\(\int \frac {\sqrt {x}}{(a+b x)^2} \, dx\) [458]

Optimal result

Integrand size = 13, antiderivative size = 46 \[ \int \frac {\sqrt {x}}{(a+b x)^2} \, dx=-\frac {\sqrt {x}}{b (a+b x)}+\frac {\arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} b^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {43, 65, 211} \[ \int \frac {\sqrt {x}}{(a+b x)^2} \, dx=\frac {\arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} b^{3/2}}-\frac {\sqrt {x}}{b (a+b x)} \]

[In]

Int[Sqrt[x]/(a + b*x)^2,x]

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps \begin{align*} \text {integral}& = -\frac {\sqrt {x}}{b (a+b x)}+\frac {\int \frac {1}{\sqrt {x} (a+b x)} \, dx}{2 b} \\ & = -\frac {\sqrt {x}}{b (a+b x)}+\frac {\text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {x}\right )}{b} \\ & = -\frac {\sqrt {x}}{b (a+b x)}+\frac {\tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} b^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {x}}{(a+b x)^2} \, dx=-\frac {\sqrt {x}}{b (a+b x)}+\frac {\arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} b^{3/2}} \]

[In]

Integrate[Sqrt[x]/(a + b*x)^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.80

method	result	size
derivativedivides	\(-\frac {\sqrt {x}}{b \left (b x +a \right )}+\frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{b \sqrt {a b}}\)	\(37\)
default	\(-\frac {\sqrt {x}}{b \left (b x +a \right )}+\frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{b \sqrt {a b}}\)	\(37\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 115, normalized size of antiderivative = 2.50 \[ \int \frac {\sqrt {x}}{(a+b x)^2} \, dx=\left [-\frac {2 \, a b \sqrt {x} + \sqrt {-a b} {\left (b x + a\right )} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right )}{2 \, {\left (a b^{3} x + a^{2} b^{2}\right )}}, -\frac {a b \sqrt {x} + \sqrt {a b} {\left (b x + a\right )} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right )}{a b^{3} x + a^{2} b^{2}}\right ] \]

[In]

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

[Out]

[-1/2*(2*a*b*sqrt(x) + sqrt(-a*b)*(b*x + a)*log((b*x - a - 2*sqrt(-a*b)*sqrt(x))/(b*x + a)))/(a*b^3*x + a^2*b^
2), -(a*b*sqrt(x) + sqrt(a*b)*(b*x + a)*arctan(sqrt(a*b)/(b*sqrt(x))))/(a*b^3*x + a^2*b^2)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 269 vs. \(2 (37) = 74\).

Time = 1.96 (sec) , antiderivative size = 269, normalized size of antiderivative = 5.85 \[ \int \frac {\sqrt {x}}{(a+b x)^2} \, dx=\begin {cases} \frac {\tilde {\infty }}{\sqrt {x}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 x^{\frac {3}{2}}}{3 a^{2}} & \text {for}\: b = 0 \\- \frac {2}{b^{2} \sqrt {x}} & \text {for}\: a = 0 \\\frac {a \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} \sqrt {- \frac {a}{b}} + 2 b^{3} x \sqrt {- \frac {a}{b}}} - \frac {a \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} \sqrt {- \frac {a}{b}} + 2 b^{3} x \sqrt {- \frac {a}{b}}} - \frac {2 b \sqrt {x} \sqrt {- \frac {a}{b}}}{2 a b^{2} \sqrt {- \frac {a}{b}} + 2 b^{3} x \sqrt {- \frac {a}{b}}} + \frac {b x \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} \sqrt {- \frac {a}{b}} + 2 b^{3} x \sqrt {- \frac {a}{b}}} - \frac {b x \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} \sqrt {- \frac {a}{b}} + 2 b^{3} x \sqrt {- \frac {a}{b}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((zoo/sqrt(x), Eq(a, 0) & Eq(b, 0)), (2*x**(3/2)/(3*a**2), Eq(b, 0)), (-2/(b**2*sqrt(x)), Eq(a, 0)),
(a*log(sqrt(x) - sqrt(-a/b))/(2*a*b**2*sqrt(-a/b) + 2*b**3*x*sqrt(-a/b)) - a*log(sqrt(x) + sqrt(-a/b))/(2*a*b*
*2*sqrt(-a/b) + 2*b**3*x*sqrt(-a/b)) - 2*b*sqrt(x)*sqrt(-a/b)/(2*a*b**2*sqrt(-a/b) + 2*b**3*x*sqrt(-a/b)) + b*
x*log(sqrt(x) - sqrt(-a/b))/(2*a*b**2*sqrt(-a/b) + 2*b**3*x*sqrt(-a/b)) - b*x*log(sqrt(x) + sqrt(-a/b))/(2*a*b
**2*sqrt(-a/b) + 2*b**3*x*sqrt(-a/b)), True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.80 \[ \int \frac {\sqrt {x}}{(a+b x)^2} \, dx=-\frac {\sqrt {x}}{b^{2} x + a b} + \frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt {x}}{(a+b x)^2} \, dx=\frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b} - \frac {\sqrt {x}}{{\left (b x + a\right )} b} \]

[In]

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

[Out]

arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b) - sqrt(x)/((b*x + a)*b)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.74 \[ \int \frac {\sqrt {x}}{(a+b x)^2} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a}\,b^{3/2}}-\frac {\sqrt {x}}{b\,\left (a+b\,x\right )} \]

[In]

int(x^(1/2)/(a + b*x)^2,x)

[Out]

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