\(\int \frac {(a+b x)^2 (A+B x)}{(d+e x)^{3/2}} \, dx\) [136]

Optimal result

Integrand size = 22, antiderivative size = 124 \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^{3/2}} \, dx=\frac {2 (b d-a e)^2 (B d-A e)}{e^4 \sqrt {d+e x}}+\frac {2 (b d-a e) (3 b B d-2 A b e-a B e) \sqrt {d+e x}}{e^4}-\frac {2 b (3 b B d-A b e-2 a B e) (d+e x)^{3/2}}{3 e^4}+\frac {2 b^2 B (d+e x)^{5/2}}{5 e^4} \] Output:

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.09 \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^{3/2}} \, dx=\frac {30 a^2 e^2 (2 B d-A e+B e x)+20 a b e \left (3 A e (2 d+e x)+B \left (-8 d^2-4 d e x+e^2 x^2\right )\right )+2 b^2 \left (5 A e \left (-8 d^2-4 d e x+e^2 x^2\right )+3 B \left (16 d^3+8 d^2 e x-2 d e^2 x^2+e^3 x^3\right )\right )}{15 e^4 \sqrt {d+e x}} \] Input:

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

Output:

(30*a^2*e^2*(2*B*d - A*e + B*e*x) + 20*a*b*e*(3*A*e*(2*d + e*x) + B*(-8*d^ 
2 - 4*d*e*x + e^2*x^2)) + 2*b^2*(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)))/(15*e^4*Sqrt[d + e*x])
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 124, 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.

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

\(\Big \downarrow \) 86

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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 = 0.29 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.98

method	result	size
pseudoelliptic	\(\frac {\left (\left (6 B \,x^{3}+10 A \,x^{2}\right ) b^{2}+60 a \left (\frac {B x}{3}+A \right ) x b -30 a^{2} \left (-B x +A \right )\right ) e^{3}+120 \left (-\frac {\left (\frac {3 B x}{10}+A \right ) x \,b^{2}}{3}+a \left (-\frac {2 B x}{3}+A \right ) b +\frac {a^{2} B}{2}\right ) d \,e^{2}-80 b \left (\left (-\frac {3 B x}{5}+A \right ) b +2 B a \right ) d^{2} e +96 b^{2} B \,d^{3}}{15 \sqrt {e x +d}\, e^{4}}\)	\(121\)
risch	\(\frac {2 \left (3 e^{2} b^{2} B \,x^{2}+5 A \,b^{2} e^{2} x +10 B a b \,e^{2} x -9 b^{2} B d e x +30 A a b \,e^{2}-25 A \,b^{2} d e +15 B \,a^{2} e^{2}-50 B a b d e +33 b^{2} B \,d^{2}\right ) \sqrt {e x +d}}{15 e^{4}}-\frac {2 \left (a^{2} A \,e^{3}-2 A a b d \,e^{2}+A \,b^{2} d^{2} e -B \,a^{2} d \,e^{2}+2 B a b \,d^{2} e -b^{2} B \,d^{3}\right )}{e^{4} \sqrt {e x +d}}\)	\(163\)
gosper	\(-\frac {2 \left (-3 b^{2} B \,x^{3} e^{3}-5 A \,x^{2} b^{2} e^{3}-10 B \,x^{2} a b \,e^{3}+6 B \,x^{2} b^{2} d \,e^{2}-30 A x a b \,e^{3}+20 A x \,b^{2} d \,e^{2}-15 B x \,a^{2} e^{3}+40 B x a b d \,e^{2}-24 B x \,b^{2} d^{2} e +15 a^{2} A \,e^{3}-60 A a b d \,e^{2}+40 A \,b^{2} d^{2} e -30 B \,a^{2} d \,e^{2}+80 B a b \,d^{2} e -48 b^{2} B \,d^{3}\right )}{15 \sqrt {e x +d}\, e^{4}}\)	\(169\)
trager	\(-\frac {2 \left (-3 b^{2} B \,x^{3} e^{3}-5 A \,x^{2} b^{2} e^{3}-10 B \,x^{2} a b \,e^{3}+6 B \,x^{2} b^{2} d \,e^{2}-30 A x a b \,e^{3}+20 A x \,b^{2} d \,e^{2}-15 B x \,a^{2} e^{3}+40 B x a b d \,e^{2}-24 B x \,b^{2} d^{2} e +15 a^{2} A \,e^{3}-60 A a b d \,e^{2}+40 A \,b^{2} d^{2} e -30 B \,a^{2} d \,e^{2}+80 B a b \,d^{2} e -48 b^{2} B \,d^{3}\right )}{15 \sqrt {e x +d}\, e^{4}}\)	\(169\)
orering	\(-\frac {2 \left (-3 b^{2} B \,x^{3} e^{3}-5 A \,x^{2} b^{2} e^{3}-10 B \,x^{2} a b \,e^{3}+6 B \,x^{2} b^{2} d \,e^{2}-30 A x a b \,e^{3}+20 A x \,b^{2} d \,e^{2}-15 B x \,a^{2} e^{3}+40 B x a b d \,e^{2}-24 B x \,b^{2} d^{2} e +15 a^{2} A \,e^{3}-60 A a b d \,e^{2}+40 A \,b^{2} d^{2} e -30 B \,a^{2} d \,e^{2}+80 B a b \,d^{2} e -48 b^{2} B \,d^{3}\right )}{15 \sqrt {e x +d}\, e^{4}}\)	\(169\)
derivativedivides	\(\frac {\frac {2 b^{2} B \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 A \,b^{2} e \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {4 B a b e \left (e x +d \right )^{\frac {3}{2}}}{3}-2 B \,b^{2} d \left (e x +d \right )^{\frac {3}{2}}+4 A a b \,e^{2} \sqrt {e x +d}-4 A \,b^{2} d e \sqrt {e x +d}+2 B \,a^{2} e^{2} \sqrt {e x +d}-8 B a b d e \sqrt {e x +d}+6 B \,b^{2} d^{2} \sqrt {e x +d}-\frac {2 \left (a^{2} A \,e^{3}-2 A a b d \,e^{2}+A \,b^{2} d^{2} e -B \,a^{2} d \,e^{2}+2 B a b \,d^{2} e -b^{2} B \,d^{3}\right )}{\sqrt {e x +d}}}{e^{4}}\)	\(200\)
default	\(\frac {\frac {2 b^{2} B \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 A \,b^{2} e \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {4 B a b e \left (e x +d \right )^{\frac {3}{2}}}{3}-2 B \,b^{2} d \left (e x +d \right )^{\frac {3}{2}}+4 A a b \,e^{2} \sqrt {e x +d}-4 A \,b^{2} d e \sqrt {e x +d}+2 B \,a^{2} e^{2} \sqrt {e x +d}-8 B a b d e \sqrt {e x +d}+6 B \,b^{2} d^{2} \sqrt {e x +d}-\frac {2 \left (a^{2} A \,e^{3}-2 A a b d \,e^{2}+A \,b^{2} d^{2} e -B \,a^{2} d \,e^{2}+2 B a b \,d^{2} e -b^{2} B \,d^{3}\right )}{\sqrt {e x +d}}}{e^{4}}\)	\(200\)

