\(\int \frac {1}{a+b \sqrt {x}} \, dx\) [2194]

Optimal result

Integrand size = 11, antiderivative size = 27 \[ \int \frac {1}{a+b \sqrt {x}} \, dx=\frac {2 \sqrt {x}}{b}-\frac {2 a \log \left (a+b \sqrt {x}\right )}{b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {196, 45} \[ \int \frac {1}{a+b \sqrt {x}} \, dx=\frac {2 \sqrt {x}}{b}-\frac {2 a \log \left (a+b \sqrt {x}\right )}{b^2} \]

[In]

Int[(a + b*Sqrt[x])^(-1),x]

[Out]

(2*Sqrt[x])/b - (2*a*Log[a + b*Sqrt[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])

Rule 196

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(1/n - 1)*(a + b*x)^p, x], x, x^n], x] /
; FreeQ[{a, b, p}, x] && FractionQ[n] && IntegerQ[1/n]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {1}{a+b \sqrt {x}} \, dx=\frac {2 \sqrt {x}}{b}-\frac {2 a \log \left (a+b \sqrt {x}\right )}{b^2} \]

[In]

Integrate[(a + b*Sqrt[x])^(-1),x]

[Out]

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

Maple [A] (verified)

Time = 5.83 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89

method	result	size
derivativedivides	\(-\frac {2 a \ln \left (a +b \sqrt {x}\right )}{b^{2}}+\frac {2 \sqrt {x}}{b}\)	\(24\)
default	\(\frac {2 \sqrt {x}}{b}+\frac {a \ln \left (b \sqrt {x}-a \right )}{b^{2}}-\frac {a \ln \left (a +b \sqrt {x}\right )}{b^{2}}-\frac {a \ln \left (b^{2} x -a^{2}\right )}{b^{2}}\)	\(57\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {1}{a+b \sqrt {x}} \, dx=-\frac {2 \, {\left (a \log \left (b \sqrt {x} + a\right ) - b \sqrt {x}\right )}}{b^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {1}{a+b \sqrt {x}} \, dx=\begin {cases} - \frac {2 a \log {\left (\frac {a}{b} + \sqrt {x} \right )}}{b^{2}} + \frac {2 \sqrt {x}}{b} & \text {for}\: b \neq 0 \\\frac {x}{a} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-2*a*log(a/b + sqrt(x))/b**2 + 2*sqrt(x)/b, Ne(b, 0)), (x/a, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {1}{a+b \sqrt {x}} \, dx=-\frac {2 \, a \log \left (b \sqrt {x} + a\right )}{b^{2}} + \frac {2 \, {\left (b \sqrt {x} + a\right )}}{b^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {1}{a+b \sqrt {x}} \, dx=-\frac {2 \, a \log \left ({\left | b \sqrt {x} + a \right |}\right )}{b^{2}} + \frac {2 \, \sqrt {x}}{b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {1}{a+b \sqrt {x}} \, dx=\frac {2\,\sqrt {x}}{b}-\frac {2\,a\,\ln \left (a+b\,\sqrt {x}\right )}{b^2} \]

[In]

int(1/(a + b*x^(1/2)),x)

[Out]

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