3.1417 \(\int \frac{a+b x}{\sqrt{c+d x}} \, dx\)

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

[Out]

(-2*(b*c - a*d)*Sqrt[c + d*x])/d^2 + (2*b*(c + d*x)^(3/2))/(3*d^2)

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Rubi [A]  time = 0.0129204, 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} \[ \frac{2 b (c+d x)^{3/2}}{3 d^2}-\frac{2 \sqrt{c+d x} (b c-a d)}{d^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)/Sqrt[c + d*x],x]

[Out]

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

Mathematica [A]  time = 0.0171093, size = 29, normalized size = 0.72 \[ \frac{2 \sqrt{c+d x} (3 a d-2 b c+b d x)}{3 d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)/Sqrt[c + d*x],x]

[Out]

(2*Sqrt[c + d*x]*(-2*b*c + 3*a*d + b*d*x))/(3*d^2)

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Maple [A]  time = 0.002, size = 26, normalized size = 0.7 \begin{align*}{\frac{2\,bdx+6\,ad-4\,bc}{3\,{d}^{2}}\sqrt{dx+c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)/(d*x+c)^(1/2),x)

[Out]

2/3*(d*x+c)^(1/2)*(b*d*x+3*a*d-2*b*c)/d^2

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Maxima [A]  time = 0.955858, size = 53, normalized size = 1.32 \begin{align*} \frac{2 \,{\left (3 \, \sqrt{d x + c} a + \frac{{\left ({\left (d x + c\right )}^{\frac{3}{2}} - 3 \, \sqrt{d x + c} c\right )} b}{d}\right )}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(3*sqrt(d*x + c)*a + ((d*x + c)^(3/2) - 3*sqrt(d*x + c)*c)*b/d)/d

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Fricas [A]  time = 1.86197, size = 63, normalized size = 1.58 \begin{align*} \frac{2 \,{\left (b d x - 2 \, b c + 3 \, a d\right )} \sqrt{d x + c}}{3 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(b*d*x - 2*b*c + 3*a*d)*sqrt(d*x + c)/d^2

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Sympy [A]  time = 3.27413, size = 121, normalized size = 3.02 \begin{align*} \begin{cases} - \frac{\frac{2 a c}{\sqrt{c + d x}} + 2 a \left (- \frac{c}{\sqrt{c + d x}} - \sqrt{c + d x}\right ) + \frac{2 b c \left (- \frac{c}{\sqrt{c + d x}} - \sqrt{c + d x}\right )}{d} + \frac{2 b \left (\frac{c^{2}}{\sqrt{c + d x}} + 2 c \sqrt{c + d x} - \frac{\left (c + d x\right )^{\frac{3}{2}}}{3}\right )}{d}}{d} & \text{for}\: d \neq 0 \\\frac{a x + \frac{b x^{2}}{2}}{\sqrt{c}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(d*x+c)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.05598, size = 53, normalized size = 1.32 \begin{align*} \frac{2 \,{\left (3 \, \sqrt{d x + c} a + \frac{{\left ({\left (d x + c\right )}^{\frac{3}{2}} - 3 \, \sqrt{d x + c} c\right )} b}{d}\right )}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(3*sqrt(d*x + c)*a + ((d*x + c)^(3/2) - 3*sqrt(d*x + c)*c)*b/d)/d