3.1960 \(\int \sqrt{1+\frac{b}{x^2}} (c x)^m \, dx\)

Optimal. Leaf size=44 \[ \frac{(c x)^{m+1} \, _2F_1\left (-\frac{1}{2},\frac{1}{2} (-m-1);\frac{1-m}{2};-\frac{b}{x^2}\right )}{c (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.0170517, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {339, 364} \[ \frac{(c x)^{m+1} \, _2F_1\left (-\frac{1}{2},\frac{1}{2} (-m-1);\frac{1-m}{2};-\frac{b}{x^2}\right )}{c (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + b/x^2]*(c*x)^m,x]

[Out]

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

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

Mathematica [A]  time = 0.0157366, size = 58, normalized size = 1.32 \[ \frac{x \sqrt{\frac{b}{x^2}+1} (c x)^m \, _2F_1\left (-\frac{1}{2},\frac{m}{2};\frac{m}{2}+1;-\frac{x^2}{b}\right )}{m \sqrt{\frac{x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 + b/x^2]*(c*x)^m,x]

[Out]

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

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Maple [F]  time = 0.009, size = 0, normalized size = 0. \begin{align*} \int \sqrt{1+{\frac{b}{{x}^{2}}}} \left ( cx \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c x\right )^{m} \sqrt{\frac{b}{x^{2}} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x)^m*sqrt(b/x^2 + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 1.92034, size = 48, normalized size = 1.09 \begin{align*} - \frac{\sqrt{b} c^{m} x^{m} \Gamma \left (- \frac{m}{2}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{m}{2} \\ \frac{m}{2} + 1 \end{matrix}\middle |{\frac{x^{2} e^{i \pi }}{b}} \right )}}{2 \Gamma \left (1 - \frac{m}{2}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(b)*c**m*x**m*gamma(-m/2)*hyper((-1/2, m/2), (m/2 + 1,), x**2*exp_polar(I*pi)/b)/(2*gamma(1 - m/2))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c x\right )^{m} \sqrt{\frac{b}{x^{2}} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x)^m*sqrt(b/x^2 + 1), x)