Integrand size = 20, antiderivative size = 133 \[ \int \frac {A+B x}{\left (a+b x+c x^2\right )^{7/2}} \, dx=-\frac {2 (A b-2 a B-(b B-2 A c) x)}{5 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{5/2}}-\frac {16 (b B-2 A c) (b+2 c x)}{15 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^{3/2}}+\frac {128 c (b B-2 A c) (b+2 c x)}{15 \left (b^2-4 a c\right )^3 \sqrt {a+b x+c x^2}} \]
-2/5*(A*b-2*B*a-(-2*A*c+B*b)*x)/(-4*a*c+b^2)/(c*x^2+b*x+a)^(5/2)-16/15*(-2 *A*c+B*b)*(2*c*x+b)/(-4*a*c+b^2)^2/(c*x^2+b*x+a)^(3/2)+128/15*c*(-2*A*c+B* b)*(2*c*x+b)/(-4*a*c+b^2)^3/(c*x^2+b*x+a)^(1/2)
Time = 2.51 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.59 \[ \int \frac {A+B x}{\left (a+b x+c x^2\right )^{7/2}} \, dx=\frac {-2 A (b+2 c x) \left (3 b^4-16 b^3 c x+64 b c^2 x \left (5 a+4 c x^2\right )+8 b^2 c \left (-5 a+14 c x^2\right )+16 c^2 \left (15 a^2+20 a c x^2+8 c^2 x^4\right )\right )+2 B \left (96 a^3 c^2+48 a^2 b c (b+5 c x)+2 a b \left (-b^3+60 b^2 c x+240 b c^2 x^2+160 c^3 x^3\right )+b x \left (-5 b^4+40 b^3 c x+240 b^2 c^2 x^2+320 b c^3 x^3+128 c^4 x^4\right )\right )}{15 \left (b^2-4 a c\right )^3 (a+x (b+c x))^{5/2}} \]
(-2*A*(b + 2*c*x)*(3*b^4 - 16*b^3*c*x + 64*b*c^2*x*(5*a + 4*c*x^2) + 8*b^2 *c*(-5*a + 14*c*x^2) + 16*c^2*(15*a^2 + 20*a*c*x^2 + 8*c^2*x^4)) + 2*B*(96 *a^3*c^2 + 48*a^2*b*c*(b + 5*c*x) + 2*a*b*(-b^3 + 60*b^2*c*x + 240*b*c^2*x ^2 + 160*c^3*x^3) + b*x*(-5*b^4 + 40*b^3*c*x + 240*b^2*c^2*x^2 + 320*b*c^3 *x^3 + 128*c^4*x^4)))/(15*(b^2 - 4*a*c)^3*(a + x*(b + c*x))^(5/2))
Time = 0.26 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.05, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {1159, 1089, 1088}
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.
| \(\displaystyle \int \frac {A+B x}{\left (a+b x+c x^2\right )^{7/2}} \, dx\) |
| \(\Big \downarrow \) 1159 |
| \(\displaystyle \frac {8 (b B-2 A c) \int \frac {1}{\left (c x^2+b x+a\right )^{5/2}}dx}{5 \left (b^2-4 a c\right )}-\frac {2 (-2 a B-x (b B-2 A c)+A b)}{5 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 1089 |
| \(\displaystyle \frac {8 (b B-2 A c) \left (-\frac {8 c \int \frac {1}{\left (c x^2+b x+a\right )^{3/2}}dx}{3 \left (b^2-4 a c\right )}-\frac {2 (b+2 c x)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\right )}{5 \left (b^2-4 a c\right )}-\frac {2 (-2 a B-x (b B-2 A c)+A b)}{5 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 1088 |
| \(\displaystyle \frac {8 \left (\frac {16 c (b+2 c x)}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}-\frac {2 (b+2 c x)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\right ) (b B-2 A c)}{5 \left (b^2-4 a c\right )}-\frac {2 (-2 a B-x (b B-2 A c)+A b)}{5 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{5/2}}\) |
(-2*(A*b - 2*a*B - (b*B - 2*A*c)*x))/(5*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(5 /2)) + (8*(b*B - 2*A*c)*((-2*(b + 2*c*x))/(3*(b^2 - 4*a*c)*(a + b*x + c*x^ 2)^(3/2)) + (16*c*(b + 2*c*x))/(3*(b^2 - 4*a*c)^2*Sqrt[a + b*x + c*x^2]))) /(5*(b^2 - 4*a*c))
3.25.93.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[-2*((b + 2*c*x)/((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2])), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]
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] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[3*p])
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]
Leaf count of result is larger than twice the leaf count of optimal. \(257\) vs. \(2(121)=242\).
