\(\int \frac {1}{\sqrt {a+b \sqrt {\frac {c}{x}}}} \, dx\) [2989]

Optimal result

Integrand size = 17, antiderivative size = 95 \[ \int \frac {1}{\sqrt {a+b \sqrt {\frac {c}{x}}}} \, dx=-\frac {3 b c \sqrt {a+b \sqrt {\frac {c}{x}}}}{2 a^2 \sqrt {\frac {c}{x}}}+\frac {\sqrt {a+b \sqrt {\frac {c}{x}}} x}{a}+\frac {3 b^2 c \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {\frac {c}{x}}}}{\sqrt {a}}\right )}{2 a^{5/2}} \]

[Out]

3/2*b^2*c*arctanh((a+b*(c/x)^(1/2))^(1/2)/a^(1/2))/a^(5/2)+x*(a+b*(c/x)^(1/2))^(1/2)/a-3/2*b*c*(a+b*(c/x)^(1/2
))^(1/2)/a^2/(c/x)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {261, 196, 44, 65, 214} \[ \int \frac {1}{\sqrt {a+b \sqrt {\frac {c}{x}}}} \, dx=\frac {3 b^2 c \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {\frac {c}{x}}}}{\sqrt {a}}\right )}{2 a^{5/2}}-\frac {3 b c \sqrt {a+b \sqrt {\frac {c}{x}}}}{2 a^2 \sqrt {\frac {c}{x}}}+\frac {x \sqrt {a+b \sqrt {\frac {c}{x}}}}{a} \]

[In]

Int[1/Sqrt[a + b*Sqrt[c/x]],x]

[Out]

(-3*b*c*Sqrt[a + b*Sqrt[c/x]])/(2*a^2*Sqrt[c/x]) + (Sqrt[a + b*Sqrt[c/x]]*x)/a + (3*b^2*c*ArcTanh[Sqrt[a + b*S
qrt[c/x]]/Sqrt[a]])/(2*a^(5/2))

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 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]

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 261

Int[((a_) + (b_.)*((c_.)*(x_)^(q_.))^(n_))^(p_.), x_Symbol] :> With[{k = Denominator[n]}, Subst[Int[(a + b*c^n
*x^(n*q))^p, x], x^(1/k), (c*x^q)^(1/k)/(c^(1/k)*(x^(1/k))^(q - 1))]] /; FreeQ[{a, b, c, p, q}, x] && Fraction
Q[n]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b \sqrt {c}}{\sqrt {x}}}} \, dx,\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right ) \\ & = -\text {Subst}\left (2 \text {Subst}\left (\int \frac {1}{x^3 \sqrt {a+b \sqrt {c} x}} \, dx,x,\frac {1}{\sqrt {x}}\right ),\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right ) \\ & = \frac {\sqrt {a+b \sqrt {\frac {c}{x}}} x}{a}+\text {Subst}\left (\frac {\left (3 b \sqrt {c}\right ) \text {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b \sqrt {c} x}} \, dx,x,\frac {1}{\sqrt {x}}\right )}{2 a},\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right ) \\ & = -\frac {3 b c \sqrt {a+b \sqrt {\frac {c}{x}}}}{2 a^2 \sqrt {\frac {c}{x}}}+\frac {\sqrt {a+b \sqrt {\frac {c}{x}}} x}{a}-\text {Subst}\left (\frac {\left (3 b^2 c\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b \sqrt {c} x}} \, dx,x,\frac {1}{\sqrt {x}}\right )}{4 a^2},\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right ) \\ & = -\frac {3 b c \sqrt {a+b \sqrt {\frac {c}{x}}}}{2 a^2 \sqrt {\frac {c}{x}}}+\frac {\sqrt {a+b \sqrt {\frac {c}{x}}} x}{a}-\text {Subst}\left (\frac {\left (3 b \sqrt {c}\right ) \text {Subst}\left (\int \frac {1}{-\frac {a}{b \sqrt {c}}+\frac {x^2}{b \sqrt {c}}} \, dx,x,\sqrt {a+\frac {b \sqrt {c}}{\sqrt {x}}}\right )}{2 a^2},\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right ) \\ & = -\frac {3 b c \sqrt {a+b \sqrt {\frac {c}{x}}}}{2 a^2 \sqrt {\frac {c}{x}}}+\frac {\sqrt {a+b \sqrt {\frac {c}{x}}} x}{a}+\frac {3 b^2 c \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {\frac {c}{x}}}}{\sqrt {a}}\right )}{2 a^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.83 \[ \int \frac {1}{\sqrt {a+b \sqrt {\frac {c}{x}}}} \, dx=\frac {\left (2 a-3 b \sqrt {\frac {c}{x}}\right ) \sqrt {a+b \sqrt {\frac {c}{x}}} x}{2 a^2}+\frac {3 b^2 c \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {\frac {c}{x}}}}{\sqrt {a}}\right )}{2 a^{5/2}} \]

