Integrand size = 30, antiderivative size = 212 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{\sqrt {c+d x}} \, dx=-\frac {2 (b c-a d) \left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \sqrt {c+d x}}{d^5}-\frac {2 \left (a d \left (2 c C d-B d^2-3 c^2 D\right )-b \left (3 c^2 C d-2 B c d^2+A d^3-4 c^3 D\right )\right ) (c+d x)^{3/2}}{3 d^5}+\frac {2 \left (a d (C d-3 c D)-b \left (3 c C d-B d^2-6 c^2 D\right )\right ) (c+d x)^{5/2}}{5 d^5}+\frac {2 (b C d-4 b c D+a d D) (c+d x)^{7/2}}{7 d^5}+\frac {2 b D (c+d x)^{9/2}}{9 d^5} \]
-2/3*(a*d*(-B*d^2+2*C*c*d-3*D*c^2)-b*(A*d^3-2*B*c*d^2+3*C*c^2*d-4*D*c^3))* (d*x+c)^(3/2)/d^5+2/5*(a*d*(C*d-3*D*c)-b*(-B*d^2+3*C*c*d-6*D*c^2))*(d*x+c) ^(5/2)/d^5+2/7*(C*b*d+D*a*d-4*D*b*c)*(d*x+c)^(7/2)/d^5+2/9*b*D*(d*x+c)^(9/ 2)/d^5-2*(-a*d+b*c)*(A*d^3-B*c*d^2+C*c^2*d-D*c^3)*(d*x+c)^(1/2)/d^5
Time = 0.13 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.87 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{\sqrt {c+d x}} \, dx=\frac {2 \sqrt {c+d x} \left (3 a d \left (-48 c^3 D+8 c^2 d (7 C+3 D x)-2 c d^2 (35 B+x (14 C+9 D x))+d^3 (105 A+x (35 B+3 x (7 C+5 D x)))\right )+b \left (128 c^4 D-16 c^3 d (9 C+4 D x)+24 c^2 d^2 (7 B+x (3 C+2 D x))+d^4 x (105 A+x (63 B+5 x (9 C+7 D x)))-2 c d^3 (105 A+x (42 B+x (27 C+20 D x)))\right )\right )}{315 d^5} \]
(2*Sqrt[c + d*x]*(3*a*d*(-48*c^3*D + 8*c^2*d*(7*C + 3*D*x) - 2*c*d^2*(35*B + x*(14*C + 9*D*x)) + d^3*(105*A + x*(35*B + 3*x*(7*C + 5*D*x)))) + b*(12 8*c^4*D - 16*c^3*d*(9*C + 4*D*x) + 24*c^2*d^2*(7*B + x*(3*C + 2*D*x)) + d^ 4*x*(105*A + x*(63*B + 5*x*(9*C + 7*D*x))) - 2*c*d^3*(105*A + x*(42*B + x* (27*C + 20*D*x))))))/(315*d^5)
Time = 0.40 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2123, 2009}
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.
