Integrand size = 22, antiderivative size = 126 \[ \int \frac {(a+b x)^2 (A+B x)}{\sqrt {d+e x}} \, dx=-\frac {2 (b d-a e)^2 (B d-A e) \sqrt {d+e x}}{e^4}+\frac {2 (b d-a e) (3 b B d-2 A b e-a B e) (d+e x)^{3/2}}{3 e^4}-\frac {2 b (3 b B d-A b e-2 a B e) (d+e x)^{5/2}}{5 e^4}+\frac {2 b^2 B (d+e x)^{7/2}}{7 e^4} \] Output:
-2*(-a*e+b*d)^2*(-A*e+B*d)*(e*x+d)^(1/2)/e^4+2/3*(-a*e+b*d)*(-2*A*b*e-B*a* e+3*B*b*d)*(e*x+d)^(3/2)/e^4-2/5*b*(-A*b*e-2*B*a*e+3*B*b*d)*(e*x+d)^(5/2)/ e^4+2/7*b^2*B*(e*x+d)^(7/2)/e^4
Time = 0.09 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.09 \[ \int \frac {(a+b x)^2 (A+B x)}{\sqrt {d+e x}} \, dx=\frac {2 \sqrt {d+e x} \left (35 a^2 e^2 (-2 B d+3 A e+B e x)+14 a b e \left (5 A e (-2 d+e x)+B \left (8 d^2-4 d e x+3 e^2 x^2\right )\right )+b^2 \left (7 A e \left (8 d^2-4 d e x+3 e^2 x^2\right )-3 B \left (16 d^3-8 d^2 e x+6 d e^2 x^2-5 e^3 x^3\right )\right )\right )}{105 e^4} \] Input:
Integrate[((a + b*x)^2*(A + B*x))/Sqrt[d + e*x],x]
Output:
(2*Sqrt[d + e*x]*(35*a^2*e^2*(-2*B*d + 3*A*e + B*e*x) + 14*a*b*e*(5*A*e*(- 2*d + e*x) + B*(8*d^2 - 4*d*e*x + 3*e^2*x^2)) + b^2*(7*A*e*(8*d^2 - 4*d*e* x + 3*e^2*x^2) - 3*B*(16*d^3 - 8*d^2*e*x + 6*d*e^2*x^2 - 5*e^3*x^3))))/(10 5*e^4)
Time = 0.29 (sec) , antiderivative size = 126, 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 = {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.
| \(\displaystyle \int \frac {(a+b x)^2 (A+B x)}{\sqrt {d+e x}} \, dx\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \int \left (\frac {b (d+e x)^{3/2} (2 a B e+A b e-3 b B d)}{e^3}+\frac {\sqrt {d+e x} (a e-b d) (a B e+2 A b e-3 b B d)}{e^3}+\frac {(a e-b d)^2 (A e-B d)}{e^3 \sqrt {d+e x}}+\frac {b^2 B (d+e x)^{5/2}}{e^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 b (d+e x)^{5/2} (-2 a B e-A b e+3 b B d)}{5 e^4}+\frac {2 (d+e x)^{3/2} (b d-a e) (-a B e-2 A b e+3 b B d)}{3 e^4}-\frac {2 \sqrt {d+e x} (b d-a e)^2 (B d-A e)}{e^4}+\frac {2 b^2 B (d+e x)^{7/2}}{7 e^4}\) |
Input:
Int[((a + b*x)^2*(A + B*x))/Sqrt[d + e*x],x]
Output:
(-2*(b*d - a*e)^2*(B*d - A*e)*Sqrt[d + e*x])/e^4 + (2*(b*d - a*e)*(3*b*B*d - 2*A*b*e - a*B*e)*(d + e*x)^(3/2))/(3*e^4) - (2*b*(3*b*B*d - A*b*e - 2*a *B*e)*(d + e*x)^(5/2))/(5*e^4) + (2*b^2*B*(d + e*x)^(7/2))/(7*e^4)
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]))))
Time = 0.68 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.93
| method | result | size |
| pseudoelliptic | \(\frac {2 \sqrt {e x +d}\, \left (\left (\frac {x^{2} \left (\frac {5 B x}{7}+A \right ) b^{2}}{5}+\frac {2 \left (\frac {3 B x}{5}+A \right ) a x b}{3}+a^{2} \left (\frac {B x}{3}+A \right )\right ) e^{3}-\frac {4 \left (\frac {\left (\frac {9 B x}{14}+A \right ) x \,b^{2}}{5}+a \left (\frac {2 B x}{5}+A \right ) b +\frac {a^{2} B}{2}\right ) d \,e^{2}}{3}+\frac {8 b \left (\left (\frac {3 B x}{7}+A \right ) b +2 B a \right ) d^{2} e}{15}-\frac {16 b^{2} B \,d^{3}}{35}\right )}{e^{4}}\) | \(117\) |
