\(\int \frac {(a+\frac {b}{\sqrt {c+\frac {d}{x}}})^p}{x} \, dx\) [308]

Optimal result

Integrand size = 21, antiderivative size = 222 \[ \int \frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p}{x} \, dx=-\frac {\sqrt {c} \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,-\frac {\sqrt {c} \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )}{b-a \sqrt {c}}\right )}{\left (b-a \sqrt {c}\right ) (1+p)}+\frac {\sqrt {c} \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {\sqrt {c} \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )}{b+a \sqrt {c}}\right )}{\left (b+a \sqrt {c}\right ) (1+p)}-\frac {2 \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b}{a \sqrt {c+\frac {d}{x}}}\right )}{a (1+p)} \] Output:

-c^(1/2)*(a+b/(c+d/x)^(1/2))^(p+1)*hypergeom([1, p+1],[2+p],-c^(1/2)*(a+b/ 
(c+d/x)^(1/2))/(b-a*c^(1/2)))/(b-a*c^(1/2))/(p+1)+c^(1/2)*(a+b/(c+d/x)^(1/ 
2))^(p+1)*hypergeom([1, p+1],[2+p],c^(1/2)*(a+b/(c+d/x)^(1/2))/(b+a*c^(1/2 
)))/(b+a*c^(1/2))/(p+1)-2*(a+b/(c+d/x)^(1/2))^(p+1)*hypergeom([1, p+1],[2+ 
p],1+b/a/(c+d/x)^(1/2))/a/(p+1)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 2.10 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.79 \[ \int \frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p}{x} \, dx=-\frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^{1+p} \left (\frac {\sqrt {c} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,-\frac {\sqrt {c} \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )}{b-a \sqrt {c}}\right )}{b-a \sqrt {c}}-\frac {\sqrt {c} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {\sqrt {c} \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )}{b+a \sqrt {c}}\right )}{b+a \sqrt {c}}+\frac {2 \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b}{a \sqrt {c+\frac {d}{x}}}\right )}{a}\right )}{1+p} \] Input:

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

Output:

-(((a + b/Sqrt[c + d/x])^(1 + p)*((Sqrt[c]*Hypergeometric2F1[1, 1 + p, 2 + 
 p, -((Sqrt[c]*(a + b/Sqrt[c + d/x]))/(b - a*Sqrt[c]))])/(b - a*Sqrt[c]) - 
 (Sqrt[c]*Hypergeometric2F1[1, 1 + p, 2 + p, (Sqrt[c]*(a + b/Sqrt[c + d/x] 
))/(b + a*Sqrt[c])])/(b + a*Sqrt[c]) + (2*Hypergeometric2F1[1, 1 + p, 2 + 
p, 1 + b/(a*Sqrt[c + d/x])])/a))/(1 + p))
 

Rubi [A] (verified)

Time = 1.08 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {7268, 25, 1894, 1803, 25, 615, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p}{x} \, dx\)

\(\Big \downarrow \) 7268

\(\displaystyle -2 \int \frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p \sqrt {c+\frac {d}{x}} x}{d}d\sqrt {c+\frac {d}{x}}\)

\(\Big \downarrow \) 25

\(\displaystyle 2 \int -\frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p \sqrt {c+\frac {d}{x}} x}{d}d\sqrt {c+\frac {d}{x}}\)

\(\Big \downarrow \) 1894

\(\displaystyle 2 \int \frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p}{\left (\frac {c}{c+\frac {d}{x}}-1\right ) \sqrt {c+\frac {d}{x}}}d\sqrt {c+\frac {d}{x}}\)

\(\Big \downarrow \) 1803

\(\displaystyle -2 \int -\frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p}{\left (1-c \left (c+\frac {d}{x}\right )\right ) \sqrt {c+\frac {d}{x}}}d\frac {1}{\sqrt {c+\frac {d}{x}}}\)

\(\Big \downarrow \) 25

\(\displaystyle 2 \int \frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p}{\left (1-c \left (c+\frac {d}{x}\right )\right ) \sqrt {c+\frac {d}{x}}}d\frac {1}{\sqrt {c+\frac {d}{x}}}\)

