3.659 \(\int \frac{x^{-1+m}}{\sqrt{a+b x^2}} \, dx\)

Optimal. Leaf size=46 \[ \frac{x^m \sqrt{a+b x^2} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+2}{2};-\frac{b x^2}{a}\right )}{a m} \]

[Out]

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

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Rubi [A]  time = 0.0186178, antiderivative size = 57, normalized size of antiderivative = 1.24, 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 = {365, 364} \[ \frac{x^m \sqrt{\frac{b x^2}{a}+1} \, _2F_1\left (\frac{1}{2},\frac{m}{2};\frac{m+2}{2};-\frac{b x^2}{a}\right )}{m \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 + m)/Sqrt[a + b*x^2],x]

[Out]

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

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

Mathematica [A]  time = 0.0118479, size = 57, normalized size = 1.24 \[ \frac{x^m \sqrt{\frac{b x^2}{a}+1} \, _2F_1\left (\frac{1}{2},\frac{m}{2};\frac{m}{2}+1;-\frac{b x^2}{a}\right )}{m \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.021, size = 0, normalized size = 0. \begin{align*} \int{{x}^{-1+m}{\frac{1}{\sqrt{b{x}^{2}+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^(m - 1)/sqrt(b*x^2 + a), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{m - 1}}{\sqrt{b x^{2} + a}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^(m - 1)/sqrt(b*x^2 + a), x)

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Sympy [C]  time = 4.7015, size = 41, normalized size = 0.89 \begin{align*} \frac{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{b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt{a} \Gamma \left (\frac{m}{2} + 1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^(m - 1)/sqrt(b*x^2 + a), x)