Integrand size = 34, antiderivative size = 140 \[ \int \frac {g+h x}{\left (a+b x+c x^2\right ) \left (a d+b d x+c d x^2\right )^2} \, dx=-\frac {b g-2 a h+(2 c g-b h) x}{2 \left (b^2-4 a c\right ) d^2 \left (a+b x+c x^2\right )^2}+\frac {3 (2 c g-b h) (b+2 c x)}{2 \left (b^2-4 a c\right )^2 d^2 \left (a+b x+c x^2\right )}-\frac {6 c (2 c g-b h) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2} d^2} \]
1/2*(-b*g+2*a*h-(-b*h+2*c*g)*x)/(-4*a*c+b^2)/d^2/(c*x^2+b*x+a)^2+3/2*(-b*h +2*c*g)*(2*c*x+b)/(-4*a*c+b^2)^2/d^2/(c*x^2+b*x+a)-6*c*(-b*h+2*c*g)*arctan h((2*c*x+b)/(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(5/2)/d^2
Time = 0.09 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.94 \[ \int \frac {g+h x}{\left (a+b x+c x^2\right ) \left (a d+b d x+c d x^2\right )^2} \, dx=\frac {\frac {\left (b^2-4 a c\right ) (-b g+2 a h-2 c g x+b h x)}{(a+x (b+c x))^2}+\frac {3 (2 c g-b h) (b+2 c x)}{a+x (b+c x)}-\frac {12 c (-2 c g+b h) \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}}{2 \left (b^2-4 a c\right )^2 d^2} \]
(((b^2 - 4*a*c)*(-(b*g) + 2*a*h - 2*c*g*x + b*h*x))/(a + x*(b + c*x))^2 + (3*(2*c*g - b*h)*(b + 2*c*x))/(a + x*(b + c*x)) - (12*c*(-2*c*g + b*h)*Arc Tan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c])/(2*(b^2 - 4*a*c)^ 2*d^2)
Time = 0.31 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {1329, 1159, 1086, 1083, 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.
| \(\displaystyle \int \frac {g+h x}{\left (a+b x+c x^2\right ) \left (a d+b d x+c d x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 1329 |
| \(\displaystyle \frac {\int \frac {g+h x}{\left (c x^2+b x+a\right )^3}dx}{d^2}\) |
| \(\Big \downarrow \) 1159 |
| \(\displaystyle \frac {-\frac {3 (2 c g-b h) \int \frac {1}{\left (c x^2+b x+a\right )^2}dx}{2 \left (b^2-4 a c\right )}-\frac {-2 a h+x (2 c g-b h)+b g}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}}{d^2}\) |
| \(\Big \downarrow \) 1086 |
| \(\displaystyle \frac {-\frac {3 (2 c g-b h) \left (-\frac {2 c \int \frac {1}{c x^2+b x+a}dx}{b^2-4 a c}-\frac {b+2 c x}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\right )}{2 \left (b^2-4 a c\right )}-\frac {-2 a h+x (2 c g-b h)+b g}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}}{d^2}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {-\frac {3 (2 c g-b h) \left (\frac {4 c \int \frac {1}{b^2-(b+2 c x)^2-4 a c}d(b+2 c x)}{b^2-4 a c}-\frac {b+2 c x}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\right )}{2 \left (b^2-4 a c\right )}-\frac {-2 a h+x (2 c g-b h)+b g}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}}{d^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {-\frac {3 (2 c g-b h) \left (\frac {4 c \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}-\frac {b+2 c x}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\right )}{2 \left (b^2-4 a c\right )}-\frac {-2 a h+x (2 c g-b h)+b g}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}}{d^2}\) |
(-1/2*(b*g - 2*a*h + (2*c*g - b*h)*x)/((b^2 - 4*a*c)*(a + b*x + c*x^2)^2) - (3*(2*c*g - b*h)*(-((b + 2*c*x)/((b^2 - 4*a*c)*(a + b*x + c*x^2))) + (4* c*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(b^2 - 4*a*c)^(3/2)))/(2*(b^2 - 4*a*c)))/d^2