| \(\displaystyle \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{\sqrt {c+d x}} \, dx\) |
| \(\Big \downarrow \) 2123 |
| \(\displaystyle \int \left (\frac {\sqrt {c+d x} \left (b \left (A d^3-2 B c d^2-4 c^3 D+3 c^2 C d\right )-a d \left (-B d^2-3 c^2 D+2 c C d\right )\right )}{d^4}+\frac {(a d-b c) \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{d^4 \sqrt {c+d x}}+\frac {(c+d x)^{3/2} \left (a d (C d-3 c D)-b \left (-B d^2-6 c^2 D+3 c C d\right )\right )}{d^4}+\frac {(c+d x)^{5/2} (a d D-4 b c D+b C d)}{d^4}+\frac {b D (c+d x)^{7/2}}{d^4}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 (c+d x)^{3/2} \left (a d \left (-B d^2-3 c^2 D+2 c C d\right )-b \left (A d^3-2 B c d^2-4 c^3 D+3 c^2 C d\right )\right )}{3 d^5}-\frac {2 \sqrt {c+d x} (b c-a d) \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{d^5}+\frac {2 (c+d x)^{5/2} \left (a d (C d-3 c D)-b \left (-B d^2-6 c^2 D+3 c C d\right )\right )}{5 d^5}+\frac {2 (c+d x)^{7/2} (a d D-4 b c D+b C d)}{7 d^5}+\frac {2 b D (c+d x)^{9/2}}{9 d^5}\) |
(-2*(b*c - a*d)*(c^2*C*d - B*c*d^2 + A*d^3 - c^3*D)*Sqrt[c + d*x])/d^5 - ( 2*(a*d*(2*c*C*d - B*d^2 - 3*c^2*D) - b*(3*c^2*C*d - 2*B*c*d^2 + A*d^3 - 4* c^3*D))*(c + d*x)^(3/2))/(3*d^5) + (2*(a*d*(C*d - 3*c*D) - b*(3*c*C*d - B* d^2 - 6*c^2*D))*(c + d*x)^(5/2))/(5*d^5) + (2*(b*C*d - 4*b*c*D + a*d*D)*(c + d*x)^(7/2))/(7*d^5) + (2*b*D*(c + d*x)^(9/2))/(9*d^5)
3.1.3.3.1 Defintions of rubi rules used
Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c , d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2])
Time = 1.65 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.76
| method | result | size |
| pseudoelliptic | \(\frac {2 \sqrt {d x +c}\, \left (\left (\frac {D b \,x^{4}}{9}+\frac {\left (C b +D a \right ) x^{3}}{7}+\frac {\left (B b +C a \right ) x^{2}}{5}+\frac {\left (A b +B a \right ) x}{3}+A a \right ) d^{4}-\frac {2 \left (\frac {4 D b \,x^{3}}{21}+\frac {9 \left (C b +D a \right ) x^{2}}{35}+\frac {2 \left (B b +C a \right ) x}{5}+A b +B a \right ) c \,d^{3}}{3}+\frac {8 c^{2} \left (\frac {2 D b \,x^{2}}{7}+\frac {3 \left (C b +D a \right ) x}{7}+B b +C a \right ) d^{2}}{15}-\frac {16 \left (\frac {4}{9} D b x +C b +D a \right ) c^{3} d}{35}+\frac {128 D b \,c^{4}}{315}\right )}{d^{5}}\) | \(162\) |
