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

Optimal. Leaf size=40 \[ \frac{2 \sqrt{a+b x} (A b-a B)}{b^2}+\frac{2 B (a+b x)^{3/2}}{3 b^2} \]

[Out]

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

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Rubi [A]  time = 0.0129595, antiderivative size = 40, 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} \[ \frac{2 \sqrt{a+b x} (A b-a B)}{b^2}+\frac{2 B (a+b x)^{3/2}}{3 b^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/Sqrt[a + b*x],x]

[Out]

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

Mathematica [A]  time = 0.0164997, size = 29, normalized size = 0.72 \[ \frac{2 \sqrt{a+b x} (-2 a B+3 A b+b B x)}{3 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/Sqrt[a + b*x],x]

[Out]

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

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Maple [A]  time = 0.003, size = 26, normalized size = 0.7 \begin{align*}{\frac{2\,bBx+6\,Ab-4\,Ba}{3\,{b}^{2}}\sqrt{bx+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.09025, size = 53, normalized size = 1.32 \begin{align*} \frac{2 \,{\left (3 \, \sqrt{b x + a} A + \frac{{\left ({\left (b x + a\right )}^{\frac{3}{2}} - 3 \, \sqrt{b x + a} a\right )} B}{b}\right )}}{3 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.49851, size = 63, normalized size = 1.58 \begin{align*} \frac{2 \,{\left (B b x - 2 \, B a + 3 \, A b\right )} \sqrt{b x + a}}{3 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 3.81616, size = 121, normalized size = 3.02 \begin{align*} \begin{cases} - \frac{\frac{2 A a}{\sqrt{a + b x}} + 2 A \left (- \frac{a}{\sqrt{a + b x}} - \sqrt{a + b x}\right ) + \frac{2 B a \left (- \frac{a}{\sqrt{a + b x}} - \sqrt{a + b x}\right )}{b} + \frac{2 B \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}}{b} & \text{for}\: b \neq 0 \\\frac{A x + \frac{B x^{2}}{2}}{\sqrt{a}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14588, size = 53, normalized size = 1.32 \begin{align*} \frac{2 \,{\left (3 \, \sqrt{b x + a} A + \frac{{\left ({\left (b x + a\right )}^{\frac{3}{2}} - 3 \, \sqrt{b x + a} a\right )} B}{b}\right )}}{3 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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