Integrand size = 24, antiderivative size = 214 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^{3/2}} \, dx=\frac {2 (B d-A e) \left (c d^2+a e^2\right )^2}{e^6 \sqrt {d+e x}}+\frac {2 \left (c d^2+a e^2\right ) \left (5 B c d^2-4 A c d e+a B e^2\right ) \sqrt {d+e x}}{e^6}-\frac {4 c \left (5 B c d^3-3 A c d^2 e+3 a B d e^2-a A e^3\right ) (d+e x)^{3/2}}{3 e^6}+\frac {4 c \left (5 B c d^2-2 A c d e+a B e^2\right ) (d+e x)^{5/2}}{5 e^6}-\frac {2 c^2 (5 B d-A e) (d+e x)^{7/2}}{7 e^6}+\frac {2 B c^2 (d+e x)^{9/2}}{9 e^6} \]
-4/3*c*(-A*a*e^3-3*A*c*d^2*e+3*B*a*d*e^2+5*B*c*d^3)*(e*x+d)^(3/2)/e^6+4/5* c*(-2*A*c*d*e+B*a*e^2+5*B*c*d^2)*(e*x+d)^(5/2)/e^6-2/7*c^2*(-A*e+5*B*d)*(e *x+d)^(7/2)/e^6+2/9*B*c^2*(e*x+d)^(9/2)/e^6+2*(-A*e+B*d)*(a*e^2+c*d^2)^2/e ^6/(e*x+d)^(1/2)+2*(a*e^2+c*d^2)*(-4*A*c*d*e+B*a*e^2+5*B*c*d^2)*(e*x+d)^(1 /2)/e^6
Time = 0.14 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.00 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^{3/2}} \, dx=\frac {-6 A e \left (105 a^2 e^4+70 a c e^2 \left (8 d^2+4 d e x-e^2 x^2\right )+3 c^2 \left (128 d^4+64 d^3 e x-16 d^2 e^2 x^2+8 d e^3 x^3-5 e^4 x^4\right )\right )+2 B \left (315 a^2 e^4 (2 d+e x)+126 a c e^2 \left (16 d^3+8 d^2 e x-2 d e^2 x^2+e^3 x^3\right )+5 c^2 \left (256 d^5+128 d^4 e x-32 d^3 e^2 x^2+16 d^2 e^3 x^3-10 d e^4 x^4+7 e^5 x^5\right )\right )}{315 e^6 \sqrt {d+e x}} \]
(-6*A*e*(105*a^2*e^4 + 70*a*c*e^2*(8*d^2 + 4*d*e*x - e^2*x^2) + 3*c^2*(128 *d^4 + 64*d^3*e*x - 16*d^2*e^2*x^2 + 8*d*e^3*x^3 - 5*e^4*x^4)) + 2*B*(315* a^2*e^4*(2*d + e*x) + 126*a*c*e^2*(16*d^3 + 8*d^2*e*x - 2*d*e^2*x^2 + e^3* x^3) + 5*c^2*(256*d^5 + 128*d^4*e*x - 32*d^3*e^2*x^2 + 16*d^2*e^3*x^3 - 10 *d*e^4*x^4 + 7*e^5*x^5)))/(315*e^6*Sqrt[d + e*x])
Time = 0.34 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {652, 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.
