\(\int (a+b \sqrt {c+\frac {d}{x}})^p \, dx\) [301]

Optimal result

Integrand size = 17, antiderivative size = 166 \[ \int \left (a+b \sqrt {c+\frac {d}{x}}\right )^p \, dx=\left (a+b \sqrt {c+\frac {d}{x}}\right )^p x-\frac {b d \left (a+b \sqrt {c+\frac {d}{x}}\right )^p \operatorname {Hypergeometric2F1}\left (1,p,1+p,\frac {a+b \sqrt {c+\frac {d}{x}}}{a-b \sqrt {c}}\right )}{2 \left (a-b \sqrt {c}\right ) \sqrt {c}}+\frac {b d \left (a+b \sqrt {c+\frac {d}{x}}\right )^p \operatorname {Hypergeometric2F1}\left (1,p,1+p,\frac {a+b \sqrt {c+\frac {d}{x}}}{a+b \sqrt {c}}\right )}{2 \left (a+b \sqrt {c}\right ) \sqrt {c}} \] Output:

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.17 \[ \int \left (a+b \sqrt {c+\frac {d}{x}}\right )^p \, dx=\frac {\left (a+b \sqrt {c+\frac {d}{x}}\right )^{1+p} \left (-b \left (a+b \sqrt {c}\right )^2 d p \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {a+b \sqrt {c+\frac {d}{x}}}{a-b \sqrt {c}}\right )+\left (a-b \sqrt {c}\right ) \left (2 \left (a+b \sqrt {c}\right ) \sqrt {c} (1+p) \left (a-b \sqrt {c+\frac {d}{x}}\right ) x+b \left (a-b \sqrt {c}\right ) d p \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {a+b \sqrt {c+\frac {d}{x}}}{a+b \sqrt {c}}\right )\right )\right )}{2 \sqrt {c} \left (a^2-b^2 c\right )^2 (1+p)} \] Input:

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

Output:

((a + b*Sqrt[c + d/x])^(1 + p)*(-(b*(a + b*Sqrt[c])^2*d*p*Hypergeometric2F 
1[1, 1 + p, 2 + p, (a + b*Sqrt[c + d/x])/(a - b*Sqrt[c])]) + (a - b*Sqrt[c 
])*(2*(a + b*Sqrt[c])*Sqrt[c]*(1 + p)*(a - b*Sqrt[c + d/x])*x + b*(a - b*S 
qrt[c])*d*p*Hypergeometric2F1[1, 1 + p, 2 + p, (a + b*Sqrt[c + d/x])/(a + 
b*Sqrt[c])])))/(2*Sqrt[c]*(a^2 - b^2*c)^2*(1 + p))
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.55, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {7268, 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.

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

\(\Big \downarrow \) 7268

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

\(\Big \downarrow \) 593

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 657

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

\(\Big \downarrow \) 2009

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

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)*x)/((a^2 - 
 b^2*c)*d) - (b*p*(-1/2*((a + b*Sqrt[c])*(a + b*Sqrt[c + d/x])^(1 + p)*Hyp 
ergeometric2F1[1, 1 + p, 2 + p, (a + b*Sqrt[c + d/x])/(a - b*Sqrt[c])])/(( 
a - b*Sqrt[c])*Sqrt[c]*(1 + p)) + ((a - b*Sqrt[c])*(a + b*Sqrt[c + d/x])^( 
1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, (a + b*Sqrt[c + d/x])/(a + b*Sqr 
t[c])])/(2*(a + b*Sqrt[c])*Sqrt[c]*(1 + p))))/(2*(a^2 - b^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 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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