Integrand size = 31, antiderivative size = 128 \[ \int (A+B x) \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=-\frac {2 (b d-a e)^2 (B d-A e) (d+e x)^{3/2}}{3 e^4}+\frac {2 (b d-a e) (3 b B d-2 A b e-a B e) (d+e x)^{5/2}}{5 e^4}-\frac {2 b (3 b B d-A b e-2 a B e) (d+e x)^{7/2}}{7 e^4}+\frac {2 b^2 B (d+e x)^{9/2}}{9 e^4} \]
-2/3*(-a*e+b*d)^2*(-A*e+B*d)*(e*x+d)^(3/2)/e^4+2/5*(-a*e+b*d)*(-2*A*b*e-B* a*e+3*B*b*d)*(e*x+d)^(5/2)/e^4-2/7*b*(-A*b*e-2*B*a*e+3*B*b*d)*(e*x+d)^(7/2 )/e^4+2/9*b^2*B*(e*x+d)^(9/2)/e^4
Time = 0.04 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.08 \[ \int (A+B x) \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {2 (d+e x)^{3/2} \left (21 a^2 e^2 (-2 B d+5 A e+3 B e x)+6 a b e \left (7 A e (-2 d+3 e x)+B \left (8 d^2-12 d e x+15 e^2 x^2\right )\right )+b^2 \left (3 A e \left (8 d^2-12 d e x+15 e^2 x^2\right )+B \left (-16 d^3+24 d^2 e x-30 d e^2 x^2+35 e^3 x^3\right )\right )\right )}{315 e^4} \]
(2*(d + e*x)^(3/2)*(21*a^2*e^2*(-2*B*d + 5*A*e + 3*B*e*x) + 6*a*b*e*(7*A*e *(-2*d + 3*e*x) + B*(8*d^2 - 12*d*e*x + 15*e^2*x^2)) + b^2*(3*A*e*(8*d^2 - 12*d*e*x + 15*e^2*x^2) + B*(-16*d^3 + 24*d^2*e*x - 30*d*e^2*x^2 + 35*e^3* x^3))))/(315*e^4)
Time = 0.27 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {1184, 27, 86, 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.
| \(\displaystyle \int \left (a^2+2 a b x+b^2 x^2\right ) (A+B x) \sqrt {d+e x} \, dx\) |
| \(\Big \downarrow \) 1184 |
| \(\displaystyle \frac {\int b^2 (a+b x)^2 (A+B x) \sqrt {d+e x}dx}{b^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int (a+b x)^2 (A+B x) \sqrt {d+e x}dx\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \int \left (\frac {b (d+e x)^{5/2} (2 a B e+A b e-3 b B d)}{e^3}+\frac {(d+e x)^{3/2} (a e-b d) (a B e+2 A b e-3 b B d)}{e^3}+\frac {\sqrt {d+e x} (a e-b d)^2 (A e-B d)}{e^3}+\frac {b^2 B (d+e x)^{7/2}}{e^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 b (d+e x)^{7/2} (-2 a B e-A b e+3 b B d)}{7 e^4}+\frac {2 (d+e x)^{5/2} (b d-a e) (-a B e-2 A b e+3 b B d)}{5 e^4}-\frac {2 (d+e x)^{3/2} (b d-a e)^2 (B d-A e)}{3 e^4}+\frac {2 b^2 B (d+e x)^{9/2}}{9 e^4}\) |
(-2*(b*d - a*e)^2*(B*d - A*e)*(d + e*x)^(3/2))/(3*e^4) + (2*(b*d - a*e)*(3 *b*B*d - 2*A*b*e - a*B*e)*(d + e*x)^(5/2))/(5*e^4) - (2*b*(3*b*B*d - A*b*e - 2*a*B*e)*(d + e*x)^(7/2))/(7*e^4) + (2*b^2*B*(d + e*x)^(9/2))/(9*e^4)
3.18.87.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^p Int[(d + e*x)^m*(f + g*x )^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && E qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 0.24 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.91
| method | result | size |
| pseudoelliptic | \(\frac {2 \left (\left (\frac {3 x^{2} \left (\frac {7 B x}{9}+A \right ) b^{2}}{7}+\frac {6 x \left (\frac {5 B x}{7}+A \right ) a b}{5}+a^{2} \left (\frac {3 B x}{5}+A \right )\right ) e^{3}-\frac {4 \left (\frac {3 x \left (\frac {5 B x}{6}+A \right ) b^{2}}{7}+a \left (\frac {6 B x}{7}+A \right ) b +\frac {B \,a^{2}}{2}\right ) d \,e^{2}}{5}+\frac {8 b \left (\left (B x +A \right ) b +2 B a \right ) d^{2} e}{35}-\frac {16 B \,b^{2} d^{3}}{105}\right ) \left (e x +d \right )^{\frac {3}{2}}}{3 e^{4}}\) | \(116\) |
