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

Optimal. Leaf size=123 \[ x \sqrt{a+\frac{b}{x}} \sqrt{c+\frac{d}{x}}+\frac{(a d+b c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+\frac{b}{x}}}{\sqrt{a} \sqrt{c+\frac{d}{x}}}\right )}{\sqrt{a} \sqrt{c}}-2 \sqrt{b} \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b} \sqrt{c+\frac{d}{x}}}\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0938878, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {375, 97, 157, 63, 217, 206, 93, 208} \[ x \sqrt{a+\frac{b}{x}} \sqrt{c+\frac{d}{x}}+\frac{(a d+b c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+\frac{b}{x}}}{\sqrt{a} \sqrt{c+\frac{d}{x}}}\right )}{\sqrt{a} \sqrt{c}}-2 \sqrt{b} \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b} \sqrt{c+\frac{d}{x}}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b/x]*Sqrt[c + d/x],x]

[Out]

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

Rule 375

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> -Subst[Int[((a + b/x^n)^p*(c +
d/x^n)^q)/x^2, x], x, 1/x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && ILtQ[n, 0]

Rule 97

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p)/(b*(m + 1)), x] - Dist[1/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n
- 1)*(e + f*x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[m
, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, 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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*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+\frac{b}{x}} \sqrt{c+\frac{d}{x}} \, dx &=-\operatorname{Subst}\left (\int \frac{\sqrt{a+b x} \sqrt{c+d x}}{x^2} \, dx,x,\frac{1}{x}\right )\\ &=\sqrt{a+\frac{b}{x}} \sqrt{c+\frac{d}{x}} x-\operatorname{Subst}\left (\int \frac{\frac{1}{2} (b c+a d)+b d x}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\frac{1}{x}\right )\\ &=\sqrt{a+\frac{b}{x}} \sqrt{c+\frac{d}{x}} x-(b d) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\frac{1}{x}\right )-\frac{1}{2} (b c+a d) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\frac{1}{x}\right )\\ &=\sqrt{a+\frac{b}{x}} \sqrt{c+\frac{d}{x}} x-(2 d) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c-\frac{a d}{b}+\frac{d x^2}{b}}} \, dx,x,\sqrt{a+\frac{b}{x}}\right )-(b c+a d) \operatorname{Subst}\left (\int \frac{1}{-a+c x^2} \, dx,x,\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{c+\frac{d}{x}}}\right )\\ &=\sqrt{a+\frac{b}{x}} \sqrt{c+\frac{d}{x}} x+\frac{(b c+a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+\frac{b}{x}}}{\sqrt{a} \sqrt{c+\frac{d}{x}}}\right )}{\sqrt{a} \sqrt{c}}-(2 d) \operatorname{Subst}\left (\int \frac{1}{1-\frac{d x^2}{b}} \, dx,x,\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{c+\frac{d}{x}}}\right )\\ &=\sqrt{a+\frac{b}{x}} \sqrt{c+\frac{d}{x}} x+\frac{(b c+a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+\frac{b}{x}}}{\sqrt{a} \sqrt{c+\frac{d}{x}}}\right )}{\sqrt{a} \sqrt{c}}-2 \sqrt{b} \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b} \sqrt{c+\frac{d}{x}}}\right )\\ \end{align*}

Mathematica [A]  time = 1.00956, size = 167, normalized size = 1.36 \[ \frac{\sqrt{a+\frac{b}{x}} (c x+d)-2 \sqrt{d} \sqrt{b c-a d} \sqrt{\frac{b c x+b d}{b c x-a d x}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )+\frac{\sqrt{c+\frac{d}{x}} (a d+b c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+\frac{b}{x}}}{\sqrt{a} \sqrt{c+\frac{d}{x}}}\right )}{\sqrt{a} \sqrt{c}}}{\sqrt{c+\frac{d}{x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b/x]*(d + c*x) - 2*Sqrt[d]*Sqrt[b*c - a*d]*Sqrt[(b*d + b*c*x)/(b*c*x - a*d*x)]*ArcSinh[(Sqrt[d]*Sqrt
[a + b/x])/Sqrt[b*c - a*d]] + ((b*c + a*d)*Sqrt[c + d/x]*ArcTanh[(Sqrt[c]*Sqrt[a + b/x])/(Sqrt[a]*Sqrt[c + d/x
])])/(Sqrt[a]*Sqrt[c]))/Sqrt[c + d/x]