| \(\displaystyle \int \frac {\left (a+c x^2\right )^2 (A+B x)}{(d+e x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 652 |
| \(\displaystyle \int \left (-\frac {2 c (d+e x)^{3/2} \left (-a B e^2+2 A c d e-5 B c d^2\right )}{e^5}+\frac {\left (a e^2+c d^2\right ) \left (a B e^2-4 A c d e+5 B c d^2\right )}{e^5 \sqrt {d+e x}}+\frac {\left (a e^2+c d^2\right )^2 (A e-B d)}{e^5 (d+e x)^{3/2}}+\frac {2 c \sqrt {d+e x} \left (a A e^3-3 a B d e^2+3 A c d^2 e-5 B c d^3\right )}{e^5}+\frac {c^2 (d+e x)^{5/2} (A e-5 B d)}{e^5}+\frac {B c^2 (d+e x)^{7/2}}{e^5}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {4 c (d+e x)^{5/2} \left (a B e^2-2 A c d e+5 B c d^2\right )}{5 e^6}+\frac {2 \sqrt {d+e x} \left (a e^2+c d^2\right ) \left (a B e^2-4 A c d e+5 B c d^2\right )}{e^6}+\frac {2 \left (a e^2+c d^2\right )^2 (B d-A e)}{e^6 \sqrt {d+e x}}-\frac {4 c (d+e x)^{3/2} \left (-a A e^3+3 a B d e^2-3 A c d^2 e+5 B c d^3\right )}{3 e^6}-\frac {2 c^2 (d+e x)^{7/2} (5 B d-A e)}{7 e^6}+\frac {2 B c^2 (d+e x)^{9/2}}{9 e^6}\) |
(2*(B*d - A*e)*(c*d^2 + a*e^2)^2)/(e^6*Sqrt[d + e*x]) + (2*(c*d^2 + a*e^2) *(5*B*c*d^2 - 4*A*c*d*e + a*B*e^2)*Sqrt[d + e*x])/e^6 - (4*c*(5*B*c*d^3 - 3*A*c*d^2*e + 3*a*B*d*e^2 - a*A*e^3)*(d + e*x)^(3/2))/(3*e^6) + (4*c*(5*B* c*d^2 - 2*A*c*d*e + a*B*e^2)*(d + e*x)^(5/2))/(5*e^6) - (2*c^2*(5*B*d - A* e)*(d + e*x)^(7/2))/(7*e^6) + (2*B*c^2*(d + e*x)^(9/2))/(9*e^6)
3.15.37.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_ )^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + c *x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && IGtQ[p, 0]
Time = 0.35 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.83
| method | result | size |
| pseudoelliptic | \(\frac {\left (\left (70 B \,x^{5}+90 A \,x^{4}\right ) c^{2}+420 \left (\frac {3 B x}{5}+A \right ) x^{2} a c -630 a^{2} \left (-B x +A \right )\right ) e^{5}-1680 d \left (\left (\frac {5}{84} B \,x^{4}+\frac {3}{35} A \,x^{3}\right ) c^{2}+a x \left (\frac {3 B x}{10}+A \right ) c -\frac {3 B \,a^{2}}{4}\right ) e^{4}-3360 c \,d^{2} \left (-\frac {3 \left (\frac {5 B x}{9}+A \right ) x^{2} c}{35}+a \left (-\frac {3 B x}{5}+A \right )\right ) e^{3}-1152 c \left (x \left (\frac {5 B x}{18}+A \right ) c -\frac {7 B a}{2}\right ) d^{3} e^{2}-2304 c^{2} \left (-\frac {5 B x}{9}+A \right ) d^{4} e +2560 B \,c^{2} d^{5}}{315 \sqrt {e x +d}\, e^{6}}\) | \(177\) |