3.1.17.3.1 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)^(-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]
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] - Simp[2*c*((2*p + 3)/((p + 1)*(b^2 - 4*a*c))) Int[(a + b*x + c*x^2)^(p + 1), x], x] /; Fre eQ[{a, b, c}, x] && ILtQ[p, -1]
Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[((b*d - 2*a*e + (2*c*d - b*e)*x)/((p + 1)*(b^2 - 4*a*c)))*(a + b* x + c*x^2)^(p + 1), x] - Simp[(2*p + 3)*((2*c*d - b*e)/((p + 1)*(b^2 - 4*a* c))) Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] & & LtQ[p, -1] && NeQ[p, -3/2]
Int[((g_.) + (h_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_ ) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp[(c/f)^p Int[(g + h *x)^m*(d + e*x + f*x^2)^(p + q), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, p , q}, x] && EqQ[c*d - a*f, 0] && EqQ[b*d - a*e, 0] && (IntegerQ[p] || GtQ[c /f, 0]) && ( !IntegerQ[q] || LeafCount[d + e*x + f*x^2] <= LeafCount[a + b* x + c*x^2])
Time = 0.90 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.01
| method | result | size |
| default | \(\frac {\frac {b g -2 a h +\left (-b h +2 c g \right ) x}{2 \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{2}}+\frac {3 \left (-b h +2 c g \right ) \left (\frac {2 c x +b}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )}+\frac {4 c \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {3}{2}}}\right )}{2 \left (4 a c -b^{2}\right )}}{d^{2}}\) | \(141\) |
| risch | \(\frac {-\frac {3 c^{2} \left (b h -2 c g \right ) x^{3}}{16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}}-\frac {9 b c \left (b h -2 c g \right ) x^{2}}{2 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}-\frac {\left (5 a b c h -10 a \,c^{2} g +b^{3} h -2 b^{2} c g \right ) x}{16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}}-\frac {8 a^{2} c h +a \,b^{2} h -10 a b c g +b^{3} g}{2 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}}{\left (c \,x^{2}+b x +a \right )^{2} d^{2}}-\frac {3 c \ln \left (\left (32 a^{2} c^{3}-16 a \,b^{2} c^{2}+2 b^{4} c \right ) x -\left (-4 a c +b^{2}\right )^{\frac {5}{2}}+16 a^{2} b \,c^{2}-8 a \,b^{3} c +b^{5}\right ) b h}{d^{2} \left (-4 a c +b^{2}\right )^{\frac {5}{2}}}+\frac {6 c^{2} \ln \left (\left (32 a^{2} c^{3}-16 a \,b^{2} c^{2}+2 b^{4} c \right ) x -\left (-4 a c +b^{2}\right )^{\frac {5}{2}}+16 a^{2} b \,c^{2}-8 a \,b^{3} c +b^{5}\right ) g}{d^{2} \left (-4 a c +b^{2}\right )^{\frac {5}{2}}}+\frac {3 c \ln \left (\left (-32 a^{2} c^{3}+16 a \,b^{2} c^{2}-2 b^{4} c \right ) x -\left (-4 a c +b^{2}\right )^{\frac {5}{2}}-16 a^{2} b \,c^{2}+8 a \,b^{3} c -b^{5}\right ) b h}{d^{2} \left (-4 a c +b^{2}\right )^{\frac {5}{2}}}-\frac {6 c^{2} \ln \left (\left (-32 a^{2} c^{3}+16 a \,b^{2} c^{2}-2 b^{4} c \right ) x -\left (-4 a c +b^{2}\right )^{\frac {5}{2}}-16 a^{2} b \,c^{2}+8 a \,b^{3} c -b^{5}\right ) g}{d^{2} \left (-4 a c +b^{2}\right )^{\frac {5}{2}}}\) | \(504\) |
1/d^2*(1/2*(b*g-2*a*h+(-b*h+2*c*g)*x)/(4*a*c-b^2)/(c*x^2+b*x+a)^2+3/2*(-b* h+2*c*g)/(4*a*c-b^2)*((2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)+4*c/(4*a*c-b^2)^ (3/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 565 vs. \(2 (132) = 264\).
