\(\int \frac {\sqrt {x}}{a+b x} \, dx\) [451]

Optimal result

Integrand size = 13, antiderivative size = 40 \[ \int \frac {\sqrt {x}}{a+b x} \, dx=\frac {2 \sqrt {x}}{b}-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 40, 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 = {52, 65, 211} \[ \int \frac {\sqrt {x}}{a+b x} \, dx=\frac {2 \sqrt {x}}{b}-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{3/2}} \]

[In]

Int[Sqrt[x]/(a + b*x),x]

[Out]

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

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {x}}{a+b x} \, dx=\frac {2 \sqrt {x}}{b}-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{3/2}} \]

[In]

Integrate[Sqrt[x]/(a + b*x),x]

[Out]

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.80

method	result	size
derivativedivides	\(\frac {2 \sqrt {x}}{b}-\frac {2 a \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{b \sqrt {a b}}\)	\(32\)
default	\(\frac {2 \sqrt {x}}{b}-\frac {2 a \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{b \sqrt {a b}}\)	\(32\)
risch	\(\frac {2 \sqrt {x}}{b}-\frac {2 a \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{b \sqrt {a b}}\)	\(32\)

[In]

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

[Out]

2*x^(1/2)/b-2*a/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 = 85, normalized size of antiderivative = 2.12 \[ \int \frac {\sqrt {x}}{a+b x} \, dx=\left [\frac {\sqrt {-\frac {a}{b}} \log \left (\frac {b x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) + 2 \, \sqrt {x}}{b}, -\frac {2 \, {\left (\sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) - \sqrt {x}\right )}}{b}\right ] \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 88 vs. \(2 (36) = 72\).

Time = 0.32 (sec) , antiderivative size = 88, normalized size of antiderivative = 2.20 \[ \int \frac {\sqrt {x}}{a+b x} \, dx=\begin {cases} \tilde {\infty } \sqrt {x} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 x^{\frac {3}{2}}}{3 a} & \text {for}\: b = 0 \\\frac {2 \sqrt {x}}{b} & \text {for}\: a = 0 \\- \frac {a \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{b^{2} \sqrt {- \frac {a}{b}}} + \frac {a \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{b^{2} \sqrt {- \frac {a}{b}}} + \frac {2 \sqrt {x}}{b} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt {x}}{a+b x} \, dx=-\frac {2 \, a \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b} + \frac {2 \, \sqrt {x}}{b} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt {x}}{a+b x} \, dx=-\frac {2 \, a \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b} + \frac {2 \, \sqrt {x}}{b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.70 \[ \int \frac {\sqrt {x}}{a+b x} \, dx=\frac {2\,\sqrt {x}}{b}-\frac {2\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{b^{3/2}} \]

[In]

int(x^(1/2)/(a + b*x),x)

[Out]

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