3.2997 \(\int \sqrt{a+\frac{b}{\sqrt{\frac{c}{x}}}} (d x)^m \, dx\)

Optimal. Leaf size=78 \[ -\frac{4 a c (d x)^m \left (a+\frac{b}{\sqrt{\frac{c}{x}}}\right )^{3/2} \left (-\frac{b}{a \sqrt{\frac{c}{x}}}\right )^{-2 m} \, _2F_1\left (\frac{3}{2},-2 m-1;\frac{5}{2};\frac{b}{a \sqrt{\frac{c}{x}}}+1\right )}{3 b^2} \]

[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))

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Rubi [A]  time = 0.0760735, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {369, 343, 341, 67, 65} \[ -\frac{4 a c (d x)^m \left (a+\frac{b}{\sqrt{\frac{c}{x}}}\right )^{3/2} \left (-\frac{b}{a \sqrt{\frac{c}{x}}}\right )^{-2 m} \, _2F_1\left (\frac{3}{2},-2 m-1;\frac{5}{2};\frac{b}{a \sqrt{\frac{c}{x}}}+1\right )}{3 b^2} \]

Antiderivative was successfully verified.

[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 369

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]

Rule 343

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

Rule 341

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 67

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 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rubi steps

\begin{align*} \int \sqrt{a+\frac{b}{\sqrt{\frac{c}{x}}}} (d x)^m \, dx &=\operatorname{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 )\\ &=\operatorname{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 )\\ &=\operatorname{Subst}\left (\left (2 x^{-m} (d x)^m\right ) \operatorname{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 )\\ &=-\operatorname{Subst}\left (\frac{\left (2 a \sqrt{c} \left (-\frac{b \sqrt{x}}{a \sqrt{c}}\right )^{-2 m} (d x)^m\right ) \operatorname{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]  time = 0.106954, size = 85, normalized size = 1.09 \[ \frac{4 x (d x)^m \sqrt{a+\frac{b}{\sqrt{\frac{c}{x}}}} \, _2F_1\left (-\frac{1}{2},-2 m-\frac{5}{2};-2 m-\frac{3}{2};-\frac{a \sqrt{\frac{c}{x}}}{b}\right )}{(4 m+5) \sqrt{\frac{a \sqrt{\frac{c}{x}}}{b}+1}} \]

Antiderivative was successfully verified.

[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])

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Maple [F]  time = 0.055, size = 0, normalized size = 0. \begin{align*} \int \left ( dx \right ) ^{m}\sqrt{a+{b{\frac{1}{\sqrt{{\frac{c}{x}}}}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d x\right )^{m} \sqrt{a + \frac{b}{\sqrt{\frac{c}{x}}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d x\right )^{m} \sqrt{a + \frac{b}{\sqrt{\frac{c}{x}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d x\right )^{m} \sqrt{a + \frac{b}{\sqrt{\frac{c}{x}}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)