Integrand size = 22, antiderivative size = 116 \[ \int (A+B x) (d+e x)^{3/2} \left (a+c x^2\right ) \, dx=-\frac {2 (B d-A e) \left (c d^2+a e^2\right ) (d+e x)^{5/2}}{5 e^4}+\frac {2 \left (3 B c d^2-2 A c d e+a B e^2\right ) (d+e x)^{7/2}}{7 e^4}-\frac {2 c (3 B d-A e) (d+e x)^{9/2}}{9 e^4}+\frac {2 B c (d+e x)^{11/2}}{11 e^4} \]
-2/5*(-A*e+B*d)*(a*e^2+c*d^2)*(e*x+d)^(5/2)/e^4+2/7*(-2*A*c*d*e+B*a*e^2+3* B*c*d^2)*(e*x+d)^(7/2)/e^4-2/9*c*(-A*e+3*B*d)*(e*x+d)^(9/2)/e^4+2/11*B*c*( e*x+d)^(11/2)/e^4
Time = 0.08 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.85 \[ \int (A+B x) (d+e x)^{3/2} \left (a+c x^2\right ) \, dx=\frac {2 (d+e x)^{5/2} \left (11 A e \left (63 a e^2+c \left (8 d^2-20 d e x+35 e^2 x^2\right )\right )-3 B \left (33 a e^2 (2 d-5 e x)+c \left (16 d^3-40 d^2 e x+70 d e^2 x^2-105 e^3 x^3\right )\right )\right )}{3465 e^4} \]
(2*(d + e*x)^(5/2)*(11*A*e*(63*a*e^2 + c*(8*d^2 - 20*d*e*x + 35*e^2*x^2)) - 3*B*(33*a*e^2*(2*d - 5*e*x) + c*(16*d^3 - 40*d^2*e*x + 70*d*e^2*x^2 - 10 5*e^3*x^3))))/(3465*e^4)
Time = 0.26 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, 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.
| \(\displaystyle \int \left (a+c x^2\right ) (A+B x) (d+e x)^{3/2} \, dx\) |
| \(\Big \downarrow \) 652 |
| \(\displaystyle \int \left (\frac {(d+e x)^{5/2} \left (a B e^2-2 A c d e+3 B c d^2\right )}{e^3}+\frac {(d+e x)^{3/2} \left (a e^2+c d^2\right ) (A e-B d)}{e^3}+\frac {c (d+e x)^{7/2} (A e-3 B d)}{e^3}+\frac {B c (d+e x)^{9/2}}{e^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 (d+e x)^{7/2} \left (a B e^2-2 A c d e+3 B c d^2\right )}{7 e^4}-\frac {2 (d+e x)^{5/2} \left (a e^2+c d^2\right ) (B d-A e)}{5 e^4}-\frac {2 c (d+e x)^{9/2} (3 B d-A e)}{9 e^4}+\frac {2 B c (d+e x)^{11/2}}{11 e^4}\) |
(-2*(B*d - A*e)*(c*d^2 + a*e^2)*(d + e*x)^(5/2))/(5*e^4) + (2*(3*B*c*d^2 - 2*A*c*d*e + a*B*e^2)*(d + e*x)^(7/2))/(7*e^4) - (2*c*(3*B*d - A*e)*(d + e *x)^(9/2))/(9*e^4) + (2*B*c*(d + e*x)^(11/2))/(11*e^4)
3.15.29.3.1 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 = 0.78 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.68
| method | result | size |
| pseudoelliptic | \(\frac {2 \left (\left (\frac {5 x^{2} \left (\frac {9 B x}{11}+A \right ) c}{9}+a \left (\frac {5 B x}{7}+A \right )\right ) e^{3}-\frac {20 d \left (x \left (\frac {21 B x}{22}+A \right ) c +\frac {9 B a}{10}\right ) e^{2}}{63}+\frac {8 c \,d^{2} \left (\frac {15 B x}{11}+A \right ) e}{63}-\frac {16 B c \,d^{3}}{231}\right ) \left (e x +d \right )^{\frac {5}{2}}}{5 e^{4}}\) | \(79\) |
