3.426 \(\int \frac {x (A+B x)}{\sqrt {a+b x}} \, dx\)

Optimal. Leaf size=65 \[ \frac {2 (a+b x)^{3/2} (A b-2 a B)}{3 b^3}-\frac {2 a \sqrt {a+b x} (A b-a B)}{b^3}+\frac {2 B (a+b x)^{5/2}}{5 b^3} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 65, 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} \[ \frac {2 (a+b x)^{3/2} (A b-2 a B)}{3 b^3}-\frac {2 a \sqrt {a+b x} (A b-a B)}{b^3}+\frac {2 B (a+b x)^{5/2}}{5 b^3} \]

Antiderivative was successfully verified.

[In]

Int[(x*(A + B*x))/Sqrt[a + b*x],x]

[Out]

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

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Mathematica [A]  time = 0.05, size = 48, normalized size = 0.74 \[ \frac {2 \sqrt {a+b x} \left (8 a^2 B-2 a b (5 A+2 B x)+b^2 x (5 A+3 B x)\right )}{15 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.74, size = 48, normalized size = 0.74 \[ \frac {2 \, {\left (3 \, B b^{2} x^{2} + 8 \, B a^{2} - 10 \, A a b - {\left (4 \, B a b - 5 \, A b^{2}\right )} x\right )} \sqrt {b x + a}}{15 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*B*b^2*x^2 + 8*B*a^2 - 10*A*a*b - (4*B*a*b - 5*A*b^2)*x)*sqrt(b*x + a)/b^3

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giac [A]  time = 1.27, size = 67, normalized size = 1.03 \[ \frac {2 \, {\left (\frac {5 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} A}{b} + \frac {{\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} B}{b^{2}}\right )}}{15 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(5*((b*x + a)^(3/2) - 3*sqrt(b*x + a)*a)*A/b + (3*(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x +
a)*a^2)*B/b^2)/b

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maple [A]  time = 0.00, size = 47, normalized size = 0.72 \[ -\frac {2 \sqrt {b x +a}\, \left (-3 B \,b^{2} x^{2}-5 A \,b^{2} x +4 B a b x +10 A a b -8 B \,a^{2}\right )}{15 b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(b*x+a)^(1/2)*(-3*B*b^2*x^2-5*A*b^2*x+4*B*a*b*x+10*A*a*b-8*B*a^2)/b^3

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maxima [A]  time = 0.87, size = 54, normalized size = 0.83 \[ \frac {2 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} B - 5 \, {\left (2 \, B a - A b\right )} {\left (b x + a\right )}^{\frac {3}{2}} + 15 \, {\left (B a^{2} - A a b\right )} \sqrt {b x + a}\right )}}{15 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.38, size = 52, normalized size = 0.80 \[ \frac {2\,\sqrt {a+b\,x}\,\left (15\,B\,a^2+3\,B\,{\left (a+b\,x\right )}^2-15\,A\,a\,b+5\,A\,b\,\left (a+b\,x\right )-10\,B\,a\,\left (a+b\,x\right )\right )}{15\,b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(A + B*x))/(a + b*x)^(1/2),x)

[Out]

(2*(a + b*x)^(1/2)*(15*B*a^2 + 3*B*(a + b*x)^2 - 15*A*a*b + 5*A*b*(a + b*x) - 10*B*a*(a + b*x)))/(15*b^3)

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sympy [A]  time = 19.47, size = 182, normalized size = 2.80 \[ \begin {cases} \frac {- \frac {2 A a \left (- \frac {a}{\sqrt {a + b x}} - \sqrt {a + b x}\right )}{b} - \frac {2 A \left (\frac {a^{2}}{\sqrt {a + b x}} + 2 a \sqrt {a + b x} - \frac {\left (a + b x\right )^{\frac {3}{2}}}{3}\right )}{b} - \frac {2 B a \left (\frac {a^{2}}{\sqrt {a + b x}} + 2 a \sqrt {a + b x} - \frac {\left (a + b x\right )^{\frac {3}{2}}}{3}\right )}{b^{2}} - \frac {2 B \left (- \frac {a^{3}}{\sqrt {a + b x}} - 3 a^{2} \sqrt {a + b x} + a \left (a + b x\right )^{\frac {3}{2}} - \frac {\left (a + b x\right )^{\frac {5}{2}}}{5}\right )}{b^{2}}}{b} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{2}}{2} + \frac {B x^{3}}{3}}{\sqrt {a}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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