Integrand size = 22, antiderivative size = 116 \[ \int (A+B x) \sqrt {d+e x} \left (a+c x^2\right ) \, dx=-\frac {2 (B d-A e) \left (c d^2+a e^2\right ) (d+e x)^{3/2}}{3 e^4}+\frac {2 \left (3 B c d^2-2 A c d e+a B e^2\right ) (d+e x)^{5/2}}{5 e^4}-\frac {2 c (3 B d-A e) (d+e x)^{7/2}}{7 e^4}+\frac {2 B c (d+e x)^{9/2}}{9 e^4} \]
-2/3*(-A*e+B*d)*(a*e^2+c*d^2)*(e*x+d)^(3/2)/e^4+2/5*(-2*A*c*d*e+B*a*e^2+3* B*c*d^2)*(e*x+d)^(5/2)/e^4-2/7*c*(-A*e+3*B*d)*(e*x+d)^(7/2)/e^4+2/9*B*c*(e *x+d)^(9/2)/e^4
Time = 0.07 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.83 \[ \int (A+B x) \sqrt {d+e x} \left (a+c x^2\right ) \, dx=\frac {2 (d+e x)^{3/2} \left (105 a A e^3+21 a B e^2 (-2 d+3 e x)+3 A c e \left (8 d^2-12 d e x+15 e^2 x^2\right )+B c \left (-16 d^3+24 d^2 e x-30 d e^2 x^2+35 e^3 x^3\right )\right )}{315 e^4} \]
(2*(d + e*x)^(3/2)*(105*a*A*e^3 + 21*a*B*e^2*(-2*d + 3*e*x) + 3*A*c*e*(8*d ^2 - 12*d*e*x + 15*e^2*x^2) + B*c*(-16*d^3 + 24*d^2*e*x - 30*d*e^2*x^2 + 3 5*e^3*x^3)))/(315*e^4)
Time = 0.25 (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.
\(\displaystyle \int \left (a+c x^2\right ) (A+B x) \sqrt {d+e x} \, dx\) |
\(\Big \downarrow \) 652 |
\(\displaystyle \int \left (\frac {(d+e x)^{3/2} \left (a B e^2-2 A c d e+3 B c d^2\right )}{e^3}+\frac {\sqrt {d+e x} \left (a e^2+c d^2\right ) (A e-B d)}{e^3}+\frac {c (d+e x)^{5/2} (A e-3 B d)}{e^3}+\frac {B c (d+e x)^{7/2}}{e^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 (d+e x)^{5/2} \left (a B e^2-2 A c d e+3 B c d^2\right )}{5 e^4}-\frac {2 (d+e x)^{3/2} \left (a e^2+c d^2\right ) (B d-A e)}{3 e^4}-\frac {2 c (d+e x)^{7/2} (3 B d-A e)}{7 e^4}+\frac {2 B c (d+e x)^{9/2}}{9 e^4}\) |
(-2*(B*d - A*e)*(c*d^2 + a*e^2)*(d + e*x)^(3/2))/(3*e^4) + (2*(3*B*c*d^2 - 2*A*c*d*e + a*B*e^2)*(d + e*x)^(5/2))/(5*e^4) - (2*c*(3*B*d - A*e)*(d + e *x)^(7/2))/(7*e^4) + (2*B*c*(d + e*x)^(9/2))/(9*e^4)
3.15.30.3.1 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 = 0.31 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.67
method | result | size |
pseudoelliptic | \(\frac {2 \left (e x +d \right )^{\frac {3}{2}} \left (\left (\frac {3 x^{2} \left (\frac {7 B x}{9}+A \right ) c}{7}+a \left (\frac {3 B x}{5}+A \right )\right ) e^{3}-\frac {12 \left (x \left (\frac {5 B x}{6}+A \right ) c +\frac {7 B a}{6}\right ) d \,e^{2}}{35}+\frac {8 c \,d^{2} \left (B x +A \right ) e}{35}-\frac {16 B c \,d^{3}}{105}\right )}{3 e^{4}}\) | \(78\) |
gosper | \(\frac {2 \left (e x +d \right )^{\frac {3}{2}} \left (35 B c \,x^{3} e^{3}+45 A c \,e^{3} x^{2}-30 B \,x^{2} c d \,e^{2}-36 A c d \,e^{2} x +63 B x a \,e^{3}+24 B c \,d^{2} e x +105 A a \,e^{3}+24 A c \,d^{2} e -42 B a d \,e^{2}-16 B c \,d^{3}\right )}{315 e^{4}}\) | \(101\) |
derivativedivides | \(\frac {\frac {2 B c \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {2 \left (\left (A e -B d \right ) c -2 B c d \right ) \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (-2 \left (A e -B d \right ) d c +B \left (e^{2} a +c \,d^{2}\right )\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 \left (A e -B d \right ) \left (e^{2} a +c \,d^{2}\right ) \left (e x +d \right )^{\frac {3}{2}}}{3}}{e^{4}}\) | \(106\) |