| derivativedivides | \(\frac {\frac {2 D b \left (d x +c \right )^{\frac {9}{2}}}{9}+\frac {2 \left (\left (a d -b c \right ) D+b \left (C d -3 D c \right )\right ) \left (d x +c \right )^{\frac {7}{2}}}{7}+\frac {2 \left (\left (a d -b c \right ) \left (C d -3 D c \right )+b \left (B \,d^{2}-2 C c d +3 D c^{2}\right )\right ) \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {2 \left (\left (a d -b c \right ) \left (B \,d^{2}-2 C c d +3 D c^{2}\right )+b \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right )\right ) \left (d x +c \right )^{\frac {3}{2}}}{3}+2 \left (a d -b c \right ) \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right ) \sqrt {d x +c}}{d^{5}}\) | \(198\) |
| default | \(\frac {\frac {2 D b \left (d x +c \right )^{\frac {9}{2}}}{9}+\frac {2 \left (\left (a d -b c \right ) D+b \left (C d -3 D c \right )\right ) \left (d x +c \right )^{\frac {7}{2}}}{7}+\frac {2 \left (\left (a d -b c \right ) \left (C d -3 D c \right )+b \left (B \,d^{2}-2 C c d +3 D c^{2}\right )\right ) \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {2 \left (\left (a d -b c \right ) \left (B \,d^{2}-2 C c d +3 D c^{2}\right )+b \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right )\right ) \left (d x +c \right )^{\frac {3}{2}}}{3}+2 \left (a d -b c \right ) \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right ) \sqrt {d x +c}}{d^{5}}\) | \(198\) |
| gosper | \(\frac {2 \sqrt {d x +c}\, \left (35 D b \,x^{4} d^{4}+45 C b \,d^{4} x^{3}+45 D a \,d^{4} x^{3}-40 D b c \,d^{3} x^{3}+63 B b \,d^{4} x^{2}+63 C a \,d^{4} x^{2}-54 C b c \,d^{3} x^{2}-54 D a c \,d^{3} x^{2}+48 D b \,c^{2} d^{2} x^{2}+105 A b \,d^{4} x +105 B a \,d^{4} x -84 B b c \,d^{3} x -84 C a c \,d^{3} x +72 C b \,c^{2} d^{2} x +72 D a \,c^{2} d^{2} x -64 D b \,c^{3} d x +315 A a \,d^{4}-210 A b c \,d^{3}-210 B a c \,d^{3}+168 B b \,c^{2} d^{2}+168 C a \,c^{2} d^{2}-144 C b \,c^{3} d -144 D a \,c^{3} d +128 D b \,c^{4}\right )}{315 d^{5}}\) | \(241\) |
| trager | \(\frac {2 \sqrt {d x +c}\, \left (35 D b \,x^{4} d^{4}+45 C b \,d^{4} x^{3}+45 D a \,d^{4} x^{3}-40 D b c \,d^{3} x^{3}+63 B b \,d^{4} x^{2}+63 C a \,d^{4} x^{2}-54 C b c \,d^{3} x^{2}-54 D a c \,d^{3} x^{2}+48 D b \,c^{2} d^{2} x^{2}+105 A b \,d^{4} x +105 B a \,d^{4} x -84 B b c \,d^{3} x -84 C a c \,d^{3} x +72 C b \,c^{2} d^{2} x +72 D a \,c^{2} d^{2} x -64 D b \,c^{3} d x +315 A a \,d^{4}-210 A b c \,d^{3}-210 B a c \,d^{3}+168 B b \,c^{2} d^{2}+168 C a \,c^{2} d^{2}-144 C b \,c^{3} d -144 D a \,c^{3} d +128 D b \,c^{4}\right )}{315 d^{5}}\) | \(241\) |
2*(d*x+c)^(1/2)*((1/9*D*b*x^4+1/7*(C*b+D*a)*x^3+1/5*(B*b+C*a)*x^2+1/3*(A*b +B*a)*x+A*a)*d^4-2/3*(4/21*D*b*x^3+9/35*(C*b+D*a)*x^2+2/5*(B*b+C*a)*x+A*b+ B*a)*c*d^3+8/15*c^2*(2/7*D*b*x^2+3/7*(C*b+D*a)*x+B*b+C*a)*d^2-16/35*(4/9*D *b*x+C*b+D*a)*c^3*d+128/315*D*b*c^4)/d^5
