\(\int \frac {x^2 (A+B x)}{(a+b x)^{5/2}} \, dx\) [446]

Optimal result

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

[Out]

-2/3*a^2*(A*b-B*a)/b^4/(b*x+a)^(3/2)+2/3*B*(b*x+a)^(3/2)/b^4+2*a*(2*A*b-3*B*a)/b^4/(b*x+a)^(1/2)+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} \[ \int \frac {x^2 (A+B x)}{(a+b x)^{5/2}} \, dx=-\frac {2 a^2 (A b-a B)}{3 b^4 (a+b x)^{3/2}}+\frac {2 a (2 A b-3 a B)}{b^4 \sqrt {a+b x}}+\frac {2 \sqrt {a+b x} (A b-3 a B)}{b^4}+\frac {2 B (a+b x)^{3/2}}{3 b^4} \]

[In]

Int[(x^2*(A + B*x))/(a + b*x)^(5/2),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.69 \[ \int \frac {x^2 (A+B x)}{(a+b x)^{5/2}} \, dx=\frac {2 \left (-16 a^3 B+8 a^2 b (A-3 B x)-6 a b^2 x (-2 A+B x)+b^3 x^2 (3 A+B x)\right )}{3 b^4 (a+b x)^{3/2}} \]

[In]

Integrate[(x^2*(A + B*x))/(a + b*x)^(5/2),x]

[Out]

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

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.68

method	result	size
pseudoelliptic	\(\frac {\left (2 x^{3} B +6 A \,x^{2}\right ) b^{3}+24 x \left (-\frac {B x}{2}+A \right ) a \,b^{2}+16 a^{2} \left (-3 B x +A \right ) b -32 a^{3} B}{3 \left (b x +a \right )^{\frac {3}{2}} b^{4}}\)	\(62\)
risch	\(\frac {2 \left (b B x +3 A b -8 B a \right ) \sqrt {b x +a}}{3 b^{4}}+\frac {2 a \left (6 A \,b^{2} x -9 B a b x +5 a b A -8 a^{2} B \right )}{3 b^{4} \left (b x +a \right )^{\frac {3}{2}}}\)	\(65\)
gosper	\(\frac {\frac {2}{3} b^{3} B \,x^{3}+2 A \,b^{3} x^{2}-4 B a \,b^{2} x^{2}+8 a \,b^{2} A x -16 a^{2} b B x +\frac {16}{3} a^{2} b A -\frac {32}{3} a^{3} B}{\left (b x +a \right )^{\frac {3}{2}} b^{4}}\)	\(70\)
trager	\(\frac {\frac {2}{3} b^{3} B \,x^{3}+2 A \,b^{3} x^{2}-4 B a \,b^{2} x^{2}+8 a \,b^{2} A x -16 a^{2} b B x +\frac {16}{3} a^{2} b A -\frac {32}{3} a^{3} B}{\left (b x +a \right )^{\frac {3}{2}} b^{4}}\)	\(70\)
derivativedivides	\(\frac {\frac {2 B \left (b x +a \right )^{\frac {3}{2}}}{3}+2 A b \sqrt {b x +a}-6 B a \sqrt {b x +a}+\frac {2 a \left (2 A b -3 B a \right )}{\sqrt {b x +a}}-\frac {2 a^{2} \left (A b -B a \right )}{3 \left (b x +a \right )^{\frac {3}{2}}}}{b^{4}}\)	\(76\)
default	\(\frac {\frac {2 B \left (b x +a \right )^{\frac {3}{2}}}{3}+2 A b \sqrt {b x +a}-6 B a \sqrt {b x +a}+\frac {2 a \left (2 A b -3 B a \right )}{\sqrt {b x +a}}-\frac {2 a^{2} \left (A b -B a \right )}{3 \left (b x +a \right )^{\frac {3}{2}}}}{b^{4}}\)	\(76\)

[In]

int(x^2*(B*x+A)/(b*x+a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/3*((2*B*x^3+6*A*x^2)*b^3+24*x*(-1/2*B*x+A)*a*b^2+16*a^2*(-3*B*x+A)*b-32*a^3*B)/(b*x+a)^(3/2)/b^4

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.01 \[ \int \frac {x^2 (A+B x)}{(a+b x)^{5/2}} \, dx=\frac {2 \, {\left (B b^{3} x^{3} - 16 \, B a^{3} + 8 \, A a^{2} b - 3 \, {\left (2 \, B a b^{2} - A b^{3}\right )} x^{2} - 12 \, {\left (2 \, B a^{2} b - A a b^{2}\right )} x\right )} \sqrt {b x + a}}{3 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} \]

[In]

integrate(x^2*(B*x+A)/(b*x+a)^(5/2),x, algorithm="fricas")

[Out]

2/3*(B*b^3*x^3 - 16*B*a^3 + 8*A*a^2*b - 3*(2*B*a*b^2 - A*b^3)*x^2 - 12*(2*B*a^2*b - A*a*b^2)*x)*sqrt(b*x + a)/
(b^6*x^2 + 2*a*b^5*x + a^2*b^4)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 299 vs. \(2 (88) = 176\).