Time = 0.47 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.94
| method | result | size |
| default | \(A \left (\frac {\frac {4 c x}{5}+\frac {2 b}{5}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}+\frac {16 c \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )}{5 \left (4 a c -b^{2}\right )}\right )+B \left (-\frac {1}{5 c \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}-\frac {b \left (\frac {\frac {4 c x}{5}+\frac {2 b}{5}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}+\frac {16 c \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )}{5 \left (4 a c -b^{2}\right )}\right )}{2 c}\right )\) | \(258\) |
| trager | \(\frac {\frac {512}{15} A \,c^{5} x^{5}-\frac {256}{15} B b \,c^{4} x^{5}+\frac {256}{3} A b \,c^{4} x^{4}-\frac {128}{3} B \,b^{2} c^{3} x^{4}+\frac {256}{3} A a \,c^{4} x^{3}+64 A \,b^{2} c^{3} x^{3}-\frac {128}{3} B a b \,c^{3} x^{3}-32 B \,b^{3} c^{2} x^{3}+128 A a b \,c^{3} x^{2}+\frac {32}{3} A \,b^{3} c^{2} x^{2}-64 B a \,b^{2} c^{2} x^{2}-\frac {16}{3} B \,b^{4} c \,x^{2}+64 A \,a^{2} c^{3} x +32 A a \,b^{2} c^{2} x -\frac {4}{3} A \,b^{4} c x -32 B \,a^{2} b \,c^{2} x -16 B a \,b^{3} c x +\frac {2}{3} B \,b^{5} x +32 A \,a^{2} b \,c^{2}-\frac {16}{3} A a \,b^{3} c +\frac {2}{5} A \,b^{5}-\frac {64}{5} B \,a^{3} c^{2}-\frac {32}{5} B \,a^{2} b^{2} c +\frac {4}{15} B a \,b^{4}}{\left (4 a c -b^{2}\right )^{3} \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}\) | \(266\) |
| gosper | \(\frac {\frac {512}{15} A \,c^{5} x^{5}-\frac {256}{15} B b \,c^{4} x^{5}+\frac {256}{3} A b \,c^{4} x^{4}-\frac {128}{3} B \,b^{2} c^{3} x^{4}+\frac {256}{3} A a \,c^{4} x^{3}+64 A \,b^{2} c^{3} x^{3}-\frac {128}{3} B a b \,c^{3} x^{3}-32 B \,b^{3} c^{2} x^{3}+128 A a b \,c^{3} x^{2}+\frac {32}{3} A \,b^{3} c^{2} x^{2}-64 B a \,b^{2} c^{2} x^{2}-\frac {16}{3} B \,b^{4} c \,x^{2}+64 A \,a^{2} c^{3} x +32 A a \,b^{2} c^{2} x -\frac {4}{3} A \,b^{4} c x -32 B \,a^{2} b \,c^{2} x -16 B a \,b^{3} c x +\frac {2}{3} B \,b^{5} x +32 A \,a^{2} b \,c^{2}-\frac {16}{3} A a \,b^{3} c +\frac {2}{5} A \,b^{5}-\frac {64}{5} B \,a^{3} c^{2}-\frac {32}{5} B \,a^{2} b^{2} c +\frac {4}{15} B a \,b^{4}}{\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} \left (64 a^{3} c^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right )}\) | \(288\) |
A*(2/5*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(5/2)+16/5*c/(4*a*c-b^2)*(2/3*( 2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)+16/3*c/(4*a*c-b^2)^2*(2*c*x+b)/(c *x^2+b*x+a)^(1/2)))+B*(-1/5/c/(c*x^2+b*x+a)^(5/2)-1/2*b/c*(2/5*(2*c*x+b)/( 4*a*c-b^2)/(c*x^2+b*x+a)^(5/2)+16/5*c/(4*a*c-b^2)*(2/3*(2*c*x+b)/(4*a*c-b^ 2)/(c*x^2+b*x+a)^(3/2)+16/3*c/(4*a*c-b^2)^2*(2*c*x+b)/(c*x^2+b*x+a)^(1/2)) ))
Leaf count of result is larger than twice the leaf count of optimal. 543 vs. \(2 (121) = 242\).
Time = 2.87 (sec) , antiderivative size = 543, normalized size of antiderivative = 4.08 \[ \int \frac {A+B x}{\left (a+b x+c x^2\right )^{7/2}} \, dx=-\frac {2 \, {\left (2 \, B a b^{4} + 3 \, A b^{5} - 128 \, {\left (B b c^{4} - 2 \, A c^{5}\right )} x^{5} - 320 \, {\left (B b^{2} c^{3} - 2 \, A b c^{4}\right )} x^{4} - 80 \, {\left (3 \, B b^{3} c^{2} - 8 \, A a c^{4} + 2 \, {\left (2 \, B a b - 3 \, A b^{2}\right )} c^{3}\right )} x^{3} - 48 \, {\left (2 \, B a^{3} - 5 \, A a^{2} b\right )} c^{2} - 40 \, {\left (B b^{4} c - 24 \, A a b c^{3} + 2 \, {\left (6 \, B a b^{2} - A b^{3}\right )} c^{2}\right )} x^{2} - 8 \, {\left (6 \, B a^{2} b^{2} + 5 \, A a b^{3}\right )} c + 5 \, {\left (B b^{5} + 96 \, A a^{2} c^{3} - 48 \, {\left (B a^{2} b - A a b^{2}\right )} c^{2} - 2 \, {\left (12 \, B a b^{3} + A b^{4}\right )} c\right )} x\right )} \sqrt {c x^{2} + b x + a}}{15 \, {\left (a^{3} b^{6} - 12 \, a^{4} b^{4} c + 48 \, a^{5} b^{2} c^{2} - 64 \, a^{6} c^{3} + {\left (b^{6} c^{3} - 12 \, a b^{4} c^{4} + 48 \, a^{2} b^{2} c^{5} - 64 \, a^{3} c^{6}\right )} x^{6} + 3 \, {\left (b^{7} c^{2} - 12 \, a b^{5} c^{3} + 48 \, a^{2} b^{3} c^{4} - 64 \, a^{3} b c^{5}\right )} x^{5} + 3 \, {\left (b^{8} c - 11 \, a b^{6} c^{2} + 36 \, a^{2} b^{4} c^{3} - 16 \, a^{3} b^{2} c^{4} - 64 \, a^{4} c^{5}\right )} x^{4} + {\left (b^{9} - 6 \, a b^{7} c - 24 \, a^{2} b^{5} c^{2} + 224 \, a^{3} b^{3} c^{3} - 384 \, a^{4} b c^{4}\right )} x^{3} + 3 \, {\left (a b^{8} - 11 \, a^{2} b^{6} c + 36 \, a^{3} b^{4} c^{2} - 16 \, a^{4} b^{2} c^{3} - 64 \, a^{5} c^{4}\right )} x^{2} + 3 \, {\left (a^{2} b^{7} - 12 \, a^{3} b^{5} c + 48 \, a^{4} b^{3} c^{2} - 64 \, a^{5} b c^{3}\right )} x\right )}} \]
-2/15*(2*B*a*b^4 + 3*A*b^5 - 128*(B*b*c^4 - 2*A*c^5)*x^5 - 320*(B*b^2*c^3 - 2*A*b*c^4)*x^4 - 80*(3*B*b^3*c^2 - 8*A*a*c^4 + 2*(2*B*a*b - 3*A*b^2)*c^3 )*x^3 - 48*(2*B*a^3 - 5*A*a^2*b)*c^2 - 40*(B*b^4*c - 24*A*a*b*c^3 + 2*(6*B *a*b^2 - A*b^3)*c^2)*x^2 - 8*(6*B*a^2*b^2 + 5*A*a*b^3)*c + 5*(B*b^5 + 96*A *a^2*c^3 - 48*(B*a^2*b - A*a*b^2)*c^2 - 2*(12*B*a*b^3 + A*b^4)*c)*x)*sqrt( c*x^2 + b*x + a)/(a^3*b^6 - 12*a^4*b^4*c + 48*a^5*b^2*c^2 - 64*a^6*c^3 + ( b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6)*x^6 + 3*(b^7*c^2 - 1 2*a*b^5*c^3 + 48*a^2*b^3*c^4 - 64*a^3*b*c^5)*x^5 + 3*(b^8*c - 11*a*b^6*c^2 + 36*a^2*b^4*c^3 - 16*a^3*b^2*c^4 - 64*a^4*c^5)*x^4 + (b^9 - 6*a*b^7*c - 24*a^2*b^5*c^2 + 224*a^3*b^3*c^3 - 384*a^4*b*c^4)*x^3 + 3*(a*b^8 - 11*a^2* b^6*c + 36*a^3*b^4*c^2 - 16*a^4*b^2*c^3 - 64*a^5*c^4)*x^2 + 3*(a^2*b^7 - 1 2*a^3*b^5*c + 48*a^4*b^3*c^2 - 64*a^5*b*c^3)*x)