| derivativedivides | \(\frac {\frac {2 b^{2} B \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (2 b \left (a e -d b \right ) B +b^{2} \left (A e -B d \right )\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 \left (\left (a e -d b \right )^{2} B +2 b \left (a e -d b \right ) \left (A e -B d \right )\right ) \left (e x +d \right )^{\frac {3}{2}}}{3}+2 \left (a e -d b \right )^{2} \left (A e -B d \right ) \sqrt {e x +d}}{e^{4}}\) | \(121\) |
| default | \(\frac {\frac {2 b^{2} B \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (2 b \left (a e -d b \right ) B +b^{2} \left (A e -B d \right )\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 \left (\left (a e -d b \right )^{2} B +2 b \left (a e -d b \right ) \left (A e -B d \right )\right ) \left (e x +d \right )^{\frac {3}{2}}}{3}+2 \left (a e -d b \right )^{2} \left (A e -B d \right ) \sqrt {e x +d}}{e^{4}}\) | \(121\) |
| gosper | \(\frac {2 \sqrt {e x +d}\, \left (15 b^{2} B \,x^{3} e^{3}+21 A \,x^{2} b^{2} e^{3}+42 B \,x^{2} a b \,e^{3}-18 B \,x^{2} b^{2} d \,e^{2}+70 A x a b \,e^{3}-28 A x \,b^{2} d \,e^{2}+35 B x \,a^{2} e^{3}-56 B x a b d \,e^{2}+24 B x \,b^{2} d^{2} e +105 a^{2} A \,e^{3}-140 A a b d \,e^{2}+56 A \,b^{2} d^{2} e -70 B \,a^{2} d \,e^{2}+112 B a b \,d^{2} e -48 b^{2} B \,d^{3}\right )}{105 e^{4}}\) | \(169\) |
| trager | \(\frac {2 \sqrt {e x +d}\, \left (15 b^{2} B \,x^{3} e^{3}+21 A \,x^{2} b^{2} e^{3}+42 B \,x^{2} a b \,e^{3}-18 B \,x^{2} b^{2} d \,e^{2}+70 A x a b \,e^{3}-28 A x \,b^{2} d \,e^{2}+35 B x \,a^{2} e^{3}-56 B x a b d \,e^{2}+24 B x \,b^{2} d^{2} e +105 a^{2} A \,e^{3}-140 A a b d \,e^{2}+56 A \,b^{2} d^{2} e -70 B \,a^{2} d \,e^{2}+112 B a b \,d^{2} e -48 b^{2} B \,d^{3}\right )}{105 e^{4}}\) | \(169\) |
| risch | \(\frac {2 \sqrt {e x +d}\, \left (15 b^{2} B \,x^{3} e^{3}+21 A \,x^{2} b^{2} e^{3}+42 B \,x^{2} a b \,e^{3}-18 B \,x^{2} b^{2} d \,e^{2}+70 A x a b \,e^{3}-28 A x \,b^{2} d \,e^{2}+35 B x \,a^{2} e^{3}-56 B x a b d \,e^{2}+24 B x \,b^{2} d^{2} e +105 a^{2} A \,e^{3}-140 A a b d \,e^{2}+56 A \,b^{2} d^{2} e -70 B \,a^{2} d \,e^{2}+112 B a b \,d^{2} e -48 b^{2} B \,d^{3}\right )}{105 e^{4}}\) | \(169\) |
| orering | \(\frac {2 \sqrt {e x +d}\, \left (15 b^{2} B \,x^{3} e^{3}+21 A \,x^{2} b^{2} e^{3}+42 B \,x^{2} a b \,e^{3}-18 B \,x^{2} b^{2} d \,e^{2}+70 A x a b \,e^{3}-28 A x \,b^{2} d \,e^{2}+35 B x \,a^{2} e^{3}-56 B x a b d \,e^{2}+24 B x \,b^{2} d^{2} e +105 a^{2} A \,e^{3}-140 A a b d \,e^{2}+56 A \,b^{2} d^{2} e -70 B \,a^{2} d \,e^{2}+112 B a b \,d^{2} e -48 b^{2} B \,d^{3}\right )}{105 e^{4}}\) | \(169\) |
Input:
int((b*x+a)^2*(B*x+A)/(e*x+d)^(1/2),x,method=_RETURNVERBOSE)
Output:
2*(e*x+d)^(1/2)*((1/5*x^2*(5/7*B*x+A)*b^2+2/3*(3/5*B*x+A)*a*x*b+a^2*(1/3*B *x+A))*e^3-4/3*(1/5*(9/14*B*x+A)*x*b^2+a*(2/5*B*x+A)*b+1/2*a^2*B)*d*e^2+8/ 15*b*((3/7*B*x+A)*b+2*B*a)*d^2*e-16/35*b^2*B*d^3)/e^4
