3.2984 \(\int \frac{\sqrt{a+b \sqrt{\frac{c}{x}}}}{x} \, dx\)

Optimal. Leaf size=51 \[ 4 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{\frac{c}{x}}}}{\sqrt{a}}\right )-4 \sqrt{a+b \sqrt{\frac{c}{x}}} \]

[Out]

-4*Sqrt[a + b*Sqrt[c/x]] + 4*Sqrt[a]*ArcTanh[Sqrt[a + b*Sqrt[c/x]]/Sqrt[a]]

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Rubi [A]  time = 0.0349198, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {369, 266, 50, 63, 208} \[ 4 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{\frac{c}{x}}}}{\sqrt{a}}\right )-4 \sqrt{a+b \sqrt{\frac{c}{x}}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Sqrt[c/x]]/x,x]

[Out]

-4*Sqrt[a + b*Sqrt[c/x]] + 4*Sqrt[a]*ArcTanh[Sqrt[a + b*Sqrt[c/x]]/Sqrt[a]]

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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

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

Mathematica [A]  time = 0.0465543, size = 51, normalized size = 1. \[ 4 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{\frac{c}{x}}}}{\sqrt{a}}\right )-4 \sqrt{a+b \sqrt{\frac{c}{x}}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*Sqrt[c/x]]/x,x]

[Out]

-4*Sqrt[a + b*Sqrt[c/x]] + 4*Sqrt[a]*ArcTanh[Sqrt[a + b*Sqrt[c/x]]/Sqrt[a]]

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Maple [B]  time = 0.022, size = 150, normalized size = 2.9 \begin{align*} 2\,{\frac{1}{bx\sqrt{a}}\sqrt{a+b\sqrt{{\frac{c}{x}}}} \left ( \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}^{3/2}ab+2\,{a}^{3/2}\sqrt{ax+b\sqrt{{\frac{c}{x}}}x}x-2\, \left ( ax+b\sqrt{{\frac{c}{x}}}x \right ) ^{3/2}\sqrt{a} \right ){\frac{1}{\sqrt{x \left ( a+b\sqrt{{\frac{c}{x}}} \right ) }}}{\frac{1}{\sqrt{{\frac{c}{x}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.4072, size = 262, normalized size = 5.14 \begin{align*} \left [2 \, \sqrt{a} \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 ) - 4 \, \sqrt{b \sqrt{\frac{c}{x}} + a}, -4 \, \sqrt{-a} \arctan \left (\frac{\sqrt{b \sqrt{\frac{c}{x}} + a} \sqrt{-a}}{a}\right ) - 4 \, \sqrt{b \sqrt{\frac{c}{x}} + a}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[2*sqrt(a)*log(2*sqrt(b*sqrt(c/x) + a)*sqrt(a)*x*sqrt(c/x) + 2*a*x*sqrt(c/x) + b*c) - 4*sqrt(b*sqrt(c/x) + a),
-4*sqrt(-a)*arctan(sqrt(b*sqrt(c/x) + a)*sqrt(-a)/a) - 4*sqrt(b*sqrt(c/x) + a)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a + b*sqrt(c/x))/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError