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

Optimal. Leaf size=38 \[ \frac{2 B \sqrt{a+b x}}{b^2}-\frac{2 (A b-a B)}{b^2 \sqrt{a+b x}} \]

[Out]

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

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Rubi [A]  time = 0.0133618, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {43} \[ \frac{2 B \sqrt{a+b x}}{b^2}-\frac{2 (A b-a B)}{b^2 \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(a + b*x)^(3/2),x]

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0159248, size = 27, normalized size = 0.71 \[ \frac{2 (2 a B-A b+b B x)}{b^2 \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.002, size = 26, normalized size = 0.7 \begin{align*} -2\,{\frac{-bBx+Ab-2\,Ba}{\sqrt{bx+a}{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 2.79799, size = 50, normalized size = 1.32 \begin{align*} \frac{2 \,{\left (\frac{\sqrt{b x + a} B}{b} + \frac{B a - A b}{\sqrt{b x + a} b}\right )}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.22688, size = 74, normalized size = 1.95 \begin{align*} \frac{2 \,{\left (B b x + 2 \, B a - A b\right )} \sqrt{b x + a}}{b^{3} x + a b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.21379, size = 60, normalized size = 1.58 \begin{align*} \begin{cases} - \frac{2 A}{b \sqrt{a + b x}} + \frac{4 B a}{b^{2} \sqrt{a + b x}} + \frac{2 B x}{b \sqrt{a + b x}} & \text{for}\: b \neq 0 \\\frac{A x + \frac{B x^{2}}{2}}{a^{\frac{3}{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.19465, size = 46, normalized size = 1.21 \begin{align*} \frac{2 \, \sqrt{b x + a} B}{b^{2}} + \frac{2 \,{\left (B a - A b\right )}}{\sqrt{b x + a} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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