Time = 0.35 (sec) , antiderivative size = 1150, normalized size of antiderivative = 8.21 \[ \int \frac {g+h x}{\left (a+b x+c x^2\right ) \left (a d+b d x+c d x^2\right )^2} \, dx=\text {Too large to display} \]
[1/2*(6*(2*(b^2*c^3 - 4*a*c^4)*g - (b^3*c^2 - 4*a*b*c^3)*h)*x^3 + 9*(2*(b^ 3*c^2 - 4*a*b*c^3)*g - (b^4*c - 4*a*b^2*c^2)*h)*x^2 - 6*(2*a^2*c^2*g - a^2 *b*c*h + (2*c^4*g - b*c^3*h)*x^4 + 2*(2*b*c^3*g - b^2*c^2*h)*x^3 + (2*(b^2 *c^2 + 2*a*c^3)*g - (b^3*c + 2*a*b*c^2)*h)*x^2 + 2*(2*a*b*c^2*g - a*b^2*c* h)*x)*sqrt(b^2 - 4*a*c)*log((2*c^2*x^2 + 2*b*c*x + b^2 - 2*a*c + sqrt(b^2 - 4*a*c)*(2*c*x + b))/(c*x^2 + b*x + a)) - (b^5 - 14*a*b^3*c + 40*a^2*b*c^ 2)*g - (a*b^4 + 4*a^2*b^2*c - 32*a^3*c^2)*h + 2*(2*(b^4*c + a*b^2*c^2 - 20 *a^2*c^3)*g - (b^5 + a*b^3*c - 20*a^2*b*c^2)*h)*x)/((b^6*c^2 - 12*a*b^4*c^ 3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)*d^2*x^4 + 2*(b^7*c - 12*a*b^5*c^2 + 48*a^ 2*b^3*c^3 - 64*a^3*b*c^4)*d^2*x^3 + (b^8 - 10*a*b^6*c + 24*a^2*b^4*c^2 + 3 2*a^3*b^2*c^3 - 128*a^4*c^4)*d^2*x^2 + 2*(a*b^7 - 12*a^2*b^5*c + 48*a^3*b^ 3*c^2 - 64*a^4*b*c^3)*d^2*x + (a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 6 4*a^5*c^3)*d^2), 1/2*(6*(2*(b^2*c^3 - 4*a*c^4)*g - (b^3*c^2 - 4*a*b*c^3)*h )*x^3 + 9*(2*(b^3*c^2 - 4*a*b*c^3)*g - (b^4*c - 4*a*b^2*c^2)*h)*x^2 - 12*( 2*a^2*c^2*g - a^2*b*c*h + (2*c^4*g - b*c^3*h)*x^4 + 2*(2*b*c^3*g - b^2*c^2 *h)*x^3 + (2*(b^2*c^2 + 2*a*c^3)*g - (b^3*c + 2*a*b*c^2)*h)*x^2 + 2*(2*a*b *c^2*g - a*b^2*c*h)*x)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c* x + b)/(b^2 - 4*a*c)) - (b^5 - 14*a*b^3*c + 40*a^2*b*c^2)*g - (a*b^4 + 4*a ^2*b^2*c - 32*a^3*c^2)*h + 2*(2*(b^4*c + a*b^2*c^2 - 20*a^2*c^3)*g - (b^5 + a*b^3*c - 20*a^2*b*c^2)*h)*x)/((b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c...
Leaf count of result is larger than twice the leaf count of optimal. 709 vs. \(2 (133) = 266\).
