\(\int \frac {(a+b x)^3 (A+B x)}{(d+e x)^{3/2}} \, dx\) [144]

Optimal result

Integrand size = 22, antiderivative size = 167 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^{3/2}} \, dx=-\frac {2 (b d-a e)^3 (B d-A e)}{e^5 \sqrt {d+e x}}-\frac {2 (b d-a e)^2 (4 b B d-3 A b e-a B e) \sqrt {d+e x}}{e^5}+\frac {2 b (b d-a e) (2 b B d-A b e-a B e) (d+e x)^{3/2}}{e^5}-\frac {2 b^2 (4 b B d-A b e-3 a B e) (d+e x)^{5/2}}{5 e^5}+\frac {2 b^3 B (d+e x)^{7/2}}{7 e^5} \] Output:

-2*(-a*e+b*d)^3*(-A*e+B*d)/e^5/(e*x+d)^(1/2)-2*(-a*e+b*d)^2*(-3*A*b*e-B*a* 
e+4*B*b*d)*(e*x+d)^(1/2)/e^5+2*b*(-a*e+b*d)*(-A*b*e-B*a*e+2*B*b*d)*(e*x+d) 
^(3/2)/e^5-2/5*b^2*(-A*b*e-3*B*a*e+4*B*b*d)*(e*x+d)^(5/2)/e^5+2/7*b^3*B*(e 
*x+d)^(7/2)/e^5
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.33 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^{3/2}} \, dx=\frac {2 \left (35 a^3 e^3 (2 B d-A e+B e x)+35 a^2 b e^2 \left (3 A e (2 d+e x)+B \left (-8 d^2-4 d e x+e^2 x^2\right )\right )+7 a b^2 e \left (5 A e \left (-8 d^2-4 d e x+e^2 x^2\right )+3 B \left (16 d^3+8 d^2 e x-2 d e^2 x^2+e^3 x^3\right )\right )+b^3 \left (7 A e \left (16 d^3+8 d^2 e x-2 d e^2 x^2+e^3 x^3\right )+B \left (-128 d^4-64 d^3 e x+16 d^2 e^2 x^2-8 d e^3 x^3+5 e^4 x^4\right )\right )\right )}{35 e^5 \sqrt {d+e x}} \] Input:

Integrate[((a + b*x)^3*(A + B*x))/(d + e*x)^(3/2),x]
 

Output:

(2*(35*a^3*e^3*(2*B*d - A*e + B*e*x) + 35*a^2*b*e^2*(3*A*e*(2*d + e*x) + B 
*(-8*d^2 - 4*d*e*x + e^2*x^2)) + 7*a*b^2*e*(5*A*e*(-8*d^2 - 4*d*e*x + e^2* 
x^2) + 3*B*(16*d^3 + 8*d^2*e*x - 2*d*e^2*x^2 + e^3*x^3)) + b^3*(7*A*e*(16* 
d^3 + 8*d^2*e*x - 2*d*e^2*x^2 + e^3*x^3) + B*(-128*d^4 - 64*d^3*e*x + 16*d 
^2*e^2*x^2 - 8*d*e^3*x^3 + 5*e^4*x^4))))/(35*e^5*Sqrt[d + e*x])
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 167, 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.

\(\displaystyle \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^{3/2}} \, dx\)

\(\Big \downarrow \) 86

\(\displaystyle \int \left (\frac {b^2 (d+e x)^{3/2} (3 a B e+A b e-4 b B d)}{e^4}-\frac {3 b \sqrt {d+e x} (b d-a e) (a B e+A b e-2 b B d)}{e^4}+\frac {(a e-b d)^2 (a B e+3 A b e-4 b B d)}{e^4 \sqrt {d+e x}}+\frac {(a e-b d)^3 (A e-B d)}{e^4 (d+e x)^{3/2}}+\frac {b^3 B (d+e x)^{5/2}}{e^4}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 b^2 (d+e x)^{5/2} (-3 a B e-A b e+4 b B d)}{5 e^5}+\frac {2 b (d+e x)^{3/2} (b d-a e) (-a B e-A b e+2 b B d)}{e^5}-\frac {2 \sqrt {d+e x} (b d-a e)^2 (-a B e-3 A b e+4 b B d)}{e^5}-\frac {2 (b d-a e)^3 (B d-A e)}{e^5 \sqrt {d+e x}}+\frac {2 b^3 B (d+e x)^{7/2}}{7 e^5}\)

