3.2986 \(\int \frac {\sqrt {a+b \sqrt {\frac {c}{x}}}}{x^3} \, dx\)

Optimal. Leaf size=116 \[ \frac {4 a^3 \left (a+b \sqrt {\frac {c}{x}}\right )^{3/2}}{3 b^4 c^2}-\frac {12 a^2 \left (a+b \sqrt {\frac {c}{x}}\right )^{5/2}}{5 b^4 c^2}-\frac {4 \left (a+b \sqrt {\frac {c}{x}}\right )^{9/2}}{9 b^4 c^2}+\frac {12 a \left (a+b \sqrt {\frac {c}{x}}\right )^{7/2}}{7 b^4 c^2} \]

[Out]

4/3*a^3*(a+b*(c/x)^(1/2))^(3/2)/b^4/c^2-12/5*a^2*(a+b*(c/x)^(1/2))^(5/2)/b^4/c^2+12/7*a*(a+b*(c/x)^(1/2))^(7/2
)/b^4/c^2-4/9*(a+b*(c/x)^(1/2))^(9/2)/b^4/c^2

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Rubi [A]  time = 0.07, antiderivative size = 116, 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 = {369, 266, 43} \[ -\frac {12 a^2 \left (a+b \sqrt {\frac {c}{x}}\right )^{5/2}}{5 b^4 c^2}+\frac {4 a^3 \left (a+b \sqrt {\frac {c}{x}}\right )^{3/2}}{3 b^4 c^2}-\frac {4 \left (a+b \sqrt {\frac {c}{x}}\right )^{9/2}}{9 b^4 c^2}+\frac {12 a \left (a+b \sqrt {\frac {c}{x}}\right )^{7/2}}{7 b^4 c^2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Sqrt[c/x]]/x^3,x]

[Out]

(4*a^3*(a + b*Sqrt[c/x])^(3/2))/(3*b^4*c^2) - (12*a^2*(a + b*Sqrt[c/x])^(5/2))/(5*b^4*c^2) + (12*a*(a + b*Sqrt
[c/x])^(7/2))/(7*b^4*c^2) - (4*(a + b*Sqrt[c/x])^(9/2))/(9*b^4*c^2)

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])

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 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]

Rubi steps

\begin {align*} \int \frac {\sqrt {a+b \sqrt {\frac {c}{x}}}}{x^3} \, dx &=\operatorname {Subst}\left (\int \frac {\sqrt {a+\frac {b \sqrt {c}}{\sqrt {x}}}}{x^3} \, dx,\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right )\\ &=-\operatorname {Subst}\left (2 \operatorname {Subst}\left (\int x^3 \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^3 \sqrt {a+b \sqrt {c} x}}{b^3 c^{3/2}}+\frac {3 a^2 \left (a+b \sqrt {c} x\right )^{3/2}}{b^3 c^{3/2}}-\frac {3 a \left (a+b \sqrt {c} x\right )^{5/2}}{b^3 c^{3/2}}+\frac {\left (a+b \sqrt {c} x\right )^{7/2}}{b^3 c^{3/2}}\right ) \, dx,x,\frac {1}{\sqrt {x}}\right ),\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right )\\ &=\frac {4 a^3 \left (a+b \sqrt {\frac {c}{x}}\right )^{3/2}}{3 b^4 c^2}-\frac {12 a^2 \left (a+b \sqrt {\frac {c}{x}}\right )^{5/2}}{5 b^4 c^2}+\frac {12 a \left (a+b \sqrt {\frac {c}{x}}\right )^{7/2}}{7 b^4 c^2}-\frac {4 \left (a+b \sqrt {\frac {c}{x}}\right )^{9/2}}{9 b^4 c^2}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 75, normalized size = 0.65 \[ \frac {4 \left (a+b \sqrt {\frac {c}{x}}\right )^{3/2} \left (16 a^3 x-24 a^2 b x \sqrt {\frac {c}{x}}+30 a b^2 c-35 b^3 c \sqrt {\frac {c}{x}}\right )}{315 b^4 c^2 x} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*Sqrt[c/x]]/x^3,x]

