\(\int x^2 \sqrt {a+b x} (A+B x) \, dx\) [389]

Optimal result

Integrand size = 18, antiderivative size = 95 \[ \int x^2 \sqrt {a+b x} (A+B x) \, dx=\frac {2 a^2 (A b-a B) (a+b x)^{3/2}}{3 b^4}-\frac {2 a (2 A b-3 a B) (a+b x)^{5/2}}{5 b^4}+\frac {2 (A b-3 a B) (a+b x)^{7/2}}{7 b^4}+\frac {2 B (a+b x)^{9/2}}{9 b^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {78} \[ \int x^2 \sqrt {a+b x} (A+B x) \, dx=\frac {2 a^2 (a+b x)^{3/2} (A b-a B)}{3 b^4}+\frac {2 (a+b x)^{7/2} (A b-3 a B)}{7 b^4}-\frac {2 a (a+b x)^{5/2} (2 A b-3 a B)}{5 b^4}+\frac {2 B (a+b x)^{9/2}}{9 b^4} \]

[In]

Int[x^2*Sqrt[a + b*x]*(A + B*x),x]

[Out]

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((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]))))

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {a^2 (-A b+a B) \sqrt {a+b x}}{b^3}+\frac {a (-2 A b+3 a B) (a+b x)^{3/2}}{b^3}+\frac {(A b-3 a B) (a+b x)^{5/2}}{b^3}+\frac {B (a+b x)^{7/2}}{b^3}\right ) \, dx \\ & = \frac {2 a^2 (A b-a B) (a+b x)^{3/2}}{3 b^4}-\frac {2 a (2 A b-3 a B) (a+b x)^{5/2}}{5 b^4}+\frac {2 (A b-3 a B) (a+b x)^{7/2}}{7 b^4}+\frac {2 B (a+b x)^{9/2}}{9 b^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.68 \[ \int x^2 \sqrt {a+b x} (A+B x) \, dx=\frac {2 (a+b x)^{3/2} \left (-16 a^3 B+24 a^2 b (A+B x)-6 a b^2 x (6 A+5 B x)+5 b^3 x^2 (9 A+7 B x)\right )}{315 b^4} \]

[In]

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

[Out]

(2*(a + b*x)^(3/2)*(-16*a^3*B + 24*a^2*b*(A + B*x) - 6*a*b^2*x*(6*A + 5*B*x) + 5*b^3*x^2*(9*A + 7*B*x)))/(315*
b^4)

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.60

method	result	size
pseudoelliptic	\(\frac {16 \left (b x +a \right )^{\frac {3}{2}} \left (\frac {15 x^{2} \left (\frac {7 B x}{9}+A \right ) b^{3}}{8}-\frac {3 x \left (\frac {5 B x}{6}+A \right ) a \,b^{2}}{2}+a^{2} \left (B x +A \right ) b -\frac {2 a^{3} B}{3}\right )}{105 b^{4}}\)	\(57\)
gosper	\(\frac {2 \left (b x +a \right )^{\frac {3}{2}} \left (35 b^{3} B \,x^{3}+45 A \,b^{3} x^{2}-30 B a \,b^{2} x^{2}-36 a \,b^{2} A x +24 a^{2} b B x +24 a^{2} b A -16 a^{3} B \right )}{315 b^{4}}\)	\(71\)
derivativedivides	\(\frac {\frac {2 B \left (b x +a \right )^{\frac {9}{2}}}{9}+\frac {2 \left (A b -3 B a \right ) \left (b x +a \right )^{\frac {7}{2}}}{7}+\frac {2 \left (a^{2} B -2 a \left (A b -B a \right )\right ) \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 a^{2} \left (A b -B a \right ) \left (b x +a \right )^{\frac {3}{2}}}{3}}{b^{4}}\)	\(80\)
default	\(\frac {\frac {2 B \left (b x +a \right )^{\frac {9}{2}}}{9}+\frac {2 \left (A b -3 B a \right ) \left (b x +a \right )^{\frac {7}{2}}}{7}+\frac {2 \left (a^{2} B -2 a \left (A b -B a \right )\right ) \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 a^{2} \left (A b -B a \right ) \left (b x +a \right )^{\frac {3}{2}}}{3}}{b^{4}}\)	\(80\)
trager	\(\frac {2 \left (35 B \,x^{4} b^{4}+45 A \,x^{3} b^{4}+5 B \,x^{3} a \,b^{3}+9 A \,x^{2} a \,b^{3}-6 B \,x^{2} a^{2} b^{2}-12 A x \,a^{2} b^{2}+8 B x \,a^{3} b +24 A \,a^{3} b -16 B \,a^{4}\right ) \sqrt {b x +a}}{315 b^{4}}\)	\(95\)
risch	\(\frac {2 \left (35 B \,x^{4} b^{4}+45 A \,x^{3} b^{4}+5 B \,x^{3} a \,b^{3}+9 A \,x^{2} a \,b^{3}-6 B \,x^{2} a^{2} b^{2}-12 A x \,a^{2} b^{2}+8 B x \,a^{3} b +24 A \,a^{3} b -16 B \,a^{4}\right ) \sqrt {b x +a}}{315 b^{4}}\)	\(95\)

