3.2983 \(\int \sqrt{a+b \sqrt{\frac{c}{x}}} \, dx\)

Optimal. Leaf size=92 \[ -\frac{b^2 c \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{\frac{c}{x}}}}{\sqrt{a}}\right )}{2 a^{3/2}}+\frac{b c \sqrt{a+b \sqrt{\frac{c}{x}}}}{2 a \sqrt{\frac{c}{x}}}+x \sqrt{a+b \sqrt{\frac{c}{x}}} \]

[Out]

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

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Rubi [A]  time = 0.0486613, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {255, 190, 47, 51, 63, 208} \[ -\frac{b^2 c \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{\frac{c}{x}}}}{\sqrt{a}}\right )}{2 a^{3/2}}+\frac{b c \sqrt{a+b \sqrt{\frac{c}{x}}}}{2 a \sqrt{\frac{c}{x}}}+x \sqrt{a+b \sqrt{\frac{c}{x}}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Sqrt[c/x]],x]

[Out]

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

Rule 255

Int[((a_) + (b_.)*((c_.)*(x_)^(q_.))^(n_))^(p_.), x_Symbol] :> With[{k = Denominator[n]}, Subst[Int[(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, p, q}, x] && Fraction
Q[n]

Rule 190

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(1/n - 1)*(a + b*x)^p, x], x, x^n], x] /
; FreeQ[{a, b, p}, x] && FractionQ[n] && IntegerQ[1/n]

Rule 47

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 \sqrt{a+b \sqrt{\frac{c}{x}}} \, dx &=\operatorname{Subst}\left (\int \sqrt{a+\frac{b \sqrt{c}}{\sqrt{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^3} \, dx,x,\frac{1}{\sqrt{x}}\right ),\sqrt{x},\frac{\sqrt{\frac{c}{x}} x}{\sqrt{c}}\right )\\ &=\sqrt{a+b \sqrt{\frac{c}{x}}} x-\operatorname{Subst}\left (\frac{1}{2} \left (b \sqrt{c}\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b \sqrt{c} x}} \, dx,x,\frac{1}{\sqrt{x}}\right ),\sqrt{x},\frac{\sqrt{\frac{c}{x}} x}{\sqrt{c}}\right )\\ &=\frac{b c \sqrt{a+b \sqrt{\frac{c}{x}}}}{2 a \sqrt{\frac{c}{x}}}+\sqrt{a+b \sqrt{\frac{c}{x}}} x+\operatorname{Subst}\left (\frac{\left (b^2 c\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b \sqrt{c} x}} \, dx,x,\frac{1}{\sqrt{x}}\right )}{4 a},\sqrt{x},\frac{\sqrt{\frac{c}{x}} x}{\sqrt{c}}\right )\\ &=\frac{b c \sqrt{a+b \sqrt{\frac{c}{x}}}}{2 a \sqrt{\frac{c}{x}}}+\sqrt{a+b \sqrt{\frac{c}{x}}} x+\operatorname{Subst}\left (\frac{\left (b \sqrt{c}\right ) \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 )}{2 a},\sqrt{x},\frac{\sqrt{\frac{c}{x}} x}{\sqrt{c}}\right )\\ &=\frac{b c \sqrt{a+b \sqrt{\frac{c}{x}}}}{2 a \sqrt{\frac{c}{x}}}+\sqrt{a+b \sqrt{\frac{c}{x}}} x-\frac{b^2 c \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{\frac{c}{x}}}}{\sqrt{a}}\right )}{2 a^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0656965, size = 79, normalized size = 0.86 \[ \frac{\sqrt{a} x \sqrt{a+b \sqrt{\frac{c}{x}}} \left (2 a+b \sqrt{\frac{c}{x}}\right )-b^2 c \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{\frac{c}{x}}}}{\sqrt{a}}\right )}{2 a^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*Sqrt[c/x]],x]

[Out]

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

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Maple [B]  time = 0.02, size = 147, normalized size = 1.6 \begin{align*}{\frac{1}{4}\sqrt{a+b\sqrt{{\frac{c}{x}}}}\sqrt{x} \left ( 2\,{a}^{3/2}\sqrt{ax+b\sqrt{{\frac{c}{x}}}x}\sqrt{{\frac{c}{x}}}\sqrt{x}b-{b}^{2}c\ln \left ({\frac{1}{2} \left ( b\sqrt{{\frac{c}{x}}}\sqrt{x}+2\,\sqrt{ax+b\sqrt{{\frac{c}{x}}}x}\sqrt{a}+2\,a\sqrt{x} \right ){\frac{1}{\sqrt{a}}}} \right ) a+4\,{a}^{5/2}\sqrt{ax+b\sqrt{{\frac{c}{x}}}x}\sqrt{x} \right ){\frac{1}{\sqrt{x \left ( a+b\sqrt{{\frac{c}{x}}} \right ) }}}{a}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.5454, size = 373, normalized size = 4.05 \begin{align*} \left [\frac{\sqrt{a} b^{2} c \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 ) + 2 \,{\left (a b x \sqrt{\frac{c}{x}} + 2 \, a^{2} x\right )} \sqrt{b \sqrt{\frac{c}{x}} + a}}{4 \, a^{2}}, \frac{\sqrt{-a} b^{2} c \arctan \left (\frac{\sqrt{b \sqrt{\frac{c}{x}} + a} \sqrt{-a}}{a}\right ) +{\left (a b x \sqrt{\frac{c}{x}} + 2 \, a^{2} x\right )} \sqrt{b \sqrt{\frac{c}{x}} + a}}{2 \, a^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(sqrt(a)*b^2*c*log(-2*sqrt(b*sqrt(c/x) + a)*sqrt(a)*x*sqrt(c/x) + 2*a*x*sqrt(c/x) + b*c) + 2*(a*b*x*sqrt(
c/x) + 2*a^2*x)*sqrt(b*sqrt(c/x) + a))/a^2, 1/2*(sqrt(-a)*b^2*c*arctan(sqrt(b*sqrt(c/x) + a)*sqrt(-a)/a) + (a*
b*x*sqrt(c/x) + 2*a^2*x)*sqrt(b*sqrt(c/x) + a))/a^2]

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

[Out]

Integral(sqrt(a + b*sqrt(c/x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError