\(\int x^3 (a+b x)^{3/2} (A+B x) \, dx\) [235]

Optimal result

Integrand size = 18, antiderivative size = 122 \[ \int x^3 (a+b x)^{3/2} (A+B x) \, dx=-\frac {2 a^3 (A b-a B) (a+b x)^{5/2}}{5 b^5}+\frac {2 a^2 (3 A b-4 a B) (a+b x)^{7/2}}{7 b^5}-\frac {2 a (A b-2 a B) (a+b x)^{9/2}}{3 b^5}+\frac {2 (A b-4 a B) (a+b x)^{11/2}}{11 b^5}+\frac {2 B (a+b x)^{13/2}}{13 b^5} \] Output:

-2/5*a^3*(A*b-B*a)*(b*x+a)^(5/2)/b^5+2/7*a^2*(3*A*b-4*B*a)*(b*x+a)^(7/2)/b 
^5-2/3*a*(A*b-2*B*a)*(b*x+a)^(9/2)/b^5+2/11*(A*b-4*B*a)*(b*x+a)^(11/2)/b^5 
+2/13*B*(b*x+a)^(13/2)/b^5
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.71 \[ \int x^3 (a+b x)^{3/2} (A+B x) \, dx=\frac {2 (a+b x)^{5/2} \left (128 a^4 B+105 b^4 x^3 (13 A+11 B x)-70 a b^3 x^2 (13 A+12 B x)+40 a^2 b^2 x (13 A+14 B x)-16 a^3 b (13 A+20 B x)\right )}{15015 b^5} \] Input:

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

Output:

(2*(a + b*x)^(5/2)*(128*a^4*B + 105*b^4*x^3*(13*A + 11*B*x) - 70*a*b^3*x^2 
*(13*A + 12*B*x) + 40*a^2*b^2*x*(13*A + 14*B*x) - 16*a^3*b*(13*A + 20*B*x) 
))/(15015*b^5)
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, 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.

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

\(\Big \downarrow \) 86

\(\displaystyle \int \left (\frac {a^3 (a+b x)^{3/2} (a B-A b)}{b^4}-\frac {a^2 (a+b x)^{5/2} (4 a B-3 A b)}{b^4}+\frac {(a+b x)^{9/2} (A b-4 a B)}{b^4}+\frac {3 a (a+b x)^{7/2} (2 a B-A b)}{b^4}+\frac {B (a+b x)^{11/2}}{b^4}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 a^3 (a+b x)^{5/2} (A b-a B)}{5 b^5}+\frac {2 a^2 (a+b x)^{7/2} (3 A b-4 a B)}{7 b^5}+\frac {2 (a+b x)^{11/2} (A b-4 a B)}{11 b^5}-\frac {2 a (a+b x)^{9/2} (A b-2 a B)}{3 b^5}+\frac {2 B (a+b x)^{13/2}}{13 b^5}\)

Input:

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

Output:

(-2*a^3*(A*b - a*B)*(a + b*x)^(5/2))/(5*b^5) + (2*a^2*(3*A*b - 4*a*B)*(a + 
 b*x)^(7/2))/(7*b^5) - (2*a*(A*b - 2*a*B)*(a + b*x)^(9/2))/(3*b^5) + (2*(A 
*b - 4*a*B)*(a + b*x)^(11/2))/(11*b^5) + (2*B*(a + b*x)^(13/2))/(13*b^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 = 0.39 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.61

