Integrand size = 22, antiderivative size = 169 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^{5/2}} \, dx=-\frac {2 (b d-a e)^3 (B d-A e)}{3 e^5 (d+e x)^{3/2}}+\frac {2 (b d-a e)^2 (4 b B d-3 A b e-a B e)}{e^5 \sqrt {d+e x}}+\frac {6 b (b d-a e) (2 b B d-A b e-a B e) \sqrt {d+e x}}{e^5}-\frac {2 b^2 (4 b B d-A b e-3 a B e) (d+e x)^{3/2}}{3 e^5}+\frac {2 b^3 B (d+e x)^{5/2}}{5 e^5} \] Output:
-2/3*(-a*e+b*d)^3*(-A*e+B*d)/e^5/(e*x+d)^(3/2)+2*(-a*e+b*d)^2*(-3*A*b*e-B* a*e+4*B*b*d)/e^5/(e*x+d)^(1/2)+6*b*(-a*e+b*d)*(-A*b*e-B*a*e+2*B*b*d)*(e*x+ d)^(1/2)/e^5-2/3*b^2*(-A*b*e-3*B*a*e+4*B*b*d)*(e*x+d)^(3/2)/e^5+2/5*b^3*B* (e*x+d)^(5/2)/e^5
Time = 0.18 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.32 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^{5/2}} \, dx=\frac {2 \left (-5 a^3 e^3 (2 B d+A e+3 B e x)+15 a^2 b e^2 \left (-A e (2 d+3 e x)+B \left (8 d^2+12 d e x+3 e^2 x^2\right )\right )+15 a b^2 e \left (A e \left (8 d^2+12 d e x+3 e^2 x^2\right )+B \left (-16 d^3-24 d^2 e x-6 d e^2 x^2+e^3 x^3\right )\right )+b^3 \left (5 A e \left (-16 d^3-24 d^2 e x-6 d e^2 x^2+e^3 x^3\right )+B \left (128 d^4+192 d^3 e x+48 d^2 e^2 x^2-8 d e^3 x^3+3 e^4 x^4\right )\right )\right )}{15 e^5 (d+e x)^{3/2}} \] Input:
Integrate[((a + b*x)^3*(A + B*x))/(d + e*x)^(5/2),x]
Output:
(2*(-5*a^3*e^3*(2*B*d + A*e + 3*B*e*x) + 15*a^2*b*e^2*(-(A*e*(2*d + 3*e*x) ) + B*(8*d^2 + 12*d*e*x + 3*e^2*x^2)) + 15*a*b^2*e*(A*e*(8*d^2 + 12*d*e*x + 3*e^2*x^2) + B*(-16*d^3 - 24*d^2*e*x - 6*d*e^2*x^2 + e^3*x^3)) + b^3*(5* A*e*(-16*d^3 - 24*d^2*e*x - 6*d*e^2*x^2 + e^3*x^3) + B*(128*d^4 + 192*d^3* e*x + 48*d^2*e^2*x^2 - 8*d*e^3*x^3 + 3*e^4*x^4))))/(15*e^5*(d + e*x)^(3/2) )
Time = 0.31 (sec) , antiderivative size = 169, 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.
