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

Optimal. Leaf size=63 \[ -\frac {2 (A b-2 a B)}{b^3 \sqrt {a+b x}}+\frac {2 a (A b-a B)}{3 b^3 (a+b x)^{3/2}}+\frac {2 B \sqrt {a+b x}}{b^3} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {77} \[ -\frac {2 (A b-2 a B)}{b^3 \sqrt {a+b x}}+\frac {2 a (A b-a B)}{3 b^3 (a+b x)^{3/2}}+\frac {2 B \sqrt {a+b x}}{b^3} \]

Antiderivative was successfully verified.

[In]

Int[(x*(A + B*x))/(a + b*x)^(5/2),x]

[Out]

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

Rule 77

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

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Mathematica [A]  time = 0.05, size = 46, normalized size = 0.73 \[ \frac {16 a^2 B-4 a b (A-6 B x)+6 b^2 x (B x-A)}{3 b^3 (a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.70, size = 69, normalized size = 1.10 \[ \frac {2 \, {\left (3 \, B b^{2} x^{2} + 8 \, B a^{2} - 2 \, A a b + 3 \, {\left (4 \, B a b - A b^{2}\right )} x\right )} \sqrt {b x + a}}{3 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.27, size = 55, normalized size = 0.87 \[ \frac {2 \, \sqrt {b x + a} B}{b^{3}} + \frac {2 \, {\left (6 \, {\left (b x + a\right )} B a - B a^{2} - 3 \, {\left (b x + a\right )} A b + A a b\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 47, normalized size = 0.75 \[ -\frac {2 \left (-3 B \,b^{2} x^{2}+3 A \,b^{2} x -12 B a b x +2 A a b -8 B \,a^{2}\right )}{3 \left (b x +a \right )^{\frac {3}{2}} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.80, size = 58, normalized size = 0.92 \[ \frac {2 \, {\left (\frac {3 \, \sqrt {b x + a} B}{b} - \frac {B a^{2} - A a b - 3 \, {\left (2 \, B a - A b\right )} {\left (b x + a\right )}}{{\left (b x + a\right )}^{\frac {3}{2}} b}\right )}}{3 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.37, size = 52, normalized size = 0.83 \[ \frac {6\,B\,{\left (a+b\,x\right )}^2-2\,B\,a^2+2\,A\,a\,b-6\,A\,b\,\left (a+b\,x\right )+12\,B\,a\,\left (a+b\,x\right )}{3\,b^3\,{\left (a+b\,x\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(A + B*x))/(a + b*x)^(5/2),x)

[Out]

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

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sympy [A]  time = 1.26, size = 211, normalized size = 3.35 \[ \begin {cases} - \frac {4 A a b}{3 a b^{3} \sqrt {a + b x} + 3 b^{4} x \sqrt {a + b x}} - \frac {6 A b^{2} x}{3 a b^{3} \sqrt {a + b x} + 3 b^{4} x \sqrt {a + b x}} + \frac {16 B a^{2}}{3 a b^{3} \sqrt {a + b x} + 3 b^{4} x \sqrt {a + b x}} + \frac {24 B a b x}{3 a b^{3} \sqrt {a + b x} + 3 b^{4} x \sqrt {a + b x}} + \frac {6 B b^{2} x^{2}}{3 a b^{3} \sqrt {a + b x} + 3 b^{4} x \sqrt {a + b x}} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{2}}{2} + \frac {B x^{3}}{3}}{a^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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