method	result	size
pseudoelliptic	\(-\frac {32 \left (-\frac {105 \left (\frac {11 B x}{13}+A \right ) x^{3} b^{4}}{16}+\frac {35 a \left (\frac {12 B x}{13}+A \right ) x^{2} b^{3}}{8}-\frac {5 a^{2} x \left (\frac {14 B x}{13}+A \right ) b^{2}}{2}+a^{3} \left (\frac {20 B x}{13}+A \right ) b -\frac {8 B \,a^{4}}{13}\right ) \left (b x +a \right )^{\frac {5}{2}}}{1155 b^{5}}\)	\(75\)
gosper	\(-\frac {2 \left (b x +a \right )^{\frac {5}{2}} \left (-1155 B \,x^{4} b^{4}-1365 A \,x^{3} b^{4}+840 B \,x^{3} a \,b^{3}+910 A \,x^{2} a \,b^{3}-560 B \,x^{2} a^{2} b^{2}-520 A x \,a^{2} b^{2}+320 B x \,a^{3} b +208 A \,a^{3} b -128 B \,a^{4}\right )}{15015 b^{5}}\)	\(95\)
orering	\(-\frac {2 \left (b x +a \right )^{\frac {5}{2}} \left (-1155 B \,x^{4} b^{4}-1365 A \,x^{3} b^{4}+840 B \,x^{3} a \,b^{3}+910 A \,x^{2} a \,b^{3}-560 B \,x^{2} a^{2} b^{2}-520 A x \,a^{2} b^{2}+320 B x \,a^{3} b +208 A \,a^{3} b -128 B \,a^{4}\right )}{15015 b^{5}}\)	\(95\)
derivativedivides	\(\frac {\frac {2 B \left (b x +a \right )^{\frac {13}{2}}}{13}+\frac {2 \left (A b -4 B a \right ) \left (b x +a \right )^{\frac {11}{2}}}{11}+\frac {2 \left (3 a^{2} B -3 a \left (A b -B a \right )\right ) \left (b x +a \right )^{\frac {9}{2}}}{9}+\frac {2 \left (-a^{3} B +3 a^{2} \left (A b -B a \right )\right ) \left (b x +a \right )^{\frac {7}{2}}}{7}-\frac {2 a^{3} \left (A b -B a \right ) \left (b x +a \right )^{\frac {5}{2}}}{5}}{b^{5}}\)	\(110\)
default	\(\frac {\frac {2 B \left (b x +a \right )^{\frac {13}{2}}}{13}-\frac {2 \left (-A b +4 B a \right ) \left (b x +a \right )^{\frac {11}{2}}}{11}-\frac {2 \left (-3 a^{2} B +3 a \left (A b -B a \right )\right ) \left (b x +a \right )^{\frac {9}{2}}}{9}-\frac {2 \left (a^{3} B -3 a^{2} \left (A b -B a \right )\right ) \left (b x +a \right )^{\frac {7}{2}}}{7}-\frac {2 a^{3} \left (A b -B a \right ) \left (b x +a \right )^{\frac {5}{2}}}{5}}{b^{5}}\)	\(110\)
trager	\(-\frac {2 \left (-1155 B \,b^{6} x^{6}-1365 A \,b^{6} x^{5}-1470 B a \,b^{5} x^{5}-1820 A a \,b^{5} x^{4}-35 B \,a^{2} b^{4} x^{4}-65 A \,a^{2} b^{4} x^{3}+40 B \,a^{3} b^{3} x^{3}+78 A \,a^{3} b^{3} x^{2}-48 B \,a^{4} b^{2} x^{2}-104 A \,a^{4} b^{2} x +64 B \,a^{5} b x +208 A \,a^{5} b -128 B \,a^{6}\right ) \sqrt {b x +a}}{15015 b^{5}}\)	\(143\)
risch	\(-\frac {2 \left (-1155 B \,b^{6} x^{6}-1365 A \,b^{6} x^{5}-1470 B a \,b^{5} x^{5}-1820 A a \,b^{5} x^{4}-35 B \,a^{2} b^{4} x^{4}-65 A \,a^{2} b^{4} x^{3}+40 B \,a^{3} b^{3} x^{3}+78 A \,a^{3} b^{3} x^{2}-48 B \,a^{4} b^{2} x^{2}-104 A \,a^{4} b^{2} x +64 B \,a^{5} b x +208 A \,a^{5} b -128 B \,a^{6}\right ) \sqrt {b x +a}}{15015 b^{5}}\)	\(143\)

Input:

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

Output:

-32/1155*(-105/16*(11/13*B*x+A)*x^3*b^4+35/8*a*(12/13*B*x+A)*x^2*b^3-5/2*a 
^2*x*(14/13*B*x+A)*b^2+a^3*(20/13*B*x+A)*b-8/13*B*a^4)*(b*x+a)^(5/2)/b^5
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.17 \[ \int x^3 (a+b x)^{3/2} (A+B x) \, dx=\frac {2 \, {\left (1155 \, B b^{6} x^{6} + 128 \, B a^{6} - 208 \, A a^{5} b + 105 \, {\left (14 \, B a b^{5} + 13 \, A b^{6}\right )} x^{5} + 35 \, {\left (B a^{2} b^{4} + 52 \, A a b^{5}\right )} x^{4} - 5 \, {\left (8 \, B a^{3} b^{3} - 13 \, A a^{2} b^{4}\right )} x^{3} + 6 \, {\left (8 \, B a^{4} b^{2} - 13 \, A a^{3} b^{3}\right )} x^{2} - 8 \, {\left (8 \, B a^{5} b - 13 \, A a^{4} b^{2}\right )} x\right )} \sqrt {b x + a}}{15015 \, b^{5}} \] Input:

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

Output:

2/15015*(1155*B*b^6*x^6 + 128*B*a^6 - 208*A*a^5*b + 105*(14*B*a*b^5 + 13*A 
*b^6)*x^5 + 35*(B*a^2*b^4 + 52*A*a*b^5)*x^4 - 5*(8*B*a^3*b^3 - 13*A*a^2*b^ 
4)*x^3 + 6*(8*B*a^4*b^2 - 13*A*a^3*b^3)*x^2 - 8*(8*B*a^5*b - 13*A*a^4*b^2) 
*x)*sqrt(b*x + a)/b^5
 

Sympy [A] (verification not implemented)

Time = 0.74 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.14 \[ \int x^3 (a+b x)^{3/2} (A+B x) \, dx=\begin {cases} \frac {2 \left (\frac {B \left (a + b x\right )^{\frac {13}{2}}}{13 b} + \frac {\left (a + b x\right )^{\frac {11}{2}} \left (A b - 4 B a\right )}{11 b} + \frac {\left (a + b x\right )^{\frac {9}{2}} \left (- 3 A a b + 6 B a^{2}\right )}{9 b} + \frac {\left (a + b x\right )^{\frac {7}{2}} \cdot \left (3 A a^{2} b - 4 B a^{3}\right )}{7 b} + \frac {\left (a + b x\right )^{\frac {5}{2}} \left (- A a^{3} b + B a^{4}\right )}{5 b}\right )}{b^{4}} & \text {for}\: b \neq 0 \\a^{\frac {3}{2}} \left (\frac {A x^{4}}{4} + \frac {B x^{5}}{5}\right ) & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((2*(B*(a + b*x)**(13/2)/(13*b) + (a + b*x)**(11/2)*(A*b - 4*B*a) 
/(11*b) + (a + b*x)**(9/2)*(-3*A*a*b + 6*B*a**2)/(9*b) + (a + b*x)**(7/2)* 
(3*A*a**2*b - 4*B*a**3)/(7*b) + (a + b*x)**(5/2)*(-A*a**3*b + B*a**4)/(5*b 
))/b**4, Ne(b, 0)), (a**(3/2)*(A*x**4/4 + B*x**5/5), True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.82 \[ \int x^3 (a+b x)^{3/2} (A+B x) \, dx=\frac {2 \, {\left (1155 \, {\left (b x + a\right )}^{\frac {13}{2}} B - 1365 \, {\left (4 \, B a - A b\right )} {\left (b x + a\right )}^{\frac {11}{2}} + 5005 \, {\left (2 \, B a^{2} - A a b\right )} {\left (b x + a\right )}^{\frac {9}{2}} - 2145 \, {\left (4 \, B a^{3} - 3 \, A a^{2} b\right )} {\left (b x + a\right )}^{\frac {7}{2}} + 3003 \, {\left (B a^{4} - A a^{3} b\right )} {\left (b x + a\right )}^{\frac {5}{2}}\right )}}{15015 \, b^{5}} \] Input:

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

Output:

2/15015*(1155*(b*x + a)^(13/2)*B - 1365*(4*B*a - A*b)*(b*x + a)^(11/2) + 5 
005*(2*B*a^2 - A*a*b)*(b*x + a)^(9/2) - 2145*(4*B*a^3 - 3*A*a^2*b)*(b*x + 
a)^(7/2) + 3003*(B*a^4 - A*a^3*b)*(b*x + a)^(5/2))/b^5
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 422 vs. \(2 (104) = 208\).

Time = 0.13 (sec) , antiderivative size = 422, normalized size of antiderivative = 3.46 \[ \int x^3 (a+b x)^{3/2} (A+B x) \, dx=\frac {2 \, {\left (\frac {1287 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} A a^{2}}{b^{3}} + \frac {143 \, {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} B a^{2}}{b^{4}} + \frac {286 \, {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} A a}{b^{3}} + \frac {130 \, {\left (63 \, {\left (b x + a\right )}^{\frac {11}{2}} - 385 \, {\left (b x + a\right )}^{\frac {9}{2}} a + 990 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{2} - 1386 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{3} + 1155 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} - 693 \, \sqrt {b x + a} a^{5}\right )} B a}{b^{4}} + \frac {65 \, {\left (63 \, {\left (b x + a\right )}^{\frac {11}{2}} - 385 \, {\left (b x + a\right )}^{\frac {9}{2}} a + 990 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{2} - 1386 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{3} + 1155 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} - 693 \, \sqrt {b x + a} a^{5}\right )} A}{b^{3}} + \frac {15 \, {\left (231 \, {\left (b x + a\right )}^{\frac {13}{2}} - 1638 \, {\left (b x + a\right )}^{\frac {11}{2}} a + 5005 \, {\left (b x + a\right )}^{\frac {9}{2}} a^{2} - 8580 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{3} + 9009 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{4} - 6006 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{5} + 3003 \, \sqrt {b x + a} a^{6}\right )} B}{b^{4}}\right )}}{45045 \, b} \] Input:

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

Output:

2/45045*(1287*(5*(b*x + a)^(7/2) - 21*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/ 
2)*a^2 - 35*sqrt(b*x + a)*a^3)*A*a^2/b^3 + 143*(35*(b*x + a)^(9/2) - 180*( 
b*x + a)^(7/2)*a + 378*(b*x + a)^(5/2)*a^2 - 420*(b*x + a)^(3/2)*a^3 + 315 
*sqrt(b*x + a)*a^4)*B*a^2/b^4 + 286*(35*(b*x + a)^(9/2) - 180*(b*x + a)^(7 
/2)*a + 378*(b*x + a)^(5/2)*a^2 - 420*(b*x + a)^(3/2)*a^3 + 315*sqrt(b*x + 
 a)*a^4)*A*a/b^3 + 130*(63*(b*x + a)^(11/2) - 385*(b*x + a)^(9/2)*a + 990* 
(b*x + a)^(7/2)*a^2 - 1386*(b*x + a)^(5/2)*a^3 + 1155*(b*x + a)^(3/2)*a^4 
- 693*sqrt(b*x + a)*a^5)*B*a/b^4 + 65*(63*(b*x + a)^(11/2) - 385*(b*x + a) 
^(9/2)*a + 990*(b*x + a)^(7/2)*a^2 - 1386*(b*x + a)^(5/2)*a^3 + 1155*(b*x 
+ a)^(3/2)*a^4 - 693*sqrt(b*x + a)*a^5)*A/b^3 + 15*(231*(b*x + a)^(13/2) - 
 1638*(b*x + a)^(11/2)*a + 5005*(b*x + a)^(9/2)*a^2 - 8580*(b*x + a)^(7/2) 
*a^3 + 9009*(b*x + a)^(5/2)*a^4 - 6006*(b*x + a)^(3/2)*a^5 + 3003*sqrt(b*x 
 + a)*a^6)*B/b^4)/b
 

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.91 \[ \int x^3 (a+b x)^{3/2} (A+B x) \, dx=\frac {\left (12\,B\,a^2-6\,A\,a\,b\right )\,{\left (a+b\,x\right )}^{9/2}}{9\,b^5}+\frac {2\,B\,{\left (a+b\,x\right )}^{13/2}}{13\,b^5}+\frac {\left (2\,A\,b-8\,B\,a\right )\,{\left (a+b\,x\right )}^{11/2}}{11\,b^5}+\frac {\left (2\,B\,a^4-2\,A\,a^3\,b\right )\,{\left (a+b\,x\right )}^{5/2}}{5\,b^5}-\frac {\left (8\,B\,a^3-6\,A\,a^2\,b\right )\,{\left (a+b\,x\right )}^{7/2}}{7\,b^5} \] Input:

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

Output:

((12*B*a^2 - 6*A*a*b)*(a + b*x)^(9/2))/(9*b^5) + (2*B*(a + b*x)^(13/2))/(1 
3*b^5) + ((2*A*b - 8*B*a)*(a + b*x)^(11/2))/(11*b^5) + ((2*B*a^4 - 2*A*a^3 
*b)*(a + b*x)^(5/2))/(5*b^5) - ((8*B*a^3 - 6*A*a^2*b)*(a + b*x)^(7/2))/(7* 
b^5)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.61 \[ \int x^3 (a+b x)^{3/2} (A+B x) \, dx=\frac {2 \sqrt {b x +a}\, \left (231 b^{6} x^{6}+567 a \,b^{5} x^{5}+371 a^{2} b^{4} x^{4}+5 a^{3} b^{3} x^{3}-6 a^{4} b^{2} x^{2}+8 a^{5} b x -16 a^{6}\right )}{3003 b^{4}} \] Input:

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

Output:

(2*sqrt(a + b*x)*( - 16*a**6 + 8*a**5*b*x - 6*a**4*b**2*x**2 + 5*a**3*b**3 
*x**3 + 371*a**2*b**4*x**4 + 567*a*b**5*x**5 + 231*b**6*x**6))/(3003*b**4)