| \(\displaystyle \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \int \left (\frac {b^2 \sqrt {d+e x} (3 a B e+A b e-4 b B d)}{e^4}-\frac {3 b (b d-a e) (a B e+A b e-2 b B d)}{e^4 \sqrt {d+e x}}+\frac {(a e-b d)^2 (a B e+3 A b e-4 b B d)}{e^4 (d+e x)^{3/2}}+\frac {(a e-b d)^3 (A e-B d)}{e^4 (d+e x)^{5/2}}+\frac {b^3 B (d+e x)^{3/2}}{e^4}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 b^2 (d+e x)^{3/2} (-3 a B e-A b e+4 b B d)}{3 e^5}+\frac {6 b \sqrt {d+e x} (b d-a e) (-a B e-A b e+2 b B d)}{e^5}+\frac {2 (b d-a e)^2 (-a B e-3 A b e+4 b B d)}{e^5 \sqrt {d+e x}}-\frac {2 (b d-a e)^3 (B d-A e)}{3 e^5 (d+e x)^{3/2}}+\frac {2 b^3 B (d+e x)^{5/2}}{5 e^5}\) |
Input:
Int[((a + b*x)^3*(A + B*x))/(d + e*x)^(5/2),x]
Output:
(-2*(b*d - a*e)^3*(B*d - A*e))/(3*e^5*(d + e*x)^(3/2)) + (2*(b*d - a*e)^2* (4*b*B*d - 3*A*b*e - a*B*e))/(e^5*Sqrt[d + e*x]) + (6*b*(b*d - a*e)*(2*b*B *d - A*b*e - a*B*e)*Sqrt[d + e*x])/e^5 - (2*b^2*(4*b*B*d - A*b*e - 3*a*B*e )*(d + e*x)^(3/2))/(3*e^5) + (2*b^3*B*(d + e*x)^(5/2))/(5*e^5)
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 = 0.33 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.06
| method | result | size |
| risch | \(\frac {2 b \left (3 e^{2} b^{2} B \,x^{2}+5 A \,b^{2} e^{2} x +15 B a b \,e^{2} x -14 b^{2} B d e x +45 A a b \,e^{2}-40 A \,b^{2} d e +45 B \,a^{2} e^{2}-120 B a b d e +73 b^{2} B \,d^{2}\right ) \sqrt {e x +d}}{15 e^{5}}-\frac {2 \left (9 A x b \,e^{2}+3 B x a \,e^{2}-12 B x b d e +A a \,e^{2}+8 A b d e +2 B a d e -11 b B \,d^{2}\right ) \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}{3 e^{5} \left (e x +d \right )^{\frac {3}{2}}}\) | \(179\) |
| pseudoelliptic | \(-\frac {2 \left (\left (\left (-\frac {3}{5} B \,x^{4}-A \,x^{3}\right ) b^{3}-9 a \left (\frac {B x}{3}+A \right ) x^{2} b^{2}+9 a^{2} x \left (-B x +A \right ) b +a^{3} \left (3 B x +A \right )\right ) e^{4}+6 \left (x^{2} \left (\frac {4 B x}{15}+A \right ) b^{3}-6 a \left (-\frac {B x}{2}+A \right ) x \,b^{2}+a^{2} \left (-6 B x +A \right ) b +\frac {a^{3} B}{3}\right ) d \,e^{3}-24 \left (\left (\frac {2}{5} x^{2} B -x A \right ) b^{2}+a \left (-3 B x +A \right ) b +a^{2} B \right ) b \,d^{2} e^{2}+16 \left (\left (-\frac {12 B x}{5}+A \right ) b +3 B a \right ) b^{2} d^{3} e -\frac {128 b^{3} B \,d^{4}}{5}\right )}{3 \left (e x +d \right )^{\frac {3}{2}} e^{5}}\) | \(194\) |
| gosper | \(-\frac {2 \left (-3 B \,x^{4} b^{3} e^{4}-5 A \,x^{3} b^{3} e^{4}-15 B \,x^{3} a \,b^{2} e^{4}+8 B \,x^{3} b^{3} d \,e^{3}-45 A \,x^{2} a \,b^{2} e^{4}+30 A \,x^{2} b^{3} d \,e^{3}-45 B \,x^{2} a^{2} b \,e^{4}+90 B \,x^{2} a \,b^{2} d \,e^{3}-48 B \,x^{2} b^{3} d^{2} e^{2}+45 A x \,a^{2} b \,e^{4}-180 A x a \,b^{2} d \,e^{3}+120 A x \,b^{3} d^{2} e^{2}+15 B x \,a^{3} e^{4}-180 B x \,a^{2} b d \,e^{3}+360 B x a \,b^{2} d^{2} e^{2}-192 B x \,b^{3} d^{3} e +5 a^{3} A \,e^{4}+30 A \,a^{2} b d \,e^{3}-120 A a \,b^{2} d^{2} e^{2}+80 A \,b^{3} d^{3} e +10 B \,a^{3} d \,e^{3}-120 B \,a^{2} b \,d^{2} e^{2}+240 B a \,b^{2} d^{3} e -128 b^{3} B \,d^{4}\right )}{15 \left (e x +d \right )^{\frac {3}{2}} e^{5}}\) | \(301\) |