Time = 1.09 (sec) , antiderivative size = 709, normalized size of antiderivative = 5.06 \[ \int \frac {g+h x}{\left (a+b x+c x^2\right ) \left (a d+b d x+c d x^2\right )^2} \, dx=\frac {3 c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b h - 2 c g\right ) \log {\left (x + \frac {- 192 a^{3} c^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b h - 2 c g\right ) + 144 a^{2} b^{2} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b h - 2 c g\right ) - 36 a b^{4} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b h - 2 c g\right ) + 3 b^{6} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b h - 2 c g\right ) + 3 b^{2} c h - 6 b c^{2} g}{6 b c^{2} h - 12 c^{3} g} \right )}}{d^{2}} - \frac {3 c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b h - 2 c g\right ) \log {\left (x + \frac {192 a^{3} c^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b h - 2 c g\right ) - 144 a^{2} b^{2} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b h - 2 c g\right ) + 36 a b^{4} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b h - 2 c g\right ) - 3 b^{6} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b h - 2 c g\right ) + 3 b^{2} c h - 6 b c^{2} g}{6 b c^{2} h - 12 c^{3} g} \right )}}{d^{2}} + \frac {- 8 a^{2} c h - a b^{2} h + 10 a b c g - b^{3} g + x^{3} \left (- 6 b c^{2} h + 12 c^{3} g\right ) + x^{2} \left (- 9 b^{2} c h + 18 b c^{2} g\right ) + x \left (- 10 a b c h + 20 a c^{2} g - 2 b^{3} h + 4 b^{2} c g\right )}{32 a^{4} c^{2} d^{2} - 16 a^{3} b^{2} c d^{2} + 2 a^{2} b^{4} d^{2} + x^{4} \cdot \left (32 a^{2} c^{4} d^{2} - 16 a b^{2} c^{3} d^{2} + 2 b^{4} c^{2} d^{2}\right ) + x^{3} \cdot \left (64 a^{2} b c^{3} d^{2} - 32 a b^{3} c^{2} d^{2} + 4 b^{5} c d^{2}\right ) + x^{2} \cdot \left (64 a^{3} c^{3} d^{2} - 12 a b^{4} c d^{2} + 2 b^{6} d^{2}\right ) + x \left (64 a^{3} b c^{2} d^{2} - 32 a^{2} b^{3} c d^{2} + 4 a b^{5} d^{2}\right )} \]
3*c*sqrt(-1/(4*a*c - b**2)**5)*(b*h - 2*c*g)*log(x + (-192*a**3*c**4*sqrt( -1/(4*a*c - b**2)**5)*(b*h - 2*c*g) + 144*a**2*b**2*c**3*sqrt(-1/(4*a*c - b**2)**5)*(b*h - 2*c*g) - 36*a*b**4*c**2*sqrt(-1/(4*a*c - b**2)**5)*(b*h - 2*c*g) + 3*b**6*c*sqrt(-1/(4*a*c - b**2)**5)*(b*h - 2*c*g) + 3*b**2*c*h - 6*b*c**2*g)/(6*b*c**2*h - 12*c**3*g))/d**2 - 3*c*sqrt(-1/(4*a*c - b**2)** 5)*(b*h - 2*c*g)*log(x + (192*a**3*c**4*sqrt(-1/(4*a*c - b**2)**5)*(b*h - 2*c*g) - 144*a**2*b**2*c**3*sqrt(-1/(4*a*c - b**2)**5)*(b*h - 2*c*g) + 36* a*b**4*c**2*sqrt(-1/(4*a*c - b**2)**5)*(b*h - 2*c*g) - 3*b**6*c*sqrt(-1/(4 *a*c - b**2)**5)*(b*h - 2*c*g) + 3*b**2*c*h - 6*b*c**2*g)/(6*b*c**2*h - 12 *c**3*g))/d**2 + (-8*a**2*c*h - a*b**2*h + 10*a*b*c*g - b**3*g + x**3*(-6* b*c**2*h + 12*c**3*g) + x**2*(-9*b**2*c*h + 18*b*c**2*g) + x*(-10*a*b*c*h + 20*a*c**2*g - 2*b**3*h + 4*b**2*c*g))/(32*a**4*c**2*d**2 - 16*a**3*b**2* c*d**2 + 2*a**2*b**4*d**2 + x**4*(32*a**2*c**4*d**2 - 16*a*b**2*c**3*d**2 + 2*b**4*c**2*d**2) + x**3*(64*a**2*b*c**3*d**2 - 32*a*b**3*c**2*d**2 + 4* b**5*c*d**2) + x**2*(64*a**3*c**3*d**2 - 12*a*b**4*c*d**2 + 2*b**6*d**2) + x*(64*a**3*b*c**2*d**2 - 32*a**2*b**3*c*d**2 + 4*a*b**5*d**2))
