3.229 \(\int \frac{\sqrt{a+\frac{b}{x}}}{(c+\frac{d}{x})^2} \, dx\)

Optimal. Leaf size=147 \[ \frac{2 d \sqrt{a+\frac{b}{x}}}{c^2 \left (c+\frac{d}{x}\right )}+\frac{\sqrt{d} (3 b c-4 a d) \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )}{c^3 \sqrt{b c-a d}}+\frac{(b c-4 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{\sqrt{a} c^3}+\frac{x \sqrt{a+\frac{b}{x}}}{c \left (c+\frac{d}{x}\right )} \]

[Out]

(2*d*Sqrt[a + b/x])/(c^2*(c + d/x)) + (Sqrt[a + b/x]*x)/(c*(c + d/x)) + (Sqrt[d]*(3*b*c - 4*a*d)*ArcTan[(Sqrt[
d]*Sqrt[a + b/x])/Sqrt[b*c - a*d]])/(c^3*Sqrt[b*c - a*d]) + ((b*c - 4*a*d)*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/(Sq
rt[a]*c^3)

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Rubi [A]  time = 0.208627, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {375, 99, 151, 156, 63, 208, 205} \[ \frac{2 d \sqrt{a+\frac{b}{x}}}{c^2 \left (c+\frac{d}{x}\right )}+\frac{\sqrt{d} (3 b c-4 a d) \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )}{c^3 \sqrt{b c-a d}}+\frac{(b c-4 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{\sqrt{a} c^3}+\frac{x \sqrt{a+\frac{b}{x}}}{c \left (c+\frac{d}{x}\right )} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b/x]/(c + d/x)^2,x]

[Out]

(2*d*Sqrt[a + b/x])/(c^2*(c + d/x)) + (Sqrt[a + b/x]*x)/(c*(c + d/x)) + (Sqrt[d]*(3*b*c - 4*a*d)*ArcTan[(Sqrt[
d]*Sqrt[a + b/x])/Sqrt[b*c - a*d]])/(c^3*Sqrt[b*c - a*d]) + ((b*c - 4*a*d)*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/(Sq
rt[a]*c^3)

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 99

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 + 1))/((m + 1)*(b*e - a*f)), x] - Dist[1/((m + 1)*(b*e - a*f)), Int[(a +
b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + p + 2)*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

Rule 156

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

Rule 205

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

Rubi steps

\begin{align*} \int \frac{\sqrt{a+\frac{b}{x}}}{\left (c+\frac{d}{x}\right )^2} \, dx &=-\operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x^2 (c+d x)^2} \, dx,x,\frac{1}{x}\right )\\ &=\frac{\sqrt{a+\frac{b}{x}} x}{c \left (c+\frac{d}{x}\right )}-\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} (b c-4 a d)-\frac{3 b d x}{2}}{x \sqrt{a+b x} (c+d x)^2} \, dx,x,\frac{1}{x}\right )}{c}\\ &=\frac{2 d \sqrt{a+\frac{b}{x}}}{c^2 \left (c+\frac{d}{x}\right )}+\frac{\sqrt{a+\frac{b}{x}} x}{c \left (c+\frac{d}{x}\right )}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{1}{2} (b c-4 a d) (b c-a d)+b d (b c-a d) x}{x \sqrt{a+b x} (c+d x)} \, dx,x,\frac{1}{x}\right )}{c^2 (b c-a d)}\\ &=\frac{2 d \sqrt{a+\frac{b}{x}}}{c^2 \left (c+\frac{d}{x}\right )}+\frac{\sqrt{a+\frac{b}{x}} x}{c \left (c+\frac{d}{x}\right )}-\frac{(b c-4 a d) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\frac{1}{x}\right )}{2 c^3}+\frac{(d (3 b c-4 a d)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x} (c+d x)} \, dx,x,\frac{1}{x}\right )}{2 c^3}\\ &=\frac{2 d \sqrt{a+\frac{b}{x}}}{c^2 \left (c+\frac{d}{x}\right )}+\frac{\sqrt{a+\frac{b}{x}} x}{c \left (c+\frac{d}{x}\right )}-\frac{(b c-4 a d) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+\frac{b}{x}}\right )}{b c^3}+\frac{(d (3 b c-4 a d)) \operatorname{Subst}\left (\int \frac{1}{c-\frac{a d}{b}+\frac{d x^2}{b}} \, dx,x,\sqrt{a+\frac{b}{x}}\right )}{b c^3}\\ &=\frac{2 d \sqrt{a+\frac{b}{x}}}{c^2 \left (c+\frac{d}{x}\right )}+\frac{\sqrt{a+\frac{b}{x}} x}{c \left (c+\frac{d}{x}\right )}+\frac{\sqrt{d} (3 b c-4 a d) \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )}{c^3 \sqrt{b c-a d}}+\frac{(b c-4 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{\sqrt{a} c^3}\\ \end{align*}

