\(\int \frac {A+B x}{(a+b x)^{3/2}} \, dx\) [438]

Optimal result

Integrand size = 15, antiderivative size = 38 \[ \int \frac {A+B x}{(a+b x)^{3/2}} \, dx=-\frac {2 (A b-a B)}{b^2 \sqrt {a+b x}}+\frac {2 B \sqrt {a+b x}}{b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {45} \[ \int \frac {A+B x}{(a+b x)^{3/2}} \, dx=\frac {2 B \sqrt {a+b x}}{b^2}-\frac {2 (A b-a B)}{b^2 \sqrt {a+b x}} \]

[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 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 (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] (verified)

Time = 0.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.71 \[ \int \frac {A+B x}{(a+b x)^{3/2}} \, dx=\frac {2 (-A b+2 a B+b B x)}{b^2 \sqrt {a+b x}} \]

[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])

Maple [A] (verified)

Time = 1.42 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.68

method	result	size
gosper	\(-\frac {2 \left (-b B x +A b -2 B a \right )}{\sqrt {b x +a}\, b^{2}}\)	\(26\)
trager	\(-\frac {2 \left (-b B x +A b -2 B a \right )}{\sqrt {b x +a}\, b^{2}}\)	\(26\)
pseudoelliptic	\(\frac {\left (2 B x -2 A \right ) b +4 B a}{\sqrt {b x +a}\, b^{2}}\)	\(27\)
derivativedivides	\(\frac {2 B \sqrt {b x +a}-\frac {2 \left (A b -B a \right )}{\sqrt {b x +a}}}{b^{2}}\)	\(33\)
default	\(\frac {2 B \sqrt {b x +a}-\frac {2 \left (A b -B a \right )}{\sqrt {b x +a}}}{b^{2}}\)	\(33\)
risch	\(-\frac {2 \left (A b -B a \right )}{b^{2} \sqrt {b x +a}}+\frac {2 B \sqrt {b x +a}}{b^{2}}\)	\(35\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.92 \[ \int \frac {A+B x}{(a+b x)^{3/2}} \, dx=\frac {2 \, {\left (B b x + 2 \, B a - A b\right )} \sqrt {b x + a}}{b^{3} x + a b^{2}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.58 \[ \int \frac {A+B x}{(a+b x)^{3/2}} \, dx=\begin {cases} - \frac {2 A}{b \sqrt {a + b x}} + \frac {4 B a}{b^{2} \sqrt {a + b x}} + \frac {2 B x}{b \sqrt {a + b x}} & \text {for}\: b \neq 0 \\\frac {A x + \frac {B x^{2}}{2}}{a^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.97 \[ \int \frac {A+B x}{(a+b x)^{3/2}} \, dx=\frac {2 \, {\left (\frac {\sqrt {b x + a} B}{b} + \frac {B a - A b}{\sqrt {b x + a} b}\right )}}{b} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89 \[ \int \frac {A+B x}{(a+b x)^{3/2}} \, dx=\frac {2 \, \sqrt {b x + a} B}{b^{2}} + \frac {2 \, {\left (B a - A b\right )}}{\sqrt {b x + a} b^{2}} \]

[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)

Mupad [B] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.66 \[ \int \frac {A+B x}{(a+b x)^{3/2}} \, dx=\frac {4\,B\,a-2\,A\,b+2\,B\,b\,x}{b^2\,\sqrt {a+b\,x}} \]

[In]

int((A + B*x)/(a + b*x)^(3/2),x)

[Out]

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