Integrand size = 26, antiderivative size = 125 \[ \int x^3 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {b (5 b B-8 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{128 c^3}-\frac {\left (5 b B-8 A c-6 B c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{48 c^2}-\frac {b^3 (5 b B-8 A c) \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{128 c^{7/2}} \]
-1/48*(-6*B*c*x^2-8*A*c+5*B*b)*(c*x^4+b*x^2)^(3/2)/c^2-1/128*b^3*(-8*A*c+5 *B*b)*arctanh(x^2*c^(1/2)/(c*x^4+b*x^2)^(1/2))/c^(7/2)+1/128*b*(-8*A*c+5*B *b)*(2*c*x^2+b)*(c*x^4+b*x^2)^(1/2)/c^3
Time = 0.89 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.18 \[ \int x^3 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (\sqrt {c} \left (15 b^3 B+8 b c^2 x^2 \left (2 A+B x^2\right )+16 c^3 x^4 \left (4 A+3 B x^2\right )-2 b^2 c \left (12 A+5 B x^2\right )\right )+\frac {6 b^3 (5 b B-8 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b}-\sqrt {b+c x^2}}\right )}{x \sqrt {b+c x^2}}\right )}{384 c^{7/2}} \]
(Sqrt[x^2*(b + c*x^2)]*(Sqrt[c]*(15*b^3*B + 8*b*c^2*x^2*(2*A + B*x^2) + 16 *c^3*x^4*(4*A + 3*B*x^2) - 2*b^2*c*(12*A + 5*B*x^2)) + (6*b^3*(5*b*B - 8*A *c)*ArcTanh[(Sqrt[c]*x)/(Sqrt[b] - Sqrt[b + c*x^2])])/(x*Sqrt[b + c*x^2])) )/(384*c^(7/2))
Time = 0.27 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1940, 1225, 1087, 1091, 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.
| \(\displaystyle \int x^3 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx\) |
| \(\Big \downarrow \) 1940 |
| \(\displaystyle \frac {1}{2} \int x^2 \left (B x^2+A\right ) \sqrt {c x^4+b x^2}dx^2\) |
| \(\Big \downarrow \) 1225 |
| \(\displaystyle \frac {1}{2} \left (\frac {b (5 b B-8 A c) \int \sqrt {c x^4+b x^2}dx^2}{16 c^2}-\frac {\left (b x^2+c x^4\right )^{3/2} \left (-8 A c+5 b B-6 B c x^2\right )}{24 c^2}\right )\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {1}{2} \left (\frac {b (5 b B-8 A c) \left (\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{4 c}-\frac {b^2 \int \frac {1}{\sqrt {c x^4+b x^2}}dx^2}{8 c}\right )}{16 c^2}-\frac {\left (b x^2+c x^4\right )^{3/2} \left (-8 A c+5 b B-6 B c x^2\right )}{24 c^2}\right )\) |
| \(\Big \downarrow \) 1091 |
| \(\displaystyle \frac {1}{2} \left (\frac {b (5 b B-8 A c) \left (\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{4 c}-\frac {b^2 \int \frac {1}{1-c x^4}d\frac {x^2}{\sqrt {c x^4+b x^2}}}{4 c}\right )}{16 c^2}-\frac {\left (b x^2+c x^4\right )^{3/2} \left (-8 A c+5 b B-6 B c x^2\right )}{24 c^2}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{2} \left (\frac {b (5 b B-8 A c) \left (\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{4 c}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{4 c^{3/2}}\right )}{16 c^2}-\frac {\left (b x^2+c x^4\right )^{3/2} \left (-8 A c+5 b B-6 B c x^2\right )}{24 c^2}\right )\) |
(-1/24*((5*b*B - 8*A*c - 6*B*c*x^2)*(b*x^2 + c*x^4)^(3/2))/c^2 + (b*(5*b*B - 8*A*c)*(((b + 2*c*x^2)*Sqrt[b*x^2 + c*x^4])/(4*c) - (b^2*ArcTanh[(Sqrt[ c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(4*c^(3/2))))/(16*c^2))/2
3.1.91.3.1 Defintions of rubi rules used
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])