Timed out. \[ \int \frac {A+B x}{\left (a+b x+c x^2\right )^{7/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {A+B x}{\left (a+b x+c x^2\right )^{7/2}} \, dx=\text {Exception raised: ValueError} \]
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
Leaf count of result is larger than twice the leaf count of optimal. 433 vs. \(2 (121) = 242\).
Time = 0.29 (sec) , antiderivative size = 433, normalized size of antiderivative = 3.26 \[ \int \frac {A+B x}{\left (a+b x+c x^2\right )^{7/2}} \, dx=\frac {2 \, {\left ({\left (8 \, {\left (2 \, {\left (4 \, {\left (\frac {2 \, {\left (B b c^{4} - 2 \, A c^{5}\right )} x}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}} + \frac {5 \, {\left (B b^{2} c^{3} - 2 \, A b c^{4}\right )}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )} x + \frac {5 \, {\left (3 \, B b^{3} c^{2} + 4 \, B a b c^{3} - 6 \, A b^{2} c^{3} - 8 \, A a c^{4}\right )}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )} x + \frac {5 \, {\left (B b^{4} c + 12 \, B a b^{2} c^{2} - 2 \, A b^{3} c^{2} - 24 \, A a b c^{3}\right )}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )} x - \frac {5 \, {\left (B b^{5} - 24 \, B a b^{3} c - 2 \, A b^{4} c - 48 \, B a^{2} b c^{2} + 48 \, A a b^{2} c^{2} + 96 \, A a^{2} c^{3}\right )}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )} x - \frac {2 \, B a b^{4} + 3 \, A b^{5} - 48 \, B a^{2} b^{2} c - 40 \, A a b^{3} c - 96 \, B a^{3} c^{2} + 240 \, A a^{2} b c^{2}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )}}{15 \, {\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}} \]
2/15*((8*(2*(4*(2*(B*b*c^4 - 2*A*c^5)*x/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3) + 5*(B*b^2*c^3 - 2*A*b*c^4)/(b^6 - 12*a*b^4*c + 48*a^2*b^2* c^2 - 64*a^3*c^3))*x + 5*(3*B*b^3*c^2 + 4*B*a*b*c^3 - 6*A*b^2*c^3 - 8*A*a* c^4)/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3))*x + 5*(B*b^4*c + 12 *B*a*b^2*c^2 - 2*A*b^3*c^2 - 24*A*a*b*c^3)/(b^6 - 12*a*b^4*c + 48*a^2*b^2* c^2 - 64*a^3*c^3))*x - 5*(B*b^5 - 24*B*a*b^3*c - 2*A*b^4*c - 48*B*a^2*b*c^ 2 + 48*A*a*b^2*c^2 + 96*A*a^2*c^3)/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64 *a^3*c^3))*x - (2*B*a*b^4 + 3*A*b^5 - 48*B*a^2*b^2*c - 40*A*a*b^3*c - 96*B *a^3*c^2 + 240*A*a^2*b*c^2)/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^ 3))/(c*x^2 + b*x + a)^(5/2)
Time = 12.01 (sec) , antiderivative size = 394, normalized size of antiderivative = 2.96 \[ \int \frac {A+B x}{\left (a+b x+c x^2\right )^{7/2}} \, dx=\frac {\frac {b\,c\,\left (256\,A\,c^2-128\,B\,b\,c\right )}{15\,\left (4\,a\,c^2-b^2\,c\right )\,{\left (4\,a\,c-b^2\right )}^2}+\frac {2\,c^2\,x\,\left (256\,A\,c^2-128\,B\,b\,c\right )}{15\,\left (4\,a\,c^2-b^2\,c\right )\,{\left (4\,a\,c-b^2\right )}^2}}{\sqrt {c\,x^2+b\,x+a}}+\frac {x\,\left (\frac {4\,A\,c^2}{5\,\left (4\,a\,c^2-b^2\,c\right )}-\frac {2\,B\,b\,c}{5\,\left (4\,a\,c^2-b^2\,c\right )}\right )+\frac {2\,A\,b\,c}{5\,\left (4\,a\,c^2-b^2\,c\right )}-\frac {4\,B\,a\,c}{5\,\left (4\,a\,c^2-b^2\,c\right )}}{{\left (c\,x^2+b\,x+a\right )}^{5/2}}+\frac {x\,\left (\frac {2\,c^2\,\left (32\,A\,c-20\,B\,b\right )}{15\,\left (4\,a\,c^2-b^2\,c\right )\,\left (4\,a\,c-b^2\right )}+\frac {8\,B\,b\,c^2}{15\,\left (4\,a\,c^2-b^2\,c\right )\,\left (4\,a\,c-b^2\right )}\right )+\frac {b\,c\,\left (32\,A\,c-20\,B\,b\right )}{15\,\left (4\,a\,c^2-b^2\,c\right )\,\left (4\,a\,c-b^2\right )}+\frac {16\,B\,a\,c^2}{15\,\left (4\,a\,c^2-b^2\,c\right )\,\left (4\,a\,c-b^2\right )}}{{\left (c\,x^2+b\,x+a\right )}^{3/2}}-\frac {4\,B}{\left (60\,a\,c-15\,b^2\right )\,{\left (c\,x^2+b\,x+a\right )}^{3/2}} \]
((b*c*(256*A*c^2 - 128*B*b*c))/(15*(4*a*c^2 - b^2*c)*(4*a*c - b^2)^2) + (2 *c^2*x*(256*A*c^2 - 128*B*b*c))/(15*(4*a*c^2 - b^2*c)*(4*a*c - b^2)^2))/(a + b*x + c*x^2)^(1/2) + (x*((4*A*c^2)/(5*(4*a*c^2 - b^2*c)) - (2*B*b*c)/(5 *(4*a*c^2 - b^2*c))) + (2*A*b*c)/(5*(4*a*c^2 - b^2*c)) - (4*B*a*c)/(5*(4*a *c^2 - b^2*c)))/(a + b*x + c*x^2)^(5/2) + (x*((2*c^2*(32*A*c - 20*B*b))/(1 5*(4*a*c^2 - b^2*c)*(4*a*c - b^2)) + (8*B*b*c^2)/(15*(4*a*c^2 - b^2*c)*(4* a*c - b^2))) + (b*c*(32*A*c - 20*B*b))/(15*(4*a*c^2 - b^2*c)*(4*a*c - b^2) ) + (16*B*a*c^2)/(15*(4*a*c^2 - b^2*c)*(4*a*c - b^2)))/(a + b*x + c*x^2)^( 3/2) - (4*B)/((60*a*c - 15*b^2)*(a + b*x + c*x^2)^(3/2))