[In]

Integrate[1/Sqrt[a + b*Sqrt[c/x]],x]

[Out]

((2*a - 3*b*Sqrt[c/x])*Sqrt[a + b*Sqrt[c/x]]*x)/(2*a^2) + (3*b^2*c*ArcTanh[Sqrt[a + b*Sqrt[c/x]]/Sqrt[a]])/(2*
a^(5/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(229\) vs. \(2(73)=146\).

Time = 4.28 (sec) , antiderivative size = 230, normalized size of antiderivative = 2.42

method	result	size
default	\(\frac {\sqrt {a +b \sqrt {\frac {c}{x}}}\, \sqrt {x}\, \left (4 a \ln \left (\frac {b \sqrt {\frac {c}{x}}\, \sqrt {x}+2 \sqrt {x \left (a +b \sqrt {\frac {c}{x}}\right )}\, \sqrt {a}+2 a \sqrt {x}}{2 \sqrt {a}}\right ) c \,b^{2}+2 a^{\frac {3}{2}} \sqrt {a x +b \sqrt {\frac {c}{x}}\, x}\, \sqrt {\frac {c}{x}}\, \sqrt {x}\, b -8 a^{\frac {3}{2}} \sqrt {x \left (a +b \sqrt {\frac {c}{x}}\right )}\, \sqrt {\frac {c}{x}}\, \sqrt {x}\, b -b^{2} c \ln \left (\frac {b \sqrt {\frac {c}{x}}\, \sqrt {x}+2 \sqrt {a x +b \sqrt {\frac {c}{x}}\, x}\, \sqrt {a}+2 a \sqrt {x}}{2 \sqrt {a}}\right ) a +4 a^{\frac {5}{2}} \sqrt {a x +b \sqrt {\frac {c}{x}}\, x}\, \sqrt {x}\right )}{4 \sqrt {x \left (a +b \sqrt {\frac {c}{x}}\right )}\, a^{\frac {7}{2}}}\)	\(230\)

[In]

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

[Out]

1/4*(a+b*(c/x)^(1/2))^(1/2)*x^(1/2)*(4*a*ln(1/2*(b*(c/x)^(1/2)*x^(1/2)+2*(x*(a+b*(c/x)^(1/2)))^(1/2)*a^(1/2)+2
*a*x^(1/2))/a^(1/2))*c*b^2+2*a^(3/2)*(a*x+b*(c/x)^(1/2)*x)^(1/2)*(c/x)^(1/2)*x^(1/2)*b-8*a^(3/2)*(x*(a+b*(c/x)
^(1/2)))^(1/2)*(c/x)^(1/2)*x^(1/2)*b-b^2*c*ln(1/2*(b*(c/x)^(1/2)*x^(1/2)+2*(a*x+b*(c/x)^(1/2)*x)^(1/2)*a^(1/2)
+2*a*x^(1/2))/a^(1/2))*a+4*a^(5/2)*(a*x+b*(c/x)^(1/2)*x)^(1/2)*x^(1/2))/(x*(a+b*(c/x)^(1/2)))^(1/2)/a^(7/2)

