\(\int \frac {\sqrt {a+b x}}{x^2} \, dx\) [289]

Optimal result

Integrand size = 13, antiderivative size = 39 \[ \int \frac {\sqrt {a+b x}}{x^2} \, dx=-\frac {\sqrt {a+b x}}{x}-\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{\sqrt {a}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, 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, 214} \[ \int \frac {\sqrt {a+b x}}{x^2} \, dx=-\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {\sqrt {a+b x}}{x} \]

[In]

Int[Sqrt[a + b*x]/x^2,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {a+b x}}{x^2} \, dx=-\frac {\sqrt {a+b x}}{x}-\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{\sqrt {a}} \]

[In]

Integrate[Sqrt[a + b*x]/x^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.82

method	result	size
risch	\(-\frac {b \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {\sqrt {b x +a}}{x}\)	\(32\)
pseudoelliptic	\(-\frac {\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b x +\sqrt {b x +a}\, \sqrt {a}}{x \sqrt {a}}\)	\(36\)
derivativedivides	\(2 b \left (-\frac {\sqrt {b x +a}}{2 b x}-\frac {\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )\)	\(37\)
default	\(2 b \left (-\frac {\sqrt {b x +a}}{2 b x}-\frac {\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )\)	\(37\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 93, normalized size of antiderivative = 2.38 \[ \int \frac {\sqrt {a+b x}}{x^2} \, dx=\left [\frac {\sqrt {a} b x \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, \sqrt {b x + a} a}{2 \, a x}, \frac {\sqrt {-a} b x \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) - \sqrt {b x + a} a}{a x}\right ] \]

[In]

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

[Out]

[1/2*(sqrt(a)*b*x*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2*sqrt(b*x + a)*a)/(a*x), (sqrt(-a)*b*x*arcta
n(sqrt(b*x + a)*sqrt(-a)/a) - sqrt(b*x + a)*a)/(a*x)]

Sympy [A] (verification not implemented)

Time = 1.17 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.13 \[ \int \frac {\sqrt {a+b x}}{x^2} \, dx=- \frac {\sqrt {b} \sqrt {\frac {a}{b x} + 1}}{\sqrt {x}} - \frac {b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{\sqrt {a}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.21 \[ \int \frac {\sqrt {a+b x}}{x^2} \, dx=\frac {b \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{2 \, \sqrt {a}} - \frac {\sqrt {b x + a}}{x} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.05 \[ \int \frac {\sqrt {a+b x}}{x^2} \, dx=\frac {\frac {b^{2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {\sqrt {b x + a} b}{x}}{b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {a+b x}}{x^2} \, dx=-\frac {\sqrt {a+b\,x}}{x}-\frac {b\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )}{\sqrt {a}} \]

[In]

int((a + b*x)^(1/2)/x^2,x)

[Out]

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