\(\int \frac {\sqrt {a+b \sqrt {x}}}{x^2} \, dx\) [2237]

Optimal result

Integrand size = 17, antiderivative size = 77 \[ \int \frac {\sqrt {a+b \sqrt {x}}}{x^2} \, dx=-\frac {\sqrt {a+b \sqrt {x}}}{x}-\frac {b \sqrt {a+b \sqrt {x}}}{2 a \sqrt {x}}+\frac {b^2 \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {x}}}{\sqrt {a}}\right )}{2 a^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {272, 43, 44, 65, 214} \[ \int \frac {\sqrt {a+b \sqrt {x}}}{x^2} \, dx=\frac {b^2 \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {x}}}{\sqrt {a}}\right )}{2 a^{3/2}}-\frac {b \sqrt {a+b \sqrt {x}}}{2 a \sqrt {x}}-\frac {\sqrt {a+b \sqrt {x}}}{x} \]

[In]

Int[Sqrt[a + b*Sqrt[x]]/x^2,x]

[Out]

-(Sqrt[a + b*Sqrt[x]]/x) - (b*Sqrt[a + b*Sqrt[x]])/(2*a*Sqrt[x]) + (b^2*ArcTanh[Sqrt[a + b*Sqrt[x]]/Sqrt[a]])/
(2*a^(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 44

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

Rule 272

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

Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int \frac {\sqrt {a+b x}}{x^3} \, dx,x,\sqrt {x}\right ) \\ & = -\frac {\sqrt {a+b \sqrt {x}}}{x}+\frac {1}{2} b \text {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,\sqrt {x}\right ) \\ & = -\frac {\sqrt {a+b \sqrt {x}}}{x}-\frac {b \sqrt {a+b \sqrt {x}}}{2 a \sqrt {x}}-\frac {b^2 \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\sqrt {x}\right )}{4 a} \\ & = -\frac {\sqrt {a+b \sqrt {x}}}{x}-\frac {b \sqrt {a+b \sqrt {x}}}{2 a \sqrt {x}}-\frac {b \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sqrt {x}}\right )}{2 a} \\ & = -\frac {\sqrt {a+b \sqrt {x}}}{x}-\frac {b \sqrt {a+b \sqrt {x}}}{2 a \sqrt {x}}+\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {x}}}{\sqrt {a}}\right )}{2 a^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.88 \[ \int \frac {\sqrt {a+b \sqrt {x}}}{x^2} \, dx=\frac {\left (-2 a-b \sqrt {x}\right ) \sqrt {a+b \sqrt {x}}}{2 a x}+\frac {b^2 \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {x}}}{\sqrt {a}}\right )}{2 a^{3/2}} \]

[In]

Integrate[Sqrt[a + b*Sqrt[x]]/x^2,x]

[Out]

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

Maple [A] (verified)

Time = 12.80 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.78

method	result	size
derivativedivides	\(4 b^{2} \left (-\frac {\frac {\left (a +b \sqrt {x}\right )^{\frac {3}{2}}}{8 a}+\frac {\sqrt {a +b \sqrt {x}}}{8}}{b^{2} x}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {a +b \sqrt {x}}}{\sqrt {a}}\right )}{8 a^{\frac {3}{2}}}\right )\)	\(60\)
default	\(4 b^{2} \left (-\frac {\frac {\left (a +b \sqrt {x}\right )^{\frac {3}{2}}}{8 a}+\frac {\sqrt {a +b \sqrt {x}}}{8}}{b^{2} x}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {a +b \sqrt {x}}}{\sqrt {a}}\right )}{8 a^{\frac {3}{2}}}\right )\)	\(60\)

[In]

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

[Out]

4*b^2*(-(1/8/a*(a+b*x^(1/2))^(3/2)+1/8*(a+b*x^(1/2))^(1/2))/b^2/x+1/8/a^(3/2)*arctanh((a+b*x^(1/2))^(1/2)/a^(1
/2)))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.73 \[ \int \frac {\sqrt {a+b \sqrt {x}}}{x^2} \, dx=\left [\frac {\sqrt {a} b^{2} x \log \left (\frac {b x + 2 \, \sqrt {b \sqrt {x} + a} \sqrt {a} \sqrt {x} + 2 \, a \sqrt {x}}{x}\right ) - 2 \, {\left (a b \sqrt {x} + 2 \, a^{2}\right )} \sqrt {b \sqrt {x} + a}}{4 \, a^{2} x}, -\frac {\sqrt {-a} b^{2} x \arctan \left (\frac {\sqrt {b \sqrt {x} + a} \sqrt {-a}}{a}\right ) + {\left (a b \sqrt {x} + 2 \, a^{2}\right )} \sqrt {b \sqrt {x} + a}}{2 \, a^{2} x}\right ] \]