default | \(\frac {\frac {2 B c \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {2 \left (\left (A e -B d \right ) c -2 B c d \right ) \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (-2 \left (A e -B d \right ) d c +B \left (e^{2} a +c \,d^{2}\right )\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 \left (A e -B d \right ) \left (e^{2} a +c \,d^{2}\right ) \left (e x +d \right )^{\frac {3}{2}}}{3}}{e^{4}}\) | \(106\) |
trager | \(\frac {2 \left (35 B \,e^{4} c \,x^{4}+45 A c \,e^{4} x^{3}+5 B c d \,e^{3} x^{3}+9 A c d \,e^{3} x^{2}+63 B a \,e^{4} x^{2}-6 B c \,d^{2} e^{2} x^{2}+105 A a \,e^{4} x -12 A c \,d^{2} e^{2} x +21 B a d \,e^{3} x +8 B c \,d^{3} e x +105 A a d \,e^{3}+24 A c \,d^{3} e -42 B a \,d^{2} e^{2}-16 B c \,d^{4}\right ) \sqrt {e x +d}}{315 e^{4}}\) | \(149\) |
risch | \(\frac {2 \left (35 B \,e^{4} c \,x^{4}+45 A c \,e^{4} x^{3}+5 B c d \,e^{3} x^{3}+9 A c d \,e^{3} x^{2}+63 B a \,e^{4} x^{2}-6 B c \,d^{2} e^{2} x^{2}+105 A a \,e^{4} x -12 A c \,d^{2} e^{2} x +21 B a d \,e^{3} x +8 B c \,d^{3} e x +105 A a d \,e^{3}+24 A c \,d^{3} e -42 B a \,d^{2} e^{2}-16 B c \,d^{4}\right ) \sqrt {e x +d}}{315 e^{4}}\) | \(149\) |
2/3*(e*x+d)^(3/2)*((3/7*x^2*(7/9*B*x+A)*c+a*(3/5*B*x+A))*e^3-12/35*(x*(5/6 *B*x+A)*c+7/6*B*a)*d*e^2+8/35*c*d^2*(B*x+A)*e-16/105*B*c*d^3)/e^4
Time = 0.28 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.23 \[ \int (A+B x) \sqrt {d+e x} \left (a+c x^2\right ) \, dx=\frac {2 \, {\left (35 \, B c e^{4} x^{4} - 16 \, B c d^{4} + 24 \, A c d^{3} e - 42 \, B a d^{2} e^{2} + 105 \, A a d e^{3} + 5 \, {\left (B c d e^{3} + 9 \, A c e^{4}\right )} x^{3} - 3 \, {\left (2 \, B c d^{2} e^{2} - 3 \, A c d e^{3} - 21 \, B a e^{4}\right )} x^{2} + {\left (8 \, B c d^{3} e - 12 \, A c d^{2} e^{2} + 21 \, B a d e^{3} + 105 \, A a e^{4}\right )} x\right )} \sqrt {e x + d}}{315 \, e^{4}} \]
2/315*(35*B*c*e^4*x^4 - 16*B*c*d^4 + 24*A*c*d^3*e - 42*B*a*d^2*e^2 + 105*A *a*d*e^3 + 5*(B*c*d*e^3 + 9*A*c*e^4)*x^3 - 3*(2*B*c*d^2*e^2 - 3*A*c*d*e^3 - 21*B*a*e^4)*x^2 + (8*B*c*d^3*e - 12*A*c*d^2*e^2 + 21*B*a*d*e^3 + 105*A*a *e^4)*x)*sqrt(e*x + d)/e^4
Time = 0.77 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.44 \[ \int (A+B x) \sqrt {d+e x} \left (a+c x^2\right ) \, dx=\begin {cases} \frac {2 \left (\frac {B c \left (d + e x\right )^{\frac {9}{2}}}{9 e^{3}} + \frac {\left (d + e x\right )^{\frac {7}{2}} \left (A c e - 3 B c d\right )}{7 e^{3}} + \frac {\left (d + e x\right )^{\frac {5}{2}} \left (- 2 A c d e + B a e^{2} + 3 B c d^{2}\right )}{5 e^{3}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (A a e^{3} + A c d^{2} e - B a d e^{2} - B c d^{3}\right )}{3 e^{3}}\right )}{e} & \text {for}\: e \neq 0 \\\sqrt {d} \left (A a x + \frac {A c x^{3}}{3} + \frac {B a x^{2}}{2} + \frac {B c x^{4}}{4}\right ) & \text {otherwise} \end {cases} \]