[Out]

(4*(a + b*Sqrt[c/x])^(3/2)*(30*a*b^2*c - 35*b^3*c*Sqrt[c/x] + 16*a^3*x - 24*a^2*b*Sqrt[c/x]*x))/(315*b^4*c^2*x
)

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fricas [A]  time = 0.91, size = 77, normalized size = 0.66 \[ -\frac {4 \, {\left (35 \, b^{4} c^{2} - 6 \, a^{2} b^{2} c x - 16 \, a^{4} x^{2} + {\left (5 \, a b^{3} c x + 8 \, a^{3} b x^{2}\right )} \sqrt {\frac {c}{x}}\right )} \sqrt {b \sqrt {\frac {c}{x}} + a}}{315 \, b^{4} c^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(c/x)^(1/2))^(1/2)/x^3,x, algorithm="fricas")

[Out]

-4/315*(35*b^4*c^2 - 6*a^2*b^2*c*x - 16*a^4*x^2 + (5*a*b^3*c*x + 8*a^3*b*x^2)*sqrt(c/x))*sqrt(b*sqrt(c/x) + a)
/(b^4*c^2*x^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(c/x)^(1/2))^(1/2)/x^3,x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.02, size = 97, normalized size = 0.84 \[ -\frac {4 \sqrt {a +\sqrt {\frac {c}{x}}\, b}\, \left (a x +\sqrt {\frac {c}{x}}\, b x \right )^{\frac {3}{2}} \left (-16 a^{3} x +24 \sqrt {\frac {c}{x}}\, a^{2} b x -30 a \,b^{2} c +35 \left (\frac {c}{x}\right )^{\frac {3}{2}} b^{3} x \right )}{315 \sqrt {\left (a +\sqrt {\frac {c}{x}}\, b \right ) x}\, b^{4} c^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/315*(a+(c/x)^(1/2)*b)^(1/2)*(a*x+(c/x)^(1/2)*b*x)^(3/2)*(35*x*(c/x)^(3/2)*b^3+24*x*(c/x)^(1/2)*a^2*b-16*a^3
*x-30*a*b^2*c)/c^2/x^2/((a+(c/x)^(1/2)*b)*x)^(1/2)/b^4

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maxima [A]  time = 0.45, size = 85, normalized size = 0.73 \[ -\frac {4 \, {\left (\frac {35 \, {\left (b \sqrt {\frac {c}{x}} + a\right )}^{\frac {9}{2}}}{b^{4}} - \frac {135 \, {\left (b \sqrt {\frac {c}{x}} + a\right )}^{\frac {7}{2}} a}{b^{4}} + \frac {189 \, {\left (b \sqrt {\frac {c}{x}} + a\right )}^{\frac {5}{2}} a^{2}}{b^{4}} - \frac {105 \, {\left (b \sqrt {\frac {c}{x}} + a\right )}^{\frac {3}{2}} a^{3}}{b^{4}}\right )}}{315 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(c/x)^(1/2))^(1/2)/x^3,x, algorithm="maxima")

[Out]

-4/315*(35*(b*sqrt(c/x) + a)^(9/2)/b^4 - 135*(b*sqrt(c/x) + a)^(7/2)*a/b^4 + 189*(b*sqrt(c/x) + a)^(5/2)*a^2/b
^4 - 105*(b*sqrt(c/x) + a)^(3/2)*a^3/b^4)/c^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {a+b\,\sqrt {\frac {c}{x}}}}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*(c/x)^(1/2))^(1/2)/x^3,x)

[Out]

int((a + b*(c/x)^(1/2))^(1/2)/x^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a + b \sqrt {\frac {c}{x}}}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(c/x)**(1/2))**(1/2)/x**3,x)

[Out]

Integral(sqrt(a + b*sqrt(c/x))/x**3, x)

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