Time = 0.35 (sec) , antiderivative size = 299, normalized size of antiderivative = 3.29 \[ \int \frac {x^2 (A+B x)}{(a+b x)^{5/2}} \, dx=\begin {cases} \frac {16 A a^{2} b}{3 a b^{4} \sqrt {a + b x} + 3 b^{5} x \sqrt {a + b x}} + \frac {24 A a b^{2} x}{3 a b^{4} \sqrt {a + b x} + 3 b^{5} x \sqrt {a + b x}} + \frac {6 A b^{3} x^{2}}{3 a b^{4} \sqrt {a + b x} + 3 b^{5} x \sqrt {a + b x}} - \frac {32 B a^{3}}{3 a b^{4} \sqrt {a + b x} + 3 b^{5} x \sqrt {a + b x}} - \frac {48 B a^{2} b x}{3 a b^{4} \sqrt {a + b x} + 3 b^{5} x \sqrt {a + b x}} - \frac {12 B a b^{2} x^{2}}{3 a b^{4} \sqrt {a + b x} + 3 b^{5} x \sqrt {a + b x}} + \frac {2 B b^{3} x^{3}}{3 a b^{4} \sqrt {a + b x} + 3 b^{5} x \sqrt {a + b x}} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{3}}{3} + \frac {B x^{4}}{4}}{a^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

[In]

integrate(x**2*(B*x+A)/(b*x+a)**(5/2),x)

[Out]

Piecewise((16*A*a**2*b/(3*a*b**4*sqrt(a + b*x) + 3*b**5*x*sqrt(a + b*x)) + 24*A*a*b**2*x/(3*a*b**4*sqrt(a + b*
x) + 3*b**5*x*sqrt(a + b*x)) + 6*A*b**3*x**2/(3*a*b**4*sqrt(a + b*x) + 3*b**5*x*sqrt(a + b*x)) - 32*B*a**3/(3*
a*b**4*sqrt(a + b*x) + 3*b**5*x*sqrt(a + b*x)) - 48*B*a**2*b*x/(3*a*b**4*sqrt(a + b*x) + 3*b**5*x*sqrt(a + b*x
)) - 12*B*a*b**2*x**2/(3*a*b**4*sqrt(a + b*x) + 3*b**5*x*sqrt(a + b*x)) + 2*B*b**3*x**3/(3*a*b**4*sqrt(a + b*x
) + 3*b**5*x*sqrt(a + b*x)), Ne(b, 0)), ((A*x**3/3 + B*x**4/4)/a**(5/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.89 \[ \int \frac {x^2 (A+B x)}{(a+b x)^{5/2}} \, dx=\frac {2 \, {\left (\frac {{\left (b x + a\right )}^{\frac {3}{2}} B - 3 \, {\left (3 \, B a - A b\right )} \sqrt {b x + a}}{b} + \frac {B a^{3} - A a^{2} b - 3 \, {\left (3 \, B a^{2} - 2 \, A a b\right )} {\left (b x + a\right )}}{{\left (b x + a\right )}^{\frac {3}{2}} b}\right )}}{3 \, b^{3}} \]

[In]

integrate(x^2*(B*x+A)/(b*x+a)^(5/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.01 \[ \int \frac {x^2 (A+B x)}{(a+b x)^{5/2}} \, dx=-\frac {2 \, {\left (9 \, {\left (b x + a\right )} B a^{2} - B a^{3} - 6 \, {\left (b x + a\right )} A a b + A a^{2} b\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{4}} + \frac {2 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} B b^{8} - 9 \, \sqrt {b x + a} B a b^{8} + 3 \, \sqrt {b x + a} A b^{9}\right )}}{3 \, b^{12}} \]

[In]

integrate(x^2*(B*x+A)/(b*x+a)^(5/2),x, algorithm="giac")

[Out]

-2/3*(9*(b*x + a)*B*a^2 - B*a^3 - 6*(b*x + a)*A*a*b + A*a^2*b)/((b*x + a)^(3/2)*b^4) + 2/3*((b*x + a)^(3/2)*B*
b^8 - 9*sqrt(b*x + a)*B*a*b^8 + 3*sqrt(b*x + a)*A*b^9)/b^12

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.87 \[ \int \frac {x^2 (A+B x)}{(a+b x)^{5/2}} \, dx=\frac {2\,B\,a^3+2\,B\,{\left (a+b\,x\right )}^3+6\,A\,b\,{\left (a+b\,x\right )}^2-18\,B\,a\,{\left (a+b\,x\right )}^2-18\,B\,a^2\,\left (a+b\,x\right )-2\,A\,a^2\,b+12\,A\,a\,b\,\left (a+b\,x\right )}{3\,b^4\,{\left (a+b\,x\right )}^{3/2}} \]

[In]

int((x^2*(A + B*x))/(a + b*x)^(5/2),x)

[Out]

(2*B*a^3 + 2*B*(a + b*x)^3 + 6*A*b*(a + b*x)^2 - 18*B*a*(a + b*x)^2 - 18*B*a^2*(a + b*x) - 2*A*a^2*b + 12*A*a*
b*(a + b*x))/(3*b^4*(a + b*x)^(3/2))