\(\int (a+\frac {b}{\sqrt {c+\frac {d}{x}}})^p \, dx\) [17]

Optimal result

Integrand size = 17, antiderivative size = 180 \[ \int \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p \, dx=\frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p (d+c x)}{c}-\frac {b d \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p \operatorname {Hypergeometric2F1}\left (1,p,1+p,-\frac {\sqrt {c} \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )}{b-a \sqrt {c}}\right )}{2 \left (b-a \sqrt {c}\right ) c}-\frac {b d \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p \operatorname {Hypergeometric2F1}\left (1,p,1+p,\frac {\sqrt {c} \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )}{b+a \sqrt {c}}\right )}{2 \left (b+a \sqrt {c}\right ) c} \] Output:

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.92 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.49, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {7268, 1894, 1803, 593, 27, 657, 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.

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

\(\Big \downarrow \) 7268

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

\(\Big \downarrow \) 1894

\(\displaystyle -2 d \int \frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p}{\left (\frac {c}{c+\frac {d}{x}}-1\right )^2 \left (c+\frac {d}{x}\right )^{3/2}}d\sqrt {c+\frac {d}{x}}\)

\(\Big \downarrow \) 1803

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

\(\Big \downarrow \) 593

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 657

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 593

Int[(x_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
 Simp[(c + d*x)^(n + 1)*(c - d*x)*((a + b*x^2)^(p + 1)/(2*(p + 1)*(b*c^2 + 
a*d^2))), x] - Simp[d/(2*(p + 1)*(b*c^2 + a*d^2))   Int[(c + d*x)^n*(a + b* 
x^2)^(p + 1)*(c*n - d*(n + 2*p + 4)*x), x], x] /; FreeQ[{a, b, c, d, n}, x] 
 && LtQ[p, -1] && NeQ[b*c^2 + a*d^2, 0]
 

rule 657

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( 
x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 
2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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