3.268 \(\int \frac {\sqrt {a+\frac {b}{x}}}{(c+\frac {d}{x})^{3/2}} \, dx\)

Optimal. Leaf size=122 \[ \frac {(b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+\frac {b}{x}}}{\sqrt {a} \sqrt {c+\frac {d}{x}}}\right )}{\sqrt {a} c^{5/2}}-\frac {\sqrt {a+\frac {b}{x}} (b c-3 a d)}{a c^2 \sqrt {c+\frac {d}{x}}}+\frac {x \left (a+\frac {b}{x}\right )^{3/2}}{a c \sqrt {c+\frac {d}{x}}} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {375, 96, 94, 93, 208} \[ -\frac {\sqrt {a+\frac {b}{x}} (b c-3 a d)}{a c^2 \sqrt {c+\frac {d}{x}}}+\frac {(b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+\frac {b}{x}}}{\sqrt {a} \sqrt {c+\frac {d}{x}}}\right )}{\sqrt {a} c^{5/2}}+\frac {x \left (a+\frac {b}{x}\right )^{3/2}}{a c \sqrt {c+\frac {d}{x}}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b/x]/(c + d/x)^(3/2),x]

[Out]

-(((b*c - 3*a*d)*Sqrt[a + b/x])/(a*c^2*Sqrt[c + d/x])) + ((a + b/x)^(3/2)*x)/(a*c*Sqrt[c + d/x]) + ((b*c - 3*a
*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b/x])/(Sqrt[a]*Sqrt[c + d/x])])/(Sqrt[a]*c^(5/2))

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 94

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[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[m, 1])

Rule 96

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[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 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, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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 {\sqrt {a+\frac {b}{x}}}{\left (c+\frac {d}{x}\right )^{3/2}} \, dx &=-\operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x^2 (c+d x)^{3/2}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {\left (a+\frac {b}{x}\right )^{3/2} x}{a c \sqrt {c+\frac {d}{x}}}+\frac {\left (-\frac {b c}{2}+\frac {3 a d}{2}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x (c+d x)^{3/2}} \, dx,x,\frac {1}{x}\right )}{a c}\\ &=-\frac {(b c-3 a d) \sqrt {a+\frac {b}{x}}}{a c^2 \sqrt {c+\frac {d}{x}}}+\frac {\left (a+\frac {b}{x}\right )^{3/2} x}{a c \sqrt {c+\frac {d}{x}}}-\frac {(b c-3 a d) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,\frac {1}{x}\right )}{2 c^2}\\ &=-\frac {(b c-3 a d) \sqrt {a+\frac {b}{x}}}{a c^2 \sqrt {c+\frac {d}{x}}}+\frac {\left (a+\frac {b}{x}\right )^{3/2} x}{a c \sqrt {c+\frac {d}{x}}}-\frac {(b c-3 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 )}{c^2}\\ &=-\frac {(b c-3 a d) \sqrt {a+\frac {b}{x}}}{a c^2 \sqrt {c+\frac {d}{x}}}+\frac {\left (a+\frac {b}{x}\right )^{3/2} x}{a c \sqrt {c+\frac {d}{x}}}+\frac {(b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+\frac {b}{x}}}{\sqrt {a} \sqrt {c+\frac {d}{x}}}\right )}{\sqrt {a} c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 87, normalized size = 0.71 \[ \frac {(b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+\frac {b}{x}}}{\sqrt {a} \sqrt {c+\frac {d}{x}}}\right )}{\sqrt {a} c^{5/2}}+\frac {\sqrt {a+\frac {b}{x}} (c x+3 d)}{c^2 \sqrt {c+\frac {d}{x}}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b/x]/(c + d/x)^(3/2),x]

[Out]

(Sqrt[a + b/x]*(3*d + c*x))/(c^2*Sqrt[c + d/x]) + ((b*c - 3*a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b/x])/(Sqrt[a]*Sqrt
[c + d/x])])/(Sqrt[a]*c^(5/2))