Time = 0.26 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.96 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{\sqrt {c+d x}} \, dx=\frac {2 \, {\left (35 \, D b d^{4} x^{4} + 128 \, D b c^{4} + 315 \, A a d^{4} + 168 \, {\left (C a + B b\right )} c^{2} d^{2} - 210 \, {\left (B a + A b\right )} c d^{3} - 5 \, {\left (8 \, D b c d^{3} - 9 \, {\left (D a + C b\right )} d^{4}\right )} x^{3} + 3 \, {\left (16 \, D b c^{2} d^{2} + 21 \, {\left (C a + B b\right )} d^{4} - 18 \, {\left (D a c + C b c\right )} d^{3}\right )} x^{2} - 144 \, {\left (D a c^{3} + C b c^{3}\right )} d - {\left (64 \, D b c^{3} d + 84 \, {\left (C a + B b\right )} c d^{3} - 105 \, {\left (B a + A b\right )} d^{4} - 72 \, {\left (D a c^{2} + C b c^{2}\right )} d^{2}\right )} x\right )} \sqrt {d x + c}}{315 \, d^{5}} \]
2/315*(35*D*b*d^4*x^4 + 128*D*b*c^4 + 315*A*a*d^4 + 168*(C*a + B*b)*c^2*d^ 2 - 210*(B*a + A*b)*c*d^3 - 5*(8*D*b*c*d^3 - 9*(D*a + C*b)*d^4)*x^3 + 3*(1 6*D*b*c^2*d^2 + 21*(C*a + B*b)*d^4 - 18*(D*a*c + C*b*c)*d^3)*x^2 - 144*(D* a*c^3 + C*b*c^3)*d - (64*D*b*c^3*d + 84*(C*a + B*b)*c*d^3 - 105*(B*a + A*b )*d^4 - 72*(D*a*c^2 + C*b*c^2)*d^2)*x)*sqrt(d*x + c)/d^5
Time = 1.25 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.51 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{\sqrt {c+d x}} \, dx=\begin {cases} \frac {2 \left (\frac {D b \left (c + d x\right )^{\frac {9}{2}}}{9 d^{4}} + \frac {\left (c + d x\right )^{\frac {7}{2}} \left (C b d + D a d - 4 D b c\right )}{7 d^{4}} + \frac {\left (c + d x\right )^{\frac {5}{2}} \left (B b d^{2} + C a d^{2} - 3 C b c d - 3 D a c d + 6 D b c^{2}\right )}{5 d^{4}} + \frac {\left (c + d x\right )^{\frac {3}{2}} \left (A b d^{3} + B a d^{3} - 2 B b c d^{2} - 2 C a c d^{2} + 3 C b c^{2} d + 3 D a c^{2} d - 4 D b c^{3}\right )}{3 d^{4}} + \frac {\sqrt {c + d x} \left (A a d^{4} - A b c d^{3} - B a c d^{3} + B b c^{2} d^{2} + C a c^{2} d^{2} - C b c^{3} d - D a c^{3} d + D b c^{4}\right )}{d^{4}}\right )}{d} & \text {for}\: d \neq 0 \\\frac {A a x + \frac {D b x^{5}}{5} + \frac {x^{4} \left (C b + D a\right )}{4} + \frac {x^{3} \left (B b + C a\right )}{3} + \frac {x^{2} \left (A b + B a\right )}{2}}{\sqrt {c}} & \text {otherwise} \end {cases} \]
Piecewise((2*(D*b*(c + d*x)**(9/2)/(9*d**4) + (c + d*x)**(7/2)*(C*b*d + D* a*d - 4*D*b*c)/(7*d**4) + (c + d*x)**(5/2)*(B*b*d**2 + C*a*d**2 - 3*C*b*c* d - 3*D*a*c*d + 6*D*b*c**2)/(5*d**4) + (c + d*x)**(3/2)*(A*b*d**3 + B*a*d* *3 - 2*B*b*c*d**2 - 2*C*a*c*d**2 + 3*C*b*c**2*d + 3*D*a*c**2*d - 4*D*b*c** 3)/(3*d**4) + sqrt(c + d*x)*(A*a*d**4 - A*b*c*d**3 - B*a*c*d**3 + B*b*c**2 *d**2 + C*a*c**2*d**2 - C*b*c**3*d - D*a*c**3*d + D*b*c**4)/d**4)/d, Ne(d, 0)), ((A*a*x + D*b*x**5/5 + x**4*(C*b + D*a)/4 + x**3*(B*b + C*a)/3 + x** 2*(A*b + B*a)/2)/sqrt(c), True))