| risch | \(-\frac {2 \left (-35 B \,c^{2} x^{4} e^{4}-45 A \,c^{2} e^{4} x^{3}+85 B \,c^{2} d \,e^{3} x^{3}+117 A \,c^{2} d \,e^{3} x^{2}-126 B a c \,e^{4} x^{2}-165 B \,c^{2} d^{2} e^{2} x^{2}-210 A a c \,e^{4} x -261 A \,c^{2} d^{2} e^{2} x +378 B a c d \,e^{3} x +325 B \,c^{2} d^{3} e x +1050 A a c d \,e^{3}+837 A \,c^{2} d^{3} e -315 B \,e^{4} a^{2}-1386 B a c \,d^{2} e^{2}-965 B \,c^{2} d^{4}\right ) \sqrt {e x +d}}{315 e^{6}}-\frac {2 \left (A \,a^{2} e^{5}+2 A a c \,d^{2} e^{3}+A \,c^{2} d^{4} e -B \,a^{2} d \,e^{4}-2 B a c \,d^{3} e^{2}-B \,c^{2} d^{5}\right )}{e^{6} \sqrt {e x +d}}\) | \(253\) |
| gosper | \(-\frac {2 \left (-35 B \,x^{5} c^{2} e^{5}-45 A \,x^{4} c^{2} e^{5}+50 B \,x^{4} c^{2} d \,e^{4}+72 A \,x^{3} c^{2} d \,e^{4}-126 B \,x^{3} a c \,e^{5}-80 B \,x^{3} c^{2} d^{2} e^{3}-210 A \,x^{2} a c \,e^{5}-144 A \,x^{2} c^{2} d^{2} e^{3}+252 B \,x^{2} a c d \,e^{4}+160 B \,x^{2} c^{2} d^{3} e^{2}+840 A x a c d \,e^{4}+576 A x \,c^{2} d^{3} e^{2}-315 B x \,a^{2} e^{5}-1008 B x a c \,d^{2} e^{3}-640 B x \,c^{2} d^{4} e +315 A \,a^{2} e^{5}+1680 A a c \,d^{2} e^{3}+1152 A \,c^{2} d^{4} e -630 B \,a^{2} d \,e^{4}-2016 B a c \,d^{3} e^{2}-1280 B \,c^{2} d^{5}\right )}{315 \sqrt {e x +d}\, e^{6}}\) | \(259\) |
| trager | \(-\frac {2 \left (-35 B \,x^{5} c^{2} e^{5}-45 A \,x^{4} c^{2} e^{5}+50 B \,x^{4} c^{2} d \,e^{4}+72 A \,x^{3} c^{2} d \,e^{4}-126 B \,x^{3} a c \,e^{5}-80 B \,x^{3} c^{2} d^{2} e^{3}-210 A \,x^{2} a c \,e^{5}-144 A \,x^{2} c^{2} d^{2} e^{3}+252 B \,x^{2} a c d \,e^{4}+160 B \,x^{2} c^{2} d^{3} e^{2}+840 A x a c d \,e^{4}+576 A x \,c^{2} d^{3} e^{2}-315 B x \,a^{2} e^{5}-1008 B x a c \,d^{2} e^{3}-640 B x \,c^{2} d^{4} e +315 A \,a^{2} e^{5}+1680 A a c \,d^{2} e^{3}+1152 A \,c^{2} d^{4} e -630 B \,a^{2} d \,e^{4}-2016 B a c \,d^{3} e^{2}-1280 B \,c^{2} d^{5}\right )}{315 \sqrt {e x +d}\, e^{6}}\) | \(259\) |
| derivativedivides | \(\frac {\frac {2 B \,c^{2} \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {2 A \,c^{2} e \left (e x +d \right )^{\frac {7}{2}}}{7}-\frac {10 B \,c^{2} d \left (e x +d \right )^{\frac {7}{2}}}{7}-\frac {8 A \,c^{2} d e \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {4 B a c \,e^{2} \left (e x +d \right )^{\frac {5}{2}}}{5}+4 B \,c^{2} d^{2} \left (e x +d \right )^{\frac {5}{2}}+\frac {4 A a c \,e^{3} \left (e x +d \right )^{\frac {3}{2}}}{3}+4 A \,c^{2} d^{2} e \left (e x +d \right )^{\frac {3}{2}}-4 B a c d \,e^{2} \left (e x +d \right )^{\frac {3}{2}}-\frac {20 B \,c^{2} d^{3} \left (e x +d \right )^{\frac {3}{2}}}{3}-8 A a c d \,e^{3} \sqrt {e x +d}-8 A \,c^{2} d^{3} e \sqrt {e x +d}+2 B \,a^{2} e^{4} \sqrt {e x +d}+12 B a c \,d^{2} e^{2} \sqrt {e x +d}+10 B \,c^{2} d^{4} \sqrt {e x +d}-\frac {2 \left (A \,a^{2} e^{5}+2 A a c \,d^{2} e^{3}+A \,c^{2} d^{4} e -B \,a^{2} d \,e^{4}-2 B a c \,d^{3} e^{2}-B \,c^{2} d^{5}\right )}{\sqrt {e x +d}}}{e^{6}}\) | \(308\) |
| default | \(\frac {\frac {2 B \,c^{2} \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {2 A \,c^{2} e \left (e x +d \right )^{\frac {7}{2}}}{7}-\frac {10 B \,c^{2} d \left (e x +d \right )^{\frac {7}{2}}}{7}-\frac {8 A \,c^{2} d e \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {4 B a c \,e^{2} \left (e x +d \right )^{\frac {5}{2}}}{5}+4 B \,c^{2} d^{2} \left (e x +d \right )^{\frac {5}{2}}+\frac {4 A a c \,e^{3} \left (e x +d \right )^{\frac {3}{2}}}{3}+4 A \,c^{2} d^{2} e \left (e x +d \right )^{\frac {3}{2}}-4 B a c d \,e^{2} \left (e x +d \right )^{\frac {3}{2}}-\frac {20 B \,c^{2} d^{3} \left (e x +d \right )^{\frac {3}{2}}}{3}-8 A a c d \,e^{3} \sqrt {e x +d}-8 A \,c^{2} d^{3} e \sqrt {e x +d}+2 B \,a^{2} e^{4} \sqrt {e x +d}+12 B a c \,d^{2} e^{2} \sqrt {e x +d}+10 B \,c^{2} d^{4} \sqrt {e x +d}-\frac {2 \left (A \,a^{2} e^{5}+2 A a c \,d^{2} e^{3}+A \,c^{2} d^{4} e -B \,a^{2} d \,e^{4}-2 B a c \,d^{3} e^{2}-B \,c^{2} d^{5}\right )}{\sqrt {e x +d}}}{e^{6}}\) | \(308\) |
1/315*(((70*B*x^5+90*A*x^4)*c^2+420*(3/5*B*x+A)*x^2*a*c-630*a^2*(-B*x+A))* e^5-1680*d*((5/84*B*x^4+3/35*A*x^3)*c^2+a*x*(3/10*B*x+A)*c-3/4*B*a^2)*e^4- 3360*c*d^2*(-3/35*(5/9*B*x+A)*x^2*c+a*(-3/5*B*x+A))*e^3-1152*c*(x*(5/18*B* x+A)*c-7/2*B*a)*d^3*e^2-2304*c^2*(-5/9*B*x+A)*d^4*e+2560*B*c^2*d^5)/(e*x+d )^(1/2)/e^6
Time = 0.34 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.20 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (35 \, B c^{2} e^{5} x^{5} + 1280 \, B c^{2} d^{5} - 1152 \, A c^{2} d^{4} e + 2016 \, B a c d^{3} e^{2} - 1680 \, A a c d^{2} e^{3} + 630 \, B a^{2} d e^{4} - 315 \, A a^{2} e^{5} - 5 \, {\left (10 \, B c^{2} d e^{4} - 9 \, A c^{2} e^{5}\right )} x^{4} + 2 \, {\left (40 \, B c^{2} d^{2} e^{3} - 36 \, A c^{2} d e^{4} + 63 \, B a c e^{5}\right )} x^{3} - 2 \, {\left (80 \, B c^{2} d^{3} e^{2} - 72 \, A c^{2} d^{2} e^{3} + 126 \, B a c d e^{4} - 105 \, A a c e^{5}\right )} x^{2} + {\left (640 \, B c^{2} d^{4} e - 576 \, A c^{2} d^{3} e^{2} + 1008 \, B a c d^{2} e^{3} - 840 \, A a c d e^{4} + 315 \, B a^{2} e^{5}\right )} x\right )} \sqrt {e x + d}}{315 \, {\left (e^{7} x + d e^{6}\right )}} \]
2/315*(35*B*c^2*e^5*x^5 + 1280*B*c^2*d^5 - 1152*A*c^2*d^4*e + 2016*B*a*c*d ^3*e^2 - 1680*A*a*c*d^2*e^3 + 630*B*a^2*d*e^4 - 315*A*a^2*e^5 - 5*(10*B*c^ 2*d*e^4 - 9*A*c^2*e^5)*x^4 + 2*(40*B*c^2*d^2*e^3 - 36*A*c^2*d*e^4 + 63*B*a *c*e^5)*x^3 - 2*(80*B*c^2*d^3*e^2 - 72*A*c^2*d^2*e^3 + 126*B*a*c*d*e^4 - 1 05*A*a*c*e^5)*x^2 + (640*B*c^2*d^4*e - 576*A*c^2*d^3*e^2 + 1008*B*a*c*d^2* e^3 - 840*A*a*c*d*e^4 + 315*B*a^2*e^5)*x)*sqrt(e*x + d)/(e^7*x + d*e^6)
Time = 5.97 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.51 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^{3/2}} \, dx=\begin {cases} \frac {2 \left (\frac {B c^{2} \left (d + e x\right )^{\frac {9}{2}}}{9 e^{5}} + \frac {\left (d + e x\right )^{\frac {7}{2}} \left (A c^{2} e - 5 B c^{2} d\right )}{7 e^{5}} + \frac {\left (d + e x\right )^{\frac {5}{2}} \left (- 4 A c^{2} d e + 2 B a c e^{2} + 10 B c^{2} d^{2}\right )}{5 e^{5}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \cdot \left (2 A a c e^{3} + 6 A c^{2} d^{2} e - 6 B a c d e^{2} - 10 B c^{2} d^{3}\right )}{3 e^{5}} + \frac {\sqrt {d + e x} \left (- 4 A a c d e^{3} - 4 A c^{2} d^{3} e + B a^{2} e^{4} + 6 B a c d^{2} e^{2} + 5 B c^{2} d^{4}\right )}{e^{5}} + \frac {\left (- A e + B d\right ) \left (a e^{2} + c d^{2}\right )^{2}}{e^{5} \sqrt {d + e x}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {A a^{2} x + \frac {2 A a c x^{3}}{3} + \frac {A c^{2} x^{5}}{5} + \frac {B a^{2} x^{2}}{2} + \frac {B a c x^{4}}{2} + \frac {B c^{2} x^{6}}{6}}{d^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((2*(B*c**2*(d + e*x)**(9/2)/(9*e**5) + (d + e*x)**(7/2)*(A*c**2* e - 5*B*c**2*d)/(7*e**5) + (d + e*x)**(5/2)*(-4*A*c**2*d*e + 2*B*a*c*e**2 + 10*B*c**2*d**2)/(5*e**5) + (d + e*x)**(3/2)*(2*A*a*c*e**3 + 6*A*c**2*d** 2*e - 6*B*a*c*d*e**2 - 10*B*c**2*d**3)/(3*e**5) + sqrt(d + e*x)*(-4*A*a*c* d*e**3 - 4*A*c**2*d**3*e + B*a**2*e**4 + 6*B*a*c*d**2*e**2 + 5*B*c**2*d**4 )/e**5 + (-A*e + B*d)*(a*e**2 + c*d**2)**2/(e**5*sqrt(d + e*x)))/e, Ne(e, 0)), ((A*a**2*x + 2*A*a*c*x**3/3 + A*c**2*x**5/5 + B*a**2*x**2/2 + B*a*c*x **4/2 + B*c**2*x**6/6)/d**(3/2), True))