Input:

Int[((a + b*x)^3*(A + B*x))/(d + e*x)^(3/2),x]
 

Output:

(-2*(b*d - a*e)^3*(B*d - A*e))/(e^5*Sqrt[d + e*x]) - (2*(b*d - a*e)^2*(4*b 
*B*d - 3*A*b*e - a*B*e)*Sqrt[d + e*x])/e^5 + (2*b*(b*d - a*e)*(2*b*B*d - A 
*b*e - a*B*e)*(d + e*x)^(3/2))/e^5 - (2*b^2*(4*b*B*d - A*b*e - 3*a*B*e)*(d 
 + e*x)^(5/2))/(5*e^5) + (2*b^3*B*(d + e*x)^(7/2))/(7*e^5)
 

Defintions of rubi rules used

rule 86

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]))))
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.13

method	result	size
pseudoelliptic	\(-\frac {2 \left (\left (-\frac {x^{3} \left (\frac {5 B x}{7}+A \right ) b^{3}}{5}-\left (\frac {3 B x}{5}+A \right ) a \,x^{2} b^{2}-3 a^{2} \left (\frac {B x}{3}+A \right ) x b +a^{3} \left (-B x +A \right )\right ) e^{4}-6 \left (-\frac {\left (\frac {4 B x}{7}+A \right ) x^{2} b^{3}}{15}-\frac {2 a \left (\frac {3 B x}{10}+A \right ) x \,b^{2}}{3}+a^{2} \left (-\frac {2 B x}{3}+A \right ) b +\frac {a^{3} B}{3}\right ) d \,e^{3}+8 b \left (-\frac {\left (\frac {2 B x}{7}+A \right ) x \,b^{2}}{5}+a \left (-\frac {3 B x}{5}+A \right ) b +a^{2} B \right ) d^{2} e^{2}-\frac {16 b^{2} \left (\left (-\frac {4 B x}{7}+A \right ) b +3 B a \right ) d^{3} e}{5}+\frac {128 b^{3} B \,d^{4}}{35}\right )}{\sqrt {e x +d}\, e^{5}}\)	\(189\)
risch	\(\frac {2 \left (5 b^{3} B \,x^{3} e^{3}+7 A \,b^{3} e^{3} x^{2}+21 B a \,b^{2} e^{3} x^{2}-13 B \,b^{3} d \,e^{2} x^{2}+35 A x a \,b^{2} e^{3}-21 A x \,b^{3} d \,e^{2}+35 B x \,a^{2} b \,e^{3}-63 B a \,b^{2} d \,e^{2} x +29 B x \,b^{3} d^{2} e +105 A \,a^{2} b \,e^{3}-175 A a \,b^{2} d \,e^{2}+77 A \,b^{3} d^{2} e +35 B \,a^{3} e^{3}-175 B \,a^{2} b d \,e^{2}+231 B a \,b^{2} d^{2} e -93 b^{3} B \,d^{3}\right ) \sqrt {e x +d}}{35 e^{5}}-\frac {2 \left (a^{3} A \,e^{4}-3 A \,a^{2} b d \,e^{3}+3 A a \,b^{2} d^{2} e^{2}-A \,b^{3} d^{3} e -B \,a^{3} d \,e^{3}+3 B \,a^{2} b \,d^{2} e^{2}-3 B a \,b^{2} d^{3} e +b^{3} B \,d^{4}\right )}{e^{5} \sqrt {e x +d}}\)	\(289\)
gosper	\(-\frac {2 \left (-5 B \,x^{4} b^{3} e^{4}-7 A \,x^{3} b^{3} e^{4}-21 B \,x^{3} a \,b^{2} e^{4}+8 B \,x^{3} b^{3} d \,e^{3}-35 A \,x^{2} a \,b^{2} e^{4}+14 A \,x^{2} b^{3} d \,e^{3}-35 B \,x^{2} a^{2} b \,e^{4}+42 B \,x^{2} a \,b^{2} d \,e^{3}-16 B \,x^{2} b^{3} d^{2} e^{2}-105 A x \,a^{2} b \,e^{4}+140 A x a \,b^{2} d \,e^{3}-56 A x \,b^{3} d^{2} e^{2}-35 B x \,a^{3} e^{4}+140 B x \,a^{2} b d \,e^{3}-168 B x a \,b^{2} d^{2} e^{2}+64 B x \,b^{3} d^{3} e +35 a^{3} A \,e^{4}-210 A \,a^{2} b d \,e^{3}+280 A a \,b^{2} d^{2} e^{2}-112 A \,b^{3} d^{3} e -70 B \,a^{3} d \,e^{3}+280 B \,a^{2} b \,d^{2} e^{2}-336 B a \,b^{2} d^{3} e +128 b^{3} B \,d^{4}\right )}{35 \sqrt {e x +d}\, e^{5}}\)	\(301\)
trager	\(-\frac {2 \left (-5 B \,x^{4} b^{3} e^{4}-7 A \,x^{3} b^{3} e^{4}-21 B \,x^{3} a \,b^{2} e^{4}+8 B \,x^{3} b^{3} d \,e^{3}-35 A \,x^{2} a \,b^{2} e^{4}+14 A \,x^{2} b^{3} d \,e^{3}-35 B \,x^{2} a^{2} b \,e^{4}+42 B \,x^{2} a \,b^{2} d \,e^{3}-16 B \,x^{2} b^{3} d^{2} e^{2}-105 A x \,a^{2} b \,e^{4}+140 A x a \,b^{2} d \,e^{3}-56 A x \,b^{3} d^{2} e^{2}-35 B x \,a^{3} e^{4}+140 B x \,a^{2} b d \,e^{3}-168 B x a \,b^{2} d^{2} e^{2}+64 B x \,b^{3} d^{3} e +35 a^{3} A \,e^{4}-210 A \,a^{2} b d \,e^{3}+280 A a \,b^{2} d^{2} e^{2}-112 A \,b^{3} d^{3} e -70 B \,a^{3} d \,e^{3}+280 B \,a^{2} b \,d^{2} e^{2}-336 B a \,b^{2} d^{3} e +128 b^{3} B \,d^{4}\right )}{35 \sqrt {e x +d}\, e^{5}}\)	\(301\)