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* p + 1))) Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*( x_)^2)^(p_), x_Symbol] :> Simp[(-(b*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x))*((a + b*x + c*x^2)^(p + 1)/(2*c^2*(p + 1)*(2*p + 3))), x] + Simp[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c , d, e, f, g, p}, x] && !LeQ[p, -1]
Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_) ^(n_))^(q_.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1) *(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] && !IntegerQ[p] && NeQ[k, j] && I ntegerQ[Simplify[j/n]] && IntegerQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]
Time = 1.81 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.12
| method | result | size |
| risch | \(-\frac {\left (-48 B \,c^{3} x^{6}-64 A \,c^{3} x^{4}-8 B b \,c^{2} x^{4}-16 A b \,c^{2} x^{2}+10 B \,b^{2} c \,x^{2}+24 b^{2} A c -15 B \,b^{3}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{384 c^{3}}+\frac {b^{3} \left (8 A c -5 B b \right ) \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{128 c^{\frac {7}{2}} x \sqrt {c \,x^{2}+b}}\) | \(140\) |
| pseudoelliptic | \(-\frac {\left (-\frac {1}{2} A \,b^{3} c +\frac {5}{16} B \,b^{4}\right ) \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right )+\left (b^{2} \left (\frac {5 x^{2} B}{12}+A \right ) c^{\frac {3}{2}}-\frac {2 x^{2} \left (\frac {x^{2} B}{2}+A \right ) b \,c^{\frac {5}{2}}}{3}+\left (-2 B \,x^{6}-\frac {8}{3} A \,x^{4}\right ) c^{\frac {7}{2}}-\frac {5 B \sqrt {c}\, b^{3}}{8}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}+\frac {\ln \left (2\right ) \left (A c -\frac {5 B b}{8}\right ) b^{3}}{2}}{16 c^{\frac {7}{2}}}\) | \(141\) |
| default | \(\frac {\sqrt {x^{4} c +b \,x^{2}}\, \left (48 B \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {5}{2}} x^{5}+64 A \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {5}{2}} x^{3}-40 B \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {3}{2}} b \,x^{3}-48 A \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {3}{2}} b x +24 A \sqrt {c \,x^{2}+b}\, c^{\frac {3}{2}} b^{2} x +30 B \left (c \,x^{2}+b \right )^{\frac {3}{2}} \sqrt {c}\, b^{2} x -15 B \sqrt {c \,x^{2}+b}\, \sqrt {c}\, b^{3} x +24 A \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{3} c -15 B \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{4}\right )}{384 x \sqrt {c \,x^{2}+b}\, c^{\frac {7}{2}}}\) | \(206\) |
-1/384*(-48*B*c^3*x^6-64*A*c^3*x^4-8*B*b*c^2*x^4-16*A*b*c^2*x^2+10*B*b^2*c *x^2+24*A*b^2*c-15*B*b^3)/c^3*(x^2*(c*x^2+b))^(1/2)+1/128*b^3*(8*A*c-5*B*b )/c^(7/2)*ln(c^(1/2)*x+(c*x^2+b)^(1/2))*(x^2*(c*x^2+b))^(1/2)/x/(c*x^2+b)^ (1/2)
Time = 0.54 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.18 \[ \int x^3 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\left [-\frac {3 \, {\left (5 \, B b^{4} - 8 \, A b^{3} c\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, {\left (48 \, B c^{4} x^{6} + 15 \, B b^{3} c - 24 \, A b^{2} c^{2} + 8 \, {\left (B b c^{3} + 8 \, A c^{4}\right )} x^{4} - 2 \, {\left (5 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{768 \, c^{4}}, \frac {3 \, {\left (5 \, B b^{4} - 8 \, A b^{3} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + {\left (48 \, B c^{4} x^{6} + 15 \, B b^{3} c - 24 \, A b^{2} c^{2} + 8 \, {\left (B b c^{3} + 8 \, A c^{4}\right )} x^{4} - 2 \, {\left (5 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{384 \, c^{4}}\right ] \]
