\(\int \frac {(a+\frac {b}{\sqrt {c+\frac {d}{x}}})^p}{x^4} \, dx\) [311]

Optimal result

Integrand size = 21, antiderivative size = 316 \[ \int \frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p}{x^4} \, dx=-\frac {b (3-p) \left (60 a^2 c-b^2 \left (20-9 p+p^2\right )\right ) \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^{1+p} \left (c+\frac {d}{x}\right )^{3/2}}{180 a^4 d^3}+\frac {\left (60 a^2 c-b^2 \left (20-9 p+p^2\right )\right ) \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^{1+p} \left (c+\frac {d}{x}\right )^2}{60 a^3 d^3}+\frac {b (5-p) \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^{1+p} \left (c+\frac {d}{x}\right )^{5/2}}{15 a^2 d^3}-\frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^{1+p} \left (c+\frac {d}{x}\right )^3}{3 a d^3}-\frac {b^2 \left (360 a^4 c^2-b^2 (2-p) (3-p) \left (60 a^2 c-b^2 \left (20-9 p+p^2\right )\right )\right ) \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^{1+p} \operatorname {Hypergeometric2F1}\left (3,1+p,2+p,1+\frac {b}{a \sqrt {c+\frac {d}{x}}}\right )}{180 a^7 d^3 (1+p)} \] Output:

-1/180*b*(3-p)*(60*a^2*c-b^2*(p^2-9*p+20))*(a+b/(c+d/x)^(1/2))^(p+1)*(c+d/ 
x)^(3/2)/a^4/d^3+1/60*(60*a^2*c-b^2*(p^2-9*p+20))*(a+b/(c+d/x)^(1/2))^(p+1 
)*(c+d/x)^2/a^3/d^3+1/15*b*(5-p)*(a+b/(c+d/x)^(1/2))^(p+1)*(c+d/x)^(5/2)/a 
^2/d^3-1/3*(a+b/(c+d/x)^(1/2))^(p+1)*(c+d/x)^3/a/d^3-1/180*b^2*(360*a^4*c^ 
2-b^2*(2-p)*(3-p)*(60*a^2*c-b^2*(p^2-9*p+20)))*(a+b/(c+d/x)^(1/2))^(p+1)*h 
ypergeom([3, p+1],[2+p],1+b/a/(c+d/x)^(1/2))/a^7/d^3/(p+1)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 4.05 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.50 \[ \int \frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p}{x^4} \, dx=-\frac {2 b^2 \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p \left (b+a \sqrt {c+\frac {d}{x}}\right ) \left (a^4 c^2 \operatorname {Hypergeometric2F1}\left (3,1+p,2+p,1+\frac {b}{a \sqrt {c+\frac {d}{x}}}\right )-2 a^2 b^2 c \operatorname {Hypergeometric2F1}\left (5,1+p,2+p,1+\frac {b}{a \sqrt {c+\frac {d}{x}}}\right )+b^4 \operatorname {Hypergeometric2F1}\left (7,1+p,2+p,1+\frac {b}{a \sqrt {c+\frac {d}{x}}}\right )\right )}{a^7 d^3 (1+p) \sqrt {c+\frac {d}{x}}} \] Input:

Integrate[(a + b/Sqrt[c + d/x])^p/x^4,x]
 

Output:

(-2*b^2*(a + b/Sqrt[c + d/x])^p*(b + a*Sqrt[c + d/x])*(a^4*c^2*Hypergeomet 
ric2F1[3, 1 + p, 2 + p, 1 + b/(a*Sqrt[c + d/x])] - 2*a^2*b^2*c*Hypergeomet 
ric2F1[5, 1 + p, 2 + p, 1 + b/(a*Sqrt[c + d/x])] + b^4*Hypergeometric2F1[7 
, 1 + p, 2 + p, 1 + b/(a*Sqrt[c + d/x])]))/(a^7*d^3*(1 + p)*Sqrt[c + d/x])
 

Rubi [A] (warning: unable to verify)

Time = 1.25 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {7268, 1894, 1803, 520, 2124, 25, 520, 87, 75}

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.

\(\displaystyle \int \frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p}{x^4} \, dx\)

\(\Big \downarrow \) 7268

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

\(\Big \downarrow \) 1894

\(\displaystyle -\frac {2 \int \left (\frac {c}{c+\frac {d}{x}}-1\right )^2 \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p \left (c+\frac {d}{x}\right )^{5/2}d\sqrt {c+\frac {d}{x}}}{d^3}\)

\(\Big \downarrow \) 1803

\(\displaystyle \frac {2 \int \frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p \left (1-c \left (c+\frac {d}{x}\right )\right )^2}{\left (c+\frac {d}{x}\right )^{7/2}}d\frac {1}{\sqrt {c+\frac {d}{x}}}}{d^3}\)

