3.2695 \(\int \frac {x^m}{\sqrt {a+b x^{-2+m}}} \, dx\)

Optimal. Leaf size=65 \[ \frac {x^{m+1} \sqrt {a+b x^{m-2}} \, _2F_1\left (1,-\frac {3 m}{2 (2-m)};\frac {1-2 m}{2-m};-\frac {b x^{m-2}}{a}\right )}{a (m+1)} \]

[Out]

x^(1+m)*hypergeom([1, -3/2*m/(2-m)],[(1-2*m)/(2-m)],-b*x^(-2+m)/a)*(a+b*x^(-2+m))^(1/2)/a/(1+m)

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Rubi [A]  time = 0.03, antiderivative size = 80, normalized size of antiderivative = 1.23, 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} \[ \frac {x^{m+1} \sqrt {\frac {b x^{m-2}}{a}+1} \, _2F_1\left (\frac {1}{2},-\frac {m+1}{2-m};\frac {1-2 m}{2-m};-\frac {b x^{m-2}}{a}\right )}{(m+1) \sqrt {a+b x^{m-2}}} \]

Antiderivative was successfully verified.

[In]

Int[x^m/Sqrt[a + b*x^(-2 + m)],x]

[Out]

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

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])

Rubi steps

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

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Mathematica [A]  time = 0.18, size = 110, normalized size = 1.69 \[ \frac {2 x \left (6 a x^2 \sqrt {\frac {a x^{2-m}}{b}+1} \, _2F_1\left (\frac {1}{2},\frac {m-8}{2 (m-2)};\frac {3 (m-4)}{2 (m-2)};-\frac {a x^{2-m}}{b}\right )+(m-8) \left (a x^2+b x^m\right )\right )}{b (m-8) (m+4) \sqrt {a+b x^{m-2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*((-8 + m)*(a*x^2 + b*x^m) + 6*a*x^2*Sqrt[1 + (a*x^(2 - m))/b]*Hypergeometric2F1[1/2, (-8 + m)/(2*(-2 + m)
), (3*(-4 + m))/(2*(-2 + m)), -((a*x^(2 - m))/b)]))/(b*(-8 + m)*(4 + m)*Sqrt[a + b*x^(-2 + m)])

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.22, size = 0, normalized size = 0.00 \[ \int \frac {x^{m}}{\sqrt {b \,x^{m -2}+a}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/(a + b*x^(m - 2))^(1/2),x)

[Out]

int(x^m/(a + b*x^(m - 2))^(1/2), x)

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sympy [C]  time = 7.84, size = 94, normalized size = 1.45 \[ \frac {x x^{m} \Gamma \left (\frac {m}{m - 2} + \frac {1}{m - 2}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{m - 2} + \frac {1}{m - 2} \\ \frac {m}{m - 2} + 1 + \frac {1}{m - 2} \end {matrix}\middle | {\frac {b x^{m} e^{i \pi }}{a x^{2}}} \right )}}{\sqrt {a} m \Gamma \left (\frac {m}{m - 2} + 1 + \frac {1}{m - 2}\right ) - 2 \sqrt {a} \Gamma \left (\frac {m}{m - 2} + 1 + \frac {1}{m - 2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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