________________________________________________________________________________________

Maple [B]  time = 0.046, size = 253, normalized size = 2.1 \begin{align*}{\frac{x}{2}\sqrt{{\frac{cx+d}{x}}}\sqrt{{\frac{ax+b}{x}}} \left ( -2\,bd\ln \left ({\frac{adx+bcx+2\,\sqrt{bd}\sqrt{ac{x}^{2}+adx+bcx+bd}+2\,bd}{x}} \right ) \sqrt{ac}+\sqrt{bd}\ln \left ({\frac{1}{2} \left ( 2\,acx+2\,\sqrt{ac{x}^{2}+adx+bcx+bd}\sqrt{ac}+ad+bc \right ){\frac{1}{\sqrt{ac}}}} \right ) ad+\sqrt{bd}\ln \left ({\frac{1}{2} \left ( 2\,acx+2\,\sqrt{ac{x}^{2}+adx+bcx+bd}\sqrt{ac}+ad+bc \right ){\frac{1}{\sqrt{ac}}}} \right ) bc+2\,\sqrt{ac{x}^{2}+adx+bcx+bd}\sqrt{ac}\sqrt{bd} \right ){\frac{1}{\sqrt{ac{x}^{2}+adx+bcx+bd}}}{\frac{1}{\sqrt{ac}}}{\frac{1}{\sqrt{bd}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a + b/x)*sqrt(c + d/x), x)

________________________________________________________________________________________

Fricas [A]  time = 5.40341, size = 1995, normalized size = 16.22 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(4*a*c*x*sqrt((a*x + b)/x)*sqrt((c*x + d)/x) + 2*sqrt(b*d)*a*c*log(-(8*b^2*d^2 + (b^2*c^2 + 6*a*b*c*d + a
^2*d^2)*x^2 - 4*(2*b*d*x + (b*c + a*d)*x^2)*sqrt(b*d)*sqrt((a*x + b)/x)*sqrt((c*x + d)/x) + 8*(b^2*c*d + a*b*d
^2)*x)/x^2) + sqrt(a*c)*(b*c + a*d)*log(-8*a^2*c^2*x^2 - b^2*c^2 - 6*a*b*c*d - a^2*d^2 - 4*(2*a*c*x^2 + (b*c +
a*d)*x)*sqrt(a*c)*sqrt((a*x + b)/x)*sqrt((c*x + d)/x) - 8*(a*b*c^2 + a^2*c*d)*x))/(a*c), 1/4*(4*a*c*x*sqrt((a
*x + b)/x)*sqrt((c*x + d)/x) + 4*sqrt(-b*d)*a*c*arctan(1/2*(2*b*d*x + (b*c + a*d)*x^2)*sqrt(-b*d)*sqrt((a*x +
b)/x)*sqrt((c*x + d)/x)/(a*b*c*d*x^2 + b^2*d^2 + (b^2*c*d + a*b*d^2)*x)) + sqrt(a*c)*(b*c + a*d)*log(-8*a^2*c^
2*x^2 - b^2*c^2 - 6*a*b*c*d - a^2*d^2 - 4*(2*a*c*x^2 + (b*c + a*d)*x)*sqrt(a*c)*sqrt((a*x + b)/x)*sqrt((c*x +
d)/x) - 8*(a*b*c^2 + a^2*c*d)*x))/(a*c), 1/2*(2*a*c*x*sqrt((a*x + b)/x)*sqrt((c*x + d)/x) + sqrt(b*d)*a*c*log(
-(8*b^2*d^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*b*d*x + (b*c + a*d)*x^2)*sqrt(b*d)*sqrt((a*x + b)/x)*
sqrt((c*x + d)/x) + 8*(b^2*c*d + a*b*d^2)*x)/x^2) - sqrt(-a*c)*(b*c + a*d)*arctan(2*sqrt(-a*c)*x*sqrt((a*x + b
)/x)*sqrt((c*x + d)/x)/(2*a*c*x + b*c + a*d)))/(a*c), 1/2*(2*a*c*x*sqrt((a*x + b)/x)*sqrt((c*x + d)/x) + 2*sqr
t(-b*d)*a*c*arctan(1/2*(2*b*d*x + (b*c + a*d)*x^2)*sqrt(-b*d)*sqrt((a*x + b)/x)*sqrt((c*x + d)/x)/(a*b*c*d*x^2
+ b^2*d^2 + (b^2*c*d + a*b*d^2)*x)) - sqrt(-a*c)*(b*c + a*d)*arctan(2*sqrt(-a*c)*x*sqrt((a*x + b)/x)*sqrt((c*
x + d)/x)/(2*a*c*x + b*c + a*d)))/(a*c)]

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a + b/x)*sqrt(c + d/x), x)