\(\Big \downarrow \) 520

\(\displaystyle \frac {2 \left (-\frac {\int \frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p \left (-6 a \left (c+\frac {d}{x}\right )^{3/2} c^2+\frac {12 a c}{\sqrt {c+\frac {d}{x}}}+b (5-p)\right )}{\left (c+\frac {d}{x}\right )^3}d\frac {1}{\sqrt {c+\frac {d}{x}}}}{6 a}-\frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^{p+1}}{6 a \left (c+\frac {d}{x}\right )^3}\right )}{d^3}\)

\(\Big \downarrow \) 2124

\(\displaystyle \frac {2 \left (-\frac {-\frac {\int -\frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p \left (60 c a^2-30 c^2 \left (c+\frac {d}{x}\right ) a^2-b^2 \left (p^2-9 p+20\right )\right )}{\left (c+\frac {d}{x}\right )^{5/2}}d\frac {1}{\sqrt {c+\frac {d}{x}}}}{5 a}-\frac {b (5-p) \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^{p+1}}{5 a \left (c+\frac {d}{x}\right )^{5/2}}}{6 a}-\frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^{p+1}}{6 a \left (c+\frac {d}{x}\right )^3}\right )}{d^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 \left (-\frac {\frac {\int \frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p \left (60 c a^2-30 c^2 \left (c+\frac {d}{x}\right ) a^2-b^2 \left (p^2-9 p+20\right )\right )}{\left (c+\frac {d}{x}\right )^{5/2}}d\frac {1}{\sqrt {c+\frac {d}{x}}}}{5 a}-\frac {b (5-p) \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^{p+1}}{5 a \left (c+\frac {d}{x}\right )^{5/2}}}{6 a}-\frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^{p+1}}{6 a \left (c+\frac {d}{x}\right )^3}\right )}{d^3}\)

\(\Big \downarrow \) 520

\(\displaystyle \frac {2 \left (-\frac {\frac {-\frac {\int \frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p \left (\frac {120 c^2 a^3}{\sqrt {c+\frac {d}{x}}}+b (3-p) \left (60 a^2 c-b^2 \left (p^2-9 p+20\right )\right )\right )}{\left (c+\frac {d}{x}\right )^2}d\frac {1}{\sqrt {c+\frac {d}{x}}}}{4 a}-\frac {\left (60 a^2 c-b^2 \left (p^2-9 p+20\right )\right ) \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^{p+1}}{4 a \left (c+\frac {d}{x}\right )^2}}{5 a}-\frac {b (5-p) \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^{p+1}}{5 a \left (c+\frac {d}{x}\right )^{5/2}}}{6 a}-\frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^{p+1}}{6 a \left (c+\frac {d}{x}\right )^3}\right )}{d^3}\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {2 \left (-\frac {\frac {-\frac {\frac {\left (360 a^4 c^2-b^2 (2-p) (3-p) \left (60 a^2 c-b^2 \left (p^2-9 p+20\right )\right )\right ) \int \frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p}{\left (c+\frac {d}{x}\right )^{3/2}}d\frac {1}{\sqrt {c+\frac {d}{x}}}}{3 a}-\frac {b (3-p) \left (60 a^2 c-b^2 \left (p^2-9 p+20\right )\right ) \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^{p+1}}{3 a \left (c+\frac {d}{x}\right )^{3/2}}}{4 a}-\frac {\left (60 a^2 c-b^2 \left (p^2-9 p+20\right )\right ) \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^{p+1}}{4 a \left (c+\frac {d}{x}\right )^2}}{5 a}-\frac {b (5-p) \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^{p+1}}{5 a \left (c+\frac {d}{x}\right )^{5/2}}}{6 a}-\frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^{p+1}}{6 a \left (c+\frac {d}{x}\right )^3}\right )}{d^3}\)

\(\Big \downarrow \) 75

\(\displaystyle \frac {2 \left (-\frac {\frac {-\frac {\left (60 a^2 c-b^2 \left (p^2-9 p+20\right )\right ) \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^{p+1}}{4 a \left (c+\frac {d}{x}\right )^2}-\frac {-\frac {b (3-p) \left (60 a^2 c-b^2 \left (p^2-9 p+20\right )\right ) \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^{p+1}}{3 a \left (c+\frac {d}{x}\right )^{3/2}}-\frac {b^2 \left (360 a^4 c^2-b^2 (2-p) (3-p) \left (60 a^2 c-b^2 \left (p^2-9 p+20\right )\right )\right ) \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^{p+1} \operatorname {Hypergeometric2F1}\left (3,p+1,p+2,\frac {b}{a \sqrt {c+\frac {d}{x}}}+1\right )}{3 a^4 (p+1)}}{4 a}}{5 a}-\frac {b (5-p) \left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^{p+1}}{5 a \left (c+\frac {d}{x}\right )^{5/2}}}{6 a}-\frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^{p+1}}{6 a \left (c+\frac {d}{x}\right )^3}\right )}{d^3}\)

Input:

Int[(a + b/Sqrt[c + d/x])^p/x^4,x]
 

Output:

