\(\int (d x)^m \sqrt {a+\frac {b}{\sqrt {c x}}} \, dx\) [2921]

Optimal result

Integrand size = 21, antiderivative size = 76 \[ \int (d x)^m \sqrt {a+\frac {b}{\sqrt {c x}}} \, dx=\frac {4 b^2 (d x)^m \left (-\frac {b}{a \sqrt {c x}}\right )^{2 m} \left (a+\frac {b}{\sqrt {c x}}\right )^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},3+2 m,\frac {5}{2},1+\frac {b}{a \sqrt {c x}}\right )}{3 a^3 c} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {374, 350, 348, 346, 69, 67} \[ \int (d x)^m \sqrt {a+\frac {b}{\sqrt {c x}}} \, dx=\frac {4 b^2 (d x)^m \left (a+\frac {b}{\sqrt {c x}}\right )^{3/2} \left (-\frac {b}{a \sqrt {c x}}\right )^{2 m} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},2 m+3,\frac {5}{2},\frac {b}{a \sqrt {c x}}+1\right )}{3 a^3 c} \]

[In]

Int[(d*x)^m*Sqrt[a + b/Sqrt[c*x]],x]

[Out]

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

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 374

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_)*(x_))^(n_))^(p_.), x_Symbol] :> Dist[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]

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

Mathematica [A] (verified)

Time = 1.36 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.78 \[ \int (d x)^m \sqrt {a+\frac {b}{\sqrt {c x}}} \, dx=\frac {4 (d x)^m \left (-\frac {a \sqrt {c x}}{b}\right )^{\frac {1}{2}-2 m} \sqrt {a+\frac {b}{\sqrt {c x}}} \left (b+a \sqrt {c x}\right ) \left (-5 b \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1}{2}-2 m,\frac {5}{2},1+\frac {a \sqrt {c x}}{b}\right )+3 \left (b+a \sqrt {c x}\right ) \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{2}-2 m,\frac {7}{2},1+\frac {a \sqrt {c x}}{b}\right )\right )}{15 a^2 c} \]

[In]

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

[Out]

(4*(d*x)^m*(-((a*Sqrt[c*x])/b))^(1/2 - 2*m)*Sqrt[a + b/Sqrt[c*x]]*(b + a*Sqrt[c*x])*(-5*b*Hypergeometric2F1[3/
2, 1/2 - 2*m, 5/2, 1 + (a*Sqrt[c*x])/b] + 3*(b + a*Sqrt[c*x])*Hypergeometric2F1[5/2, 1/2 - 2*m, 7/2, 1 + (a*Sq
rt[c*x])/b]))/(15*a^2*c)

Maple [F]

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

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

[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]

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

[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]

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

[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(-2)]

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

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

[In]

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

[Out]

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