3.2995 \(\int \sqrt{a+b (\frac{c}{x})^{3/2}} (d x)^m \, dx\)

Optimal. Leaf size=102 \[ \frac{x (d x)^m \sqrt{a+\frac{b c^3}{x^3 \left (\frac{c}{x}\right )^{3/2}}} \, _2F_1\left (-\frac{1}{2},-\frac{2}{3} (m+1);\frac{1}{3} (1-2 m);-\frac{b c^3}{a \left (\frac{c}{x}\right )^{3/2} x^3}\right )}{(m+1) \sqrt{\frac{b c^3}{a x^3 \left (\frac{c}{x}\right )^{3/2}}+1}} \]

[Out]

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

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Rubi [A]  time = 0.105256, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {369, 343, 341, 339, 365, 364} \[ \frac{x (d x)^m \sqrt{a+\frac{b c^3}{x^3 \left (\frac{c}{x}\right )^{3/2}}} \, _2F_1\left (-\frac{1}{2},-\frac{2}{3} (m+1);\frac{1}{3} (1-2 m);-\frac{b c^3}{a \left (\frac{c}{x}\right )^{3/2} x^3}\right )}{(m+1) \sqrt{\frac{b c^3}{a x^3 \left (\frac{c}{x}\right )^{3/2}}+1}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*(c/x)^(3/2)]*(d*x)^m,x]

[Out]

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

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 339

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Dist[((c*x)^(m + 1)*(1/x)^(m + 1))/c, 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 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

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

Mathematica [F]  time = 0.126196, size = 0, normalized size = 0. \[ \int \sqrt{a+b \left (\frac{c}{x}\right )^{3/2}} (d x)^m \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Sqrt[a + b*(c/x)^(3/2)]*(d*x)^m,x]

[Out]

Integrate[Sqrt[a + b*(c/x)^(3/2)]*(d*x)^m, x]

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Maple [F]  time = 0.05, size = 0, normalized size = 0. \begin{align*} \int \left ( dx \right ) ^{m}\sqrt{a+b \left ({\frac{c}{x}} \right ) ^{{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m*(a+b*(c/x)^(3/2))^(1/2),x)

[Out]

int((d*x)^m*(a+b*(c/x)^(3/2))^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \left (\frac{c}{x}\right )^{\frac{3}{2}} + a} \left (d x\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*(a+b*(c/x)^(3/2))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*(c/x)^(3/2) + a)*(d*x)^m, 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)^(3/2))^(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 + b \left (\frac{c}{x}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)**m*(a+b*(c/x)**(3/2))**(1/2),x)

[Out]

Integral((d*x)**m*sqrt(a + b*(c/x)**(3/2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \left (\frac{c}{x}\right )^{\frac{3}{2}} + a} \left (d x\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*(a+b*(c/x)^(3/2))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*(c/x)^(3/2) + a)*(d*x)^m, x)