orering	\(-\frac {2 \left (-5 B \,x^{4} b^{3} e^{4}-7 A \,x^{3} b^{3} e^{4}-21 B \,x^{3} a \,b^{2} e^{4}+8 B \,x^{3} b^{3} d \,e^{3}-35 A \,x^{2} a \,b^{2} e^{4}+14 A \,x^{2} b^{3} d \,e^{3}-35 B \,x^{2} a^{2} b \,e^{4}+42 B \,x^{2} a \,b^{2} d \,e^{3}-16 B \,x^{2} b^{3} d^{2} e^{2}-105 A x \,a^{2} b \,e^{4}+140 A x a \,b^{2} d \,e^{3}-56 A x \,b^{3} d^{2} e^{2}-35 B x \,a^{3} e^{4}+140 B x \,a^{2} b d \,e^{3}-168 B x a \,b^{2} d^{2} e^{2}+64 B x \,b^{3} d^{3} e +35 a^{3} A \,e^{4}-210 A \,a^{2} b d \,e^{3}+280 A a \,b^{2} d^{2} e^{2}-112 A \,b^{3} d^{3} e -70 B \,a^{3} d \,e^{3}+280 B \,a^{2} b \,d^{2} e^{2}-336 B a \,b^{2} d^{3} e +128 b^{3} B \,d^{4}\right )}{35 \sqrt {e x +d}\, e^{5}}\)	\(301\)
derivativedivides	\(\frac {\frac {2 b^{3} B \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 A \,b^{3} e \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {6 B a \,b^{2} e \left (e x +d \right )^{\frac {5}{2}}}{5}-\frac {8 B \,b^{3} d \left (e x +d \right )^{\frac {5}{2}}}{5}+2 A a \,b^{2} e^{2} \left (e x +d \right )^{\frac {3}{2}}-2 A \,b^{3} d e \left (e x +d \right )^{\frac {3}{2}}+2 B \,a^{2} b \,e^{2} \left (e x +d \right )^{\frac {3}{2}}-6 B a \,b^{2} d e \left (e x +d \right )^{\frac {3}{2}}+4 B \,b^{3} d^{2} \left (e x +d \right )^{\frac {3}{2}}+6 A \,a^{2} b \,e^{3} \sqrt {e x +d}-12 A a \,b^{2} d \,e^{2} \sqrt {e x +d}+6 A \,b^{3} d^{2} e \sqrt {e x +d}+2 B \,a^{3} e^{3} \sqrt {e x +d}-12 B \,a^{2} b d \,e^{2} \sqrt {e x +d}+18 B a \,b^{2} d^{2} e \sqrt {e x +d}-8 B \,b^{3} d^{3} \sqrt {e x +d}-\frac {2 \left (a^{3} A \,e^{4}-3 A \,a^{2} b d \,e^{3}+3 A a \,b^{2} d^{2} e^{2}-A \,b^{3} d^{3} e -B \,a^{3} d \,e^{3}+3 B \,a^{2} b \,d^{2} e^{2}-3 B a \,b^{2} d^{3} e +b^{3} B \,d^{4}\right )}{\sqrt {e x +d}}}{e^{5}}\)	\(355\)
default	\(\frac {\frac {2 b^{3} B \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 A \,b^{3} e \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {6 B a \,b^{2} e \left (e x +d \right )^{\frac {5}{2}}}{5}-\frac {8 B \,b^{3} d \left (e x +d \right )^{\frac {5}{2}}}{5}+2 A a \,b^{2} e^{2} \left (e x +d \right )^{\frac {3}{2}}-2 A \,b^{3} d e \left (e x +d \right )^{\frac {3}{2}}+2 B \,a^{2} b \,e^{2} \left (e x +d \right )^{\frac {3}{2}}-6 B a \,b^{2} d e \left (e x +d \right )^{\frac {3}{2}}+4 B \,b^{3} d^{2} \left (e x +d \right )^{\frac {3}{2}}+6 A \,a^{2} b \,e^{3} \sqrt {e x +d}-12 A a \,b^{2} d \,e^{2} \sqrt {e x +d}+6 A \,b^{3} d^{2} e \sqrt {e x +d}+2 B \,a^{3} e^{3} \sqrt {e x +d}-12 B \,a^{2} b d \,e^{2} \sqrt {e x +d}+18 B a \,b^{2} d^{2} e \sqrt {e x +d}-8 B \,b^{3} d^{3} \sqrt {e x +d}-\frac {2 \left (a^{3} A \,e^{4}-3 A \,a^{2} b d \,e^{3}+3 A a \,b^{2} d^{2} e^{2}-A \,b^{3} d^{3} e -B \,a^{3} d \,e^{3}+3 B \,a^{2} b \,d^{2} e^{2}-3 B a \,b^{2} d^{3} e +b^{3} B \,d^{4}\right )}{\sqrt {e x +d}}}{e^{5}}\)	\(355\)

