\(\int \frac {x^2 (A+B x)}{(a+b x)^{3/2}} \, dx\) [436]

Optimal result

Integrand size = 18, antiderivative size = 91 \[ \int \frac {x^2 (A+B x)}{(a+b x)^{3/2}} \, dx=-\frac {2 a^2 (A b-a B)}{b^4 \sqrt {a+b x}}-\frac {2 a (2 A b-3 a B) \sqrt {a+b x}}{b^4}+\frac {2 (A b-3 a B) (a+b x)^{3/2}}{3 b^4}+\frac {2 B (a+b x)^{5/2}}{5 b^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 91, 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} \[ \int \frac {x^2 (A+B x)}{(a+b x)^{3/2}} \, dx=-\frac {2 a^2 (A b-a B)}{b^4 \sqrt {a+b x}}+\frac {2 (a+b x)^{3/2} (A b-3 a B)}{3 b^4}-\frac {2 a \sqrt {a+b x} (2 A b-3 a B)}{b^4}+\frac {2 B (a+b x)^{5/2}}{5 b^4} \]

[In]

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

[Out]

(-2*a^2*(A*b - a*B))/(b^4*Sqrt[a + b*x]) - (2*a*(2*A*b - 3*a*B)*Sqrt[a + b*x])/b^4 + (2*(A*b - 3*a*B)*(a + b*x
)^(3/2))/(3*b^4) + (2*B*(a + b*x)^(5/2))/(5*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)}{b^3 (a+b x)^{3/2}}+\frac {a (-2 A b+3 a B)}{b^3 \sqrt {a+b x}}+\frac {(A b-3 a B) \sqrt {a+b x}}{b^3}+\frac {B (a+b x)^{3/2}}{b^3}\right ) \, dx \\ & = -\frac {2 a^2 (A b-a B)}{b^4 \sqrt {a+b x}}-\frac {2 a (2 A b-3 a B) \sqrt {a+b x}}{b^4}+\frac {2 (A b-3 a B) (a+b x)^{3/2}}{3 b^4}+\frac {2 B (a+b x)^{5/2}}{5 b^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.74 \[ \int \frac {x^2 (A+B x)}{(a+b x)^{3/2}} \, dx=\frac {2 \left (48 a^3 B-8 a^2 b (5 A-3 B x)+b^3 x^2 (5 A+3 B x)-2 a b^2 x (10 A+3 B x)\right )}{15 b^4 \sqrt {a+b x}} \]

[In]

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

[Out]

(2*(48*a^3*B - 8*a^2*b*(5*A - 3*B*x) + b^3*x^2*(5*A + 3*B*x) - 2*a*b^2*x*(10*A + 3*B*x)))/(15*b^4*Sqrt[a + b*x
])

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.64

method	result	size
pseudoelliptic	\(-\frac {16 \left (-\frac {\left (\frac {3 B x}{5}+A \right ) x^{2} b^{3}}{8}+\frac {x \left (\frac {3 B x}{10}+A \right ) a \,b^{2}}{2}+a^{2} \left (-\frac {3 B x}{5}+A \right ) b -\frac {6 a^{3} B}{5}\right )}{3 \sqrt {b x +a}\, b^{4}}\)	\(58\)
gosper	\(-\frac {2 \left (-3 b^{3} B \,x^{3}-5 A \,b^{3} x^{2}+6 B a \,b^{2} x^{2}+20 a \,b^{2} A x -24 a^{2} b B x +40 a^{2} b A -48 a^{3} B \right )}{15 \sqrt {b x +a}\, b^{4}}\)	\(71\)
trager	\(-\frac {2 \left (-3 b^{3} B \,x^{3}-5 A \,b^{3} x^{2}+6 B a \,b^{2} x^{2}+20 a \,b^{2} A x -24 a^{2} b B x +40 a^{2} b A -48 a^{3} B \right )}{15 \sqrt {b x +a}\, b^{4}}\)	\(71\)
risch	\(-\frac {2 \left (-3 b^{2} B \,x^{2}-5 A \,b^{2} x +9 B a b x +25 a b A -33 a^{2} B \right ) \sqrt {b x +a}}{15 b^{4}}-\frac {2 a^{2} \left (A b -B a \right )}{b^{4} \sqrt {b x +a}}\)	\(71\)
derivativedivides	\(\frac {\frac {2 B \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 A b \left (b x +a \right )^{\frac {3}{2}}}{3}-2 B a \left (b x +a \right )^{\frac {3}{2}}-4 A a b \sqrt {b x +a}+6 B \,a^{2} \sqrt {b x +a}-\frac {2 a^{2} \left (A b -B a \right )}{\sqrt {b x +a}}}{b^{4}}\)	\(84\)
default	\(\frac {\frac {2 B \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 A b \left (b x +a \right )^{\frac {3}{2}}}{3}-2 B a \left (b x +a \right )^{\frac {3}{2}}-4 A a b \sqrt {b x +a}+6 B \,a^{2} \sqrt {b x +a}-\frac {2 a^{2} \left (A b -B a \right )}{\sqrt {b x +a}}}{b^{4}}\)	\(84\)

