3.2958 \(\int (d x)^m \sqrt{a+\frac{b}{(c x^2)^{3/2}}} \, dx\)

Optimal. Leaf size=90 \[ \frac{(d x)^{m+1} \sqrt{a+\frac{b}{\left (c x^2\right )^{3/2}}} \, _2F_1\left (-\frac{1}{2},\frac{1}{3} (-m-1);\frac{2-m}{3};-\frac{b}{a \left (c x^2\right )^{3/2}}\right )}{d (m+1) \sqrt{\frac{b}{a \left (c x^2\right )^{3/2}}+1}} \]

[Out]

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

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Rubi [A]  time = 0.0836774, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {368, 339, 365, 364} \[ \frac{(d x)^{m+1} \sqrt{a+\frac{b}{\left (c x^2\right )^{3/2}}} \, _2F_1\left (-\frac{1}{2},\frac{1}{3} (-m-1);\frac{2-m}{3};-\frac{b}{a \left (c x^2\right )^{3/2}}\right )}{d (m+1) \sqrt{\frac{b}{a \left (c x^2\right )^{3/2}}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 368

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbol] :> Dist[(d*x)^(m + 1)/(d*((c*x^q
)^(1/q))^(m + 1)), Subst[Int[x^m*(a + b*x^(n*q))^p, x], x, (c*x^q)^(1/q)], x] /; FreeQ[{a, b, c, d, m, n, p, q
}, x] && IntegerQ[n*q] && NeQ[x, (c*x^q)^(1/q)]

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

Mathematica [F]  time = 0.112051, size = 0, normalized size = 0. \[ \int (d x)^m \sqrt{a+\frac{b}{\left (c x^2\right )^{3/2}}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.055, size = 0, normalized size = 0. \begin{align*} \int \left ( dx \right ) ^{m}\sqrt{a+{b \left ( c{x}^{2} \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^2)^(3/2))^(1/2),x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (d x\right )^{m} \sqrt{\frac{a c^{2} x^{4} + \sqrt{c x^{2}} b}{c^{2} x^{4}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((d*x)^m*sqrt((a*c^2*x^4 + sqrt(c*x^2)*b)/(c^2*x^4)), x)

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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}{\left (c x^{2}\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**2)**(3/2))**(1/2),x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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