Input:

int((b*x+a)^3*(B*x+A)/(e*x+d)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-2*((-1/5*x^3*(5/7*B*x+A)*b^3-(3/5*B*x+A)*a*x^2*b^2-3*a^2*(1/3*B*x+A)*x*b+ 
a^3*(-B*x+A))*e^4-6*(-1/15*(4/7*B*x+A)*x^2*b^3-2/3*a*(3/10*B*x+A)*x*b^2+a^ 
2*(-2/3*B*x+A)*b+1/3*a^3*B)*d*e^3+8*b*(-1/5*(2/7*B*x+A)*x*b^2+a*(-3/5*B*x+ 
A)*b+a^2*B)*d^2*e^2-16/5*b^2*((-4/7*B*x+A)*b+3*B*a)*d^3*e+128/35*b^3*B*d^4 
)/(e*x+d)^(1/2)/e^5
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.63 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (5 \, B b^{3} e^{4} x^{4} - 128 \, B b^{3} d^{4} - 35 \, A a^{3} e^{4} + 112 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e - 280 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} + 70 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} - {\left (8 \, B b^{3} d e^{3} - 7 \, {\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} + {\left (16 \, B b^{3} d^{2} e^{2} - 14 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} + 35 \, {\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} - {\left (64 \, B b^{3} d^{3} e - 56 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 140 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{3} - 35 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x\right )} \sqrt {e x + d}}{35 \, {\left (e^{6} x + d e^{5}\right )}} \] Input:

integrate((b*x+a)^3*(B*x+A)/(e*x+d)^(3/2),x, algorithm="fricas")
 

