2 x)^3 \left (2+3 x^2\right )^{3/2}} \, dx=\text {Timed out} \] Input:

integrate((4*x**2+3*x+1)/(1+2*x)**3/(3*x**2+2)**(3/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.28 \[ \int \frac {1+3 x+4 x^2}{(1+2 x)^3 \left (2+3 x^2\right )^{3/2}} \, dx=\frac {322}{14641} \, \sqrt {11} \operatorname {arsinh}\left (\frac {\sqrt {6} x}{2 \, {\left | 2 \, x + 1 \right |}} - \frac {2 \, \sqrt {6}}{3 \, {\left | 2 \, x + 1 \right |}}\right ) + \frac {357 \, x}{2662 \, \sqrt {3 \, x^{2} + 2}} + \frac {161}{1331 \, \sqrt {3 \, x^{2} + 2}} - \frac {1}{22 \, {\left (4 \, \sqrt {3 \, x^{2} + 2} x^{2} + 4 \, \sqrt {3 \, x^{2} + 2} x + \sqrt {3 \, x^{2} + 2}\right )}} + \frac {7}{242 \, {\left (2 \, \sqrt {3 \, x^{2} + 2} x + \sqrt {3 \, x^{2} + 2}\right )}} \] Input:

integrate((4*x^2+3*x+1)/(1+2*x)^3/(3*x^2+2)^(3/2),x, algorithm="maxima")
 

Output:

322/14641*sqrt(11)*arcsinh(1/2*sqrt(6)*x/abs(2*x + 1) - 2/3*sqrt(6)/abs(2* 
x + 1)) + 357/2662*x/sqrt(3*x^2 + 2) + 161/1331/sqrt(3*x^2 + 2) - 1/22/(4* 
sqrt(3*x^2 + 2)*x^2 + 4*sqrt(3*x^2 + 2)*x + sqrt(3*x^2 + 2)) + 7/242/(2*sq 
rt(3*x^2 + 2)*x + sqrt(3*x^2 + 2))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 196 vs. \(2 (78) = 156\).

Time = 0.28 (sec) , antiderivative size = 196, normalized size of antiderivative = 2.02 \[ \int \frac {1+3 x+4 x^2}{(1+2 x)^3 \left (2+3 x^2\right )^{3/2}} \, dx=\frac {322}{14641} \, \sqrt {11} \log \left (-\frac {{\left | -2 \, \sqrt {3} x - \sqrt {11} - \sqrt {3} + 2 \, \sqrt {3 \, x^{2} + 2} \right |}}{2 \, \sqrt {3} x - \sqrt {11} + \sqrt {3} - 2 \, \sqrt {3 \, x^{2} + 2}}\right ) + \frac {351 \, x + 358}{2662 \, \sqrt {3 \, x^{2} + 2}} + \frac {36 \, {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 2}\right )}^{3} - \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 2}\right )}^{2} + 48 \, \sqrt {3} x + 8 \, \sqrt {3} - 48 \, \sqrt {3 \, x^{2} + 2}}{1331 \, {\left ({\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 2}\right )}^{2} + \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 2}\right )} - 2\right )}^{2}} \] Input:

integrate((4*x^2+3*x+1)/(1+2*x)^3/(3*x^2+2)^(3/2),x, algorithm="giac")
 

Output:

322/14641*sqrt(11)*log(-abs(-2*sqrt(3)*x - sqrt(11) - sqrt(3) + 2*sqrt(3*x 
^2 + 2))/(2*sqrt(3)*x - sqrt(11) + sqrt(3) - 2*sqrt(3*x^2 + 2))) + 1/2662* 
(351*x + 358)/sqrt(3*x^2 + 2) + 1/1331*(36*(sqrt(3)*x - sqrt(3*x^2 + 2))^3 
 - sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 + 2))^2 + 48*sqrt(3)*x + 8*sqrt(3) - 48 
*sqrt(3*x^2 + 2))/((sqrt(3)*x - sqrt(3*x^2 + 2))^2 + sqrt(3)*(sqrt(3)*x - 
sqrt(3*x^2 + 2)) - 2)^2
 

Mupad [B] (verification not implemented)

Time = 16.69 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.86 \[ \int \frac {1+3 x+4 x^2}{(1+2 x)^3 \left (2+3 x^2\right )^{3/2}} \, dx=\frac {322\,\sqrt {11}\,\ln \left (x+\frac {1}{2}\right )}{14641}-\frac {322\,\sqrt {11}\,\ln \left (x-\frac {\sqrt {3}\,\sqrt {11}\,\sqrt {x^2+\frac {2}{3}}}{3}-\frac {4}{3}\right )}{14641}+\frac {117\,\sqrt {3}\,\sqrt {x^2+\frac {2}{3}}}{5324\,\left (x-\frac {\sqrt {6}\,1{}\mathrm {i}}{3}\right )}+\frac {117\,\sqrt {3}\,\sqrt {x^2+\frac {2}{3}}}{5324\,\left (x+\frac {\sqrt {6}\,1{}\mathrm {i}}{3}\right )}-\frac {\sqrt {3}\,\sqrt {x^2+\frac {2}{3}}}{242\,\left (x^2+x+\frac {1}{4}\right )}+\frac {\sqrt {3}\,\sqrt {x^2+\frac {2}{3}}}{1331\,\left (x+\frac {1}{2}\right )}-\frac {\sqrt {3}\,\sqrt {6}\,\sqrt {x^2+\frac {2}{3}}\,179{}\mathrm {i}}{15972\,\left (x-\frac {\sqrt {6}\,1{}\mathrm {i}}{3}\right )}+\frac {\sqrt {3}\,\sqrt {6}\,\sqrt {x^2+\frac {2}{3}}\,179{}\mathrm {i}}{15972\,\left (x+\frac {\sqrt {6}\,1{}\mathrm {i}}{3}\right )} \] Input:

int((3*x + 4*x^2 + 1)/((2*x + 1)^3*(3*x^2 + 2)^(3/2)),x)
 

Output:

(322*11^(1/2)*log(x + 1/2))/14641 - (322*11^(1/2)*log(x - (3^(1/2)*11^(1/2 
)*(x^2 + 2/3)^(1/2))/3 - 4/3))/14641 + (117*3^(1/2)*(x^2 + 2/3)^(1/2))/(53 
24*(x - (6^(1/2)*1i)/3)) + (117*3^(1/2)*(x^2 + 2/3)^(1/2))/(5324*(x + (6^( 
1/2)*1i)/3)) - (3^(1/2)*(x^2 + 2/3)^(1/2))/(242*(x + x^2 + 1/4)) + (3^(1/2 
)*(x^2 + 2/3)^(1/2))/(1331*(x + 1/2)) - (3^(1/2)*6^(1/2)*(x^2 + 2/3)^(1/2) 
*179i)/(15972*(x - (6^(1/2)*1i)/3)) + (3^(1/2)*6^(1/2)*(x^2 + 2/3)^(1/2)*1 
79i)/(15972*(x + (6^(1/2)*1i)/3))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 246, normalized size of antiderivative = 2.54 \[ \int \frac {1+3 x+4 x^2}{(1+2 x)^3 \left (2+3 x^2\right )^{3/2}} \, dx=\frac {15708 \sqrt {3 x^{2}+2}\, x^{3}+29876 \sqrt {3 x^{2}+2}\, x^{2}+19789 \sqrt {3 x^{2}+2}\, x +3058 \sqrt {3 x^{2}+2}+7728 \sqrt {11}\, \mathrm {log}\left (\sqrt {3 x^{2}+2}\, \sqrt {11}+3 x -4\right ) x^{4}+7728 \sqrt {11}\, \mathrm {log}\left (\sqrt {3 x^{2}+2}\, \sqrt {11}+3 x -4\right ) x^{3}+7084 \sqrt {11}\, \mathrm {log}\left (\sqrt {3 x^{2}+2}\, \sqrt {11}+3 x -4\right ) x^{2}+5152 \sqrt {11}\, \mathrm {log}\left (\sqrt {3 x^{2}+2}\, \sqrt {11}+3 x -4\right ) x +1288 \sqrt {11}\, \mathrm {log}\left (\sqrt {3 x^{2}+2}\, \sqrt {11}+3 x -4\right )-7728 \sqrt {11}\, \mathrm {log}\left (2 x +1\right ) x^{4}-7728 \sqrt {11}\, \mathrm {log}\left (2 x +1\right ) x^{3}-7084 \sqrt {11}\, \mathrm {log}\left (2 x +1\right ) x^{2}-5152 \sqrt {11}\, \mathrm {log}\left (2 x +1\right ) x -1288 \sqrt {11}\, \mathrm {log}\left (2 x +1\right )}{351384 x^{4}+351384 x^{3}+322102 x^{2}+234256 x +58564} \] Input:

int((4*x^2+3*x+1)/(1+2*x)^3/(3*x^2+2)^(3/2),x)
 

Output:

(15708*sqrt(3*x**2 + 2)*x**3 + 29876*sqrt(3*x**2 + 2)*x**2 + 19789*sqrt(3* 
x**2 + 2)*x + 3058*sqrt(3*x**2 + 2) + 7728*sqrt(11)*log(sqrt(3*x**2 + 2)*s 
qrt(11) + 3*x - 4)*x**4 + 7728*sqrt(11)*log(sqrt(3*x**2 + 2)*sqrt(11) + 3* 
x - 4)*x**3 + 7084*sqrt(11)*log(sqrt(3*x**2 + 2)*sqrt(11) + 3*x - 4)*x**2 
+ 5152*sqrt(11)*log(sqrt(3*x**2 + 2)*sqrt(11) + 3*x - 4)*x + 1288*sqrt(11) 
*log(sqrt(3*x**2 + 2)*sqrt(11) + 3*x - 4) - 7728*sqrt(11)*log(2*x + 1)*x** 
4 - 7728*sqrt(11)*log(2*x + 1)*x**3 - 7084*sqrt(11)*log(2*x + 1)*x**2 - 51 
52*sqrt(11)*log(2*x + 1)*x - 1288*sqrt(11)*log(2*x + 1))/(29282*(12*x**4 + 
 12*x**3 + 11*x**2 + 8*x + 2))