| trager | \(-\frac {2 \left (-3 B \,x^{4} b^{3} e^{4}-5 A \,x^{3} b^{3} e^{4}-15 B \,x^{3} a \,b^{2} e^{4}+8 B \,x^{3} b^{3} d \,e^{3}-45 A \,x^{2} a \,b^{2} e^{4}+30 A \,x^{2} b^{3} d \,e^{3}-45 B \,x^{2} a^{2} b \,e^{4}+90 B \,x^{2} a \,b^{2} d \,e^{3}-48 B \,x^{2} b^{3} d^{2} e^{2}+45 A x \,a^{2} b \,e^{4}-180 A x a \,b^{2} d \,e^{3}+120 A x \,b^{3} d^{2} e^{2}+15 B x \,a^{3} e^{4}-180 B x \,a^{2} b d \,e^{3}+360 B x a \,b^{2} d^{2} e^{2}-192 B x \,b^{3} d^{3} e +5 a^{3} A \,e^{4}+30 A \,a^{2} b d \,e^{3}-120 A a \,b^{2} d^{2} e^{2}+80 A \,b^{3} d^{3} e +10 B \,a^{3} d \,e^{3}-120 B \,a^{2} b \,d^{2} e^{2}+240 B a \,b^{2} d^{3} e -128 b^{3} B \,d^{4}\right )}{15 \left (e x +d \right )^{\frac {3}{2}} e^{5}}\) | \(301\) |
| orering | \(-\frac {2 \left (-3 B \,x^{4} b^{3} e^{4}-5 A \,x^{3} b^{3} e^{4}-15 B \,x^{3} a \,b^{2} e^{4}+8 B \,x^{3} b^{3} d \,e^{3}-45 A \,x^{2} a \,b^{2} e^{4}+30 A \,x^{2} b^{3} d \,e^{3}-45 B \,x^{2} a^{2} b \,e^{4}+90 B \,x^{2} a \,b^{2} d \,e^{3}-48 B \,x^{2} b^{3} d^{2} e^{2}+45 A x \,a^{2} b \,e^{4}-180 A x a \,b^{2} d \,e^{3}+120 A x \,b^{3} d^{2} e^{2}+15 B x \,a^{3} e^{4}-180 B x \,a^{2} b d \,e^{3}+360 B x a \,b^{2} d^{2} e^{2}-192 B x \,b^{3} d^{3} e +5 a^{3} A \,e^{4}+30 A \,a^{2} b d \,e^{3}-120 A a \,b^{2} d^{2} e^{2}+80 A \,b^{3} d^{3} e +10 B \,a^{3} d \,e^{3}-120 B \,a^{2} b \,d^{2} e^{2}+240 B a \,b^{2} d^{3} e -128 b^{3} B \,d^{4}\right )}{15 \left (e x +d \right )^{\frac {3}{2}} e^{5}}\) | \(301\) |
| derivativedivides | \(\frac {\frac {2 b^{3} B \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 A \,b^{3} e \left (e x +d \right )^{\frac {3}{2}}}{3}+2 B a \,b^{2} e \left (e x +d \right )^{\frac {3}{2}}-\frac {8 B \,b^{3} d \left (e x +d \right )^{\frac {3}{2}}}{3}+6 A a \,b^{2} e^{2} \sqrt {e x +d}-6 A \,b^{3} d e \sqrt {e x +d}+6 B \,a^{2} b \,e^{2} \sqrt {e x +d}-18 B a \,b^{2} d e \sqrt {e x +d}+12 B \,b^{3} d^{2} \sqrt {e x +d}-\frac {2 \left (3 A \,a^{2} b \,e^{3}-6 A a \,b^{2} d \,e^{2}+3 A \,b^{3} d^{2} e +B \,a^{3} e^{3}-6 B \,a^{2} b d \,e^{2}+9 B a \,b^{2} d^{2} e -4 b^{3} B \,d^{3}\right )}{\sqrt {e x +d}}-\frac {2 \left (a^{3} A \,e^{4}-3 A \,a^{2} b d \,e^{3}+3 A a \,b^{2} d^{2} e^{2}-A \,b^{3} d^{3} e -B \,a^{3} d \,e^{3}+3 B \,a^{2} b \,d^{2} e^{2}-3 B a \,b^{2} d^{3} e +b^{3} B \,d^{4}\right )}{3 \left (e x +d \right )^{\frac {3}{2}}}}{e^{5}}\) | \(317\) |