Time = 0.21 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.93 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{\sqrt {c+d x}} \, dx=\frac {2 \, {\left (35 \, {\left (d x + c\right )}^{\frac {9}{2}} D b - 45 \, {\left (4 \, D b c - {\left (D a + C b\right )} d\right )} {\left (d x + c\right )}^{\frac {7}{2}} + 63 \, {\left (6 \, D b c^{2} - 3 \, {\left (D a + C b\right )} c d + {\left (C a + B b\right )} d^{2}\right )} {\left (d x + c\right )}^{\frac {5}{2}} - 105 \, {\left (4 \, D b c^{3} - 3 \, {\left (D a + C b\right )} c^{2} d + 2 \, {\left (C a + B b\right )} c d^{2} - {\left (B a + A b\right )} d^{3}\right )} {\left (d x + c\right )}^{\frac {3}{2}} + 315 \, {\left (D b c^{4} + A a d^{4} - {\left (D a + C b\right )} c^{3} d + {\left (C a + B b\right )} c^{2} d^{2} - {\left (B a + A b\right )} c d^{3}\right )} \sqrt {d x + c}\right )}}{315 \, d^{5}} \]
2/315*(35*(d*x + c)^(9/2)*D*b - 45*(4*D*b*c - (D*a + C*b)*d)*(d*x + c)^(7/ 2) + 63*(6*D*b*c^2 - 3*(D*a + C*b)*c*d + (C*a + B*b)*d^2)*(d*x + c)^(5/2) - 105*(4*D*b*c^3 - 3*(D*a + C*b)*c^2*d + 2*(C*a + B*b)*c*d^2 - (B*a + A*b) *d^3)*(d*x + c)^(3/2) + 315*(D*b*c^4 + A*a*d^4 - (D*a + C*b)*c^3*d + (C*a + B*b)*c^2*d^2 - (B*a + A*b)*c*d^3)*sqrt(d*x + c))/d^5
Time = 0.29 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.46 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{\sqrt {c+d x}} \, dx=\frac {2 \, {\left (315 \, \sqrt {d x + c} A a + \frac {105 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} - 3 \, \sqrt {d x + c} c\right )} B a}{d} + \frac {105 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} - 3 \, \sqrt {d x + c} c\right )} A b}{d} + \frac {21 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {d x + c} c^{2}\right )} C a}{d^{2}} + \frac {21 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {d x + c} c^{2}\right )} B b}{d^{2}} + \frac {9 \, {\left (5 \, {\left (d x + c\right )}^{\frac {7}{2}} - 21 \, {\left (d x + c\right )}^{\frac {5}{2}} c + 35 \, {\left (d x + c\right )}^{\frac {3}{2}} c^{2} - 35 \, \sqrt {d x + c} c^{3}\right )} D a}{d^{3}} + \frac {9 \, {\left (5 \, {\left (d x + c\right )}^{\frac {7}{2}} - 21 \, {\left (d x + c\right )}^{\frac {5}{2}} c + 35 \, {\left (d x + c\right )}^{\frac {3}{2}} c^{2} - 35 \, \sqrt {d x + c} c^{3}\right )} C b}{d^{3}} + \frac {{\left (35 \, {\left (d x + c\right )}^{\frac {9}{2}} - 180 \, {\left (d x + c\right )}^{\frac {7}{2}} c + 378 \, {\left (d x + c\right )}^{\frac {5}{2}} c^{2} - 420 \, {\left (d x + c\right )}^{\frac {3}{2}} c^{3} + 315 \, \sqrt {d x + c} c^{4}\right )} D b}{d^{4}}\right )}}{315 \, d} \]
