\(\int \frac {1}{\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}} \, dx\) [3063]

Optimal result

Integrand size = 22, antiderivative size = 135 \[ \int \frac {1}{\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}} \, dx=-\frac {3 b d \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}{2 a^2 \sqrt {\frac {d}{x}}}+\frac {\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x}{a}-\frac {\left (4 a c-3 b^2 d\right ) \text {arctanh}\left (\frac {2 a+b \sqrt {\frac {d}{x}}}{2 \sqrt {a} \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}\right )}{4 a^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1993, 1371, 758, 820, 738, 212} \[ \int \frac {1}{\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}} \, dx=-\frac {\left (4 a c-3 b^2 d\right ) \text {arctanh}\left (\frac {2 a+b \sqrt {\frac {d}{x}}}{2 \sqrt {a} \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}\right )}{4 a^{5/2}}-\frac {3 b d \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}{2 a^2 \sqrt {\frac {d}{x}}}+\frac {x \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}{a} \]

[In]

Int[1/Sqrt[a + b*Sqrt[d/x] + c/x],x]

[Out]

(-3*b*d*Sqrt[a + b*Sqrt[d/x] + c/x])/(2*a^2*Sqrt[d/x]) + (Sqrt[a + b*Sqrt[d/x] + c/x]*x)/a - ((4*a*c - 3*b^2*d
)*ArcTanh[(2*a + b*Sqrt[d/x])/(2*Sqrt[a]*Sqrt[a + b*Sqrt[d/x] + c/x])])/(4*a^(5/2))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 738

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 758

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m + 1)*
((a + b*x + c*x^2)^(p + 1)/((m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Dist[1/((m + 1)*(c*d^2 - b*d*e + a*e^2)),
Int[(d + e*x)^(m + 1)*Simp[c*d*(m + 1) - b*e*(m + p + 2) - c*e*(m + 2*p + 3)*x, x]*(a + b*x + c*x^2)^p, x], x]
/; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e
, 0] && NeQ[m, -1] && ((LtQ[m, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]) || (SumSimplerQ[m, 1] && IntegerQ
[p]) || ILtQ[Simplify[m + 2*p + 3], 0])

Rule 820

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2))), x] - Dist[
(b*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x]
, x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[S
implify[m + 2*p + 3], 0]

Rule 1371

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && NeQ[
b^2 - 4*a*c, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1993

Int[((a_.) + (b_.)*((d_.)/(x_))^(n_) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[-d, Subst[Int[(a + b*x^n + (
c/d^(2*n))*x^(2*n))^p/x^2, x], x, d/x], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[n2, -2*n] && IntegerQ[2*n]

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

Mathematica [A] (verified)

Time = 0.79 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.29 \[ \int \frac {1}{\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}} \, dx=\frac {\sqrt {a} d \left (2 a-3 b \sqrt {\frac {d}{x}}\right ) \left (c+\left (a+b \sqrt {\frac {d}{x}}\right ) x\right )+\sqrt {d} \left (4 a c-3 b^2 d\right ) \sqrt {\frac {d \left (c+\left (a+b \sqrt {\frac {d}{x}}\right ) x\right )}{x}} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {\frac {d}{x}}-\sqrt {\frac {d \left (c+a x+b \sqrt {\frac {d}{x}} x\right )}{x}}}{\sqrt {a} \sqrt {d}}\right )}{2 a^{5/2} d \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}} \]

[In]

Integrate[1/Sqrt[a + b*Sqrt[d/x] + c/x],x]

[Out]

(Sqrt[a]*d*(2*a - 3*b*Sqrt[d/x])*(c + (a + b*Sqrt[d/x])*x) + Sqrt[d]*(4*a*c - 3*b^2*d)*Sqrt[(d*(c + (a + b*Sqr
t[d/x])*x))/x]*ArcTanh[(Sqrt[c]*Sqrt[d/x] - Sqrt[(d*(c + a*x + b*Sqrt[d/x]*x))/x])/(Sqrt[a]*Sqrt[d])])/(2*a^(5
/2)*d*Sqrt[a + b*Sqrt[d/x] + c/x])