| derivativedivides | \(\frac {\frac {2 B \,b^{2} \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {2 \left (\left (A e -B d \right ) b^{2}+B \left (2 a b e -2 b^{2} d \right )\right ) \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (\left (A e -B d \right ) \left (2 a b e -2 b^{2} d \right )+B \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right )\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 \left (A e -B d \right ) \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right ) \left (e x +d \right )^{\frac {3}{2}}}{3}}{e^{4}}\) | \(148\) |
| default | \(\frac {\frac {2 B \,b^{2} \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {2 \left (\left (A e -B d \right ) b^{2}+B \left (2 a b e -2 b^{2} d \right )\right ) \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (\left (A e -B d \right ) \left (2 a b e -2 b^{2} d \right )+B \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right )\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 \left (A e -B d \right ) \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right ) \left (e x +d \right )^{\frac {3}{2}}}{3}}{e^{4}}\) | \(148\) |
| gosper | \(\frac {2 \left (e x +d \right )^{\frac {3}{2}} \left (35 B \,b^{2} x^{3} e^{3}+45 A \,b^{2} e^{3} x^{2}+90 B \,x^{2} a b \,e^{3}-30 B \,x^{2} b^{2} d \,e^{2}+126 A x a b \,e^{3}-36 A x \,b^{2} d \,e^{2}+63 B x \,a^{2} e^{3}-72 B x a b d \,e^{2}+24 B x \,b^{2} d^{2} e +105 A \,a^{2} e^{3}-84 A a b d \,e^{2}+24 A \,b^{2} d^{2} e -42 B \,a^{2} d \,e^{2}+48 B a b \,d^{2} e -16 B \,b^{2} d^{3}\right )}{315 e^{4}}\) | \(169\) |
| trager | \(\frac {2 \left (35 B \,b^{2} e^{4} x^{4}+45 A \,b^{2} e^{4} x^{3}+90 B a b \,e^{4} x^{3}+5 B \,b^{2} e^{3} d \,x^{3}+126 A a b \,e^{4} x^{2}+9 A \,b^{2} d \,e^{3} x^{2}+63 B \,a^{2} e^{4} x^{2}+18 B a b d \,e^{3} x^{2}-6 B \,b^{2} d^{2} e^{2} x^{2}+105 A \,a^{2} e^{4} x +42 A a b d \,e^{3} x -12 A \,b^{2} d^{2} e^{2} x +21 B \,a^{2} d \,e^{3} x -24 B a b \,d^{2} e^{2} x +8 B \,b^{2} d^{3} e x +105 A \,a^{2} d \,e^{3}-84 A a b \,d^{2} e^{2}+24 A \,b^{2} d^{3} e -42 B \,a^{2} d^{2} e^{2}+48 B a b \,d^{3} e -16 B \,b^{2} d^{4}\right ) \sqrt {e x +d}}{315 e^{4}}\) | \(253\) |
| risch | \(\frac {2 \left (35 B \,b^{2} e^{4} x^{4}+45 A \,b^{2} e^{4} x^{3}+90 B a b \,e^{4} x^{3}+5 B \,b^{2} e^{3} d \,x^{3}+126 A a b \,e^{4} x^{2}+9 A \,b^{2} d \,e^{3} x^{2}+63 B \,a^{2} e^{4} x^{2}+18 B a b d \,e^{3} x^{2}-6 B \,b^{2} d^{2} e^{2} x^{2}+105 A \,a^{2} e^{4} x +42 A a b d \,e^{3} x -12 A \,b^{2} d^{2} e^{2} x +21 B \,a^{2} d \,e^{3} x -24 B a b \,d^{2} e^{2} x +8 B \,b^{2} d^{3} e x +105 A \,a^{2} d \,e^{3}-84 A a b \,d^{2} e^{2}+24 A \,b^{2} d^{3} e -42 B \,a^{2} d^{2} e^{2}+48 B a b \,d^{3} e -16 B \,b^{2} d^{4}\right ) \sqrt {e x +d}}{315 e^{4}}\) | \(253\) |
2/3*((3/7*x^2*(7/9*B*x+A)*b^2+6/5*x*(5/7*B*x+A)*a*b+a^2*(3/5*B*x+A))*e^3-4 /5*(3/7*x*(5/6*B*x+A)*b^2+a*(6/7*B*x+A)*b+1/2*B*a^2)*d*e^2+8/35*b*((B*x+A) *b+2*B*a)*d^2*e-16/105*B*b^2*d^3)*(e*x+d)^(3/2)/e^4