2/315*(315*sqrt(d*x + c)*A*a + 105*((d*x + c)^(3/2) - 3*sqrt(d*x + c)*c)*B *a/d + 105*((d*x + c)^(3/2) - 3*sqrt(d*x + c)*c)*A*b/d + 21*(3*(d*x + c)^( 5/2) - 10*(d*x + c)^(3/2)*c + 15*sqrt(d*x + c)*c^2)*C*a/d^2 + 21*(3*(d*x + c)^(5/2) - 10*(d*x + c)^(3/2)*c + 15*sqrt(d*x + c)*c^2)*B*b/d^2 + 9*(5*(d *x + c)^(7/2) - 21*(d*x + c)^(5/2)*c + 35*(d*x + c)^(3/2)*c^2 - 35*sqrt(d* x + c)*c^3)*D*a/d^3 + 9*(5*(d*x + c)^(7/2) - 21*(d*x + c)^(5/2)*c + 35*(d* x + c)^(3/2)*c^2 - 35*sqrt(d*x + c)*c^3)*C*b/d^3 + (35*(d*x + c)^(9/2) - 1 80*(d*x + c)^(7/2)*c + 378*(d*x + c)^(5/2)*c^2 - 420*(d*x + c)^(3/2)*c^3 + 315*sqrt(d*x + c)*c^4)*D*b/d^4)/d
Time = 3.71 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.66 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{\sqrt {c+d x}} \, dx=\frac {2\,A\,b\,{\left (c+d\,x\right )}^{3/2}-6\,A\,b\,c\,\sqrt {c+d\,x}}{3\,d^2}+\frac {2\,B\,a\,{\left (c+d\,x\right )}^{3/2}-6\,B\,a\,c\,\sqrt {c+d\,x}}{3\,d^2}+\frac {6\,B\,b\,{\left (c+d\,x\right )}^{5/2}+30\,B\,b\,c^2\,\sqrt {c+d\,x}-20\,B\,b\,c\,{\left (c+d\,x\right )}^{3/2}}{15\,d^3}+\frac {6\,C\,a\,{\left (c+d\,x\right )}^{5/2}+30\,C\,a\,c^2\,\sqrt {c+d\,x}-20\,C\,a\,c\,{\left (c+d\,x\right )}^{3/2}}{15\,d^3}+\frac {2\,A\,a\,\sqrt {c+d\,x}}{d}+\frac {2\,C\,b\,{\left (c+d\,x\right )}^{7/2}}{7\,d^4}-\frac {2\,a\,\sqrt {c+d\,x}\,D\,\left (6\,c\,{\left (c+d\,x\right )}^2-20\,c^2\,\left (c+d\,x\right )+30\,c^3-5\,d^3\,x^3\right )}{35\,d^4}+\frac {2\,b\,x^4\,\sqrt {c+d\,x}\,D}{9\,d}-\frac {6\,C\,b\,c\,{\left (c+d\,x\right )}^{5/2}}{5\,d^4}-\frac {8\,b\,c\,D\,\left (\frac {2\,{\left (c+d\,x\right )}^{7/2}}{7\,d^4}-\frac {2\,c^3\,\sqrt {c+d\,x}}{d^4}+\frac {2\,c^2\,{\left (c+d\,x\right )}^{3/2}}{d^4}-\frac {6\,c\,{\left (c+d\,x\right )}^{5/2}}{5\,d^4}\right )}{9\,d}-\frac {2\,C\,b\,c^3\,\sqrt {c+d\,x}}{d^4}+\frac {2\,C\,b\,c^2\,{\left (c+d\,x\right )}^{3/2}}{d^4} \]
(2*A*b*(c + d*x)^(3/2) - 6*A*b*c*(c + d*x)^(1/2))/(3*d^2) + (2*B*a*(c + d* x)^(3/2) - 6*B*a*c*(c + d*x)^(1/2))/(3*d^2) + (6*B*b*(c + d*x)^(5/2) + 30* B*b*c^2*(c + d*x)^(1/2) - 20*B*b*c*(c + d*x)^(3/2))/(15*d^3) + (6*C*a*(c + d*x)^(5/2) + 30*C*a*c^2*(c + d*x)^(1/2) - 20*C*a*c*(c + d*x)^(3/2))/(15*d ^3) + (2*A*a*(c + d*x)^(1/2))/d + (2*C*b*(c + d*x)^(7/2))/(7*d^4) - (2*a*( c + d*x)^(1/2)*D*(6*c*(c + d*x)^2 - 20*c^2*(c + d*x) + 30*c^3 - 5*d^3*x^3) )/(35*d^4) + (2*b*x^4*(c + d*x)^(1/2)*D)/(9*d) - (6*C*b*c*(c + d*x)^(5/2)) /(5*d^4) - (8*b*c*D*((2*(c + d*x)^(7/2))/(7*d^4) - (2*c^3*(c + d*x)^(1/2)) /d^4 + (2*c^2*(c + d*x)^(3/2))/d^4 - (6*c*(c + d*x)^(5/2))/(5*d^4)))/(9*d) - (2*C*b*c^3*(c + d*x)^(1/2))/d^4 + (2*C*b*c^2*(c + d*x)^(3/2))/d^4