| default | \(\frac {\frac {2 b^{3} B \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 A \,b^{3} e \left (e x +d \right )^{\frac {3}{2}}}{3}+2 B a \,b^{2} e \left (e x +d \right )^{\frac {3}{2}}-\frac {8 B \,b^{3} d \left (e x +d \right )^{\frac {3}{2}}}{3}+6 A a \,b^{2} e^{2} \sqrt {e x +d}-6 A \,b^{3} d e \sqrt {e x +d}+6 B \,a^{2} b \,e^{2} \sqrt {e x +d}-18 B a \,b^{2} d e \sqrt {e x +d}+12 B \,b^{3} d^{2} \sqrt {e x +d}-\frac {2 \left (3 A \,a^{2} b \,e^{3}-6 A a \,b^{2} d \,e^{2}+3 A \,b^{3} d^{2} e +B \,a^{3} e^{3}-6 B \,a^{2} b d \,e^{2}+9 B a \,b^{2} d^{2} e -4 b^{3} B \,d^{3}\right )}{\sqrt {e x +d}}-\frac {2 \left (a^{3} A \,e^{4}-3 A \,a^{2} b d \,e^{3}+3 A a \,b^{2} d^{2} e^{2}-A \,b^{3} d^{3} e -B \,a^{3} d \,e^{3}+3 B \,a^{2} b \,d^{2} e^{2}-3 B a \,b^{2} d^{3} e +b^{3} B \,d^{4}\right )}{3 \left (e x +d \right )^{\frac {3}{2}}}}{e^{5}}\) | \(317\) |
Input:
int((b*x+a)^3*(B*x+A)/(e*x+d)^(5/2),x,method=_RETURNVERBOSE)
Output:
2/15*b*(3*B*b^2*e^2*x^2+5*A*b^2*e^2*x+15*B*a*b*e^2*x-14*B*b^2*d*e*x+45*A*a *b*e^2-40*A*b^2*d*e+45*B*a^2*e^2-120*B*a*b*d*e+73*B*b^2*d^2)*(e*x+d)^(1/2) /e^5-2/3*(9*A*b*e^2*x+3*B*a*e^2*x-12*B*b*d*e*x+A*a*e^2+8*A*b*d*e+2*B*a*d*e -11*B*b*d^2)*(a^2*e^2-2*a*b*d*e+b^2*d^2)/e^5/(e*x+d)^(3/2)
Time = 0.10 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.68 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (3 \, B b^{3} e^{4} x^{4} + 128 \, B b^{3} d^{4} - 5 \, A a^{3} e^{4} - 80 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 120 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} - 10 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} - {\left (8 \, B b^{3} d e^{3} - 5 \, {\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} + 3 \, {\left (16 \, B b^{3} d^{2} e^{2} - 10 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} + 15 \, {\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} + 3 \, {\left (64 \, B b^{3} d^{3} e - 40 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 60 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{3} - 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} e^{5}\right )}} \] Input:
integrate((b*x+a)^3*(B*x+A)/(e*x+d)^(5/2),x, algorithm="fricas")
Output:
2/15*(3*B*b^3*e^4*x^4 + 128*B*b^3*d^4 - 5*A*a^3*e^4 - 80*(3*B*a*b^2 + A*b^ 3)*d^3*e + 120*(B*a^2*b + A*a*b^2)*d^2*e^2 - 10*(B*a^3 + 3*A*a^2*b)*d*e^3 - (8*B*b^3*d*e^3 - 5*(3*B*a*b^2 + A*b^3)*e^4)*x^3 + 3*(16*B*b^3*d^2*e^2 - 10*(3*B*a*b^2 + A*b^3)*d*e^3 + 15*(B*a^2*b + A*a*b^2)*e^4)*x^2 + 3*(64*B*b ^3*d^3*e - 40*(3*B*a*b^2 + A*b^3)*d^2*e^2 + 60*(B*a^2*b + A*a*b^2)*d*e^3 - 5*(B*a^3 + 3*A*a^2*b)*e^4)*x)*sqrt(e*x + d)/(e^7*x^2 + 2*d*e^6*x + d^2*e^ 5)