Time = 0.32 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.72 \[ \int (A+B x) \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {2 \, {\left (35 \, B b^{2} e^{4} x^{4} - 16 \, B b^{2} d^{4} + 105 \, A a^{2} d e^{3} + 24 \, {\left (2 \, B a b + A b^{2}\right )} d^{3} e - 42 \, {\left (B a^{2} + 2 \, A a b\right )} d^{2} e^{2} + 5 \, {\left (B b^{2} d e^{3} + 9 \, {\left (2 \, B a b + A b^{2}\right )} e^{4}\right )} x^{3} - 3 \, {\left (2 \, B b^{2} d^{2} e^{2} - 3 \, {\left (2 \, B a b + A b^{2}\right )} d e^{3} - 21 \, {\left (B a^{2} + 2 \, A a b\right )} e^{4}\right )} x^{2} + {\left (8 \, B b^{2} d^{3} e + 105 \, A a^{2} e^{4} - 12 \, {\left (2 \, B a b + A b^{2}\right )} d^{2} e^{2} + 21 \, {\left (B a^{2} + 2 \, A a b\right )} d e^{3}\right )} x\right )} \sqrt {e x + d}}{315 \, e^{4}} \]
2/315*(35*B*b^2*e^4*x^4 - 16*B*b^2*d^4 + 105*A*a^2*d*e^3 + 24*(2*B*a*b + A *b^2)*d^3*e - 42*(B*a^2 + 2*A*a*b)*d^2*e^2 + 5*(B*b^2*d*e^3 + 9*(2*B*a*b + A*b^2)*e^4)*x^3 - 3*(2*B*b^2*d^2*e^2 - 3*(2*B*a*b + A*b^2)*d*e^3 - 21*(B* a^2 + 2*A*a*b)*e^4)*x^2 + (8*B*b^2*d^3*e + 105*A*a^2*e^4 - 12*(2*B*a*b + A *b^2)*d^2*e^2 + 21*(B*a^2 + 2*A*a*b)*d*e^3)*x)*sqrt(e*x + d)/e^4
Time = 1.05 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.01 \[ \int (A+B x) \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\begin {cases} \frac {2 \left (\frac {B b^{2} \left (d + e x\right )^{\frac {9}{2}}}{9 e^{3}} + \frac {\left (d + e x\right )^{\frac {7}{2}} \left (A b^{2} e + 2 B a b e - 3 B b^{2} d\right )}{7 e^{3}} + \frac {\left (d + e x\right )^{\frac {5}{2}} \cdot \left (2 A a b e^{2} - 2 A b^{2} d e + B a^{2} e^{2} - 4 B a b d e + 3 B b^{2} d^{2}\right )}{5 e^{3}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (A a^{2} e^{3} - 2 A a b d e^{2} + A b^{2} d^{2} e - B a^{2} d e^{2} + 2 B a b d^{2} e - B b^{2} d^{3}\right )}{3 e^{3}}\right )}{e} & \text {for}\: e \neq 0 \\\sqrt {d} \left (A a^{2} x + \frac {B b^{2} x^{4}}{4} + \frac {x^{3} \left (A b^{2} + 2 B a b\right )}{3} + \frac {x^{2} \cdot \left (2 A a b + B a^{2}\right )}{2}\right ) & \text {otherwise} \end {cases} \]
Piecewise((2*(B*b**2*(d + e*x)**(9/2)/(9*e**3) + (d + e*x)**(7/2)*(A*b**2* e + 2*B*a*b*e - 3*B*b**2*d)/(7*e**3) + (d + e*x)**(5/2)*(2*A*a*b*e**2 - 2* A*b**2*d*e + B*a**2*e**2 - 4*B*a*b*d*e + 3*B*b**2*d**2)/(5*e**3) + (d + e* x)**(3/2)*(A*a**2*e**3 - 2*A*a*b*d*e**2 + A*b**2*d**2*e - B*a**2*d*e**2 + 2*B*a*b*d**2*e - B*b**2*d**3)/(3*e**3))/e, Ne(e, 0)), (sqrt(d)*(A*a**2*x + B*b**2*x**4/4 + x**3*(A*b**2 + 2*B*a*b)/3 + x**2*(2*A*a*b + B*a**2)/2), T rue))
Time = 0.19 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.24 \[ \int (A+B x) \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {2 \, {\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} B b^{2} - 45 \, {\left (3 \, B b^{2} d - {\left (2 \, B a b + A b^{2}\right )} e\right )} {\left (e x + d\right )}^{\frac {7}{2}} + 63 \, {\left (3 \, B b^{2} d^{2} - 2 \, {\left (2 \, B a b + A b^{2}\right )} d e + {\left (B a^{2} + 2 \, A a b\right )} e^{2}\right )} {\left (e x + d\right )}^{\frac {5}{2}} - 105 \, {\left (B b^{2} d^{3} - A a^{2} e^{3} - {\left (2 \, B a b + A b^{2}\right )} d^{2} e + {\left (B a^{2} + 2 \, A a b\right )} d e^{2}\right )} {\left (e x + d\right )}^{\frac {3}{2}}\right )}}{315 \, e^{4}} \]
