\(\int \frac {A+B x}{\sqrt {a+b x}} \, dx\) [163]

Optimal result

Integrand size = 15, antiderivative size = 44 \[ \int \frac {A+B x}{\sqrt {a+b x}} \, dx=\frac {2 A \sqrt {a+b x}}{b}+\frac {2 B \sqrt {a+b x} \left (-a+\frac {1}{3} (a+b x)\right )}{b^2} \]

[Out]

2*alpha*(b*x+a)^(1/2)/b+2*beta*(1/3*b*x-2/3*a)*(b*x+a)^(1/2)/b^2

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.91, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {45} \[ \int \frac {A+B x}{\sqrt {a+b x}} \, dx=\frac {2 \sqrt {a+b x} (A b-a B)}{b^2}+\frac {2 B (a+b x)^{3/2}}{3 b^2} \]

[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 45

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*} \text {integral}& = \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] (verified)

Time = 0.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.66 \[ \int \frac {A+B x}{\sqrt {a+b x}} \, dx=\frac {2 \sqrt {a+b x} (3 A b-2 a B+b B x)}{3 b^2} \]

[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)

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.61

method	result	size
gosper	\(-\frac {2 \sqrt {b x +a}\, \left (-b \beta x +2 a \beta -3 \alpha b \right )}{3 b^{2}}\)	\(27\)
trager	\(-\frac {2 \sqrt {b x +a}\, \left (-b \beta x +2 a \beta -3 \alpha b \right )}{3 b^{2}}\)	\(27\)
risch	\(-\frac {2 \sqrt {b x +a}\, \left (-b \beta x +2 a \beta -3 \alpha b \right )}{3 b^{2}}\)	\(27\)
pseudoelliptic	\(-\frac {2 \sqrt {b x +a}\, \left (-b \beta x +2 a \beta -3 \alpha b \right )}{3 b^{2}}\)	\(27\)
derivativedivides	\(\frac {\frac {2 \beta \left (b x +a \right )^{\frac {3}{2}}}{3}-2 a \beta \sqrt {b x +a}+2 \alpha b \sqrt {b x +a}}{b^{2}}\)	\(38\)
default	\(\frac {\frac {2 \beta \left (b x +a \right )^{\frac {3}{2}}}{3}-2 a \beta \sqrt {b x +a}+2 \alpha b \sqrt {b x +a}}{b^{2}}\)	\(38\)

[In]

int((beta*x+alpha)/(b*x+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/3*(b*x+a)^(1/2)*(-b*beta*x+2*a*beta-3*alpha*b)/b^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.57 \[ \int \frac {A+B x}{\sqrt {a+b x}} \, dx=\frac {2 \, {\left (b \beta x + 3 \, \alpha b - 2 \, a \beta \right )} \sqrt {b x + a}}{3 \, b^{2}} \]

[In]

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

[Out]

2/3*(b*beta*x + 3*alpha*b - 2*a*beta)*sqrt(b*x + a)/b^2

Sympy [A] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.20 \[ \int \frac {A+B x}{\sqrt {a+b x}} \, dx=\begin {cases} \frac {2 A \sqrt {a + b x} + \frac {2 B \left (- a \sqrt {a + b x} + \frac {\left (a + b x\right )^{\frac {3}{2}}}{3}\right )}{b}}{b} & \text {for}\: b \neq 0 \\\frac {A x + \frac {B x^{2}}{2}}{\sqrt {a}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.89 \[ \int \frac {A+B x}{\sqrt {a+b x}} \, dx=\frac {2 \, {\left (3 \, \sqrt {b x + a} \alpha + \frac {{\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} \beta }{b}\right )}}{3 \, b} \]

[In]

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

[Out]

2/3*(3*sqrt(b*x + a)*alpha + ((b*x + a)^(3/2) - 3*sqrt(b*x + a)*a)*beta/b)/b

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.89 \[ \int \frac {A+B x}{\sqrt {a+b x}} \, dx=\frac {2 \, {\left (3 \, \sqrt {b x + a} \alpha + \frac {{\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} \beta }{b}\right )}}{3 \, b} \]

[In]

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

[Out]

2/3*(3*sqrt(b*x + a)*alpha + ((b*x + a)^(3/2) - 3*sqrt(b*x + a)*a)*beta/b)/b

Mupad [B] (verification not implemented)

Time = 16.96 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.64 \[ \int \frac {A+B x}{\sqrt {a+b x}} \, dx=\frac {2\,\sqrt {a+b\,x}\,\left (3\,\alpha \,b+\left (a+b\,x\right )\,\beta -3\,a\,\beta \right )}{3\,b^2} \]

[In]

int((alpha + x*beta)/(a + b*x)^(1/2),x)

[Out]

(2*(a + b*x)^(1/2)*(3*alpha*b + (a + b*x)*beta - 3*a*beta))/(3*b^2)