Time = 0.20 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.20 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (\frac {35 \, {\left (e x + d\right )}^{\frac {9}{2}} B c^{2} - 45 \, {\left (5 \, B c^{2} d - A c^{2} e\right )} {\left (e x + d\right )}^{\frac {7}{2}} + 126 \, {\left (5 \, B c^{2} d^{2} - 2 \, A c^{2} d e + B a c e^{2}\right )} {\left (e x + d\right )}^{\frac {5}{2}} - 210 \, {\left (5 \, B c^{2} d^{3} - 3 \, A c^{2} d^{2} e + 3 \, B a c d e^{2} - A a c e^{3}\right )} {\left (e x + d\right )}^{\frac {3}{2}} + 315 \, {\left (5 \, B c^{2} d^{4} - 4 \, A c^{2} d^{3} e + 6 \, B a c d^{2} e^{2} - 4 \, A a c d e^{3} + B a^{2} e^{4}\right )} \sqrt {e x + d}}{e^{5}} + \frac {315 \, {\left (B c^{2} d^{5} - A c^{2} d^{4} e + 2 \, B a c d^{3} e^{2} - 2 \, A a c d^{2} e^{3} + B a^{2} d e^{4} - A a^{2} e^{5}\right )}}{\sqrt {e x + d} e^{5}}\right )}}{315 \, e} \]
2/315*((35*(e*x + d)^(9/2)*B*c^2 - 45*(5*B*c^2*d - A*c^2*e)*(e*x + d)^(7/2 ) + 126*(5*B*c^2*d^2 - 2*A*c^2*d*e + B*a*c*e^2)*(e*x + d)^(5/2) - 210*(5*B *c^2*d^3 - 3*A*c^2*d^2*e + 3*B*a*c*d*e^2 - A*a*c*e^3)*(e*x + d)^(3/2) + 31 5*(5*B*c^2*d^4 - 4*A*c^2*d^3*e + 6*B*a*c*d^2*e^2 - 4*A*a*c*d*e^3 + B*a^2*e ^4)*sqrt(e*x + d))/e^5 + 315*(B*c^2*d^5 - A*c^2*d^4*e + 2*B*a*c*d^3*e^2 - 2*A*a*c*d^2*e^3 + B*a^2*d*e^4 - A*a^2*e^5)/(sqrt(e*x + d)*e^5))/e
Time = 0.28 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.57 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (B c^{2} d^{5} - A c^{2} d^{4} e + 2 \, B a c d^{3} e^{2} - 2 \, A a c d^{2} e^{3} + B a^{2} d e^{4} - A a^{2} e^{5}\right )}}{\sqrt {e x + d} e^{6}} + \frac {2 \, {\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} B c^{2} e^{48} - 225 \, {\left (e x + d\right )}^{\frac {7}{2}} B c^{2} d e^{48} + 630 \, {\left (e x + d\right )}^{\frac {5}{2}} B c^{2} d^{2} e^{48} - 1050 \, {\left (e x + d\right )}^{\frac {3}{2}} B c^{2} d^{3} e^{48} + 1575 \, \sqrt {e x + d} B c^{2} d^{4} e^{48} + 45 \, {\left (e x + d\right )}^{\frac {7}{2}} A c^{2} e^{49} - 252 \, {\left (e x + d\right )}^{\frac {5}{2}} A c^{2} d