Piecewise((2*(B*c*(d + e*x)**(9/2)/(9*e**3) + (d + e*x)**(7/2)*(A*c*e - 3* B*c*d)/(7*e**3) + (d + e*x)**(5/2)*(-2*A*c*d*e + B*a*e**2 + 3*B*c*d**2)/(5 *e**3) + (d + e*x)**(3/2)*(A*a*e**3 + A*c*d**2*e - B*a*d*e**2 - B*c*d**3)/ (3*e**3))/e, Ne(e, 0)), (sqrt(d)*(A*a*x + A*c*x**3/3 + B*a*x**2/2 + B*c*x* *4/4), True))
Time = 0.20 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.90 \[ \int (A+B x) \sqrt {d+e x} \left (a+c x^2\right ) \, dx=\frac {2 \, {\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} B c - 45 \, {\left (3 \, B c d - A c e\right )} {\left (e x + d\right )}^{\frac {7}{2}} + 63 \, {\left (3 \, B c d^{2} - 2 \, A c d e + B a e^{2}\right )} {\left (e x + d\right )}^{\frac {5}{2}} - 105 \, {\left (B c d^{3} - A c d^{2} e + B a d e^{2} - A a e^{3}\right )} {\left (e x + d\right )}^{\frac {3}{2}}\right )}}{315 \, e^{4}} \]
2/315*(35*(e*x + d)^(9/2)*B*c - 45*(3*B*c*d - A*c*e)*(e*x + d)^(7/2) + 63* (3*B*c*d^2 - 2*A*c*d*e + B*a*e^2)*(e*x + d)^(5/2) - 105*(B*c*d^3 - A*c*d^2 *e + B*a*d*e^2 - A*a*e^3)*(e*x + d)^(3/2))/e^4
Leaf count of result is larger than twice the leaf count of optimal. 310 vs. \(2 (100) = 200\).
Time = 0.29 (sec) , antiderivative size = 310, normalized size of antiderivative = 2.67 \[ \int (A+B x) \sqrt {d+e x} \left (a+c x^2\right ) \, dx=\frac {2 \, {\left (315 \, \sqrt {e x + d} A a d + 105 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} A a + \frac {105 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} B a d}{e} + \frac {21 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} A c d}{e^{2}} + \frac {21 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} B a}{e} + \frac {9 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} B c d}{e^{3}} + \frac {9 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} A c}{e^{2}} + \frac {{\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} - 180 \, {\left (e x + d\right )}^{\frac {7}{2}} d + 378 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2} - 420 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3} + 315 \, \sqrt {e x + d} d^{4}\right )} B c}{e^{3}}\right )}}{315 \, e} \]
2/315*(315*sqrt(e*x + d)*A*a*d + 105*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d) *A*a + 105*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*B*a*d/e + 21*(3*(e*x + d) ^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*A*c*d/e^2 + 21*(3*(e *x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*B*a/e + 9*(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/e^3 + 9*(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/e^2 + (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/e^3)/e
Time = 10.46 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.86 \[ \int (A+B x) \sqrt {d+e x} \left (a+c x^2\right ) \, dx=\frac {{\left (d+e\,x\right )}^{5/2}\,\left (6\,B\,c\,d^2-4\,A\,c\,d\,e+2\,B\,a\,e^2\right )}{5\,e^4}+\frac {2\,B\,c\,{\left (d+e\,x\right )}^{9/2}}{9\,e^4}+\frac {2\,c\,\left (A\,e-3\,B\,d\right )\,{\left (d+e\,x\right )}^{7/2}}{7\,e^4}+\frac {2\,\left (c\,d^2+a\,e^2\right )\,\left (A\,e-B\,d\right )\,{\left (d+e\,x\right )}^{3/2}}{3\,e^4} \]