[-1/768*(3*(5*B*b^4 - 8*A*b^3*c)*sqrt(c)*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) - 2*(48*B*c^4*x^6 + 15*B*b^3*c - 24*A*b^2*c^2 + 8*(B*b*c^ 3 + 8*A*c^4)*x^4 - 2*(5*B*b^2*c^2 - 8*A*b*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/c ^4, 1/384*(3*(5*B*b^4 - 8*A*b^3*c)*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sqr t(-c)/(c*x^2 + b)) + (48*B*c^4*x^6 + 15*B*b^3*c - 24*A*b^2*c^2 + 8*(B*b*c^ 3 + 8*A*c^4)*x^4 - 2*(5*B*b^2*c^2 - 8*A*b*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/c ^4]
Time = 0.68 (sec) , antiderivative size = 287, normalized size of antiderivative = 2.30 \[ \int x^3 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {A \left (\begin {cases} \frac {b^{3} \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x^{2} + c x^{4}} + 2 c x^{2} \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x^{2}\right ) \log {\left (\frac {b}{2 c} + x^{2} \right )}}{\sqrt {c \left (\frac {b}{2 c} + x^{2}\right )^{2}}} & \text {otherwise} \end {cases}\right )}{16 c^{2}} + \sqrt {b x^{2} + c x^{4}} \left (- \frac {b^{2}}{8 c^{2}} + \frac {b x^{2}}{12 c} + \frac {x^{4}}{3}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (b x^{2}\right )^{\frac {5}{2}}}{5 b^{2}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases}\right )}{2} + \frac {B \left (\begin {cases} - \frac {5 b^{4} \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x^{2} + c x^{4}} + 2 c x^{2} \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x^{2}\right ) \log {\left (\frac {b}{2 c} + x^{2} \right )}}{\sqrt {c \left (\frac {b}{2 c} + x^{2}\right )^{2}}} & \text {otherwise} \end {cases}\right )}{128 c^{3}} + \sqrt {b x^{2} + c x^{4}} \cdot \left (\frac {5 b^{3}}{64 c^{3}} - \frac {5 b^{2} x^{2}}{96 c^{2}} + \frac {b x^{4}}{24 c} + \frac {x^{6}}{4}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (b x^{2}\right )^{\frac {7}{2}}}{7 b^{3}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases}\right )}{2} \]
A*Piecewise((b**3*Piecewise((log(b + 2*sqrt(c)*sqrt(b*x**2 + c*x**4) + 2*c *x**2)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c) + x**2)*log(b/(2*c) + x**2)/sqrt (c*(b/(2*c) + x**2)**2), True))/(16*c**2) + sqrt(b*x**2 + c*x**4)*(-b**2/( 8*c**2) + b*x**2/(12*c) + x**4/3), Ne(c, 0)), (2*(b*x**2)**(5/2)/(5*b**2), Ne(b, 0)), (0, True))/2 + B*Piecewise((-5*b**4*Piecewise((log(b + 2*sqrt( c)*sqrt(b*x**2 + c*x**4) + 2*c*x**2)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c) + x**2)*log(b/(2*c) + x**2)/sqrt(c*(b/(2*c) + x**2)**2), True))/(128*c**3) + sqrt(b*x**2 + c*x**4)*(5*b**3/(64*c**3) - 5*b**2*x**2/(96*c**2) + b*x**4/ (24*c) + x**6/4), Ne(c, 0)), (2*(b*x**2)**(7/2)/(7*b**3), Ne(b, 0)), (0, T rue))/2
Leaf count of result is larger than twice the leaf count of optimal. 225 vs. \(2 (109) = 218\).
