\(\int \frac {x}{(a+b x)^{3/2}} \, dx\) [346]

Optimal result

Integrand size = 11, antiderivative size = 30 \[ \int \frac {x}{(a+b x)^{3/2}} \, dx=\frac {2 a}{b^2 \sqrt {a+b x}}+\frac {2 \sqrt {a+b x}}{b^2} \]

[Out]

2*a/b^2/(b*x+a)^(1/2)+2*(b*x+a)^(1/2)/b^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {45} \[ \int \frac {x}{(a+b x)^{3/2}} \, dx=\frac {2 a}{b^2 \sqrt {a+b x}}+\frac {2 \sqrt {a+b x}}{b^2} \]

[In]

Int[x/(a + b*x)^(3/2),x]

[Out]

(2*a)/(b^2*Sqrt[a + b*x]) + (2*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 x)^{3/2}}+\frac {1}{b \sqrt {a+b x}}\right ) \, dx \\ & = \frac {2 a}{b^2 \sqrt {a+b x}}+\frac {2 \sqrt {a+b x}}{b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70 \[ \int \frac {x}{(a+b x)^{3/2}} \, dx=\frac {2 (2 a+b x)}{b^2 \sqrt {a+b x}} \]

[In]

Integrate[x/(a + b*x)^(3/2),x]

[Out]

(2*(2*a + b*x))/(b^2*Sqrt[a + b*x])

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67

method	result	size
gosper	\(\frac {2 b x +4 a}{b^{2} \sqrt {b x +a}}\)	\(20\)
trager	\(\frac {2 b x +4 a}{b^{2} \sqrt {b x +a}}\)	\(20\)
pseudoelliptic	\(\frac {2 b x +4 a}{b^{2} \sqrt {b x +a}}\)	\(20\)
derivativedivides	\(\frac {2 \sqrt {b x +a}+\frac {2 a}{\sqrt {b x +a}}}{b^{2}}\)	\(23\)
default	\(\frac {2 \sqrt {b x +a}+\frac {2 a}{\sqrt {b x +a}}}{b^{2}}\)	\(23\)
risch	\(\frac {2 a}{b^{2} \sqrt {b x +a}}+\frac {2 \sqrt {b x +a}}{b^{2}}\)	\(27\)

[In]

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

[Out]

2/(b*x+a)^(1/2)*(b*x+2*a)/b^2

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {x}{(a+b x)^{3/2}} \, dx=\frac {2 \, {\left (b x + 2 \, a\right )} \sqrt {b x + a}}{b^{3} x + a b^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {x}{(a+b x)^{3/2}} \, dx=\begin {cases} \frac {4 a}{b^{2} \sqrt {a + b x}} + \frac {2 x}{b \sqrt {a + b x}} & \text {for}\: b \neq 0 \\\frac {x^{2}}{2 a^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {x}{(a+b x)^{3/2}} \, dx=\frac {2 \, \sqrt {b x + a}}{b^{2}} + \frac {2 \, a}{\sqrt {b x + a} b^{2}} \]

[In]

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

[Out]

2*sqrt(b*x + a)/b^2 + 2*a/(sqrt(b*x + a)*b^2)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {x}{(a+b x)^{3/2}} \, dx=\frac {2 \, {\left (\frac {\sqrt {b x + a}}{b} + \frac {a}{\sqrt {b x + a} b}\right )}}{b} \]

[In]

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

[Out]

2*(sqrt(b*x + a)/b + a/(sqrt(b*x + a)*b))/b

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.63 \[ \int \frac {x}{(a+b x)^{3/2}} \, dx=\frac {4\,a+2\,b\,x}{b^2\,\sqrt {a+b\,x}} \]

[In]

int(x/(a + b*x)^(3/2),x)

[Out]

(4*a + 2*b*x)/(b^2*(a + b*x)^(1/2))