Integrand size = 17, antiderivative size = 92 \[ \int \sqrt {a+b \sqrt {\frac {c}{x}}} \, dx=\frac {b c \sqrt {a+b \sqrt {\frac {c}{x}}}}{2 a \sqrt {\frac {c}{x}}}+\sqrt {a+b \sqrt {\frac {c}{x}}} x-\frac {b^2 c \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {\frac {c}{x}}}}{\sqrt {a}}\right )}{2 a^{3/2}} \]
-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)
Time = 0.14 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.85 \[ \int \sqrt {a+b \sqrt {\frac {c}{x}}} \, dx=\frac {\sqrt {a+b \sqrt {\frac {c}{x}}} \left (2 a+b \sqrt {\frac {c}{x}}\right ) x}{2 a}-\frac {b^2 c \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {\frac {c}{x}}}}{\sqrt {a}}\right )}{2 a^{3/2}} \]
(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))
Time = 0.24 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.26, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {787, 774, 798, 51, 52, 73, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sqrt {a+b \sqrt {\frac {c}{x}}} \, dx\) |
\(\Big \downarrow \) 787 |
\(\displaystyle \int \sqrt {a+\frac {b \sqrt {c}}{\sqrt {x}}}dx\) |
\(\Big \downarrow \) 774 |
\(\displaystyle 2 \int \frac {\sqrt {a+\frac {b c}{\sqrt {\frac {c}{x}} x}} \sqrt {\frac {c}{x}} x}{\sqrt {c}}d\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle -2 \int \frac {c^{3/2} \sqrt {a+b \sqrt {\frac {c}{x}} x}}{\left (\frac {c}{x}\right )^{3/2} x^3}d\frac {\sqrt {c}}{\sqrt {\frac {c}{x}} x}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle -2 \left (\frac {1}{4} b \sqrt {c} \int \frac {1}{x \sqrt {a+b \sqrt {\frac {c}{x}} x}}d\frac {\sqrt {c}}{\sqrt {\frac {c}{x}} x}-\frac {\sqrt {a+b x \sqrt {\frac {c}{x}}}}{2 x}\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle -2 \left (\frac {1}{4} b \sqrt {c} \left (-\frac {b \sqrt {c} \int \frac {\sqrt {c}}{\sqrt {\frac {c}{x}} x \sqrt {a+b \sqrt {\frac {c}{x}} x}}d\frac {\sqrt {c}}{\sqrt {\frac {c}{x}} x}}{2 a}-\frac {\sqrt {c} \sqrt {a+b x \sqrt {\frac {c}{x}}}}{a x \sqrt {\frac {c}{x}}}\right )-\frac {\sqrt {a+b x \sqrt {\frac {c}{x}}}}{2 x}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -2 \left (\frac {1}{4} b \sqrt {c} \left (-\frac {\int \frac {1}{\frac {x}{b \sqrt {c}}-\frac {a}{b \sqrt {c}}}d\sqrt {a+b \sqrt {\frac {c}{x}} x}}{a}-\frac {\sqrt {c} \sqrt {a+b x \sqrt {\frac {c}{x}}}}{a x \sqrt {\frac {c}{x}}}\right )-\frac {\sqrt {a+b x \sqrt {\frac {c}{x}}}}{2 x}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -2 \left (\frac {1}{4} b \sqrt {c} \left (\frac {b \sqrt {c} \text {arctanh}\left (\frac {\sqrt {a+b x \sqrt {\frac {c}{x}}}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\sqrt {c} \sqrt {a+b x \sqrt {\frac {c}{x}}}}{a x \sqrt {\frac {c}{x}}}\right )-\frac {\sqrt {a+b x \sqrt {\frac {c}{x}}}}{2 x}\right )\) |
-2*(-1/2*Sqrt[a + b*Sqrt[c/x]*x]/x + (b*Sqrt[c]*(-((Sqrt[c]*Sqrt[a + b*Sqr t[c/x]*x])/(a*Sqrt[c/x]*x)) + (b*Sqrt[c]*ArcTanh[Sqrt[a + b*Sqrt[c/x]*x]/S qrt[a]])/a^(3/2)))/4)
3.30.83.3.1 Defintions of rubi rules used
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] - Simp[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 ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
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] - Simp[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] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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]
