\(\int \sqrt {a+b \sqrt {\frac {c}{x}}} (d x)^m \, dx\) [2996]

Optimal result

Integrand size = 23, antiderivative size = 60 \[ \int \sqrt {a+b \sqrt {\frac {c}{x}}} (d x)^m \, dx=\frac {4 \left (a+b \sqrt {\frac {c}{x}}\right )^{3/2} x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (-1-4 m),\frac {5}{2},\frac {a+b \sqrt {\frac {c}{x}}}{a}\right )}{3 a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.33, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {376, 350, 348, 346, 69, 67} \[ \int \sqrt {a+b \sqrt {\frac {c}{x}}} (d x)^m \, dx=\frac {4 b^2 c (d x)^m \left (a+b \sqrt {\frac {c}{x}}\right )^{3/2} \left (-\frac {b \sqrt {\frac {c}{x}}}{a}\right )^{2 m} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},2 m+3,\frac {5}{2},\frac {\sqrt {\frac {c}{x}} b}{a}+1\right )}{3 a^3} \]

[In]

Int[Sqrt[a + b*Sqrt[c/x]]*(d*x)^m,x]

[Out]

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

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 346

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(-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 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 {c}}{\sqrt {x}}} (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 {c}}{\sqrt {x}}} 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 \sqrt {a+\frac {b \sqrt {c}}{x}} x^{-1+2 (1+m)} \, dx,x,\sqrt {x}\right ),\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+b \sqrt {c} x} \, dx,x,\frac {1}{\sqrt {x}}\right ),\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right ) \\ & = \text {Subst}\left (\frac {\left (2 b^3 c^{3/2} \left (-\frac {b \sqrt {c}}{a \sqrt {x}}\right )^{2 m} (d x)^m\right ) \text {Subst}\left (\int \left (-\frac {b \sqrt {c} x}{a}\right )^{-1-2 (1+m)} \sqrt {a+b \sqrt {c} x} \, dx,x,\frac {1}{\sqrt {x}}\right )}{a^3},\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right ) \\ & = \frac {4 b^2 c \left (a+b \sqrt {\frac {c}{x}}\right )^{3/2} \left (-\frac {b \sqrt {\frac {c}{x}}}{a}\right )^{2 m} (d x)^m \, _2F_1\left (\frac {3}{2},3+2 m;\frac {5}{2};1+\frac {b \sqrt {\frac {c}{x}}}{a}\right )}{3 a^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.13 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.30 \[ \int \sqrt {a+b \sqrt {\frac {c}{x}}} (d x)^m \, dx=\frac {\sqrt {a+b \sqrt {\frac {c}{x}}} x (d x)^m \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-2 (1+m),-1-2 m,-\frac {b \sqrt {\frac {c}{x}}}{a}\right )}{(1+m) \sqrt {1+\frac {b \sqrt {\frac {c}{x}}}{a}}} \]

[In]

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

[Out]

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

Maple [F]

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

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

[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]

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

[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]

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

[In]

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

[Out]

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

Giac [F(-2)]

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

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{1,[0,1,1,0]%%%} / %%%{1,[0,0,0,1]%%%} Error: Bad Argumen
t Value

Mupad [F(-1)]

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

[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)