[In]

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

[Out]

-16/3*(-1/8*(3/5*B*x+A)*x^2*b^3+1/2*x*(3/10*B*x+A)*a*b^2+a^2*(-3/5*B*x+A)*b-6/5*a^3*B)/(b*x+a)^(1/2)/b^4

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.90 \[ \int \frac {x^2 (A+B x)}{(a+b x)^{3/2}} \, dx=\frac {2 \, {\left (3 \, B b^{3} x^{3} + 48 \, B a^{3} - 40 \, A a^{2} b - {\left (6 \, B a b^{2} - 5 \, A b^{3}\right )} x^{2} + 4 \, {\left (6 \, B a^{2} b - 5 \, A a b^{2}\right )} x\right )} \sqrt {b x + a}}{15 \, {\left (b^{5} x + a b^{4}\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.20 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.15 \[ \int \frac {x^2 (A+B x)}{(a+b x)^{3/2}} \, dx=\begin {cases} \frac {2 \left (\frac {B \left (a + b x\right )^{\frac {5}{2}}}{5 b} + \frac {a^{2} \left (- A b + B a\right )}{b \sqrt {a + b x}} + \frac {\left (a + b x\right )^{\frac {3}{2}} \left (A b - 3 B a\right )}{3 b} + \frac {\sqrt {a + b x} \left (- 2 A a b + 3 B a^{2}\right )}{b}\right )}{b^{3}} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{3}}{3} + \frac {B x^{4}}{4}}{a^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.93 \[ \int \frac {x^2 (A+B x)}{(a+b x)^{3/2}} \, dx=\frac {2 \, {\left (\frac {3 \, {\left (b x + a\right )}^{\frac {5}{2}} B - 5 \, {\left (3 \, B a - A b\right )} {\left (b x + a\right )}^{\frac {3}{2}} + 15 \, {\left (3 \, B a^{2} - 2 \, A a b\right )} \sqrt {b x + a}}{b} + \frac {15 \, {\left (B a^{3} - A a^{2} b\right )}}{\sqrt {b x + a} b}\right )}}{15 \, b^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.12 \[ \int \frac {x^2 (A+B x)}{(a+b x)^{3/2}} \, dx=\frac {2 \, {\left (B a^{3} - A a^{2} b\right )}}{\sqrt {b x + a} b^{4}} + \frac {2 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} B b^{16} - 15 \, {\left (b x + a\right )}^{\frac {3}{2}} B a b^{16} + 45 \, \sqrt {b x + a} B a^{2} b^{16} + 5 \, {\left (b x + a\right )}^{\frac {3}{2}} A b^{17} - 30 \, \sqrt {b x + a} A a b^{17}\right )}}{15 \, b^{20}} \]

[In]

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

[Out]

2*(B*a^3 - A*a^2*b)/(sqrt(b*x + a)*b^4) + 2/15*(3*(b*x + a)^(5/2)*B*b^16 - 15*(b*x + a)^(3/2)*B*a*b^16 + 45*sq
rt(b*x + a)*B*a^2*b^16 + 5*(b*x + a)^(3/2)*A*b^17 - 30*sqrt(b*x + a)*A*a*b^17)/b^20

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.91 \[ \int \frac {x^2 (A+B x)}{(a+b x)^{3/2}} \, dx=\frac {\left (6\,B\,a^2-4\,A\,a\,b\right )\,\sqrt {a+b\,x}}{b^4}+\frac {2\,B\,{\left (a+b\,x\right )}^{5/2}}{5\,b^4}+\frac {\left (2\,A\,b-6\,B\,a\right )\,{\left (a+b\,x\right )}^{3/2}}{3\,b^4}+\frac {2\,B\,a^3-2\,A\,a^2\,b}{b^4\,\sqrt {a+b\,x}} \]

[In]

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

[Out]

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