Time = 0.21 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.80 \[ \int x^3 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=-\frac {1}{96} \, {\left (\frac {12 \, \sqrt {c x^{4} + b x^{2}} b x^{2}}{c} - \frac {3 \, b^{3} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {5}{2}}} + \frac {6 \, \sqrt {c x^{4} + b x^{2}} b^{2}}{c^{2}} - \frac {16 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{c}\right )} A + \frac {1}{768} \, {\left (\frac {60 \, \sqrt {c x^{4} + b x^{2}} b^{2} x^{2}}{c^{2}} + \frac {96 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} x^{2}}{c} - \frac {15 \, b^{4} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {7}{2}}} + \frac {30 \, \sqrt {c x^{4} + b x^{2}} b^{3}}{c^{3}} - \frac {80 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b}{c^{2}}\right )} B \]
-1/96*(12*sqrt(c*x^4 + b*x^2)*b*x^2/c - 3*b^3*log(2*c*x^2 + b + 2*sqrt(c*x ^4 + b*x^2)*sqrt(c))/c^(5/2) + 6*sqrt(c*x^4 + b*x^2)*b^2/c^2 - 16*(c*x^4 + b*x^2)^(3/2)/c)*A + 1/768*(60*sqrt(c*x^4 + b*x^2)*b^2*x^2/c^2 + 96*(c*x^4 + b*x^2)^(3/2)*x^2/c - 15*b^4*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqr t(c))/c^(7/2) + 30*sqrt(c*x^4 + b*x^2)*b^3/c^3 - 80*(c*x^4 + b*x^2)^(3/2)* b/c^2)*B
Time = 0.30 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.42 \[ \int x^3 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {1}{384} \, {\left (2 \, {\left (4 \, {\left (6 \, B x^{2} \mathrm {sgn}\left (x\right ) + \frac {B b c^{5} \mathrm {sgn}\left (x\right ) + 8 \, A c^{6} \mathrm {sgn}\left (x\right )}{c^{6}}\right )} x^{2} - \frac {5 \, B b^{2} c^{4} \mathrm {sgn}\left (x\right ) - 8 \, A b c^{5} \mathrm {sgn}\left (x\right )}{c^{6}}\right )} x^{2} + \frac {3 \, {\left (5 \, B b^{3} c^{3} \mathrm {sgn}\left (x\right ) - 8 \, A b^{2} c^{4} \mathrm {sgn}\left (x\right )\right )}}{c^{6}}\right )} \sqrt {c x^{2} + b} x + \frac {{\left (5 \, B b^{4} \mathrm {sgn}\left (x\right ) - 8 \, A b^{3} c \mathrm {sgn}\left (x\right )\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{128 \, c^{\frac {7}{2}}} - \frac {{\left (5 \, B b^{4} \log \left ({\left | b \right |}\right ) - 8 \, A b^{3} c \log \left ({\left | b \right |}\right )\right )} \mathrm {sgn}\left (x\right )}{256 \, c^{\frac {7}{2}}} \]
1/384*(2*(4*(6*B*x^2*sgn(x) + (B*b*c^5*sgn(x) + 8*A*c^6*sgn(x))/c^6)*x^2 - (5*B*b^2*c^4*sgn(x) - 8*A*b*c^5*sgn(x))/c^6)*x^2 + 3*(5*B*b^3*c^3*sgn(x) - 8*A*b^2*c^4*sgn(x))/c^6)*sqrt(c*x^2 + b)*x + 1/128*(5*B*b^4*sgn(x) - 8*A *b^3*c*sgn(x))*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))/c^(7/2) - 1/256*(5*B *b^4*log(abs(b)) - 8*A*b^3*c*log(abs(b)))*sgn(x)/c^(7/2)
Time = 9.62 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.42 \[ \int x^3 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx=\frac {B\,x^2\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{8\,c}-\frac {5\,B\,b\,\left (\frac {b^3\,\ln \left (b+2\,c\,x^2+2\,\sqrt {c}\,\left |x\right |\,\sqrt {c\,x^2+b}\right )}{16\,c^{5/2}}+\frac {\sqrt {c\,x^4+b\,x^2}\,\left (-3\,b^2+2\,b\,c\,x^2+8\,c^2\,x^4\right )}{24\,c^2}\right )}{16\,c}+\frac {A\,b^3\,\ln \left (b+2\,c\,x^2+2\,\sqrt {c}\,\left |x\right |\,\sqrt {c\,x^2+b}\right )}{32\,c^{5/2}}+\frac {A\,\sqrt {c\,x^4+b\,x^2}\,\left (-3\,b^2+2\,b\,c\,x^2+8\,c^2\,x^4\right )}{48\,c^2} \]
(B*x^2*(b*x^2 + c*x^4)^(3/2))/(8*c) - (5*B*b*((b^3*log(b + 2*c*x^2 + 2*c^( 1/2)*abs(x)*(b + c*x^2)^(1/2)))/(16*c^(5/2)) + ((b*x^2 + c*x^4)^(1/2)*(8*c ^2*x^4 - 3*b^2 + 2*b*c*x^2))/(24*c^2)))/(16*c) + (A*b^3*log(b + 2*c*x^2 + 2*c^(1/2)*abs(x)*(b + c*x^2)^(1/2)))/(32*c^(5/2)) + (A*(b*x^2 + c*x^4)^(1/ 2)*(8*c^2*x^4 - 3*b^2 + 2*b*c*x^2))/(48*c^2)