| gosper | \(\frac {2 \left (e x +d \right )^{\frac {5}{2}} \left (315 B c \,x^{3} e^{3}+385 A c \,e^{3} x^{2}-210 B \,x^{2} c d \,e^{2}-220 A c d \,e^{2} x +495 B x a \,e^{3}+120 B c \,d^{2} e x +693 A a \,e^{3}+88 A c \,d^{2} e -198 B a d \,e^{2}-48 B c \,d^{3}\right )}{3465 e^{4}}\) | \(101\) |
| derivativedivides | \(\frac {\frac {2 B c \left (e x +d \right )^{\frac {11}{2}}}{11}+\frac {2 \left (\left (A e -B d \right ) c -2 B c d \right ) \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {2 \left (-2 \left (A e -B d \right ) d c +B \left (e^{2} a +c \,d^{2}\right )\right ) \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (A e -B d \right ) \left (e^{2} a +c \,d^{2}\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}}{e^{4}}\) | \(106\) |
| default | \(\frac {\frac {2 B c \left (e x +d \right )^{\frac {11}{2}}}{11}+\frac {2 \left (\left (A e -B d \right ) c -2 B c d \right ) \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {2 \left (-2 \left (A e -B d \right ) d c +B \left (e^{2} a +c \,d^{2}\right )\right ) \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (A e -B d \right ) \left (e^{2} a +c \,d^{2}\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}}{e^{4}}\) | \(106\) |
| trager | \(\frac {2 \left (315 B \,e^{5} c \,x^{5}+385 A c \,e^{5} x^{4}+420 B c d \,e^{4} x^{4}+550 A c d \,e^{4} x^{3}+495 B \,e^{5} a \,x^{3}+15 B c \,d^{2} e^{3} x^{3}+693 A a \,e^{5} x^{2}+33 A c \,d^{2} e^{3} x^{2}+792 B a d \,e^{4} x^{2}-18 B c \,d^{3} e^{2} x^{2}+1386 A a d \,e^{4} x -44 A c \,d^{3} e^{2} x +99 B a \,d^{2} e^{3} x +24 B c \,d^{4} e x +693 A a \,d^{2} e^{3}+88 A c \,d^{4} e -198 B a \,d^{3} e^{2}-48 B c \,d^{5}\right ) \sqrt {e x +d}}{3465 e^{4}}\) | \(201\) |
| risch | \(\frac {2 \left (315 B \,e^{5} c \,x^{5}+385 A c \,e^{5} x^{4}+420 B c d \,e^{4} x^{4}+550 A c d \,e^{4} x^{3}+495 B \,e^{5} a \,x^{3}+15 B c \,d^{2} e^{3} x^{3}+693 A a \,e^{5} x^{2}+33 A c \,d^{2} e^{3} x^{2}+792 B a d \,e^{4} x^{2}-18 B c \,d^{3} e^{2} x^{2}+1386 A a d \,e^{4} x -44 A c \,d^{3} e^{2} x +99 B a \,d^{2} e^{3} x +24 B c \,d^{4} e x +693 A a \,d^{2} e^{3}+88 A c \,d^{4} e -198 B a \,d^{3} e^{2}-48 B c \,d^{5}\right ) \sqrt {e x +d}}{3465 e^{4}}\) | \(201\) |
2/5*((5/9*x^2*(9/11*B*x+A)*c+a*(5/7*B*x+A))*e^3-20/63*d*(x*(21/22*B*x+A)*c +9/10*B*a)*e^2+8/63*c*d^2*(15/11*B*x+A)*e-16/231*B*c*d^3)*(e*x+d)^(5/2)/e^ 4
