3.30.85 \(\int \frac {\sqrt {a+b \sqrt {\frac {c}{x}}}}{x^2} \, dx\) [2985]

Optimal. Leaf size=56 \[ \frac {4 a \left (a+b \sqrt {\frac {c}{x}}\right )^{3/2}}{3 b^2 c}-\frac {4 \left (a+b \sqrt {\frac {c}{x}}\right )^{5/2}}{5 b^2 c} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 56, normalized size of antiderivative = 1.00, 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 = {376, 272, 45} \begin {gather*} \frac {4 a \left (a+b \sqrt {\frac {c}{x}}\right )^{3/2}}{3 b^2 c}-\frac {4 \left (a+b \sqrt {\frac {c}{x}}\right )^{5/2}}{5 b^2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Sqrt[c/x]]/x^2,x]

[Out]

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

Rule 45

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])

Rule 272

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 376

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]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 54, normalized size = 0.96 \begin {gather*} -\frac {4 \sqrt {a+b \sqrt {\frac {c}{x}}} \left (-2 a^2+a b \sqrt {\frac {c}{x}}+\frac {3 b^2 c}{x}\right )}{15 b^2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Sqrt[a + b*Sqrt[c/x]]*(-2*a^2 + a*b*Sqrt[c/x] + (3*b^2*c)/x))/(15*b^2*c)

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Maple [A]
time = 0.47, size = 70, normalized size = 1.25

method	result	size
derivativedivides	\(-\frac {4 \left (\frac {\left (a +b \sqrt {\frac {c}{x}}\right )^{\frac {5}{2}}}{5}-\frac {a \left (a +b \sqrt {\frac {c}{x}}\right )^{\frac {3}{2}}}{3}\right )}{c \,b^{2}}\)	\(41\)
default	\(-\frac {4 \sqrt {a +b \sqrt {\frac {c}{x}}}\, \left (a x +b \sqrt {\frac {c}{x}}\, x \right )^{\frac {3}{2}} \left (3 b \sqrt {\frac {c}{x}}-2 a \right )}{15 c x \sqrt {x \left (a +b \sqrt {\frac {c}{x}}\right )}\, b^{2}}\)	\(70\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/15*(a+b*(c/x)^(1/2))^(1/2)*(a*x+b*(c/x)^(1/2)*x)^(3/2)/c/x*(3*b*(c/x)^(1/2)-2*a)/(x*(a+b*(c/x)^(1/2)))^(1/2
)/b^2

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Maxima [A]
time = 0.28, size = 43, normalized size = 0.77 \begin {gather*} -\frac {4 \, {\left (\frac {3 \, {\left (b \sqrt {\frac {c}{x}} + a\right )}^{\frac {5}{2}}}{b^{2}} - \frac {5 \, {\left (b \sqrt {\frac {c}{x}} + a\right )}^{\frac {3}{2}} a}{b^{2}}\right )}}{15 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/15*(3*(b*sqrt(c/x) + a)^(5/2)/b^2 - 5*(b*sqrt(c/x) + a)^(3/2)*a/b^2)/c

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Fricas [A]
time = 0.41, size = 48, normalized size = 0.86 \begin {gather*} -\frac {4 \, {\left (a b x \sqrt {\frac {c}{x}} + 3 \, b^{2} c - 2 \, a^{2} x\right )} \sqrt {b \sqrt {\frac {c}{x}} + a}}{15 \, b^{2} c x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/15*(a*b*x*sqrt(c/x) + 3*b^2*c - 2*a^2*x)*sqrt(b*sqrt(c/x) + a)/(b^2*c*x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a + b \sqrt {\frac {c}{x}}}}{x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

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Mupad [B]
time = 1.39, size = 52, normalized size = 0.93 \begin {gather*} -\frac {\sqrt {a+b\,\sqrt {\frac {c}{x}}}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{2},2;\ 3;\ -\frac {b\,\sqrt {\frac {c}{x}}}{a}\right )}{x\,\sqrt {\frac {b\,\sqrt {\frac {c}{x}}}{a}+1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*(c/x)^(1/2))^(1/2)/x^2,x)

[Out]

-((a + b*(c/x)^(1/2))^(1/2)*hypergeom([-1/2, 2], 3, -(b*(c/x)^(1/2))/a))/(x*((b*(c/x)^(1/2))/a + 1)^(1/2))

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