Mathematica [A]  time = 0.311993, size = 122, normalized size = 0.83 \[ \frac{\frac{c x \sqrt{a+\frac{b}{x}} (c x+2 d)}{c x+d}+\frac{\sqrt{d} (3 b c-4 a d) \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )}{\sqrt{b c-a d}}+\frac{(b c-4 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{\sqrt{a}}}{c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b/x]/(c + d/x)^2,x]

[Out]

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

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Maple [B]  time = 0.013, size = 943, normalized size = 6.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a + \frac{b}{x}}}{{\left (c + \frac{d}{x}\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a + b/x)/(c + d/x)^2, x)

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Fricas [A]  time = 1.50225, size = 1760, normalized size = 11.97 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*((b*c*d - 4*a*d^2 + (b*c^2 - 4*a*c*d)*x)*sqrt(a)*log(2*a*x - 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + (3*a*b
*c*d - 4*a^2*d^2 + (3*a*b*c^2 - 4*a^2*c*d)*x)*sqrt(-d/(b*c - a*d))*log(-(2*(b*c - a*d)*x*sqrt(-d/(b*c - a*d))*
sqrt((a*x + b)/x) - b*d + (b*c - 2*a*d)*x)/(c*x + d)) - 2*(a*c^2*x^2 + 2*a*c*d*x)*sqrt((a*x + b)/x))/(a*c^4*x
+ a*c^3*d), -1/2*(2*(b*c*d - 4*a*d^2 + (b*c^2 - 4*a*c*d)*x)*sqrt(-a)*arctan(sqrt(-a)*sqrt((a*x + b)/x)/a) + (3
*a*b*c*d - 4*a^2*d^2 + (3*a*b*c^2 - 4*a^2*c*d)*x)*sqrt(-d/(b*c - a*d))*log(-(2*(b*c - a*d)*x*sqrt(-d/(b*c - a*
d))*sqrt((a*x + b)/x) - b*d + (b*c - 2*a*d)*x)/(c*x + d)) - 2*(a*c^2*x^2 + 2*a*c*d*x)*sqrt((a*x + b)/x))/(a*c^
4*x + a*c^3*d), 1/2*(2*(3*a*b*c*d - 4*a^2*d^2 + (3*a*b*c^2 - 4*a^2*c*d)*x)*sqrt(d/(b*c - a*d))*arctan(-(b*c -
a*d)*x*sqrt(d/(b*c - a*d))*sqrt((a*x + b)/x)/(a*d*x + b*d)) - (b*c*d - 4*a*d^2 + (b*c^2 - 4*a*c*d)*x)*sqrt(a)*
log(2*a*x - 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 2*(a*c^2*x^2 + 2*a*c*d*x)*sqrt((a*x + b)/x))/(a*c^4*x + a*c^3
*d), ((3*a*b*c*d - 4*a^2*d^2 + (3*a*b*c^2 - 4*a^2*c*d)*x)*sqrt(d/(b*c - a*d))*arctan(-(b*c - a*d)*x*sqrt(d/(b*
c - a*d))*sqrt((a*x + b)/x)/(a*d*x + b*d)) - (b*c*d - 4*a*d^2 + (b*c^2 - 4*a*c*d)*x)*sqrt(-a)*arctan(sqrt(-a)*
sqrt((a*x + b)/x)/a) + (a*c^2*x^2 + 2*a*c*d*x)*sqrt((a*x + b)/x))/(a*c^4*x + a*c^3*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2} \sqrt{a + \frac{b}{x}}}{\left (c x + d\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: TypeError