3.2991 \(\int \frac{1}{\sqrt{a+b \sqrt{\frac{c}{x}}} x^2} \, dx\)

Optimal. Leaf size=54 \[ \frac{4 a \sqrt{a+b \sqrt{\frac{c}{x}}}}{b^2 c}-\frac{4 \left (a+b \sqrt{\frac{c}{x}}\right )^{3/2}}{3 b^2 c} \]

[Out]

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

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Rubi [A]  time = 0.0441508, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {369, 266, 43} \[ \frac{4 a \sqrt{a+b \sqrt{\frac{c}{x}}}}{b^2 c}-\frac{4 \left (a+b \sqrt{\frac{c}{x}}\right )^{3/2}}{3 b^2 c} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a + b*Sqrt[c/x]]*x^2),x]

[Out]

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

Rule 369

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbol] :> With[{k = Denominator[n]}, Su
bst[Int[(d*x)^m*(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, d, m, p, q}, x] && FractionQ[n]

Rule 266

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]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.033207, size = 42, normalized size = 0.78 \[ -\frac{4 \left (b \sqrt{\frac{c}{x}}-2 a\right ) \sqrt{a+b \sqrt{\frac{c}{x}}}}{3 b^2 c} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[a + b*Sqrt[c/x]]*x^2),x]

[Out]

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

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Maple [C]  time = 0.046, size = 274, normalized size = 5.1 \begin{align*} -{\frac{1}{3\,{b}^{3}}\sqrt{a+b\sqrt{{\frac{c}{x}}}} \left ( 6\,\sqrt{x \left ( a+b\sqrt{{\frac{c}{x}}} \right ) }{a}^{5/2}{x}^{3/2}+6\,\sqrt{ax+b\sqrt{{\frac{c}{x}}}x}{a}^{5/2}{x}^{3/2}+3\,\ln \left ( 1/2\,{\frac{1}{\sqrt{a}} \left ( b\sqrt{{\frac{c}{x}}}\sqrt{x}+2\,\sqrt{ax+b\sqrt{{\frac{c}{x}}}x}\sqrt{a}+2\,a\sqrt{x} \right ) } \right ) \sqrt{{\frac{c}{x}}}{x}^{2}{a}^{2}b-3\,\ln \left ( 1/2\,{\frac{1}{\sqrt{a}} \left ( b\sqrt{{\frac{c}{x}}}\sqrt{x}+2\,\sqrt{x \left ( a+b\sqrt{{\frac{c}{x}}} \right ) }\sqrt{a}+2\,a\sqrt{x} \right ) } \right ) \sqrt{{\frac{c}{x}}}{x}^{2}{a}^{2}b+4\, \left ( ax+b\sqrt{{\frac{c}{x}}}x \right ) ^{3/2}\sqrt{{\frac{c}{x}}}\sqrt{a}\sqrt{x}b-12\, \left ( ax+b\sqrt{{\frac{c}{x}}}x \right ) ^{3/2}{a}^{3/2}\sqrt{x} \right ){x}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{x \left ( a+b\sqrt{{\frac{c}{x}}} \right ) }}} \left ({\frac{c}{x}} \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(a+b*(c/x)^(1/2))^(1/2)*(6*(x*(a+b*(c/x)^(1/2)))^(1/2)*a^(5/2)*x^(3/2)+6*(a*x+b*(c/x)^(1/2)*x)^(1/2)*a^(5
/2)*x^(3/2)+3*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))*(c/x)^
(1/2)*x^2*a^2*b-3*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
/x)^(1/2)*x^2*a^2*b+4*(a*x+b*(c/x)^(1/2)*x)^(3/2)*(c/x)^(1/2)*a^(1/2)*x^(1/2)*b-12*(a*x+b*(c/x)^(1/2)*x)^(3/2)
*a^(3/2)*x^(1/2))/x^(5/2)/(x*(a+b*(c/x)^(1/2)))^(1/2)/b^3/(c/x)^(3/2)/a^(1/2)

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Maxima [A]  time = 0.925253, size = 57, normalized size = 1.06 \begin{align*} -\frac{4 \,{\left (\frac{{\left (b \sqrt{\frac{c}{x}} + a\right )}^{\frac{3}{2}}}{b^{2}} - \frac{3 \, \sqrt{b \sqrt{\frac{c}{x}} + a} a}{b^{2}}\right )}}{3 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.4377, size = 76, normalized size = 1.41 \begin{align*} -\frac{4 \, \sqrt{b \sqrt{\frac{c}{x}} + a}{\left (b \sqrt{\frac{c}{x}} - 2 \, a\right )}}{3 \, b^{2} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{2} \sqrt{a + b \sqrt{\frac{c}{x}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.15511, size = 51, normalized size = 0.94 \begin{align*} -\frac{4 \,{\left ({\left (b \sqrt{\frac{c}{x}} + a\right )}^{\frac{3}{2}} - 3 \, \sqrt{b \sqrt{\frac{c}{x}} + a} a\right )}}{3 \, b^{2} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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