Time = 9.61 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.68 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^{5/2}} \, dx=\begin {cases} \frac {2 \left (\frac {B b^{3} \left (d + e x\right )^{\frac {5}{2}}}{5 e^{4}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (A b^{3} e + 3 B a b^{2} e - 4 B b^{3} d\right )}{3 e^{4}} + \frac {\sqrt {d + e x} \left (3 A a b^{2} e^{2} - 3 A b^{3} d e + 3 B a^{2} b e^{2} - 9 B a b^{2} d e + 6 B b^{3} d^{2}\right )}{e^{4}} - \frac {\left (a e - b d\right )^{2} \cdot \left (3 A b e + B a e - 4 B b d\right )}{e^{4} \sqrt {d + e x}} + \frac {\left (- A e + B d\right ) \left (a e - b d\right )^{3}}{3 e^{4} \left (d + e x\right )^{\frac {3}{2}}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {A a^{3} x + \frac {B b^{3} x^{5}}{5} + \frac {x^{4} \left (A b^{3} + 3 B a b^{2}\right )}{4} + \frac {x^{3} \cdot \left (3 A a b^{2} + 3 B a^{2} b\right )}{3} + \frac {x^{2} \cdot \left (3 A a^{2} b + B a^{3}\right )}{2}}{d^{\frac {5}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate((b*x+a)**3*(B*x+A)/(e*x+d)**(5/2),x)
Output:
Piecewise((2*(B*b**3*(d + e*x)**(5/2)/(5*e**4) + (d + e*x)**(3/2)*(A*b**3* e + 3*B*a*b**2*e - 4*B*b**3*d)/(3*e**4) + sqrt(d + e*x)*(3*A*a*b**2*e**2 - 3*A*b**3*d*e + 3*B*a**2*b*e**2 - 9*B*a*b**2*d*e + 6*B*b**3*d**2)/e**4 - ( a*e - b*d)**2*(3*A*b*e + B*a*e - 4*B*b*d)/(e**4*sqrt(d + e*x)) + (-A*e + B *d)*(a*e - b*d)**3/(3*e**4*(d + e*x)**(3/2)))/e, Ne(e, 0)), ((A*a**3*x + B *b**3*x**5/5 + x**4*(A*b**3 + 3*B*a*b**2)/4 + x**3*(3*A*a*b**2 + 3*B*a**2* b)/3 + x**2*(3*A*a**2*b + B*a**3)/2)/d**(5/2), True))
Time = 0.03 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.60 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (\frac {3 \, {\left (e x + d\right )}^{\frac {5}{2}} B b^{3} - 5 \, {\left (4 \, B b^{3} d - {\left (3 \, B a b^{2} + A b^{3}\right )} e\right )} {\left (e x + d\right )}^{\frac {3}{2}} + 45 \, {\left (2 \, B b^{3} d^{2} - {\left (3 \, B a b^{2} + A b^{3}\right )} d e + {\left (B a^{2} b + A a b^{2}\right )} e^{2}\right )} \sqrt {e x + d}}{e^{4}} - \frac {5 \, {\left (B b^{3} d^{4} + A a^{3} e^{4} - {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 3 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} - {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} - 3 \, {\left (4 \, B b^{3} d^{3} - 3 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e + 6 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{2} - {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{3}\right )} {\left (e x + d\right )}\right )}}{{\left (e x + d\right )}^{\frac {3}{2}} e^{4}}\right )}}{15 \, e} \] Input:
integrate((b*x+a)^3*(B*x+A)/(e*x+d)^(5/2),x, algorithm="maxima")
Output:
2/15*((3*(e*x + d)^(5/2)*B*b^3 - 5*(4*B*b^3*d - (3*B*a*b^2 + A*b^3)*e)*(e* x + d)^(3/2) + 45*(2*B*b^3*d^2 - (3*B*a*b^2 + A*b^3)*d*e + (B*a^2*b + A*a* b^2)*e^2)*sqrt(e*x + d))/e^4 - 5*(B*b^3*d^4 + A*a^3*e^4 - (3*B*a*b^2 + A*b ^3)*d^3*e + 3*(B*a^2*b + A*a*b^2)*d^2*e^2 - (B*a^3 + 3*A*a^2*b)*d*e^3 - 3* (4*B*b^3*d^3 - 3*(3*B*a*b^2 + A*b^3)*d^2*e + 6*(B*a^2*b + A*a*b^2)*d*e^2 - (B*a^3 + 3*A*a^2*b)*e^3)*(e*x + d))/((e*x + d)^(3/2)*e^4))/e
Leaf count of result is larger than twice the leaf count of optimal. 364 vs. \(2 (153) = 306\).