Exception generated. \[ \int \frac {g+h x}{\left (a+b x+c x^2\right ) \left (a d+b d x+c d x^2\right )^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for more deta
Time = 0.28 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.56 \[ \int \frac {g+h x}{\left (a+b x+c x^2\right ) \left (a d+b d x+c d x^2\right )^2} \, dx=\frac {6 \, {\left (2 \, c^{2} g - b c h\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{4} d^{2} - 8 \, a b^{2} c d^{2} + 16 \, a^{2} c^{2} d^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {12 \, c^{3} g x^{3} - 6 \, b c^{2} h x^{3} + 18 \, b c^{2} g x^{2} - 9 \, b^{2} c h x^{2} + 4 \, b^{2} c g x + 20 \, a c^{2} g x - 2 \, b^{3} h x - 10 \, a b c h x - b^{3} g + 10 \, a b c g - a b^{2} h - 8 \, a^{2} c h}{2 \, {\left (b^{4} d^{2} - 8 \, a b^{2} c d^{2} + 16 \, a^{2} c^{2} d^{2}\right )} {\left (c x^{2} + b x + a\right )}^{2}} \]
6*(2*c^2*g - b*c*h)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((b^4*d^2 - 8*a *b^2*c*d^2 + 16*a^2*c^2*d^2)*sqrt(-b^2 + 4*a*c)) + 1/2*(12*c^3*g*x^3 - 6*b *c^2*h*x^3 + 18*b*c^2*g*x^2 - 9*b^2*c*h*x^2 + 4*b^2*c*g*x + 20*a*c^2*g*x - 2*b^3*h*x - 10*a*b*c*h*x - b^3*g + 10*a*b*c*g - a*b^2*h - 8*a^2*c*h)/((b^ 4*d^2 - 8*a*b^2*c*d^2 + 16*a^2*c^2*d^2)*(c*x^2 + b*x + a)^2)
Time = 0.42 (sec) , antiderivative size = 395, normalized size of antiderivative = 2.82 \[ \int \frac {g+h x}{\left (a+b x+c x^2\right ) \left (a d+b d x+c d x^2\right )^2} \, dx=\frac {6\,c\,\mathrm {atan}\left (\frac {d^2\,\left (\frac {6\,c^2\,x\,\left (b\,h-2\,c\,g\right )}{d^2\,{\left (4\,a\,c-b^2\right )}^{5/2}}+\frac {3\,c\,\left (b\,h-2\,c\,g\right )\,\left (16\,a^2\,b\,c^2\,d^2-8\,a\,b^3\,c\,d^2+b^5\,d^2\right )}{d^4\,{\left (4\,a\,c-b^2\right )}^{5/2}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}\right )\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}{6\,c^2\,g-3\,b\,c\,h}\right )\,\left (b\,h-2\,c\,g\right )}{d^2\,{\left (4\,a\,c-b^2\right )}^{5/2}}-\frac {\frac {8\,c\,h\,a^2+h\,a\,b^2-10\,c\,g\,a\,b+g\,b^3}{2\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {x\,\left (b^2+5\,a\,c\right )\,\left (b\,h-2\,c\,g\right )}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}+\frac {3\,c^2\,x^3\,\left (b\,h-2\,c\,g\right )}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}+\frac {9\,b\,c\,x^2\,\left (b\,h-2\,c\,g\right )}{2\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}}{x^2\,\left (b^2\,d^2+2\,a\,c\,d^2\right )+a^2\,d^2+c^2\,d^2\,x^4+2\,a\,b\,d^2\,x+2\,b\,c\,d^2\,x^3} \]
(6*c*atan((d^2*((6*c^2*x*(b*h - 2*c*g))/(d^2*(4*a*c - b^2)^(5/2)) + (3*c*( b*h - 2*c*g)*(b^5*d^2 + 16*a^2*b*c^2*d^2 - 8*a*b^3*c*d^2))/(d^4*(4*a*c - b ^2)^(5/2)*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)))*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) /(6*c^2*g - 3*b*c*h))*(b*h - 2*c*g))/(d^2*(4*a*c - b^2)^(5/2)) - ((b^3*g + a*b^2*h + 8*a^2*c*h - 10*a*b*c*g)/(2*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (x *(5*a*c + b^2)*(b*h - 2*c*g))/(b^4 + 16*a^2*c^2 - 8*a*b^2*c) + (3*c^2*x^3* (b*h - 2*c*g))/(b^4 + 16*a^2*c^2 - 8*a*b^2*c) + (9*b*c*x^2*(b*h - 2*c*g))/ (2*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)))/(x^2*(b^2*d^2 + 2*a*c*d^2) + a^2*d^2 + c^2*d^2*x^4 + 2*a*b*d^2*x + 2*b*c*d^2*x^3)