\(\int \sqrt {a+\frac {b}{\sqrt {\frac {c}{x}}}} (d x)^m \, dx\) [2997]

Optimal result

Integrand size = 23, antiderivative size = 78 \[ \int \sqrt {a+\frac {b}{\sqrt {\frac {c}{x}}}} (d x)^m \, dx=-\frac {4 a c \left (a+\frac {b}{\sqrt {\frac {c}{x}}}\right )^{3/2} \left (-\frac {b}{a \sqrt {\frac {c}{x}}}\right )^{-2 m} (d x)^m \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-1-2 m,\frac {5}{2},1+\frac {b}{a \sqrt {\frac {c}{x}}}\right )}{3 b^2} \]

[Out]

-4/3*a*c*(d*x)^m*hypergeom([3/2, -1-2*m],[5/2],1+b/a/(c/x)^(1/2))*(a+b/(c/x)^(1/2))^(3/2)/b^2/((-b/a/(c/x)^(1/
2))^(2*m))

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 78, 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} \[ \int \sqrt {a+\frac {b}{\sqrt {\frac {c}{x}}}} (d x)^m \, dx=-\frac {4 a c (d x)^m \left (a+\frac {b}{\sqrt {\frac {c}{x}}}\right )^{3/2} \left (-\frac {b}{a \sqrt {\frac {c}{x}}}\right )^{-2 m} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-2 m-1,\frac {5}{2},\frac {b}{a \sqrt {\frac {c}{x}}}+1\right )}{3 b^2} \]

[In]

Int[Sqrt[a + b/Sqrt[c/x]]*(d*x)^m,x]

[Out]

(-4*a*c*(a + b/Sqrt[c/x])^(3/2)*(d*x)^m*Hypergeometric2F1[3/2, -1 - 2*m, 5/2, 1 + b/(a*Sqrt[c/x])])/(3*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 \sqrt {a+\frac {b \sqrt {x}}{\sqrt {c}}} (d x)^m \, dx,\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right ) \\ & = \text {Subst}\left (\left (x^{-m} (d x)^m\right ) \int \sqrt {a+\frac {b \sqrt {x}}{\sqrt {c}}} x^m \, 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 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 \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 \left (a+\frac {b}{\sqrt {\frac {c}{x}}}\right )^{3/2} \left (-\frac {b}{a \sqrt {\frac {c}{x}}}\right )^{-2 m} (d x)^m \, _2F_1\left (\frac {3}{2},-1-2 m;\frac {5}{2};1+\frac {b}{a \sqrt {\frac {c}{x}}}\right )}{3 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.72 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.09 \[ \int \sqrt {a+\frac {b}{\sqrt {\frac {c}{x}}}} (d x)^m \, dx=\frac {4 \sqrt {a+\frac {b}{\sqrt {\frac {c}{x}}}} x (d x)^m \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {5}{2}-2 m,-\frac {3}{2}-2 m,-\frac {a \sqrt {\frac {c}{x}}}{b}\right )}{(5+4 m) \sqrt {1+\frac {a \sqrt {\frac {c}{x}}}{b}}} \]

[In]

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

[Out]

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

Maple [F]

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

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

[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]

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

[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]

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

[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]

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

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

[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)