\(\int \frac {\sqrt {a+b x}}{x} \, dx\) [288]

Optimal result

Integrand size = 13, antiderivative size = 35 \[ \int \frac {\sqrt {a+b x}}{x} \, dx=2 \sqrt {a+b x}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, 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, 214} \[ \int \frac {\sqrt {a+b x}}{x} \, dx=2 \sqrt {a+b x}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]

[In]

Int[Sqrt[a + b*x]/x,x]

[Out]

2*Sqrt[a + b*x] - 2*Sqrt[a]*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]

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 214

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

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {a+b x}}{x} \, dx=2 \sqrt {a+b x}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]

[In]

Integrate[Sqrt[a + b*x]/x,x]

[Out]

2*Sqrt[a + b*x] - 2*Sqrt[a]*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.80

method	result	size
derivativedivides	\(-2 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {a}+2 \sqrt {b x +a}\)	\(28\)
default	\(-2 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {a}+2 \sqrt {b x +a}\)	\(28\)
pseudoelliptic	\(-2 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {a}+2 \sqrt {b x +a}\)	\(28\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.09 \[ \int \frac {\sqrt {a+b x}}{x} \, dx=\left [\sqrt {a} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, \sqrt {b x + a}, 2 \, \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + 2 \, \sqrt {b x + a}\right ] \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (31) = 62\).

Time = 0.84 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.94 \[ \int \frac {\sqrt {a+b x}}{x} \, dx=- 2 \sqrt {a} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )} + \frac {2 a}{\sqrt {b} \sqrt {x} \sqrt {\frac {a}{b x} + 1}} + \frac {2 \sqrt {b} \sqrt {x}}{\sqrt {\frac {a}{b x} + 1}} \]

[In]

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

[Out]

-2*sqrt(a)*asinh(sqrt(a)/(sqrt(b)*sqrt(x))) + 2*a/(sqrt(b)*sqrt(x)*sqrt(a/(b*x) + 1)) + 2*sqrt(b)*sqrt(x)/sqrt
(a/(b*x) + 1)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.20 \[ \int \frac {\sqrt {a+b x}}{x} \, dx=\sqrt {a} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right ) + 2 \, \sqrt {b x + a} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.91 \[ \int \frac {\sqrt {a+b x}}{x} \, dx=\frac {2 \, a \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + 2 \, \sqrt {b x + a} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77 \[ \int \frac {\sqrt {a+b x}}{x} \, dx=2\,\sqrt {a+b\,x}-2\,\sqrt {a}\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right ) \]

[In]

int((a + b*x)^(1/2)/x,x)

[Out]

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