3.249 \(\int \frac {1}{\sqrt {a+\frac {b}{x}} (c+\frac {d}{x})} \, dx\)

Optimal. Leaf size=108 \[ -\frac {(2 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{a^{3/2} c^2}-\frac {2 d^{3/2} \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {a+\frac {b}{x}}}{\sqrt {b c-a d}}\right )}{c^2 \sqrt {b c-a d}}+\frac {x \sqrt {a+\frac {b}{x}}}{a c} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {375, 103, 156, 63, 208, 205} \[ -\frac {(2 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{a^{3/2} c^2}-\frac {2 d^{3/2} \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {a+\frac {b}{x}}}{\sqrt {b c-a d}}\right )}{c^2 \sqrt {b c-a d}}+\frac {x \sqrt {a+\frac {b}{x}}}{a c} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a + b/x]*(c + d/x)),x]

[Out]

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

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 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(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*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

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 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]

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 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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.25, size = 104, normalized size = 0.96 \[ \frac {-\frac {(2 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {2 d^{3/2} \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {a+\frac {b}{x}}}{\sqrt {b c-a d}}\right )}{\sqrt {b c-a d}}+\frac {c x \sqrt {a+\frac {b}{x}}}{a}}{c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[a + b/x]*(c + d/x)),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.99, size = 542, normalized size = 5.02 \[ \left [\frac {2 \, a^{2} d \sqrt {-\frac {d}{b c - a d}} \log \left (-\frac {2 \, {\left (b c - a d\right )} x \sqrt {-\frac {d}{b c - a d}} \sqrt {\frac {a x + b}{x}} - b d + {\left (b c - 2 \, a d\right )} x}{c x + d}\right ) + 2 \, a c x \sqrt {\frac {a x + b}{x}} + {\left (b c + 2 \, a d\right )} \sqrt {a} \log \left (2 \, a x - 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right )}{2 \, a^{2} c^{2}}, \frac {a^{2} d \sqrt {-\frac {d}{b c - a d}} \log \left (-\frac {2 \, {\left (b c - a d\right )} x \sqrt {-\frac {d}{b c - a d}} \sqrt {\frac {a x + b}{x}} - b d + {\left (b c - 2 \, a d\right )} x}{c x + d}\right ) + a c x \sqrt {\frac {a x + b}{x}} + {\left (b c + 2 \, a d\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right )}{a^{2} c^{2}}, -\frac {4 \, a^{2} d \sqrt {\frac {d}{b c - a d}} \arctan \left (-\frac {{\left (b c - a d\right )} x \sqrt {\frac {d}{b c - a d}} \sqrt {\frac {a x + b}{x}}}{a d x + b d}\right ) - 2 \, a c x \sqrt {\frac {a x + b}{x}} - {\left (b c + 2 \, a d\right )} \sqrt {a} \log \left (2 \, a x - 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right )}{2 \, a^{2} c^{2}}, -\frac {2 \, a^{2} d \sqrt {\frac {d}{b c - a d}} \arctan \left (-\frac {{\left (b c - a d\right )} x \sqrt {\frac {d}{b c - a d}} \sqrt {\frac {a x + b}{x}}}{a d x + b d}\right ) - a c x \sqrt {\frac {a x + b}{x}} - {\left (b c + 2 \, a d\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right )}{a^{2} c^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.18, size = 134, normalized size = 1.24 \[ -b^{2} {\left (\frac {2 \, d^{2} \arctan \left (\frac {d \sqrt {\frac {a x + b}{x}}}{\sqrt {b c d - a d^{2}}}\right )}{\sqrt {b c d - a d^{2}} b^{2} c^{2}} + \frac {\sqrt {\frac {a x + b}{x}}}{{\left (a - \frac {a x + b}{x}\right )} a b c} - \frac {{\left (b c + 2 \, a d\right )} \arctan \left (\frac {\sqrt {\frac {a x + b}{x}}}{\sqrt {-a}}\right )}{\sqrt {-a} a b^{2} c^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b^2*(2*d^2*arctan(d*sqrt((a*x + b)/x)/sqrt(b*c*d - a*d^2))/(sqrt(b*c*d - a*d^2)*b^2*c^2) + sqrt((a*x + b)/x)/
((a - (a*x + b)/x)*a*b*c) - (b*c + 2*a*d)*arctan(sqrt((a*x + b)/x)/sqrt(-a))/(sqrt(-a)*a*b^2*c^2))