Output:

2/35*(5*B*b^3*e^4*x^4 - 128*B*b^3*d^4 - 35*A*a^3*e^4 + 112*(3*B*a*b^2 + A* 
b^3)*d^3*e - 280*(B*a^2*b + A*a*b^2)*d^2*e^2 + 70*(B*a^3 + 3*A*a^2*b)*d*e^ 
3 - (8*B*b^3*d*e^3 - 7*(3*B*a*b^2 + A*b^3)*e^4)*x^3 + (16*B*b^3*d^2*e^2 - 
14*(3*B*a*b^2 + A*b^3)*d*e^3 + 35*(B*a^2*b + A*a*b^2)*e^4)*x^2 - (64*B*b^3 
*d^3*e - 56*(3*B*a*b^2 + A*b^3)*d^2*e^2 + 140*(B*a^2*b + A*a*b^2)*d*e^3 - 
35*(B*a^3 + 3*A*a^2*b)*e^4)*x)*sqrt(e*x + d)/(e^6*x + d*e^5)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 340 vs. \(2 (165) = 330\).

Time = 9.55 (sec) , antiderivative size = 340, normalized size of antiderivative = 2.04 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^{3/2}} \, dx=\begin {cases} \frac {2 \left (\frac {B b^{3} \left (d + e x\right )^{\frac {7}{2}}}{7 e^{4}} + \frac {\left (d + e x\right )^{\frac {5}{2}} \left (A b^{3} e + 3 B a b^{2} e - 4 B b^{3} d\right )}{5 e^{4}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \cdot \left (3 A a b^{2} e^{2} - 3 A b^{3} d e + 3 B a^{2} b e^{2} - 9 B a b^{2} d e + 6 B b^{3} d^{2}\right )}{3 e^{4}} + \frac {\sqrt {d + e x} \left (3 A a^{2} b e^{3} - 6 A a b^{2} d e^{2} + 3 A b^{3} d^{2} e + B a^{3} e^{3} - 6 B a^{2} b d e^{2} + 9 B a b^{2} d^{2} e - 4 B b^{3} d^{3}\right )}{e^{4}} + \frac {\left (- A e + B d\right ) \left (a e - b d\right )^{3}}{e^{4} \sqrt {d + e x}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {A a^{3} x + \frac {B b^{3} x^{5}}{5} + \frac {x^{4} \left (A b^{3} + 3 B a b^{2}\right )}{4} + \frac {x^{3} \cdot \left (3 A a b^{2} + 3 B a^{2} b\right )}{3} + \frac {x^{2} \cdot \left (3 A a^{2} b + B a^{3}\right )}{2}}{d^{\frac {3}{2}}} & \text {otherwise} \end {cases} \] Input:

integrate((b*x+a)**3*(B*x+A)/(e*x+d)**(3/2),x)
 

Output:

Piecewise((2*(B*b**3*(d + e*x)**(7/2)/(7*e**4) + (d + e*x)**(5/2)*(A*b**3* 
e + 3*B*a*b**2*e - 4*B*b**3*d)/(5*e**4) + (d + e*x)**(3/2)*(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)/(3*e* 
*4) + sqrt(d + e*x)*(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*b**3*d**3)/e* 
*4 + (-A*e + B*d)*(a*e - b*d)**3/(e**4*sqrt(d + e*x)))/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**(3/2), True))
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.63 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (\frac {5 \, {\left (e x + d\right )}^{\frac {7}{2}} B b^{3} - 7 \, {\left (4 \, B b^{3} d - {\left (3 \, B a b^{2} + A b^{3}\right )} e\right )} {\left (e x + d\right )}^{\frac {5}{2}} + 35 \, {\left (2 \, B b^{3} d^{2} - {\left (3 \, B a b^{2} + A b^{3}\right )} d e + {\left (B a^{2} b + A a b^{2}\right )} e^{2}\right )} {\left (e x + d\right )}^{\frac {3}{2}} - 35 \, {\left (4 \, B b^{3} d^{3} - 3 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e + 6 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{2} - {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{3}\right )} \sqrt {e x + d}}{e^{4}} - \frac {35 \, {\left (B b^{3} d^{4} + A a^{3} e^{4} - {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 3 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} - {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3}\right )}}{\sqrt {e x + d} e^{4}}\right )}}{35 \, e} \] Input:

integrate((b*x+a)^3*(B*x+A)/(e*x+d)^(3/2),x, algorithm="maxima")
 

