\(\int (d x)^m \sqrt {a+\frac {b}{(c x)^{3/2}}} \, dx\) [16]

Optimal result

Integrand size = 21, antiderivative size = 82 \[ \int (d x)^m \sqrt {a+\frac {b}{(c x)^{3/2}}} \, dx=\frac {(d x)^{1+m} \sqrt {a+\frac {b}{(c x)^{3/2}}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {2}{3} (1+m),\frac {1}{3} (1-2 m),-\frac {b}{a (c x)^{3/2}}\right )}{d (1+m) \sqrt {1+\frac {b}{a (c x)^{3/2}}}} \] Output:

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

Mathematica [A] (verified)

Time = 1.79 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.02 \[ \int (d x)^m \sqrt {a+\frac {b}{(c x)^{3/2}}} \, dx=\frac {4 x (d x)^m \sqrt {a+\frac {b}{(c x)^{3/2}}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{6} (1+4 m),\frac {1}{6} (7+4 m),-\frac {a (c x)^{3/2}}{b}\right )}{(1+4 m) \sqrt {\frac {b+a (c x)^{3/2}}{b}}} \] Input:

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

Output:

(4*x*(d*x)^m*Sqrt[a + b/(c*x)^(3/2)]*Hypergeometric2F1[-1/2, (1 + 4*m)/6, 
(7 + 4*m)/6, -((a*(c*x)^(3/2))/b)])/((1 + 4*m)*Sqrt[(b + a*(c*x)^(3/2))/b] 
)
 

Rubi [A] (warning: unable to verify)

Time = 0.49 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.18, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {891, 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.

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

\(\Big \downarrow \) 891

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

\(\Big \downarrow \) 866

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

\(\Big \downarrow \) 864

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

\(\Big \downarrow \) 862

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

\(\Big \downarrow \) 889

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

\(\Big \downarrow \) 888

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

Input:

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

Output:

((1/(c*x))^(2*m)*(c*x)^(-1 - m)*(d*x)^m*Sqrt[a + b*c^3*x^3]*Hypergeometric 
2F1[-1/2, (-2*(1 + m))/3, (1 - 2*m)/3, -((b*c^3*x^3)/a)])/(c*(1 + m)*Sqrt[ 
1 + (b*c^3*x^3)/a])
 

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 891

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_)*(x_))^(n_))^(p_.), x_Symbol] :> 
Simp[1/c   Subst[Int[(d*(x/c))^m*(a + b*x^n)^p, x], x, c*x], x] /; FreeQ[{a 
, b, c, d, m, n, p}, x]
 

Maple [F]

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

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

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

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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