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.58

method	result	size
default	\(\frac {\sqrt {\frac {b \sqrt {\frac {d}{x}}\, x +a x +c}{x}}\, \sqrt {x}\, \left (4 a^{\frac {5}{2}} \sqrt {b \sqrt {\frac {d}{x}}\, x +a x +c}\, \sqrt {x}-6 a^{\frac {3}{2}} \sqrt {b \sqrt {\frac {d}{x}}\, x +a x +c}\, \sqrt {\frac {d}{x}}\, \sqrt {x}\, b +3 \ln \left (\frac {\sqrt {\frac {d}{x}}\, \sqrt {x}\, b +2 \sqrt {b \sqrt {\frac {d}{x}}\, x +a x +c}\, \sqrt {a}+2 a \sqrt {x}}{2 \sqrt {a}}\right ) d a \,b^{2}-4 \ln \left (\frac {\sqrt {\frac {d}{x}}\, \sqrt {x}\, b +2 \sqrt {b \sqrt {\frac {d}{x}}\, x +a x +c}\, \sqrt {a}+2 a \sqrt {x}}{2 \sqrt {a}}\right ) a^{2} c \right )}{4 \sqrt {b \sqrt {\frac {d}{x}}\, x +a x +c}\, a^{\frac {7}{2}}}\)	\(213\)

[In]

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

[Out]

1/4*((b*(d/x)^(1/2)*x+a*x+c)/x)^(1/2)*x^(1/2)*(4*a^(5/2)*(b*(d/x)^(1/2)*x+a*x+c)^(1/2)*x^(1/2)-6*a^(3/2)*(b*(d
/x)^(1/2)*x+a*x+c)^(1/2)*(d/x)^(1/2)*x^(1/2)*b+3*ln(1/2*((d/x)^(1/2)*x^(1/2)*b+2*(b*(d/x)^(1/2)*x+a*x+c)^(1/2)
*a^(1/2)+2*a*x^(1/2))/a^(1/2))*d*a*b^2-4*ln(1/2*((d/x)^(1/2)*x^(1/2)*b+2*(b*(d/x)^(1/2)*x+a*x+c)^(1/2)*a^(1/2)
+2*a*x^(1/2))/a^(1/2))*a^2*c)/(b*(d/x)^(1/2)*x+a*x+c)^(1/2)/a^(7/2)

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F]

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

[In]

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

[Out]

Integral(1/sqrt(a + b*sqrt(d/x) + c/x), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.61 \[ \int \frac {1}{\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}} \, dx=-\frac {2 \, \sqrt {a d^{2} x + \sqrt {d x} b d^{2} + c d^{2}} {\left (\frac {3 \, b}{a^{2}} - \frac {2 \, \sqrt {d x}}{a d}\right )} + \frac {{\left (3 \, b^{2} d^{2} - 4 \, a c d\right )} \log \left ({\left | -b d^{2} - 2 \, \sqrt {a d} {\left (\sqrt {a d} \sqrt {d x} - \sqrt {a d^{2} x + \sqrt {d x} b d^{2} + c d^{2}}\right )} \right |}\right )}{\sqrt {a d} a^{2}} - \frac {3 \, b^{2} d^{2} \log \left ({\left | -b d^{2} + 2 \, \sqrt {c d^{2}} \sqrt {a d} \right |}\right ) - 4 \, a c d \log \left ({\left | -b d^{2} + 2 \, \sqrt {c d^{2}} \sqrt {a d} \right |}\right ) + 6 \, \sqrt {c d^{2}} \sqrt {a d} b}{\sqrt {a d} a^{2}}}{4 \, \sqrt {d} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

-1/4*(2*sqrt(a*d^2*x + sqrt(d*x)*b*d^2 + c*d^2)*(3*b/a^2 - 2*sqrt(d*x)/(a*d)) + (3*b^2*d^2 - 4*a*c*d)*log(abs(
-b*d^2 - 2*sqrt(a*d)*(sqrt(a*d)*sqrt(d*x) - sqrt(a*d^2*x + sqrt(d*x)*b*d^2 + c*d^2))))/(sqrt(a*d)*a^2) - (3*b^
2*d^2*log(abs(-b*d^2 + 2*sqrt(c*d^2)*sqrt(a*d))) - 4*a*c*d*log(abs(-b*d^2 + 2*sqrt(c*d^2)*sqrt(a*d))) + 6*sqrt
(c*d^2)*sqrt(a*d)*b)/(sqrt(a*d)*a^2))/(sqrt(d)*sgn(x))

Mupad [F(-1)]

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

[In]

int(1/(a + c/x + b*(d/x)^(1/2))^(1/2),x)

[Out]

int(1/(a + c/x + b*(d/x)^(1/2))^(1/2), x)