[In]

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

[Out]

[1/4*(sqrt(a)*b^2*x*log((b*x + 2*sqrt(b*sqrt(x) + a)*sqrt(a)*sqrt(x) + 2*a*sqrt(x))/x) - 2*(a*b*sqrt(x) + 2*a^
2)*sqrt(b*sqrt(x) + a))/(a^2*x), -1/2*(sqrt(-a)*b^2*x*arctan(sqrt(b*sqrt(x) + a)*sqrt(-a)/a) + (a*b*sqrt(x) +
2*a^2)*sqrt(b*sqrt(x) + a))/(a^2*x)]

Sympy [A] (verification not implemented)

Time = 2.07 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.36 \[ \int \frac {\sqrt {a+b \sqrt {x}}}{x^2} \, dx=- \frac {a}{\sqrt {b} x^{\frac {5}{4}} \sqrt {\frac {a}{b \sqrt {x}} + 1}} - \frac {3 \sqrt {b}}{2 x^{\frac {3}{4}} \sqrt {\frac {a}{b \sqrt {x}} + 1}} - \frac {b^{\frac {3}{2}}}{2 a \sqrt [4]{x} \sqrt {\frac {a}{b \sqrt {x}} + 1}} + \frac {b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt [4]{x}} \right )}}{2 a^{\frac {3}{2}}} \]

[In]

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

[Out]

-a/(sqrt(b)*x**(5/4)*sqrt(a/(b*sqrt(x)) + 1)) - 3*sqrt(b)/(2*x**(3/4)*sqrt(a/(b*sqrt(x)) + 1)) - b**(3/2)/(2*a
*x**(1/4)*sqrt(a/(b*sqrt(x)) + 1)) + b**2*asinh(sqrt(a)/(sqrt(b)*x**(1/4)))/(2*a**(3/2))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.30 \[ \int \frac {\sqrt {a+b \sqrt {x}}}{x^2} \, dx=-\frac {b^{2} \log \left (\frac {\sqrt {b \sqrt {x} + a} - \sqrt {a}}{\sqrt {b \sqrt {x} + a} + \sqrt {a}}\right )}{4 \, a^{\frac {3}{2}}} - \frac {{\left (b \sqrt {x} + a\right )}^{\frac {3}{2}} b^{2} + \sqrt {b \sqrt {x} + a} a b^{2}}{2 \, {\left ({\left (b \sqrt {x} + a\right )}^{2} a - 2 \, {\left (b \sqrt {x} + a\right )} a^{2} + a^{3}\right )}} \]

[In]

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

[Out]

-1/4*b^2*log((sqrt(b*sqrt(x) + a) - sqrt(a))/(sqrt(b*sqrt(x) + a) + sqrt(a)))/a^(3/2) - 1/2*((b*sqrt(x) + a)^(
3/2)*b^2 + sqrt(b*sqrt(x) + a)*a*b^2)/((b*sqrt(x) + a)^2*a - 2*(b*sqrt(x) + a)*a^2 + a^3)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {a+b \sqrt {x}}}{x^2} \, dx=-\frac {\frac {b^{3} \arctan \left (\frac {\sqrt {b \sqrt {x} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {{\left (b \sqrt {x} + a\right )}^{\frac {3}{2}} b^{3} + \sqrt {b \sqrt {x} + a} a b^{3}}{a b^{2} x}}{2 \, b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.22 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.70 \[ \int \frac {\sqrt {a+b \sqrt {x}}}{x^2} \, dx=\frac {b^2\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,\sqrt {x}}}{\sqrt {a}}\right )}{2\,a^{3/2}}-\frac {\sqrt {a+b\,\sqrt {x}}}{2\,x}-\frac {{\left (a+b\,\sqrt {x}\right )}^{3/2}}{2\,a\,x} \]

[In]

int((a + b*x^(1/2))^(1/2)/x^2,x)

[Out]

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