3.7.60 \(\int \frac {x^{-2+m}}{\sqrt {a+b x^2}} \, dx\) [660]

Optimal. Leaf size=51 \[ -\frac {x^{-1+m} \sqrt {a+b x^2} \, _2F_1\left (1,\frac {m}{2};\frac {1+m}{2};-\frac {b x^2}{a}\right )}{a (1-m)} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 66, normalized size of antiderivative = 1.29, 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 = {372, 371} \begin {gather*} -\frac {x^{m-1} \sqrt {\frac {b x^2}{a}+1} \, _2F_1\left (\frac {1}{2},\frac {m-1}{2};\frac {m+1}{2};-\frac {b x^2}{a}\right )}{(1-m) \sqrt {a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(-2 + m)/Sqrt[a + b*x^2],x]

[Out]

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

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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])

Rubi steps

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

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Mathematica [A]
time = 0.17, size = 65, normalized size = 1.27 \begin {gather*} \frac {x^{-1+m} \sqrt {1+\frac {b x^2}{a}} \, _2F_1\left (\frac {1}{2},\frac {1}{2} (-1+m);1+\frac {1}{2} (-1+m);-\frac {b x^2}{a}\right )}{(-1+m) \sqrt {a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {x^{-2+m}}{\sqrt {b \,x^{2}+a}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 4.14, size = 53, normalized size = 1.04 \begin {gather*} \frac {x^{m} \Gamma \left (\frac {m}{2} - \frac {1}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{2} - \frac {1}{2} \\ \frac {m}{2} + \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} x \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^{m-2}}{\sqrt {b\,x^2+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(m - 2)/(a + b*x^2)^(1/2),x)

[Out]

int(x^(m - 2)/(a + b*x^2)^(1/2), x)

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