Output:

2/35*((5*(e*x + d)^(7/2)*B*b^3 - 7*(4*B*b^3*d - (3*B*a*b^2 + A*b^3)*e)*(e* 
x + d)^(5/2) + 35*(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)*(e*x + d)^(3/2) - 35*(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)*sqrt(e*x + d))/e^ 
4 - 35*(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)/(sqrt(e*x + d)*e^4))/e
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 385 vs. \(2 (153) = 306\).

Time = 0.13 (sec) , antiderivative size = 385, normalized size of antiderivative = 2.31 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^{3/2}} \, dx=-\frac {2 \, {\left (B b^{3} d^{4} - 3 \, B a b^{2} d^{3} e - A b^{3} d^{3} e + 3 \, B a^{2} b d^{2} e^{2} + 3 \, A a b^{2} d^{2} e^{2} - B a^{3} d e^{3} - 3 \, A a^{2} b d e^{3} + A a^{3} e^{4}\right )}}{\sqrt {e x + d} e^{5}} + \frac {2 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} B b^{3} e^{30} - 28 \, {\left (e x + d\right )}^{\frac {5}{2}} B b^{3} d e^{30} + 70 \, {\left (e x + d\right )}^{\frac {3}{2}} B b^{3} d^{2} e^{30} - 140 \, \sqrt {e x + d} B b^{3} d^{3} e^{30} + 21 \, {\left (e x + d\right )}^{\frac {5}{2}} B a b^{2} e^{31} + 7 \, {\left (e x + d\right )}^{\frac {5}{2}} A b^{3} e^{31} - 105 \, {\left (e x + d\right )}^{\frac {3}{2}} B a b^{2} d e^{31} - 35 \, {\left (e x + d\right )}^{\frac {3}{2}} A b^{3} d e^{31} + 315 \, \sqrt {e x + d} B a b^{2} d^{2} e^{31} + 105 \, \sqrt {e x + d} A b^{3} d^{2} e^{31} + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} B a^{2} b e^{32} + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} A a b^{2} e^{32} - 210 \, \sqrt {e x + d} B a^{2} b d e^{32} - 210 \, \sqrt {e x + d} A a b^{2} d e^{32} + 35 \, \sqrt {e x + d} B a^{3} e^{33} + 105 \, \sqrt {e x + d} A a^{2} b e^{33}\right )}}{35 \, e^{35}} \] Input:

integrate((b*x+a)^3*(B*x+A)/(e*x+d)^(3/2),x, algorithm="giac")
 

Output:

-2*(B*b^3*d^4 - 3*B*a*b^2*d^3*e - A*b^3*d^3*e + 3*B*a^2*b*d^2*e^2 + 3*A*a* 
b^2*d^2*e^2 - B*a^3*d*e^3 - 3*A*a^2*b*d*e^3 + A*a^3*e^4)/(sqrt(e*x + d)*e^ 
5) + 2/35*(5*(e*x + d)^(7/2)*B*b^3*e^30 - 28*(e*x + d)^(5/2)*B*b^3*d*e^30 
+ 70*(e*x + d)^(3/2)*B*b^3*d^2*e^30 - 140*sqrt(e*x + d)*B*b^3*d^3*e^30 + 2 
1*(e*x + d)^(5/2)*B*a*b^2*e^31 + 7*(e*x + d)^(5/2)*A*b^3*e^31 - 105*(e*x + 
 d)^(3/2)*B*a*b^2*d*e^31 - 35*(e*x + d)^(3/2)*A*b^3*d*e^31 + 315*sqrt(e*x 
+ d)*B*a*b^2*d^2*e^31 + 105*sqrt(e*x + d)*A*b^3*d^2*e^31 + 35*(e*x + d)^(3 
/2)*B*a^2*b*e^32 + 35*(e*x + d)^(3/2)*A*a*b^2*e^32 - 210*sqrt(e*x + d)*B*a 
^2*b*d*e^32 - 210*sqrt(e*x + d)*A*a*b^2*d*e^32 + 35*sqrt(e*x + d)*B*a^3*e^ 
33 + 105*sqrt(e*x + d)*A*a^2*b*e^33)/e^35
 