[In]

int(x^2*(B*x+A)*(b*x+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

16/105*(b*x+a)^(3/2)*(15/8*x^2*(7/9*B*x+A)*b^3-3/2*x*(5/6*B*x+A)*a*b^2+a^2*(B*x+A)*b-2/3*a^3*B)/b^4

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00 \[ \int x^2 \sqrt {a+b x} (A+B x) \, dx=\frac {2 \, {\left (35 \, B b^{4} x^{4} - 16 \, B a^{4} + 24 \, A a^{3} b + 5 \, {\left (B a b^{3} + 9 \, A b^{4}\right )} x^{3} - 3 \, {\left (2 \, B a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{2} + 4 \, {\left (2 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} x\right )} \sqrt {b x + a}}{315 \, b^{4}} \]

[In]

integrate(x^2*(B*x+A)*(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

2/315*(35*B*b^4*x^4 - 16*B*a^4 + 24*A*a^3*b + 5*(B*a*b^3 + 9*A*b^4)*x^3 - 3*(2*B*a^2*b^2 - 3*A*a*b^3)*x^2 + 4*
(2*B*a^3*b - 3*A*a^2*b^2)*x)*sqrt(b*x + a)/b^4

Sympy [A] (verification not implemented)

Time = 0.63 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.16 \[ \int x^2 \sqrt {a+b x} (A+B x) \, dx=\begin {cases} \frac {2 \left (\frac {B \left (a + b x\right )^{\frac {9}{2}}}{9 b} + \frac {\left (a + b x\right )^{\frac {7}{2}} \left (A b - 3 B a\right )}{7 b} + \frac {\left (a + b x\right )^{\frac {5}{2}} \left (- 2 A a b + 3 B a^{2}\right )}{5 b} + \frac {\left (a + b x\right )^{\frac {3}{2}} \left (A a^{2} b - B a^{3}\right )}{3 b}\right )}{b^{3}} & \text {for}\: b \neq 0 \\\sqrt {a} \left (\frac {A x^{3}}{3} + \frac {B x^{4}}{4}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate(x**2*(B*x+A)*(b*x+a)**(1/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.81 \[ \int x^2 \sqrt {a+b x} (A+B x) \, dx=\frac {2 \, {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} B - 45 \, {\left (3 \, B a - A b\right )} {\left (b x + a\right )}^{\frac {7}{2}} + 63 \, {\left (3 \, B a^{2} - 2 \, A a b\right )} {\left (b x + a\right )}^{\frac {5}{2}} - 105 \, {\left (B a^{3} - A a^{2} b\right )} {\left (b x + a\right )}^{\frac {3}{2}}\right )}}{315 \, b^{4}} \]

[In]

integrate(x^2*(B*x+A)*(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

2/315*(35*(b*x + a)^(9/2)*B - 45*(3*B*a - A*b)*(b*x + a)^(7/2) + 63*(3*B*a^2 - 2*A*a*b)*(b*x + a)^(5/2) - 105*
(B*a^3 - A*a^2*b)*(b*x + a)^(3/2))/b^4

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 207 vs. \(2 (80) = 160\).

Time = 0.28 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.18 \[ \int x^2 \sqrt {a+b x} (A+B x) \, dx=\frac {2 \, {\left (\frac {21 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} A a}{b^{2}} + \frac {9 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} B a}{b^{3}} + \frac {9 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} A}{b^{2}} + \frac {{\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} B}{b^{3}}\right )}}{315 \, b} \]

[In]

integrate(x^2*(B*x+A)*(b*x+a)^(1/2),x, algorithm="giac")

[Out]

2/315*(21*(3*(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2)*A*a/b^2 + 9*(5*(b*x + a)^(7/2) - 2
1*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2 - 35*sqrt(b*x + a)*a^3)*B*a/b^3 + 9*(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/b^2 + (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/b^3)/b

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.89 \[ \int x^2 \sqrt {a+b x} (A+B x) \, dx=\frac {\left (6\,B\,a^2-4\,A\,a\,b\right )\,{\left (a+b\,x\right )}^{5/2}}{5\,b^4}+\frac {2\,B\,{\left (a+b\,x\right )}^{9/2}}{9\,b^4}+\frac {\left (2\,A\,b-6\,B\,a\right )\,{\left (a+b\,x\right )}^{7/2}}{7\,b^4}-\frac {\left (2\,B\,a^3-2\,A\,a^2\,b\right )\,{\left (a+b\,x\right )}^{3/2}}{3\,b^4} \]

[In]

int(x^2*(A + B*x)*(a + b*x)^(1/2),x)

[Out]

((6*B*a^2 - 4*A*a*b)*(a + b*x)^(5/2))/(5*b^4) + (2*B*(a + b*x)^(9/2))/(9*b^4) + ((2*A*b - 6*B*a)*(a + b*x)^(7/
2))/(7*b^4) - ((2*B*a^3 - 2*A*a^2*b)*(a + b*x)^(3/2))/(3*b^4)