Time = 0.13 (sec) , antiderivative size = 364, normalized size of antiderivative = 2.15 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (12 \, {\left (e x + d\right )} B b^{3} d^{3} - B b^{3} d^{4} - 27 \, {\left (e x + d\right )} B a b^{2} d^{2} e - 9 \, {\left (e x + d\right )} A b^{3} d^{2} e + 3 \, B a b^{2} d^{3} e + A b^{3} d^{3} e + 18 \, {\left (e x + d\right )} B a^{2} b d e^{2} + 18 \, {\left (e x + d\right )} A a b^{2} d e^{2} - 3 \, B a^{2} b d^{2} e^{2} - 3 \, A a b^{2} d^{2} e^{2} - 3 \, {\left (e x + d\right )} B a^{3} e^{3} - 9 \, {\left (e x + d\right )} A a^{2} b e^{3} + B a^{3} d e^{3} + 3 \, A a^{2} b d e^{3} - A a^{3} e^{4}\right )}}{3 \, {\left (e x + d\right )}^{\frac {3}{2}} e^{5}} + \frac {2 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} B b^{3} e^{20} - 20 \, {\left (e x + d\right )}^{\frac {3}{2}} B b^{3} d e^{20} + 90 \, \sqrt {e x + d} B b^{3} d^{2} e^{20} + 15 \, {\left (e x + d\right )}^{\frac {3}{2}} B a b^{2} e^{21} + 5 \, {\left (e x + d\right )}^{\frac {3}{2}} A b^{3} e^{21} - 135 \, \sqrt {e x + d} B a b^{2} d e^{21} - 45 \, \sqrt {e x + d} A b^{3} d e^{21} + 45 \, \sqrt {e x + d} B a^{2} b e^{22} + 45 \, \sqrt {e x + d} A a b^{2} e^{22}\right )}}{15 \, e^{25}} \] Input:
integrate((b*x+a)^3*(B*x+A)/(e*x+d)^(5/2),x, algorithm="giac")
Output:
2/3*(12*(e*x + d)*B*b^3*d^3 - B*b^3*d^4 - 27*(e*x + d)*B*a*b^2*d^2*e - 9*( e*x + d)*A*b^3*d^2*e + 3*B*a*b^2*d^3*e + A*b^3*d^3*e + 18*(e*x + d)*B*a^2* b*d*e^2 + 18*(e*x + d)*A*a*b^2*d*e^2 - 3*B*a^2*b*d^2*e^2 - 3*A*a*b^2*d^2*e ^2 - 3*(e*x + d)*B*a^3*e^3 - 9*(e*x + d)*A*a^2*b*e^3 + B*a^3*d*e^3 + 3*A*a ^2*b*d*e^3 - A*a^3*e^4)/((e*x + d)^(3/2)*e^5) + 2/15*(3*(e*x + d)^(5/2)*B* b^3*e^20 - 20*(e*x + d)^(3/2)*B*b^3*d*e^20 + 90*sqrt(e*x + d)*B*b^3*d^2*e^ 20 + 15*(e*x + d)^(3/2)*B*a*b^2*e^21 + 5*(e*x + d)^(3/2)*A*b^3*e^21 - 135* sqrt(e*x + d)*B*a*b^2*d*e^21 - 45*sqrt(e*x + d)*A*b^3*d*e^21 + 45*sqrt(e*x + d)*B*a^2*b*e^22 + 45*sqrt(e*x + d)*A*a*b^2*e^22)/e^25