Time = 0.41 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.64 \[ \int (A+B x) (d+e x)^{3/2} \left (a+c x^2\right ) \, dx=\frac {2 \, {\left (315 \, B c e^{5} x^{5} - 48 \, B c d^{5} + 88 \, A c d^{4} e - 198 \, B a d^{3} e^{2} + 693 \, A a d^{2} e^{3} + 35 \, {\left (12 \, B c d e^{4} + 11 \, A c e^{5}\right )} x^{4} + 5 \, {\left (3 \, B c d^{2} e^{3} + 110 \, A c d e^{4} + 99 \, B a e^{5}\right )} x^{3} - 3 \, {\left (6 \, B c d^{3} e^{2} - 11 \, A c d^{2} e^{3} - 264 \, B a d e^{4} - 231 \, A a e^{5}\right )} x^{2} + {\left (24 \, B c d^{4} e - 44 \, A c d^{3} e^{2} + 99 \, B a d^{2} e^{3} + 1386 \, A a d e^{4}\right )} x\right )} \sqrt {e x + d}}{3465 \, e^{4}} \]
2/3465*(315*B*c*e^5*x^5 - 48*B*c*d^5 + 88*A*c*d^4*e - 198*B*a*d^3*e^2 + 69 3*A*a*d^2*e^3 + 35*(12*B*c*d*e^4 + 11*A*c*e^5)*x^4 + 5*(3*B*c*d^2*e^3 + 11 0*A*c*d*e^4 + 99*B*a*e^5)*x^3 - 3*(6*B*c*d^3*e^2 - 11*A*c*d^2*e^3 - 264*B* a*d*e^4 - 231*A*a*e^5)*x^2 + (24*B*c*d^4*e - 44*A*c*d^3*e^2 + 99*B*a*d^2*e ^3 + 1386*A*a*d*e^4)*x)*sqrt(e*x + d)/e^4
Time = 0.85 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.44 \[ \int (A+B x) (d+e x)^{3/2} \left (a+c x^2\right ) \, dx=\begin {cases} \frac {2 \left (\frac {B c \left (d + e x\right )^{\frac {11}{2}}}{11 e^{3}} + \frac {\left (d + e x\right )^{\frac {9}{2}} \left (A c e - 3 B c d\right )}{9 e^{3}} + \frac {\left (d + e x\right )^{\frac {7}{2}} \left (- 2 A c d e + B a e^{2} + 3 B c d^{2}\right )}{7 e^{3}} + \frac {\left (d + e x\right )^{\frac {5}{2}} \left (A a e^{3} + A c d^{2} e - B a d e^{2} - B c d^{3}\right )}{5 e^{3}}\right )}{e} & \text {for}\: e \neq 0 \\d^{\frac {3}{2}} \left (A a x + \frac {A c x^{3}}{3} + \frac {B a x^{2}}{2} + \frac {B c x^{4}}{4}\right ) & \text {otherwise} \end {cases} \]
Piecewise((2*(B*c*(d + e*x)**(11/2)/(11*e**3) + (d + e*x)**(9/2)*(A*c*e - 3*B*c*d)/(9*e**3) + (d + e*x)**(7/2)*(-2*A*c*d*e + B*a*e**2 + 3*B*c*d**2)/ (7*e**3) + (d + e*x)**(5/2)*(A*a*e**3 + A*c*d**2*e - B*a*d*e**2 - B*c*d**3 )/(5*e**3))/e, Ne(e, 0)), (d**(3/2)*(A*a*x + A*c*x**3/3 + B*a*x**2/2 + B*c *x**4/4), True))
Time = 0.19 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.90 \[ \int (A+B x) (d+e x)^{3/2} \left (a+c x^2\right ) \, dx=\frac {2 \, {\left (315 \, {\left (e x + d\right )}^{\frac {11}{2}} B c - 385 \, {\left (3 \, B c d - A c e\right )} {\left (e x + d\right )}^{\frac {9}{2}} + 495 \, {\left (3 \, B c d^{2} - 2 \, A c d e + B a e^{2}\right )} {\left (e x + d\right )}^{\frac {7}{2}} - 693 \, {\left (B c d^{3} - A c d^{2} e + B a d e^{2} - A a e^{3}\right )} {\left (e x + d\right )}^{\frac {5}{2}}\right )}}{3465 \, e^{4}} \]
2/3465*(315*(e*x + d)^(11/2)*B*c - 385*(3*B*c*d - A*c*e)*(e*x + d)^(9/2) + 495*(3*B*c*d^2 - 2*A*c*d*e + B*a*e^2)*(e*x + d)^(7/2) - 693*(B*c*d^3 - A* c*d^2*e + B*a*d*e^2 - A*a*e^3)*(e*x + d)^(5/2))/e^4
Leaf count of result is larger than twice the leaf count of optimal. 548 vs. \(2 (100) = 200\).
