\(\int \frac {(d x)^m}{\sqrt {a+b (\frac {c}{x})^{3/2}}} \, dx\) [97]

Optimal result

Integrand size = 23, antiderivative size = 88 \[ \int \frac {(d x)^m}{\sqrt {a+b \left (\frac {c}{x}\right )^{3/2}}} \, dx=\frac {\sqrt {1+\frac {b \left (\frac {c}{x}\right )^{3/2}}{a}} (d x)^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {2}{3} (1+m),\frac {1}{3} (1-2 m),-\frac {b \left (\frac {c}{x}\right )^{3/2}}{a}\right )}{d (1+m) \sqrt {a+b \left (\frac {c}{x}\right )^{3/2}}} \] Output:

(1+b*(c/x)^(3/2)/a)^(1/2)*(d*x)^(1+m)*hypergeom([1/2, -2/3-2/3*m],[1/3-2/3 
*m],-b*(c/x)^(3/2)/a)/d/(1+m)/(a+b*(c/x)^(3/2))^(1/2)
                                                                                    
                                                                                    
 

Mathematica [F]

\[ \int \frac {(d x)^m}{\sqrt {a+b \left (\frac {c}{x}\right )^{3/2}}} \, dx=\int \frac {(d x)^m}{\sqrt {a+b \left (\frac {c}{x}\right )^{3/2}}} \, dx \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.33 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.65, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {893, 866, 864, 862, 889, 888}

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.

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

\(\Big \downarrow \) 893

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

\(\Big \downarrow \) 866

\(\displaystyle x^{-m} (d x)^m \int \frac {x^m}{\sqrt {\frac {b c^{3/2}}{x^{3/2}}+a}}dx\)

\(\Big \downarrow \) 864

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

\(\Big \downarrow \) 862

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

\(\Big \downarrow \) 889

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

\(\Big \downarrow \) 888

\(\displaystyle \frac {x^{-m} \left (\frac {\sqrt {c}}{x \sqrt {\frac {c}{x}}}\right )^{2 m-2 (m+1)} \left (\frac {x \sqrt {\frac {c}{x}}}{\sqrt {c}}\right )^{2 m} (d x)^m \sqrt {\frac {b x^3 \left (\frac {c}{x}\right )^{3/2}}{a}+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {2}{3} (m+1),\frac {1}{3} (1-2 m),-\frac {b \left (\frac {c}{x}\right )^{3/2} x^3}{a}\right )}{(m+1) \sqrt {a+b x^3 \left (\frac {c}{x}\right )^{3/2}}}\)

Input:

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

Output:

((Sqrt[c]/(Sqrt[c/x]*x))^(2*m - 2*(1 + m))*(d*x)^m*((Sqrt[c/x]*x)/Sqrt[c]) 
^(2*m)*Sqrt[1 + (b*(c/x)^(3/2)*x^3)/a]*Hypergeometric2F1[1/2, (-2*(1 + m)) 
/3, (1 - 2*m)/3, -((b*(c/x)^(3/2)*x^3)/a)])/((1 + m)*x^m*Sqrt[a + b*(c/x)^ 
(3/2)*x^3])
 

Defintions of rubi rules used

rule 862

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-c^ 
(-1))*(c*x)^(m + 1)*(1/x)^(m + 1)   Subst[Int[(a + b/x^n)^p/x^(m + 2), x], 
x, 1/x], x] /; FreeQ[{a, b, c, m, p}, x] && ILtQ[n, 0] &&  !RationalQ[m]
 

rule 864

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denomi 
nator[n]}, Simp[k   Subst[Int[x^(k*(m + 1) - 1)*(a + b*x^(k*n))^p, x], x, x 
^(1/k)], x]] /; FreeQ[{a, b, m, p}, x] && FractionQ[n]
 

rule 866

Int[((c_)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^Int 
Part[m]*((c*x)^FracPart[m]/x^FracPart[m])   Int[x^m*(a + b*x^n)^p, x], x] / 
; FreeQ[{a, b, c, m, p}, x] && FractionQ[n]
 

rule 888

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p 
*((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 
, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] && (ILt 
Q[p, 0] || GtQ[a, 0])
 

rule 889

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^I 
ntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p])   Int[(c*x) 
^m*(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0 
] &&  !(ILtQ[p, 0] || GtQ[a, 0])
 

rule 893

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbo 
l] :> With[{k = Denominator[n]}, Subst[Int[(d*x)^m*(a + b*c^n*x^(n*q))^p, x 
], x^(1/k), (c*x^q)^(1/k)/(c^(1/k)*(x^(1/k))^(q - 1))]] /; FreeQ[{a, b, c, 
d, m, p, q}, x] && FractionQ[n]
 

Maple [F]

\[\int \frac {\left (d x \right )^{m}}{\sqrt {a +b \left (\frac {c}{x}\right )^{\frac {3}{2}}}}d x\]

Input:

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

Output:

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

Fricas [F(-2)]

Exception generated. \[ \int \frac {(d x)^m}{\sqrt {a+b \left (\frac {c}{x}\right )^{3/2}}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >>  Error detected within library code:   algl 
ogextint: unimplemented
 

Sympy [F]

\[ \int \frac {(d x)^m}{\sqrt {a+b \left (\frac {c}{x}\right )^{3/2}}} \, dx=\int \frac {\left (d x\right )^{m}}{\sqrt {a + b \left (\frac {c}{x}\right )^{\frac {3}{2}}}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {(d x)^m}{\sqrt {a+b \left (\frac {c}{x}\right )^{3/2}}} \, dx=\int { \frac {\left (d x\right )^{m}}{\sqrt {b \left (\frac {c}{x}\right )^{\frac {3}{2}} + a}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {(d x)^m}{\sqrt {a+b \left (\frac {c}{x}\right )^{3/2}}} \, dx=\int { \frac {\left (d x\right )^{m}}{\sqrt {b \left (\frac {c}{x}\right )^{\frac {3}{2}} + a}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(d x)^m}{\sqrt {a+b \left (\frac {c}{x}\right )^{3/2}}} \, dx=\int \frac {{\left (d\,x\right )}^m}{\sqrt {a+b\,{\left (\frac {c}{x}\right )}^{3/2}}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {(d x)^m}{\sqrt {a+b \left (\frac {c}{x}\right )^{3/2}}} \, dx=\frac {d^{m} \left (4 x^{m +\frac {1}{4}} \sqrt {\sqrt {c}\, b c +\sqrt {x}\, a x}-4 \left (\int \frac {x^{m +\frac {1}{4}} \sqrt {\sqrt {c}\, b c +\sqrt {x}\, a x}}{x}d x \right ) m -\left (\int \frac {x^{m +\frac {1}{4}} \sqrt {\sqrt {c}\, b c +\sqrt {x}\, a x}}{x}d x \right )\right )}{3 a} \] Input:

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

Output:

(d**m*(4*x**((4*m + 1)/4)*sqrt(sqrt(c)*b*c + sqrt(x)*a*x) - 4*int((x**((4* 
m + 1)/4)*sqrt(sqrt(c)*b*c + sqrt(x)*a*x))/x,x)*m - int((x**((4*m + 1)/4)* 
sqrt(sqrt(c)*b*c + sqrt(x)*a*x))/x,x)))/(3*a)