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Simp[k Subst[Int[x^(k - 1)*(a + b*x^(k*n))^p, x], x, x^(1/k)], x]] /; Fre eQ[{a, b, p}, x] && FractionQ[n]
Int[((a_) + (b_.)*((c_.)*(x_)^(q_.))^(n_))^(p_), x_Symbol] :> With[{k = Den ominator[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] && FractionQ[n]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[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]]
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 | \(\frac {\sqrt {a +b \sqrt {\frac {c}{x}}}\, \sqrt {x}\, \left (2 a^{\frac {3}{2}} \sqrt {a x +b \sqrt {\frac {c}{x}}\, x}\, \sqrt {\frac {c}{x}}\, \sqrt {x}\, b -b^{2} c \ln \left (\frac {b \sqrt {\frac {c}{x}}\, \sqrt {x}+2 \sqrt {a x +b \sqrt {\frac {c}{x}}\, x}\, \sqrt {a}+2 a \sqrt {x}}{2 \sqrt {a}}\right ) a +4 a^{\frac {5}{2}} \sqrt {a x +b \sqrt {\frac {c}{x}}\, x}\, \sqrt {x}\right )}{4 \sqrt {x \left (a +b \sqrt {\frac {c}{x}}\right )}\, a^{\frac {5}{2}}}\) | \(147\) |
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)
Time = 0.32 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.75 \[ \int \sqrt {a+b \sqrt {\frac {c}{x}}} \, dx=\left [\frac {\sqrt {a} b^{2} c \log \left (-2 \, \sqrt {b \sqrt {\frac {c}{x}} + a} \sqrt {a} x \sqrt {\frac {c}{x}} + 2 \, a x \sqrt {\frac {c}{x}} + b c\right ) + 2 \, {\left (a b x \sqrt {\frac {c}{x}} + 2 \, a^{2} x\right )} \sqrt {b \sqrt {\frac {c}{x}} + a}}{4 \, a^{2}}, \frac {\sqrt {-a} b^{2} c \arctan \left (\frac {\sqrt {b \sqrt {\frac {c}{x}} + a} \sqrt {-a}}{a}\right ) + {\left (a b x \sqrt {\frac {c}{x}} + 2 \, a^{2} x\right )} \sqrt {b \sqrt {\frac {c}{x}} + a}}{2 \, a^{2}}\right ] \]
[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]
\[ \int \sqrt {a+b \sqrt {\frac {c}{x}}} \, dx=\int \sqrt {a + b \sqrt {\frac {c}{x}}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.37 \[ \int \sqrt {a+b \sqrt {\frac {c}{x}}} \, dx=\frac {1}{4} \, {\left (\frac {b^{2} \log \left (\frac {\sqrt {b \sqrt {\frac {c}{x}} + a} - \sqrt {a}}{\sqrt {b \sqrt {\frac {c}{x}} + a} + \sqrt {a}}\right )}{a^{\frac {3}{2}}} + \frac {2 \, {\left ({\left (b \sqrt {\frac {c}{x}} + a\right )}^{\frac {3}{2}} b^{2} + \sqrt {b \sqrt {\frac {c}{x}} + a} a b^{2}\right )}}{{\left (b \sqrt {\frac {c}{x}} + a\right )}^{2} a - 2 \, {\left (b \sqrt {\frac {c}{x}} + a\right )} a^{2} + a^{3}}\right )} c \]
1/4*(b^2*log((sqrt(b*sqrt(c/x) + a) - sqrt(a))/(sqrt(b*sqrt(c/x) + a) + sq rt(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
Time = 0.34 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.41 \[ \int \sqrt {a+b \sqrt {\frac {c}{x}}} \, dx=-\frac {{\left (\frac {b^{2} c^{3} \log \left (c^{2} {\left | b \right |}\right )}{\sqrt {a c} a} - \frac {b^{2} c^{3} \log \left ({\left | -b c^{2} - 2 \, \sqrt {a c} {\left (\sqrt {a c} \sqrt {c x} - \sqrt {a c^{2} x + \sqrt {c x} b c^{2}}\right )} \right |}\right )}{\sqrt {a c} a} - 2 \, \sqrt {a c^{2} x + \sqrt {c x} b c^{2}} {\left (\frac {b c}{a} + 2 \, \sqrt {c x}\right )}\right )} \mathrm {sgn}\left (x\right )}{4 \, c^{\frac {3}{2}}} \]
-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*s qrt(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)
Timed out. \[ \int \sqrt {a+b \sqrt {\frac {c}{x}}} \, dx=\int \sqrt {a+b\,\sqrt {\frac {c}{x}}} \,d x \]