\(\int \frac {(d x)^m}{\sqrt {a+\frac {b}{\sqrt {\frac {c}{x}}}}} \, dx\) [3001]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F(-2)]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 23, antiderivative size = 76 \[ \int \frac {(d x)^m}{\sqrt {a+\frac {b}{\sqrt {\frac {c}{x}}}}} \, dx=-\frac {4 a c \sqrt {a+\frac {b}{\sqrt {\frac {c}{x}}}} \left (-\frac {b}{a \sqrt {\frac {c}{x}}}\right )^{-2 m} (d x)^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-1-2 m,\frac {3}{2},1+\frac {b}{a \sqrt {\frac {c}{x}}}\right )}{b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {376, 350, 348, 69, 67} \[ \int \frac {(d x)^m}{\sqrt {a+\frac {b}{\sqrt {\frac {c}{x}}}}} \, dx=-\frac {4 a c (d x)^m \sqrt {a+\frac {b}{\sqrt {\frac {c}{x}}}} \left (-\frac {b}{a \sqrt {\frac {c}{x}}}\right )^{-2 m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-2 m-1,\frac {3}{2},\frac {b}{a \sqrt {\frac {c}{x}}}+1\right )}{b^2} \]

[In]

Int[(d*x)^m/Sqrt[a + b/Sqrt[c/x]],x]

[Out]

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

Rule 67

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] && (Intege
rQ[m] || GtQ[-d/(b*c), 0])

Rule 69

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

Rule 348

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Dist[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 350

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

Rule 376

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbol] :> With[{k = Denominator[n]}, Su
bst[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]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {(d x)^m}{\sqrt {a+\frac {b \sqrt {x}}{\sqrt {c}}}} \, dx,\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right ) \\ & = \text {Subst}\left (\left (x^{-m} (d x)^m\right ) \int \frac {x^m}{\sqrt {a+\frac {b \sqrt {x}}{\sqrt {c}}}} \, dx,\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right ) \\ & = \text {Subst}\left (\left (2 x^{-m} (d x)^m\right ) \text {Subst}\left (\int \frac {x^{-1+2 (1+m)}}{\sqrt {a+\frac {b x}{\sqrt {c}}}} \, dx,x,\sqrt {x}\right ),\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right ) \\ & = -\text {Subst}\left (\frac {\left (2 a \sqrt {c} \left (-\frac {b \sqrt {x}}{a \sqrt {c}}\right )^{-2 m} (d x)^m\right ) \text {Subst}\left (\int \frac {\left (-\frac {b x}{a \sqrt {c}}\right )^{-1+2 (1+m)}}{\sqrt {a+\frac {b x}{\sqrt {c}}}} \, dx,x,\sqrt {x}\right )}{b},\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right ) \\ & = -\frac {4 a c \sqrt {a+\frac {b}{\sqrt {\frac {c}{x}}}} \left (-\frac {b}{a \sqrt {\frac {c}{x}}}\right )^{-2 m} (d x)^m \, _2F_1\left (\frac {1}{2},-1-2 m;\frac {3}{2};1+\frac {b}{a \sqrt {\frac {c}{x}}}\right )}{b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.02 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.53 \[ \int \frac {(d x)^m}{\sqrt {a+\frac {b}{\sqrt {\frac {c}{x}}}}} \, dx=\frac {a^2 c \left (\frac {a \sqrt {\frac {c}{x}}}{b+a \sqrt {\frac {c}{x}}}\right )^{-\frac {1}{2}+2 m} (d x)^m \operatorname {Hypergeometric2F1}\left (2+2 m,\frac {5}{2}+2 m,3+2 m,\frac {b}{b+a \sqrt {\frac {c}{x}}}\right )}{(1+m) \sqrt {a+\frac {b}{\sqrt {\frac {c}{x}}}} \left (b+a \sqrt {\frac {c}{x}}\right )^2} \]

[In]

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

[Out]

(a^2*c*((a*Sqrt[c/x])/(b + a*Sqrt[c/x]))^(-1/2 + 2*m)*(d*x)^m*Hypergeometric2F1[2 + 2*m, 5/2 + 2*m, 3 + 2*m, b
/(b + a*Sqrt[c/x])])/((1 + m)*Sqrt[a + b/Sqrt[c/x]]*(b + a*Sqrt[c/x])^2)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F(-2)]

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

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   alglogextint: unimplemented

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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