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fricas [A]  time = 1.30, size = 319, normalized size = 2.61 \[ \left [-\frac {{\left (b c d - 3 \, a d^{2} + {\left (b c^{2} - 3 \, a c d\right )} x\right )} \sqrt {a c} \log \left (-8 \, a^{2} c^{2} x^{2} - b^{2} c^{2} - 6 \, a b c d - a^{2} d^{2} + 4 \, {\left (2 \, a c x^{2} + {\left (b c + a d\right )} x\right )} \sqrt {a c} \sqrt {\frac {a x + b}{x}} \sqrt {\frac {c x + d}{x}} - 8 \, {\left (a b c^{2} + a^{2} c d\right )} x\right ) - 4 \, {\left (a c^{2} x^{2} + 3 \, a c d x\right )} \sqrt {\frac {a x + b}{x}} \sqrt {\frac {c x + d}{x}}}{4 \, {\left (a c^{4} x + a c^{3} d\right )}}, -\frac {{\left (b c d - 3 \, a d^{2} + {\left (b c^{2} - 3 \, a c d\right )} x\right )} \sqrt {-a c} \arctan \left (\frac {2 \, \sqrt {-a c} x \sqrt {\frac {a x + b}{x}} \sqrt {\frac {c x + d}{x}}}{2 \, a c x + b c + a d}\right ) - 2 \, {\left (a c^{2} x^{2} + 3 \, a c d x\right )} \sqrt {\frac {a x + b}{x}} \sqrt {\frac {c x + d}{x}}}{2 \, {\left (a c^{4} x + a c^{3} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^(1/2)/(c+d/x)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^(1/2)/(c+d/x)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(x)]Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):
Check [abs(t_nostep)]Unable to divide, perhaps due to rounding error%%%{%%%{1,[1]%%%},[2,1,2]%%%}+%%%{%%{[-2,0
]:[1,0,%%%{-1,[1]%%%}]%%},[1,1,3]%%%}+%%%{1,[0,1,4]%%%} / %%%{%%%{1,[2]%%%},[2,0,0]%%%}+%%%{%%{[%%%{-2,[1]%%%}
,0]:[1,0,%%%{-1,[1]%%%}]%%},[1,0,1]%%%}+%%%{%%%{1,[1]%%%},[0,0,2]%%%} Error: Bad Argument Value

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maple [B]  time = 0.08, size = 280, normalized size = 2.30 \[ \frac {\sqrt {\frac {a x +b}{x}}\, \sqrt {\frac {c x +d}{x}}\, \left (-3 a c d x \ln \left (\frac {2 a c x +a d +b c +2 \sqrt {\left (a x +b \right ) \left (c x +d \right )}\, \sqrt {a c}}{2 \sqrt {a c}}\right )+b \,c^{2} x \ln \left (\frac {2 a c x +a d +b c +2 \sqrt {\left (a x +b \right ) \left (c x +d \right )}\, \sqrt {a c}}{2 \sqrt {a c}}\right )-3 a \,d^{2} \ln \left (\frac {2 a c x +a d +b c +2 \sqrt {\left (a x +b \right ) \left (c x +d \right )}\, \sqrt {a c}}{2 \sqrt {a c}}\right )+b c d \ln \left (\frac {2 a c x +a d +b c +2 \sqrt {\left (a x +b \right ) \left (c x +d \right )}\, \sqrt {a c}}{2 \sqrt {a c}}\right )+2 \sqrt {\left (a x +b \right ) \left (c x +d \right )}\, \sqrt {a c}\, c x +6 \sqrt {\left (a x +b \right ) \left (c x +d \right )}\, \sqrt {a c}\, d \right ) x}{2 \sqrt {a c}\, \left (c x +d \right ) \sqrt {\left (a x +b \right ) \left (c x +d \right )}\, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x)^(1/2)/(c+d/x)^(3/2),x)

[Out]

1/2*((a*x+b)/x)^(1/2)*x*((c*x+d)/x)^(1/2)*(-3*ln(1/2*(2*a*c*x+a*d+b*c+2*((a*x+b)*(c*x+d))^(1/2)*(a*c)^(1/2))/(
a*c)^(1/2))*x*a*c*d+ln(1/2*(2*a*c*x+a*d+b*c+2*((a*x+b)*(c*x+d))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2))*x*b*c^2+2*x*c*
((a*x+b)*(c*x+d))^(1/2)*(a*c)^(1/2)-3*ln(1/2*(2*a*c*x+a*d+b*c+2*((a*x+b)*(c*x+d))^(1/2)*(a*c)^(1/2))/(a*c)^(1/
2))*a*d^2+ln(1/2*(2*a*c*x+a*d+b*c+2*((a*x+b)*(c*x+d))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2))*b*c*d+6*d*((a*x+b)*(c*x+
d))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2)/(c*x+d)/((a*x+b)*(c*x+d))^(1/2)/c^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a + \frac {b}{x}}}{{\left (c + \frac {d}{x}\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^(1/2)/(c+d/x)^(3/2),x, algorithm="maxima")

[Out]

integrate(sqrt(a + b/x)/(c + d/x)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {a+\frac {b}{x}}}{{\left (c+\frac {d}{x}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x)^(1/2)/(c + d/x)^(3/2),x)

[Out]

int((a + b/x)^(1/2)/(c + d/x)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a + \frac {b}{x}}}{\left (c + \frac {d}{x}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)**(1/2)/(c+d/x)**(3/2),x)

[Out]

Integral(sqrt(a + b/x)/(c + d/x)**(3/2), x)

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