Fricas [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.74 \[ \int \frac {1}{\sqrt {a+b \sqrt {\frac {c}{x}}}} \, dx=\left [\frac {3 \, \sqrt {a} b^{2} c \log \left (2 \, \sqrt {b \sqrt {\frac {c}{x}} + a} \sqrt {a} x \sqrt {\frac {c}{x}} + 2 \, a x \sqrt {\frac {c}{x}} + b c\right ) - 2 \, {\left (3 \, a b x \sqrt {\frac {c}{x}} - 2 \, a^{2} x\right )} \sqrt {b \sqrt {\frac {c}{x}} + a}}{4 \, a^{3}}, -\frac {3 \, \sqrt {-a} b^{2} c \arctan \left (\frac {\sqrt {b \sqrt {\frac {c}{x}} + a} \sqrt {-a}}{a}\right ) + {\left (3 \, a b x \sqrt {\frac {c}{x}} - 2 \, a^{2} x\right )} \sqrt {b \sqrt {\frac {c}{x}} + a}}{2 \, a^{3}}\right ] \]

[In]

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

[Out]

[1/4*(3*sqrt(a)*b^2*c*log(2*sqrt(b*sqrt(c/x) + a)*sqrt(a)*x*sqrt(c/x) + 2*a*x*sqrt(c/x) + b*c) - 2*(3*a*b*x*sq
rt(c/x) - 2*a^2*x)*sqrt(b*sqrt(c/x) + a))/a^3, -1/2*(3*sqrt(-a)*b^2*c*arctan(sqrt(b*sqrt(c/x) + a)*sqrt(-a)/a)
+ (3*a*b*x*sqrt(c/x) - 2*a^2*x)*sqrt(b*sqrt(c/x) + a))/a^3]

Sympy [F]

\[ \int \frac {1}{\sqrt {a+b \sqrt {\frac {c}{x}}}} \, dx=\int \frac {1}{\sqrt {a + b \sqrt {\frac {c}{x}}}}\, dx \]

[In]

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

[Out]

Integral(1/sqrt(a + b*sqrt(c/x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.38 \[ \int \frac {1}{\sqrt {a+b \sqrt {\frac {c}{x}}}} \, dx=-\frac {1}{4} \, c {\left (\frac {3 \, b^{2} \log \left (\frac {\sqrt {b \sqrt {\frac {c}{x}} + a} - \sqrt {a}}{\sqrt {b \sqrt {\frac {c}{x}} + a} + \sqrt {a}}\right )}{a^{\frac {5}{2}}} + \frac {2 \, {\left (3 \, {\left (b \sqrt {\frac {c}{x}} + a\right )}^{\frac {3}{2}} b^{2} - 5 \, \sqrt {b \sqrt {\frac {c}{x}} + a} a b^{2}\right )}}{{\left (b \sqrt {\frac {c}{x}} + a\right )}^{2} a^{2} - 2 \, {\left (b \sqrt {\frac {c}{x}} + a\right )} a^{3} + a^{4}}\right )} \]

[In]

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

[Out]

-1/4*c*(3*b^2*log((sqrt(b*sqrt(c/x) + a) - sqrt(a))/(sqrt(b*sqrt(c/x) + a) + sqrt(a)))/a^(5/2) + 2*(3*(b*sqrt(
c/x) + a)^(3/2)*b^2 - 5*sqrt(b*sqrt(c/x) + a)*a*b^2)/((b*sqrt(c/x) + a)^2*a^2 - 2*(b*sqrt(c/x) + a)*a^3 + a^4)
)

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.46 \[ \int \frac {1}{\sqrt {a+b \sqrt {\frac {c}{x}}}} \, dx=\frac {\frac {3 \, b^{2} c^{2} \log \left (c^{2} {\left | b \right |}\right )}{\sqrt {a c} a^{2}} - \frac {3 \, b^{2} c^{2} \log \left ({\left | -b c^{2} - 2 \, \sqrt {a c} {\left (\sqrt {a c} \sqrt {c x} - \sqrt {a c^{2} x + \sqrt {c x} b c^{2}}\right )} \right |}\right )}{\sqrt {a c} a^{2}} - 2 \, \sqrt {a c^{2} x + \sqrt {c x} b c^{2}} {\left (\frac {3 \, b}{a^{2}} - \frac {2 \, \sqrt {c x}}{a c}\right )}}{4 \, \sqrt {c} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {a+b \sqrt {\frac {c}{x}}}} \, dx=\int \frac {1}{\sqrt {a+b\,\sqrt {\frac {c}{x}}}} \,d x \]

[In]

int(1/(a + b*(c/x)^(1/2))^(1/2),x)

[Out]

int(1/(a + b*(c/x)^(1/2))^(1/2), x)