\(\Big \downarrow \) 615

\(\displaystyle 2 \int \left (\frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p}{\sqrt {c+\frac {d}{x}}}-\frac {c \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p}{\left (c \left (c+\frac {d}{x}\right )-1\right ) \sqrt {c+\frac {d}{x}}}\right )d\frac {1}{\sqrt {c+\frac {d}{x}}}\)

\(\Big \downarrow \) 2009

\(\displaystyle -2 \left (\frac {\sqrt {c} \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,-\frac {\sqrt {c} \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )}{b-a \sqrt {c}}\right )}{2 (p+1) \left (b-a \sqrt {c}\right )}-\frac {\sqrt {c} \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {\sqrt {c} \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )}{\sqrt {c} a+b}\right )}{2 (p+1) \left (a \sqrt {c}+b\right )}+\frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b}{a \sqrt {c+\frac {d}{x}}}+1\right )}{a (p+1)}\right )\)

Input:

Int[(a + b/Sqrt[c + d/x])^p/x,x]
 

Output:

-2*((Sqrt[c]*(a + b/Sqrt[c + d/x])^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + 
 p, -((Sqrt[c]*(a + b/Sqrt[c + d/x]))/(b - a*Sqrt[c]))])/(2*(b - a*Sqrt[c] 
)*(1 + p)) - (Sqrt[c]*(a + b/Sqrt[c + d/x])^(1 + p)*Hypergeometric2F1[1, 1 
 + p, 2 + p, (Sqrt[c]*(a + b/Sqrt[c + d/x]))/(b + a*Sqrt[c])])/(2*(b + a*S 
qrt[c])*(1 + p)) + ((a + b/Sqrt[c + d/x])^(1 + p)*Hypergeometric2F1[1, 1 + 
 p, 2 + p, 1 + b/(a*Sqrt[c + d/x])])/(a*(1 + p)))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 615

Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), 
 x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] 
 /; FreeQ[{a, b, c, d, e, m, n}, x] && ILtQ[p, 0]
 

rule 1803

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q 
_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(d + e*x 
)^q*(a + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, c, d, e, m, n, p, q}, x] && 
 EqQ[n2, 2*n] && IntegerQ[Simplify[(m + 1)/n]]
 

rule 1894

Int[(x_)^(m_.)*((a_.) + (c_.)*(x_)^(mn2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_.)) 
^(q_.), x_Symbol] :> Int[x^(m - 2*n*p)*(d + e*x^n)^q*(c + a*x^(2*n))^p, x] 
/; FreeQ[{a, c, d, e, m, n, q}, x] && EqQ[mn2, -2*n] && IntegerQ[p]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7268

Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfQuotientOfLinears 
[u, x]}, Simp[lst[[2]]*lst[[4]]   Subst[Int[lst[[1]], x], x, lst[[3]]^(1/ls 
t[[2]])], x] /;  !FalseQ[lst]]
 

Maple [F]

\[\int \frac {\left (a +\frac {b}{\sqrt {c +\frac {d}{x}}}\right )^{p}}{x}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int \frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p}{x} \, dx=\int { \frac {{\left (a + \frac {b}{\sqrt {c + \frac {d}{x}}}\right )}^{p}}{x} \,d x } \] Input:

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

Output:

integral(((a*c*x + b*x*sqrt((c*x + d)/x) + a*d)/(c*x + d))^p/x, x)
 

Sympy [F]

\[ \int \frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p}{x} \, dx=\int \frac {\left (a + \frac {b}{\sqrt {c + \frac {d}{x}}}\right )^{p}}{x}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p}{x} \, dx=\int { \frac {{\left (a + \frac {b}{\sqrt {c + \frac {d}{x}}}\right )}^{p}}{x} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p}{x} \, dx=\int { \frac {{\left (a + \frac {b}{\sqrt {c + \frac {d}{x}}}\right )}^{p}}{x} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p}{x} \, dx=\int \frac {{\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )}^p}{x} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p}{x} \, dx=\int \frac {\left (\sqrt {c x +d}\, a +\sqrt {x}\, b \right )^{p}}{\left (c x +d \right )^{\frac {p}{2}} x}d x \] Input:

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

Output:

int((sqrt(c*x + d)*a + sqrt(x)*b)**p/((c*x + d)**(p/2)*x),x)