\(\int \sqrt {a+b \sqrt {\frac {c}{x}}} \, dx\) [2983]

Optimal result

Integrand size = 17, antiderivative size = 92 \[ \int \sqrt {a+b \sqrt {\frac {c}{x}}} \, dx=\frac {b c \sqrt {a+b \sqrt {\frac {c}{x}}}}{2 a \sqrt {\frac {c}{x}}}+\sqrt {a+b \sqrt {\frac {c}{x}}} x-\frac {b^2 c \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {\frac {c}{x}}}}{\sqrt {a}}\right )}{2 a^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {261, 196, 43, 44, 65, 214} \[ \int \sqrt {a+b \sqrt {\frac {c}{x}}} \, dx=-\frac {b^2 c \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {\frac {c}{x}}}}{\sqrt {a}}\right )}{2 a^{3/2}}+\frac {b c \sqrt {a+b \sqrt {\frac {c}{x}}}}{2 a \sqrt {\frac {c}{x}}}+x \sqrt {a+b \sqrt {\frac {c}{x}}} \]

[In]

Int[Sqrt[a + b*Sqrt[c/x]],x]

[Out]

(b*c*Sqrt[a + b*Sqrt[c/x]])/(2*a*Sqrt[c/x]) + Sqrt[a + b*Sqrt[c/x]]*x - (b^2*c*ArcTanh[Sqrt[a + b*Sqrt[c/x]]/S
qrt[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 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 \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 {\sqrt {a+b \sqrt {c} x}}{x^3} \, dx,x,\frac {1}{\sqrt {x}}\right ),\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right ) \\ & = \sqrt {a+b \sqrt {\frac {c}{x}}} x-\text {Subst}\left (\frac {1}{2} \left (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 ),\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right ) \\ & = \frac {b c \sqrt {a+b \sqrt {\frac {c}{x}}}}{2 a \sqrt {\frac {c}{x}}}+\sqrt {a+b \sqrt {\frac {c}{x}}} x+\text {Subst}\left (\frac {\left (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},\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right ) \\ & = \frac {b c \sqrt {a+b \sqrt {\frac {c}{x}}}}{2 a \sqrt {\frac {c}{x}}}+\sqrt {a+b \sqrt {\frac {c}{x}}} x+\text {Subst}\left (\frac {\left (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},\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right ) \\ & = \frac {b c \sqrt {a+b \sqrt {\frac {c}{x}}}}{2 a \sqrt {\frac {c}{x}}}+\sqrt {a+b \sqrt {\frac {c}{x}}} x-\frac {b^2 c \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {\frac {c}{x}}}}{\sqrt {a}}\right )}{2 a^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.85 \[ \int \sqrt {a+b \sqrt {\frac {c}{x}}} \, dx=\frac {\sqrt {a+b \sqrt {\frac {c}{x}}} \left (2 a+b \sqrt {\frac {c}{x}}\right ) x}{2 a}-\frac {b^2 c \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {\frac {c}{x}}}}{\sqrt {a}}\right )}{2 a^{3/2}} \]

[In]

Integrate[Sqrt[a + b*Sqrt[c/x]],x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(146\) vs. \(2(70)=140\).

Time = 3.96 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.60

method	result	size
default	\(\frac {\sqrt {a +b \sqrt {\frac {c}{x}}}\, \sqrt {x}\, \left (2 a^{\frac {3}{2}} \sqrt {a x +b \sqrt {\frac {c}{x}}\, x}\, \sqrt {\frac {c}{x}}\, \sqrt {x}\, b -b^{2} c \ln \left (\frac {b \sqrt {\frac {c}{x}}\, \sqrt {x}+2 \sqrt {a x +b \sqrt {\frac {c}{x}}\, x}\, \sqrt {a}+2 a \sqrt {x}}{2 \sqrt {a}}\right ) a +4 a^{\frac {5}{2}} \sqrt {a x +b \sqrt {\frac {c}{x}}\, x}\, \sqrt {x}\right )}{4 \sqrt {x \left (a +b \sqrt {\frac {c}{x}}\right )}\, a^{\frac {5}{2}}}\)	\(147\)

[In]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.75 \[ \int \sqrt {a+b \sqrt {\frac {c}{x}}} \, dx=\left [\frac {\sqrt {a} b^{2} c \log \left (-2 \, \sqrt {b \sqrt {\frac {c}{x}} + a} \sqrt {a} x \sqrt {\frac {c}{x}} + 2 \, a x \sqrt {\frac {c}{x}} + b c\right ) + 2 \, {\left (a b x \sqrt {\frac {c}{x}} + 2 \, a^{2} x\right )} \sqrt {b \sqrt {\frac {c}{x}} + a}}{4 \, a^{2}}, \frac {\sqrt {-a} b^{2} c \arctan \left (\frac {\sqrt {b \sqrt {\frac {c}{x}} + a} \sqrt {-a}}{a}\right ) + {\left (a b x \sqrt {\frac {c}{x}} + 2 \, a^{2} x\right )} \sqrt {b \sqrt {\frac {c}{x}} + a}}{2 \, a^{2}}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(sqrt(a + b*sqrt(c/x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.37 \[ \int \sqrt {a+b \sqrt {\frac {c}{x}}} \, dx=\frac {1}{4} \, {\left (\frac {b^{2} \log \left (\frac {\sqrt {b \sqrt {\frac {c}{x}} + a} - \sqrt {a}}{\sqrt {b \sqrt {\frac {c}{x}} + a} + \sqrt {a}}\right )}{a^{\frac {3}{2}}} + \frac {2 \, {\left ({\left (b \sqrt {\frac {c}{x}} + a\right )}^{\frac {3}{2}} b^{2} + \sqrt {b \sqrt {\frac {c}{x}} + a} a b^{2}\right )}}{{\left (b \sqrt {\frac {c}{x}} + a\right )}^{2} a - 2 \, {\left (b \sqrt {\frac {c}{x}} + a\right )} a^{2} + a^{3}}\right )} c \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.41 \[ \int \sqrt {a+b \sqrt {\frac {c}{x}}} \, dx=-\frac {{\left (\frac {b^{2} c^{3} \log \left (c^{2} {\left | b \right |}\right )}{\sqrt {a c} a} - \frac {b^{2} c^{3} \log \left ({\left | -b c^{2} - 2 \, \sqrt {a c} {\left (\sqrt {a c} \sqrt {c x} - \sqrt {a c^{2} x + \sqrt {c x} b c^{2}}\right )} \right |}\right )}{\sqrt {a c} a} - 2 \, \sqrt {a c^{2} x + \sqrt {c x} b c^{2}} {\left (\frac {b c}{a} + 2 \, \sqrt {c x}\right )}\right )} \mathrm {sgn}\left (x\right )}{4 \, c^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((a + b*(c/x)^(1/2))^(1/2),x)

[Out]

int((a + b*(c/x)^(1/2))^(1/2), x)