________________________________________________________________________________________

maple [B]  time = 0.06, size = 228, normalized size = 2.11 \[ -\frac {\left (2 a^{\frac {3}{2}} d^{2} \ln \left (\frac {-2 a d x +b c x -b d +2 \sqrt {\frac {\left (a d -b c \right ) d}{c^{2}}}\, \sqrt {\left (a x +b \right ) x}\, c}{c x +d}\right )+2 \sqrt {\frac {\left (a d -b c \right ) d}{c^{2}}}\, a c d \ln \left (\frac {2 a x +b +2 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}}{2 \sqrt {a}}\right )+\sqrt {\frac {\left (a d -b c \right ) d}{c^{2}}}\, b \,c^{2} \ln \left (\frac {2 a x +b +2 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}}{2 \sqrt {a}}\right )-2 \sqrt {\left (a x +b \right ) x}\, \sqrt {\frac {\left (a d -b c \right ) d}{c^{2}}}\, \sqrt {a}\, c^{2}\right ) \sqrt {\frac {a x +b}{x}}\, x}{2 \sqrt {\frac {\left (a d -b c \right ) d}{c^{2}}}\, \sqrt {\left (a x +b \right ) x}\, a^{\frac {3}{2}} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a + \frac {b}{x}} {\left (c + \frac {d}{x}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.98, size = 1183, normalized size = 10.95 \[ \frac {x\,\sqrt {a+\frac {b}{x}}}{a\,c}-\frac {\mathrm {atanh}\left (\frac {12\,b^4\,d^4\,\sqrt {a+\frac {b}{x}}}{\sqrt {a^3}\,\left (\frac {12\,b^4\,d^4}{a}+\frac {10\,b^5\,c\,d^3}{a^2}+\frac {2\,b^6\,c^2\,d^2}{a^3}\right )}+\frac {10\,b^5\,d^3\,\sqrt {a+\frac {b}{x}}}{\sqrt {a^3}\,\left (\frac {10\,b^5\,d^3}{a}+\frac {12\,b^4\,d^4}{c}+\frac {2\,b^6\,c\,d^2}{a^2}\right )}+\frac {2\,b^6\,d^2\,\sqrt {a+\frac {b}{x}}}{\sqrt {a^3}\,\left (\frac {2\,b^6\,d^2}{a}+\frac {10\,b^5\,d^3}{c}+\frac {12\,a\,b^4\,d^4}{c^2}\right )}\right )\,\left (2\,a\,d+b\,c\right )}{c^2\,\sqrt {a^3}}-\frac {\mathrm {atan}\left (\frac {\frac {\left (\frac {\left (\frac {2\,\left (2\,a^2\,b^3\,c^4\,d^3+2\,a\,b^4\,c^5\,d^2\right )}{a^2\,c^3}-\frac {2\,\left (4\,a^2\,b^3\,c^5\,d^2-8\,a^3\,b^2\,c^4\,d^3\right )\,\sqrt {a+\frac {b}{x}}\,\sqrt {a\,d^4-b\,c\,d^3}}{a^2\,c^2\,\left (b\,c^3-a\,c^2\,d\right )}\right )\,\sqrt {a\,d^4-b\,c\,d^3}}{b\,c^3-a\,c^2\,d}-\frac {2\,\sqrt {a+\frac {b}{x}}\,\left (8\,a^2\,b^2\,d^5+4\,a\,b^3\,c\,d^4+b^4\,c^2\,d^3\right )}{a^2\,c^2}\right )\,\sqrt {a\,d^4-b\,c\,d^3}\,1{}\mathrm {i}}{b\,c^3-a\,c^2\,d}-\frac {\left (\frac {\left (\frac {2\,\left (2\,a^2\,b^3\,c^4\,d^3+2\,a\,b^4\,c^5\,d^2\right )}{a^2\,c^3}+\frac {2\,\left (4\,a^2\,b^3\,c^5\,d^2-8\,a^3\,b^2\,c^4\,d^3\right )\,\sqrt {a+\frac {b}{x}}\,\sqrt {a\,d^4-b\,c\,d^3}}{a^2\,c^2\,\left (b\,c^3-a\,c^2\,d\right )}\right )\,\sqrt {a\,d^4-b\,c\,d^3}}{b\,c^3-a\,c^2\,d}+\frac {2\,\sqrt {a+\frac {b}{x}}\,\left (8\,a^2\,b^2\,d^5+4\,a\,b^3\,c\,d^4+b^4\,c^2\,d^3\right )}{a^2\,c^2}\right )\,\sqrt {a\,d^4-b\,c\,d^3}\,1{}\mathrm {i}}{b\,c^3-a\,c^2\,d}}{\frac {\left (\frac {\left (\frac {2\,\left (2\,a^2\,b^3\,c^4\,d^3+2\,a\,b^4\,c^5\,d^2\right )}{a^2\,c^3}-\frac {2\,\left (4\,a^2\,b^3\,c^5\,d^2-8\,a^3\,b^2\,c^4\,d^3\right )\,\sqrt {a+\frac {b}{x}}\,\sqrt {a\,d^4-b\,c\,d^3}}{a^2\,c^2\,\left (b\,c^3-a\,c^2\,d\right )}\right )\,\sqrt {a\,d^4-b\,c\,d^3}}{b\,c^3-a\,c^2\,d}-\frac {2\,\sqrt {a+\frac {b}{x}}\,\left (8\,a^2\,b^2\,d^5+4\,a\,b^3\,c\,d^4+b^4\,c^2\,d^3\right )}{a^2\,c^2}\right )\,\sqrt {a\,d^4-b\,c\,d^3}}{b\,c^3-a\,c^2\,d}-\frac {4\,\left (c\,b^4\,d^4+2\,a\,b^3\,d^5\right )}{a^2\,c^3}+\frac {\left (\frac {\left (\frac {2\,\left (2\,a^2\,b^3\,c^4\,d^3+2\,a\,b^4\,c^5\,d^2\right )}{a^2\,c^3}+\frac {2\,\left (4\,a^2\,b^3\,c^5\,d^2-8\,a^3\,b^2\,c^4\,d^3\right )\,\sqrt {a+\frac {b}{x}}\,\sqrt {a\,d^4-b\,c\,d^3}}{a^2\,c^2\,\left (b\,c^3-a\,c^2\,d\right )}\right )\,\sqrt {a\,d^4-b\,c\,d^3}}{b\,c^3-a\,c^2\,d}+\frac {2\,\sqrt {a+\frac {b}{x}}\,\left (8\,a^2\,b^2\,d^5+4\,a\,b^3\,c\,d^4+b^4\,c^2\,d^3\right )}{a^2\,c^2}\right )\,\sqrt {a\,d^4-b\,c\,d^3}}{b\,c^3-a\,c^2\,d}}\right )\,\sqrt {a\,d^4-b\,c\,d^3}\,2{}\mathrm {i}}{b\,c^3-a\,c^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b/x)^(1/2)*(c + d/x)),x)

[Out]