Time = 0.28 (sec) , antiderivative size = 548, normalized size of antiderivative = 4.72 \[ \int (A+B x) (d+e x)^{3/2} \left (a+c x^2\right ) \, dx=\frac {2 \, {\left (3465 \, \sqrt {e x + d} A a d^{2} + 2310 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} A a d + \frac {1155 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} B a d^{2}}{e} + 231 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} A a + \frac {231 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} A c d^{2}}{e^{2}} + \frac {462 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} B a d}{e} + \frac {99 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} B c d^{2}}{e^{3}} + \frac {198 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} A c d}{e^{2}} + \frac {99 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} B a}{e} + \frac {22 \, {\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} - 180 \, {\left (e x + d\right )}^{\frac {7}{2}} d + 378 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2} - 420 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3} + 315 \, \sqrt {e x + d} d^{4}\right )} B c d}{e^{3}} + \frac {11 \, {\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} - 180 \, {\left (e x + d\right )}^{\frac {7}{2}} d + 378 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2} - 420 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3} + 315 \, \sqrt {e x + d} d^{4}\right )} A c}{e^{2}} + \frac {5 \, {\left (63 \, {\left (e x + d\right )}^{\frac {11}{2}} - 385 \, {\left (e x + d\right )}^{\frac {9}{2}} d + 990 \, {\left (e x + d\right )}^{\frac {7}{2}} d^{2} - 1386 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{3} + 1155 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{4} - 693 \, \sqrt {e x + d} d^{5}\right )} B c}{e^{3}}\right )}}{3465 \, e} \]
2/3465*(3465*sqrt(e*x + d)*A*a*d^2 + 2310*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*A*a*d + 1155*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*B*a*d^2/e + 231*( 3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*A*a + 231 *(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*A*c*d^2 /e^2 + 462*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^ 2)*B*a*d/e + 99*(5*(e*x + d)^(7/2) - 21*(e*x + d)^(5/2)*d + 35*(e*x + d)^( 3/2)*d^2 - 35*sqrt(e*x + d)*d^3)*B*c*d^2/e^3 + 198*(5*(e*x + d)^(7/2) - 21 *(e*x + d)^(5/2)*d + 35*(e*x + d)^(3/2)*d^2 - 35*sqrt(e*x + d)*d^3)*A*c*d/ e^2 + 99*(5*(e*x + d)^(7/2) - 21*(e*x + d)^(5/2)*d + 35*(e*x + d)^(3/2)*d^ 2 - 35*sqrt(e*x + d)*d^3)*B*a/e + 22*(35*(e*x + d)^(9/2) - 180*(e*x + d)^( 7/2)*d + 378*(e*x + d)^(5/2)*d^2 - 420*(e*x + d)^(3/2)*d^3 + 315*sqrt(e*x + d)*d^4)*B*c*d/e^3 + 11*(35*(e*x + d)^(9/2) - 180*(e*x + d)^(7/2)*d + 378 *(e*x + d)^(5/2)*d^2 - 420*(e*x + d)^(3/2)*d^3 + 315*sqrt(e*x + d)*d^4)*A* c/e^2 + 5*(63*(e*x + d)^(11/2) - 385*(e*x + d)^(9/2)*d + 990*(e*x + d)^(7/ 2)*d^2 - 1386*(e*x + d)^(5/2)*d^3 + 1155*(e*x + d)^(3/2)*d^4 - 693*sqrt(e* x + d)*d^5)*B*c/e^3)/e
Time = 0.10 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.86 \[ \int (A+B x) (d+e x)^{3/2} \left (a+c x^2\right ) \, dx=\frac {{\left (d+e\,x\right )}^{7/2}\,\left (6\,B\,c\,d^2-4\,A\,c\,d\,e+2\,B\,a\,e^2\right )}{7\,e^4}+\frac {2\,B\,c\,{\left (d+e\,x\right )}^{11/2}}{11\,e^4}+\frac {2\,c\,\left (A\,e-3\,B\,d\right )\,{\left (d+e\,x\right )}^{9/2}}{9\,e^4}+\frac {2\,\left (c\,d^2+a\,e^2\right )\,\left (A\,e-B\,d\right )\,{\left (d+e\,x\right )}^{5/2}}{5\,e^4} \]