e^{49} + 630 \, {\left (e x + d\right )}^{\frac {3}{2}} A c^{2} d^{2} e^{49} - 1260 \, \sqrt {e x + d} A c^{2} d^{3} e^{49} + 126 \, {\left (e x + d\right )}^{\frac {5}{2}} B a c e^{50} - 630 \, {\left (e x + d\right )}^{\frac {3}{2}} B a c d e^{50} + 1890 \, \sqrt {e x + d} B a c d^{2} e^{50} + 210 \, {\left (e x + d\right )}^{\frac {3}{2}} A a c e^{51} - 1260 \, \sqrt {e x + d} A a c d e^{51} + 315 \, \sqrt {e x + d} B a^{2} e^{52}\right )}}{315 \, e^{54}} \]
2*(B*c^2*d^5 - A*c^2*d^4*e + 2*B*a*c*d^3*e^2 - 2*A*a*c*d^2*e^3 + B*a^2*d*e ^4 - A*a^2*e^5)/(sqrt(e*x + d)*e^6) + 2/315*(35*(e*x + d)^(9/2)*B*c^2*e^48 - 225*(e*x + d)^(7/2)*B*c^2*d*e^48 + 630*(e*x + d)^(5/2)*B*c^2*d^2*e^48 - 1050*(e*x + d)^(3/2)*B*c^2*d^3*e^48 + 1575*sqrt(e*x + d)*B*c^2*d^4*e^48 + 45*(e*x + d)^(7/2)*A*c^2*e^49 - 252*(e*x + d)^(5/2)*A*c^2*d*e^49 + 630*(e *x + d)^(3/2)*A*c^2*d^2*e^49 - 1260*sqrt(e*x + d)*A*c^2*d^3*e^49 + 126*(e* x + d)^(5/2)*B*a*c*e^50 - 630*(e*x + d)^(3/2)*B*a*c*d*e^50 + 1890*sqrt(e*x + d)*B*a*c*d^2*e^50 + 210*(e*x + d)^(3/2)*A*a*c*e^51 - 1260*sqrt(e*x + d) *A*a*c*d*e^51 + 315*sqrt(e*x + d)*B*a^2*e^52)/e^54
Time = 0.07 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.11 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^{3/2}} \, dx=\frac {{\left (d+e\,x\right )}^{5/2}\,\left (20\,B\,c^2\,d^2-8\,A\,c^2\,d\,e+4\,B\,a\,c\,e^2\right )}{5\,e^6}-\frac {-2\,B\,a^2\,d\,e^4+2\,A\,a^2\,e^5-4\,B\,a\,c\,d^3\,e^2+4\,A\,a\,c\,d^2\,e^3-2\,B\,c^2\,d^5+2\,A\,c^2\,d^4\,e}{e^6\,\sqrt {d+e\,x}}+\frac {4\,c\,{\left (d+e\,x\right )}^{3/2}\,\left (-5\,B\,c\,d^3+3\,A\,c\,d^2\,e-3\,B\,a\,d\,e^2+A\,a\,e^3\right )}{3\,e^6}+\frac {2\,B\,c^2\,{\left (d+e\,x\right )}^{9/2}}{9\,e^6}+\frac {2\,\left (c\,d^2+a\,e^2\right )\,\sqrt {d+e\,x}\,\left (5\,B\,c\,d^2-4\,A\,c\,d\,e+B\,a\,e^2\right )}{e^6}+\frac {2\,c^2\,\left (A\,e-5\,B\,d\right )\,{\left (d+e\,x\right )}^{7/2}}{7\,e^6} \]
((d + e*x)^(5/2)*(20*B*c^2*d^2 + 4*B*a*c*e^2 - 8*A*c^2*d*e))/(5*e^6) - (2* A*a^2*e^5 - 2*B*c^2*d^5 - 2*B*a^2*d*e^4 + 2*A*c^2*d^4*e + 4*A*a*c*d^2*e^3 - 4*B*a*c*d^3*e^2)/(e^6*(d + e*x)^(1/2)) + (4*c*(d + e*x)^(3/2)*(A*a*e^3 - 5*B*c*d^3 - 3*B*a*d*e^2 + 3*A*c*d^2*e))/(3*e^6) + (2*B*c^2*(d + e*x)^(9/2 ))/(9*e^6) + (2*(a*e^2 + c*d^2)*(d + e*x)^(1/2)*(B*a*e^2 + 5*B*c*d^2 - 4*A *c*d*e))/e^6 + (2*c^2*(A*e - 5*B*d)*(d + e*x)^(7/2))/(7*e^6)