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.34 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^{3/2}} \, dx=\frac {{\left (d+e\,x\right )}^{5/2}\,\left (2\,A\,b^3\,e-8\,B\,b^3\,d+6\,B\,a\,b^2\,e\right )}{5\,e^5}-\frac {-2\,B\,a^3\,d\,e^3+2\,A\,a^3\,e^4+6\,B\,a^2\,b\,d^2\,e^2-6\,A\,a^2\,b\,d\,e^3-6\,B\,a\,b^2\,d^3\,e+6\,A\,a\,b^2\,d^2\,e^2+2\,B\,b^3\,d^4-2\,A\,b^3\,d^3\,e}{e^5\,\sqrt {d+e\,x}}+\frac {2\,{\left (a\,e-b\,d\right )}^2\,\sqrt {d+e\,x}\,\left (3\,A\,b\,e+B\,a\,e-4\,B\,b\,d\right )}{e^5}+\frac {2\,B\,b^3\,{\left (d+e\,x\right )}^{7/2}}{7\,e^5}+\frac {2\,b\,\left (a\,e-b\,d\right )\,{\left (d+e\,x\right )}^{3/2}\,\left (A\,b\,e+B\,a\,e-2\,B\,b\,d\right )}{e^5} \] Input:

int(((A + B*x)*(a + b*x)^3)/(d + e*x)^(3/2),x)
 

Output:

((d + e*x)^(5/2)*(2*A*b^3*e - 8*B*b^3*d + 6*B*a*b^2*e))/(5*e^5) - (2*A*a^3 
*e^4 + 2*B*b^3*d^4 - 2*A*b^3*d^3*e - 2*B*a^3*d*e^3 + 6*A*a*b^2*d^2*e^2 + 6 
*B*a^2*b*d^2*e^2 - 6*A*a^2*b*d*e^3 - 6*B*a*b^2*d^3*e)/(e^5*(d + e*x)^(1/2) 
) + (2*(a*e - b*d)^2*(d + e*x)^(1/2)*(3*A*b*e + B*a*e - 4*B*b*d))/e^5 + (2 
*B*b^3*(d + e*x)^(7/2))/(7*e^5) + (2*b*(a*e - b*d)*(d + e*x)^(3/2)*(A*b*e 
+ B*a*e - 2*B*b*d))/e^5
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.11 \[ \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^{3/2}} \, dx=\frac {\frac {2}{7} b^{4} e^{4} x^{4}+\frac {8}{5} a \,b^{3} e^{4} x^{3}-\frac {16}{35} b^{4} d \,e^{3} x^{3}+4 a^{2} b^{2} e^{4} x^{2}-\frac {16}{5} a \,b^{3} d \,e^{3} x^{2}+\frac {32}{35} b^{4} d^{2} e^{2} x^{2}+8 a^{3} b \,e^{4} x -16 a^{2} b^{2} d \,e^{3} x +\frac {64}{5} a \,b^{3} d^{2} e^{2} x -\frac {128}{35} b^{4} d^{3} e x -2 a^{4} e^{4}+16 a^{3} b d \,e^{3}-32 a^{2} b^{2} d^{2} e^{2}+\frac {128}{5} a \,b^{3} d^{3} e -\frac {256}{35} b^{4} d^{4}}{\sqrt {e x +d}\, e^{5}} \] Input:

int((b*x+a)^3*(B*x+A)/(e*x+d)^(3/2),x)
 

Output:

(2*( - 35*a**4*e**4 + 280*a**3*b*d*e**3 + 140*a**3*b*e**4*x - 560*a**2*b** 
2*d**2*e**2 - 280*a**2*b**2*d*e**3*x + 70*a**2*b**2*e**4*x**2 + 448*a*b**3 
*d**3*e + 224*a*b**3*d**2*e**2*x - 56*a*b**3*d*e**3*x**2 + 28*a*b**3*e**4* 
x**3 - 128*b**4*d**4 - 64*b**4*d**3*e*x + 16*b**4*d**2*e**2*x**2 - 8*b**4* 
d*e**3*x**3 + 5*b**4*e**4*x**4))/(35*sqrt(d + e*x)*e**5)