(2*(-1/6*(a + b/Sqrt[c + d/x])^(1 + p)/(a*(c + d/x)^3) - (-1/5*(b*(5 - p)* 
(a + b/Sqrt[c + d/x])^(1 + p))/(a*(c + d/x)^(5/2)) + (-1/4*((60*a^2*c - b^ 
2*(20 - 9*p + p^2))*(a + b/Sqrt[c + d/x])^(1 + p))/(a*(c + d/x)^2) - (-1/3 
*(b*(3 - p)*(60*a^2*c - b^2*(20 - 9*p + p^2))*(a + b/Sqrt[c + d/x])^(1 + p 
))/(a*(c + d/x)^(3/2)) - (b^2*(360*a^4*c^2 - b^2*(2 - p)*(3 - p)*(60*a^2*c 
 - b^2*(20 - 9*p + p^2)))*(a + b/Sqrt[c + d/x])^(1 + p)*Hypergeometric2F1[ 
3, 1 + p, 2 + p, 1 + b/(a*Sqrt[c + d/x])])/(3*a^4*(1 + p)))/(4*a))/(5*a))/ 
(6*a)))/d^3
 

Defintions of rubi rules used

rule 25

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

rule 75

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x 
)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))*Hypergeometric2F1[-m, n + 1, n + 2, 1 + 
 d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (IntegerQ[m] 
 || GtQ[-d/(b*c), 0])
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 520

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), 
 x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2)^p, e*x, x], R = Pol 
ynomialRemainder[(a + b*x^2)^p, e*x, x]}, Simp[R*(e*x)^(m + 1)*((c + d*x)^( 
n + 1)/((m + 1)*(e*c))), x] + Simp[1/((m + 1)*(e*c))   Int[(e*x)^(m + 1)*(c 
 + d*x)^n*ExpandToSum[(m + 1)*(e*c)*Qx - d*R*(m + n + 2), x], x], x]] /; Fr 
eeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[m, -1] &&  !IntegerQ[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 2124

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] : 
> With[{Qx = PolynomialQuotient[Px, a + b*x, x], R = PolynomialRemainder[Px 
, a + b*x, x]}, Simp[R*(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((m + 1)*(b*c - 
 a*d))), x] + Simp[1/((m + 1)*(b*c - a*d))   Int[(a + b*x)^(m + 1)*(c + d*x 
)^n*ExpandToSum[(m + 1)*(b*c - a*d)*Qx - d*R*(m + n + 2), x], x], x]] /; Fr 
eeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && LtQ[m, -1] && (IntegerQ[m] ||  ! 
ILtQ[n, -1])
 

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]

\[\int \frac {\left (a +\frac {b}{\sqrt {c +\frac {d}{x}}}\right )^{p}}{x^{4}}d x\]

Input:

int((a+b/(c+d/x)^(1/2))^p/x^4,x)
 

Output:

int((a+b/(c+d/x)^(1/2))^p/x^4,x)
 

Fricas [F]

\[ \int \frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p}{x^4} \, dx=\int { \frac {{\left (a + \frac {b}{\sqrt {c + \frac {d}{x}}}\right )}^{p}}{x^{4}} \,d x } \] Input:

integrate((a+b/(c+d/x)^(1/2))^p/x^4,x, algorithm="fricas")
 

Output:

integral(((a*c*x + b*x*sqrt((c*x + d)/x) + a*d)/(c*x + d))^p/x^4, x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p}{x^4} \, dx=\text {Timed out} \] Input:

integrate((a+b/(c+d/x)**(1/2))**p/x**4,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p}{x^4} \, dx=\int { \frac {{\left (a + \frac {b}{\sqrt {c + \frac {d}{x}}}\right )}^{p}}{x^{4}} \,d x } \] Input:

integrate((a+b/(c+d/x)^(1/2))^p/x^4,x, algorithm="maxima")
 

Output:

integrate((a + b/sqrt(c + d/x))^p/x^4, x)
                                                                                    
                                                                                    
 

Giac [F]

\[ \int \frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p}{x^4} \, dx=\int { \frac {{\left (a + \frac {b}{\sqrt {c + \frac {d}{x}}}\right )}^{p}}{x^{4}} \,d x } \] Input:

integrate((a+b/(c+d/x)^(1/2))^p/x^4,x, algorithm="giac")
 

Output:

integrate((a + b/sqrt(c + d/x))^p/x^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p}{x^4} \, dx=\int \frac {{\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )}^p}{x^4} \,d x \] Input:

int((a + b/(c + d/x)^(1/2))^p/x^4,x)
 

Output:

int((a + b/(c + d/x)^(1/2))^p/x^4, x)
 

Reduce [F]

\[ \int \frac {\left (a+\frac {b}{\sqrt {c+\frac {d}{x}}}\right )^p}{x^4} \, dx=\int \frac {\left (\sqrt {c x +d}\, a +\sqrt {x}\, b \right )^{p}}{\left (c x +d \right )^{\frac {p}{2}} x^{4}}d x \] Input:

int((a+b/(c+d/x)^(1/2))^p/x^4,x)
 

Output:

int((sqrt(c*x + d)*a + sqrt(x)*b)**p/((c*x + d)**(p/2)*x**4),x)