Time = 0.07 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.23 \[ \int \frac {(a+b x)^2 (A+B x)}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (15 \, B b^{2} e^{3} x^{3} - 48 \, B b^{2} d^{3} + 105 \, A a^{2} e^{3} + 56 \, {\left (2 \, B a b + A b^{2}\right )} d^{2} e - 70 \, {\left (B a^{2} + 2 \, A a b\right )} d e^{2} - 3 \, {\left (6 \, B b^{2} d e^{2} - 7 \, {\left (2 \, B a b + A b^{2}\right )} e^{3}\right )} x^{2} + {\left (24 \, B b^{2} d^{2} e - 28 \, {\left (2 \, B a b + A b^{2}\right )} d e^{2} + 35 \, {\left (B a^{2} + 2 \, A a b\right )} e^{3}\right )} x\right )} \sqrt {e x + d}}{105 \, e^{4}} \] Input:
integrate((b*x+a)^2*(B*x+A)/(e*x+d)^(1/2),x, algorithm="fricas")
Output:
2/105*(15*B*b^2*e^3*x^3 - 48*B*b^2*d^3 + 105*A*a^2*e^3 + 56*(2*B*a*b + A*b ^2)*d^2*e - 70*(B*a^2 + 2*A*a*b)*d*e^2 - 3*(6*B*b^2*d*e^2 - 7*(2*B*a*b + A *b^2)*e^3)*x^2 + (24*B*b^2*d^2*e - 28*(2*B*a*b + A*b^2)*d*e^2 + 35*(B*a^2 + 2*A*a*b)*e^3)*x)*sqrt(e*x + d)/e^4
Time = 0.97 (sec) , antiderivative size = 255, normalized size of antiderivative = 2.02 \[ \int \frac {(a+b x)^2 (A+B x)}{\sqrt {d+e x}} \, dx=\begin {cases} \frac {2 \left (\frac {B b^{2} \left (d + e x\right )^{\frac {7}{2}}}{7 e^{3}} + \frac {\left (d + e x\right )^{\frac {5}{2}} \left (A b^{2} e + 2 B a b e - 3 B b^{2} d\right )}{5 e^{3}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \cdot \left (2 A a b e^{2} - 2 A b^{2} d e + B a^{2} e^{2} - 4 B a b d e + 3 B b^{2} d^{2}\right )}{3 e^{3}} + \frac {\sqrt {d + e x} \left (A a^{2} e^{3} - 2 A a b d e^{2} + A b^{2} d^{2} e - B a^{2} d e^{2} + 2 B a b d^{2} e - B b^{2} d^{3}\right )}{e^{3}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {A a^{2} x + \frac {B b^{2} x^{4}}{4} + \frac {x^{3} \left (A b^{2} + 2 B a b\right )}{3} + \frac {x^{2} \cdot \left (2 A a b + B a^{2}\right )}{2}}{\sqrt {d}} & \text {otherwise} \end {cases} \] Input:
integrate((b*x+a)**2*(B*x+A)/(e*x+d)**(1/2),x)
Output:
Piecewise((2*(B*b**2*(d + e*x)**(7/2)/(7*e**3) + (d + e*x)**(5/2)*(A*b**2* e + 2*B*a*b*e - 3*B*b**2*d)/(5*e**3) + (d + e*x)**(3/2)*(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)/(3*e**3) + sqrt(d + e*x)*(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)/e**3)/e, Ne(e, 0)), ((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)/sqrt(d), True))
Time = 0.04 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.26 \[ \int \frac {(a+b x)^2 (A+B x)}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (15 \, {\left (e x + d\right )}^{\frac {7}{2}} B b^{2} - 21 \, {\left (3 \, B b^{2} d - {\left (2 \, B a b + A b^{2}\right )} e\right )} {\left (e x + d\right )}^{\frac {5}{2}} + 35 \, {\left (3 \, B b^{2} d^{2} - 2 \, {\left (2 \, B a b + A b^{2}\right )} d e + {\left (B a^{2} + 2 \, A a b\right )} e^{2}\right )} {\left (e x + d\right )}^{\frac {3}{2}} - 105 \, {\left (B b^{2} d^{3} - A a^{2} e^{3} - {\left (2 \, B a b + A b^{2}\right )} d^{2} e + {\left (B a^{2} + 2 \, A a b\right )} d e^{2}\right )} \sqrt {e x + d}\right )}}{105 \, e^{4}} \] Input:
integrate((b*x+a)^2*(B*x+A)/(e*x+d)^(1/2),x, algorithm="maxima")
Output:
2/105*(15*(e*x + d)^(7/2)*B*b^2 - 21*(3*B*b^2*d - (2*B*a*b + A*b^2)*e)*(e* x + d)^(5/2) + 35*(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)^(3/2) - 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)*sqrt(e*x + d))/e^4
Time = 0.13 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.63 \[ \int \frac {(a+b x)^2 (A+B x)}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (105 \, \sqrt {e x + d} A a^{2} + \frac {35 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} B a^{2}}{e} + \frac {70 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} A a b}{e} + \frac {14 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} B a b}{e^{2}} + \frac {7 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} A b^{2}}{e^{2}} + \frac {3 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} B b^{2}}{e^{3}}\right )}}{105 \, e} \] Input:
integrate((b*x+a)^2*(B*x+A)/(e*x+d)^(1/2),x, algorithm="giac")
Output:
2/105*(105*sqrt(e*x + d)*A*a^2 + 35*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)* B*a^2/e + 70*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*A*a*b/e + 14*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*B*a*b/e^2 + 7*(3*( e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*A*b^2/e^2 + 3*(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*b^2/e^3)/e
Time = 0.04 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.91 \[ \int \frac {(a+b x)^2 (A+B x)}{\sqrt {d+e x}} \, dx=\frac {{\left (d+e\,x\right )}^{5/2}\,\left (2\,A\,b^2\,e-6\,B\,b^2\,d+4\,B\,a\,b\,e\right )}{5\,e^4}+\frac {2\,B\,b^2\,{\left (d+e\,x\right )}^{7/2}}{7\,e^4}+\frac {2\,\left (a\,e-b\,d\right )\,{\left (d+e\,x\right )}^{3/2}\,\left (2\,A\,b\,e+B\,a\,e-3\,B\,b\,d\right )}{3\,e^4}+\frac {2\,\left (A\,e-B\,d\right )\,{\left (a\,e-b\,d\right )}^2\,\sqrt {d+e\,x}}{e^4} \] Input:
int(((A + B*x)*(a + b*x)^2)/(d + e*x)^(1/2),x)
Output:
((d + e*x)^(5/2)*(2*A*b^2*e - 6*B*b^2*d + 4*B*a*b*e))/(5*e^4) + (2*B*b^2*( d + e*x)^(7/2))/(7*e^4) + (2*(a*e - b*d)*(d + e*x)^(3/2)*(2*A*b*e + B*a*e - 3*B*b*d))/(3*e^4) + (2*(A*e - B*d)*(a*e - b*d)^2*(d + e*x)^(1/2))/e^4
Time = 0.16 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.90 \[ \int \frac {(a+b x)^2 (A+B x)}{\sqrt {d+e x}} \, dx=\frac {2 \sqrt {e x +d}\, \left (5 b^{3} e^{3} x^{3}+21 a \,b^{2} e^{3} x^{2}-6 b^{3} d \,e^{2} x^{2}+35 a^{2} b \,e^{3} x -28 a \,b^{2} d \,e^{2} x +8 b^{3} d^{2} e x +35 a^{3} e^{3}-70 a^{2} b d \,e^{2}+56 a \,b^{2} d^{2} e -16 b^{3} d^{3}\right )}{35 e^{4}} \] Input:
int((b*x+a)^2*(B*x+A)/(e*x+d)^(1/2),x)
Output:
(2*sqrt(d + e*x)*(35*a**3*e**3 - 70*a**2*b*d*e**2 + 35*a**2*b*e**3*x + 56* a*b**2*d**2*e - 28*a*b**2*d*e**2*x + 21*a*b**2*e**3*x**2 - 16*b**3*d**3 + 8*b**3*d**2*e*x - 6*b**3*d*e**2*x**2 + 5*b**3*e**3*x**3))/(35*e**4)