2/315*(35*(e*x + d)^(9/2)*B*b^2 - 45*(3*B*b^2*d - (2*B*a*b + A*b^2)*e)*(e* x + d)^(7/2) + 63*(3*B*b^2*d^2 - 2*(2*B*a*b + A*b^2)*d*e + (B*a^2 + 2*A*a* b)*e^2)*(e*x + d)^(5/2) - 105*(B*b^2*d^3 - A*a^2*e^3 - (2*B*a*b + A*b^2)*d ^2*e + (B*a^2 + 2*A*a*b)*d*e^2)*(e*x + d)^(3/2))/e^4
Leaf count of result is larger than twice the leaf count of optimal. 486 vs. \(2 (112) = 224\).
Time = 0.32 (sec) , antiderivative size = 486, normalized size of antiderivative = 3.80 \[ \int (A+B x) \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {2 \, {\left (315 \, \sqrt {e x + d} A a^{2} d + 105 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} A a^{2} + \frac {105 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} B a^{2} d}{e} + \frac {210 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} A a b d}{e} + \frac {42 \, {\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 b 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 )} A b^{2} 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^{2}}{e} + \frac {42 \, {\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 b}{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 b^{2} d}{e^{3}} + \frac {18 \, {\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 b}{e^{2}} + \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 b^{2}}{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 b^{2}}{e^{3}}\right )}}{315 \, e} \]
2/315*(315*sqrt(e*x + d)*A*a^2*d + 105*((e*x + d)^(3/2) - 3*sqrt(e*x + d)* d)*A*a^2 + 105*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*B*a^2*d/e + 210*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*A*a*b*d/e + 42*(3*(e*x + d)^(5/2) - 10*(e *x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*B*a*b*d/e^2 + 21*(3*(e*x + d)^(5/2 ) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*A*b^2*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^2/e + 42*(3* (e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*A*a*b/e + 9 *(5*(e*x + d)^(7/2) - 21*(e*x + d)^(5/2)*d + 35*(e*x + d)^(3/2)*d^2 - 35*s qrt(e*x + d)*d^3)*B*b^2*d/e^3 + 18*(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*b/e^2 + 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*b^2/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*b^2/ e^3)/e
Time = 0.08 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.90 \[ \int (A+B x) \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {{\left (d+e\,x\right )}^{7/2}\,\left (2\,A\,b^2\,e-6\,B\,b^2\,d+4\,B\,a\,b\,e\right )}{7\,e^4}+\frac {2\,B\,b^2\,{\left (d+e\,x\right )}^{9/2}}{9\,e^4}+\frac {2\,\left (a\,e-b\,d\right )\,{\left (d+e\,x\right )}^{5/2}\,\left (2\,A\,b\,e+B\,a\,e-3\,B\,b\,d\right )}{5\,e^4}+\frac {2\,\left (A\,e-B\,d\right )\,{\left (a\,e-b\,d\right )}^2\,{\left (d+e\,x\right )}^{3/2}}{3\,e^4} \]