(x*(a + b/x)^(1/2))/(a*c) - (atan(((((((2*(2*a*b^4*c^5*d^2 + 2*a^2*b^3*c^4*d^3))/(a^2*c^3) - (2*(4*a^2*b^3*c^5
*d^2 - 8*a^3*b^2*c^4*d^3)*(a + b/x)^(1/2)*(a*d^4 - b*c*d^3)^(1/2))/(a^2*c^2*(b*c^3 - a*c^2*d)))*(a*d^4 - b*c*d
^3)^(1/2))/(b*c^3 - a*c^2*d) - (2*(a + b/x)^(1/2)*(8*a^2*b^2*d^5 + b^4*c^2*d^3 + 4*a*b^3*c*d^4))/(a^2*c^2))*(a
*d^4 - b*c*d^3)^(1/2)*1i)/(b*c^3 - a*c^2*d) - (((((2*(2*a*b^4*c^5*d^2 + 2*a^2*b^3*c^4*d^3))/(a^2*c^3) + (2*(4*
a^2*b^3*c^5*d^2 - 8*a^3*b^2*c^4*d^3)*(a + b/x)^(1/2)*(a*d^4 - b*c*d^3)^(1/2))/(a^2*c^2*(b*c^3 - a*c^2*d)))*(a*
d^4 - b*c*d^3)^(1/2))/(b*c^3 - a*c^2*d) + (2*(a + b/x)^(1/2)*(8*a^2*b^2*d^5 + b^4*c^2*d^3 + 4*a*b^3*c*d^4))/(a
^2*c^2))*(a*d^4 - b*c*d^3)^(1/2)*1i)/(b*c^3 - a*c^2*d))/((((((2*(2*a*b^4*c^5*d^2 + 2*a^2*b^3*c^4*d^3))/(a^2*c^
3) - (2*(4*a^2*b^3*c^5*d^2 - 8*a^3*b^2*c^4*d^3)*(a + b/x)^(1/2)*(a*d^4 - b*c*d^3)^(1/2))/(a^2*c^2*(b*c^3 - a*c
^2*d)))*(a*d^4 - b*c*d^3)^(1/2))/(b*c^3 - a*c^2*d) - (2*(a + b/x)^(1/2)*(8*a^2*b^2*d^5 + b^4*c^2*d^3 + 4*a*b^3
*c*d^4))/(a^2*c^2))*(a*d^4 - b*c*d^3)^(1/2))/(b*c^3 - a*c^2*d) - (4*(2*a*b^3*d^5 + b^4*c*d^4))/(a^2*c^3) + (((
((2*(2*a*b^4*c^5*d^2 + 2*a^2*b^3*c^4*d^3))/(a^2*c^3) + (2*(4*a^2*b^3*c^5*d^2 - 8*a^3*b^2*c^4*d^3)*(a + b/x)^(1
/2)*(a*d^4 - b*c*d^3)^(1/2))/(a^2*c^2*(b*c^3 - a*c^2*d)))*(a*d^4 - b*c*d^3)^(1/2))/(b*c^3 - a*c^2*d) + (2*(a +
b/x)^(1/2)*(8*a^2*b^2*d^5 + b^4*c^2*d^3 + 4*a*b^3*c*d^4))/(a^2*c^2))*(a*d^4 - b*c*d^3)^(1/2))/(b*c^3 - a*c^2*
d)))*(a*d^4 - b*c*d^3)^(1/2)*2i)/(b*c^3 - a*c^2*d) - (atanh((12*b^4*d^4*(a + b/x)^(1/2))/((a^3)^(1/2)*((12*b^4
*d^4)/a + (10*b^5*c*d^3)/a^2 + (2*b^6*c^2*d^2)/a^3)) + (10*b^5*d^3*(a + b/x)^(1/2))/((a^3)^(1/2)*((10*b^5*d^3)
/a + (12*b^4*d^4)/c + (2*b^6*c*d^2)/a^2)) + (2*b^6*d^2*(a + b/x)^(1/2))/((a^3)^(1/2)*((2*b^6*d^2)/a + (10*b^5*
d^3)/c + (12*a*b^4*d^4)/c^2)))*(2*a*d + b*c))/(c^2*(a^3)^(1/2))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\sqrt {a + \frac {b}{x}} \left (c x + d\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________