Time = 0.97 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.56 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^{5/2}} \, dx=\frac {{\left (d+e\,x\right )}^{3/2}\,\left (2\,A\,b^3\,e-8\,B\,b^3\,d+6\,B\,a\,b^2\,e\right )}{3\,e^5}-\frac {\left (d+e\,x\right )\,\left (2\,B\,a^3\,e^3-12\,B\,a^2\,b\,d\,e^2+6\,A\,a^2\,b\,e^3+18\,B\,a\,b^2\,d^2\,e-12\,A\,a\,b^2\,d\,e^2-8\,B\,b^3\,d^3+6\,A\,b^3\,d^2\,e\right )+\frac {2\,A\,a^3\,e^4}{3}+\frac {2\,B\,b^3\,d^4}{3}-\frac {2\,A\,b^3\,d^3\,e}{3}-\frac {2\,B\,a^3\,d\,e^3}{3}+2\,A\,a\,b^2\,d^2\,e^2+2\,B\,a^2\,b\,d^2\,e^2-2\,A\,a^2\,b\,d\,e^3-2\,B\,a\,b^2\,d^3\,e}{e^5\,{\left (d+e\,x\right )}^{3/2}}+\frac {2\,B\,b^3\,{\left (d+e\,x\right )}^{5/2}}{5\,e^5}+\frac {6\,b\,\left (a\,e-b\,d\right )\,\sqrt {d+e\,x}\,\left (A\,b\,e+B\,a\,e-2\,B\,b\,d\right )}{e^5} \] Input:
int(((A + B*x)*(a + b*x)^3)/(d + e*x)^(5/2),x)
Output:
((d + e*x)^(3/2)*(2*A*b^3*e - 8*B*b^3*d + 6*B*a*b^2*e))/(3*e^5) - ((d + e* x)*(2*B*a^3*e^3 - 8*B*b^3*d^3 + 6*A*a^2*b*e^3 + 6*A*b^3*d^2*e - 12*A*a*b^2 *d*e^2 + 18*B*a*b^2*d^2*e - 12*B*a^2*b*d*e^2) + (2*A*a^3*e^4)/3 + (2*B*b^3 *d^4)/3 - (2*A*b^3*d^3*e)/3 - (2*B*a^3*d*e^3)/3 + 2*A*a*b^2*d^2*e^2 + 2*B* a^2*b*d^2*e^2 - 2*A*a^2*b*d*e^3 - 2*B*a*b^2*d^3*e)/(e^5*(d + e*x)^(3/2)) + (2*B*b^3*(d + e*x)^(5/2))/(5*e^5) + (6*b*(a*e - b*d)*(d + e*x)^(1/2)*(A*b *e + B*a*e - 2*B*b*d))/e^5
Time = 0.15 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.14 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^{5/2}} \, dx=\frac {\frac {2}{5} b^{4} e^{4} x^{4}+\frac {8}{3} a \,b^{3} e^{4} x^{3}-\frac {16}{15} b^{4} d \,e^{3} x^{3}+12 a^{2} b^{2} e^{4} x^{2}-16 a \,b^{3} d \,e^{3} x^{2}+\frac {32}{5} b^{4} d^{2} e^{2} x^{2}-8 a^{3} b \,e^{4} x +48 a^{2} b^{2} d \,e^{3} x -64 a \,b^{3} d^{2} e^{2} x +\frac {128}{5} b^{4} d^{3} e x -\frac {2}{3} a^{4} e^{4}-\frac {16}{3} a^{3} b d \,e^{3}+32 a^{2} b^{2} d^{2} e^{2}-\frac {128}{3} a \,b^{3} d^{3} e +\frac {256}{15} b^{4} d^{4}}{\sqrt {e x +d}\, e^{5} \left (e x +d \right )} \] Input:
int((b*x+a)^3*(B*x+A)/(e*x+d)^(5/2),x)
Output:
(2*( - 5*a**4*e**4 - 40*a**3*b*d*e**3 - 60*a**3*b*e**4*x + 240*a**2*b**2*d **2*e**2 + 360*a**2*b**2*d*e**3*x + 90*a**2*b**2*e**4*x**2 - 320*a*b**3*d* *3*e - 480*a*b**3*d**2*e**2*x - 120*a*b**3*d*e**3*x**2 + 20*a*b**3*e**4*x* *3 + 128*b**4*d**4 + 192*b**4*d**3*e*x + 48*b**4*d**2*e**2*x**2 - 8*b**4*d *e**3*x**3 + 3*b**4*e**4*x**4))/(15*sqrt(d + e*x)*e**5*(d + e*x))