Input:

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

Output:

1/15*(((6*B*x^3+10*A*x^2)*b^2+60*a*(1/3*B*x+A)*x*b-30*a^2*(-B*x+A))*e^3+12 
0*(-1/3*(3/10*B*x+A)*x*b^2+a*(-2/3*B*x+A)*b+1/2*a^2*B)*d*e^2-80*b*((-3/5*B 
*x+A)*b+2*B*a)*d^2*e+96*b^2*B*d^3)/(e*x+d)^(1/2)/e^4
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.33 \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (3 \, B b^{2} e^{3} x^{3} + 48 \, B b^{2} d^{3} - 15 \, A a^{2} e^{3} - 40 \, {\left (2 \, B a b + A b^{2}\right )} d^{2} e + 30 \, {\left (B a^{2} + 2 \, A a b\right )} d e^{2} - {\left (6 \, B b^{2} d e^{2} - 5 \, {\left (2 \, B a b + A b^{2}\right )} e^{3}\right )} x^{2} + {\left (24 \, B b^{2} d^{2} e - 20 \, {\left (2 \, B a b + A b^{2}\right )} d e^{2} + 15 \, {\left (B a^{2} + 2 \, A a b\right )} e^{3}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (e^{5} x + d e^{4}\right )}} \] Input:

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

Output:

2/15*(3*B*b^2*e^3*x^3 + 48*B*b^2*d^3 - 15*A*a^2*e^3 - 40*(2*B*a*b + A*b^2) 
*d^2*e + 30*(B*a^2 + 2*A*a*b)*d*e^2 - (6*B*b^2*d*e^2 - 5*(2*B*a*b + A*b^2) 
*e^3)*x^2 + (24*B*b^2*d^2*e - 20*(2*B*a*b + A*b^2)*d*e^2 + 15*(B*a^2 + 2*A 
*a*b)*e^3)*x)*sqrt(e*x + d)/(e^5*x + d*e^4)
 

Sympy [A] (verification not implemented)

Time = 4.22 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.67 \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^{3/2}} \, dx=\begin {cases} \frac {2 \left (\frac {B b^{2} \left (d + e x\right )^{\frac {5}{2}}}{5 e^{3}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (A b^{2} e + 2 B a b e - 3 B b^{2} d\right )}{3 e^{3}} + \frac {\sqrt {d + e x} \left (2 A a b e^{2} - 2 A b^{2} d e + B a^{2} e^{2} - 4 B a b d e + 3 B b^{2} d^{2}\right )}{e^{3}} + \frac {\left (- A e + B d\right ) \left (a e - b d\right )^{2}}{e^{3} \sqrt {d + e x}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {A a^{2} x + \frac {B b^{2} x^{4}}{4} + \frac {x^{3} \left (A b^{2} + 2 B a b\right )}{3} + \frac {x^{2} \cdot \left (2 A a b + B a^{2}\right )}{2}}{d^{\frac {3}{2}}} & \text {otherwise} \end {cases} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.35 \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (\frac {3 \, {\left (e x + d\right )}^{\frac {5}{2}} B b^{2} - 5 \, {\left (3 \, B b^{2} d - {\left (2 \, B a b + A b^{2}\right )} e\right )} {\left (e x + d\right )}^{\frac {3}{2}} + 15 \, {\left (3 \, B b^{2} d^{2} - 2 \, {\left (2 \, B a b + A b^{2}\right )} d e + {\left (B a^{2} + 2 \, A a b\right )} e^{2}\right )} \sqrt {e x + d}}{e^{3}} + \frac {15 \, {\left (B b^{2} d^{3} - A a^{2} e^{3} - {\left (2 \, B a b + A b^{2}\right )} d^{2} e + {\left (B a^{2} + 2 \, A a b\right )} d e^{2}\right )}}{\sqrt {e x + d} e^{3}}\right )}}{15 \, e} \] Input:

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

Output:

2/15*((3*(e*x + d)^(5/2)*B*b^2 - 5*(3*B*b^2*d - (2*B*a*b + A*b^2)*e)*(e*x 
+ d)^(3/2) + 15*(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)*sqrt(e*x + d))/e^3 + 15*(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^3))/e
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.78 \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (B b^{2} d^{3} - 2 \, B a b d^{2} e - A b^{2} d^{2} e + B a^{2} d e^{2} + 2 \, A a b d e^{2} - A a^{2} e^{3}\right )}}{\sqrt {e x + d} e^{4}} + \frac {2 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} B b^{2} e^{16} - 15 \, {\left (e x + d\right )}^{\frac {3}{2}} B b^{2} d e^{16} + 45 \, \sqrt {e x + d} B b^{2} d^{2} e^{16} + 10 \, {\left (e x + d\right )}^{\frac {3}{2}} B a b e^{17} + 5 \, {\left (e x + d\right )}^{\frac {3}{2}} A b^{2} e^{17} - 60 \, \sqrt {e x + d} B a b d e^{17} - 30 \, \sqrt {e x + d} A b^{2} d e^{17} + 15 \, \sqrt {e x + d} B a^{2} e^{18} + 30 \, \sqrt {e x + d} A a b e^{18}\right )}}{15 \, e^{20}} \] Input:

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

Output:

2*(B*b^2*d^3 - 2*B*a*b*d^2*e - A*b^2*d^2*e + B*a^2*d*e^2 + 2*A*a*b*d*e^2 - 
 A*a^2*e^3)/(sqrt(e*x + d)*e^4) + 2/15*(3*(e*x + d)^(5/2)*B*b^2*e^16 - 15* 
(e*x + d)^(3/2)*B*b^2*d*e^16 + 45*sqrt(e*x + d)*B*b^2*d^2*e^16 + 10*(e*x + 
 d)^(3/2)*B*a*b*e^17 + 5*(e*x + d)^(3/2)*A*b^2*e^17 - 60*sqrt(e*x + d)*B*a 
*b*d*e^17 - 30*sqrt(e*x + d)*A*b^2*d*e^17 + 15*sqrt(e*x + d)*B*a^2*e^18 + 
30*sqrt(e*x + d)*A*a*b*e^18)/e^20
 

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.24 \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^{3/2}} \, dx=\frac {{\left (d+e\,x\right )}^{3/2}\,\left (2\,A\,b^2\,e-6\,B\,b^2\,d+4\,B\,a\,b\,e\right )}{3\,e^4}-\frac {-2\,B\,a^2\,d\,e^2+2\,A\,a^2\,e^3+4\,B\,a\,b\,d^2\,e-4\,A\,a\,b\,d\,e^2-2\,B\,b^2\,d^3+2\,A\,b^2\,d^2\,e}{e^4\,\sqrt {d+e\,x}}+\frac {2\,B\,b^2\,{\left (d+e\,x\right )}^{5/2}}{5\,e^4}+\frac {2\,\left (a\,e-b\,d\right )\,\sqrt {d+e\,x}\,\left (2\,A\,b\,e+B\,a\,e-3\,B\,b\,d\right )}{e^4} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.93 \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^{3/2}} \, dx=\frac {\frac {2}{5} b^{3} e^{3} x^{3}+2 a \,b^{2} e^{3} x^{2}-\frac {4}{5} b^{3} d \,e^{2} x^{2}+6 a^{2} b \,e^{3} x -8 a \,b^{2} d \,e^{2} x +\frac {16}{5} b^{3} d^{2} e x -2 a^{3} e^{3}+12 a^{2} b d \,e^{2}-16 a \,b^{2} d^{2} e +\frac {32}{5} b^{3} d^{3}}{\sqrt {e x +d}\, e^{4}} \] Input:

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

Output:

(2*( - 5*a**3*e**3 + 30*a**2*b*d*e**2 + 15*a**2*b*e**3*x - 40*a*b**2*d**2* 
e - 20*a*b**2*d*e**2*x + 5*a*b**2*e**3*x**2 + 16*b**3*d**3 + 8*b**3*d**2*e 
*x - 2*b**3*d*e**2*x**2 + b**3*e**3*x**3))/(5*sqrt(d + e*x)*e**4)