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

Optimal. Leaf size=50 \[ \frac {x^{m+3} \sqrt {a+b x^2} \, _2F_1\left (1,\frac {m+4}{2};\frac {m+5}{2};-\frac {b x^2}{a}\right )}{a (m+3)} \]

[Out]

x^(3+m)*hypergeom([1, 2+1/2*m],[5/2+1/2*m],-b*x^2/a)*(b*x^2+a)^(1/2)/a/(3+m)

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Rubi [A]  time = 0.02, antiderivative size = 63, normalized size of antiderivative = 1.26, 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} \[ \frac {x^{m+3} \sqrt {\frac {b x^2}{a}+1} \, _2F_1\left (\frac {1}{2},\frac {m+3}{2};\frac {m+5}{2};-\frac {b x^2}{a}\right )}{(m+3) \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^(2 + m)/Sqrt[a + b*x^2],x]

[Out]

(x^(3 + m)*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[1/2, (3 + m)/2, (5 + m)/2, -((b*x^2)/a)])/((3 + m)*Sqrt[a + b
*x^2])

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^{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^{3+m} \sqrt {1+\frac {b x^2}{a}} \, _2F_1\left (\frac {1}{2},\frac {3+m}{2};\frac {5+m}{2};-\frac {b x^2}{a}\right )}{(3+m) \sqrt {a+b x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 65, normalized size = 1.30 \[ \frac {x^{m+3} \sqrt {\frac {b x^2}{a}+1} \, _2F_1\left (\frac {1}{2},\frac {m+3}{2};\frac {m+3}{2}+1;-\frac {b x^2}{a}\right )}{(m+3) \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.89, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{m + 2}}{\sqrt {b x^{2} + a}}, x\right ) \]

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

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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maple [F]  time = 0.26, size = 0, normalized size = 0.00 \[ \int \frac {x^{m +2}}{\sqrt {b \,x^{2}+a}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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

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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sympy [C]  time = 2.64, size = 54, normalized size = 1.08 \[ \frac {x^{3} x^{m} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{2} + \frac {3}{2} \\ \frac {m}{2} + \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**3*x**m*gamma(m/2 + 3/2)*hyper((1/2, m/2 + 3/2), (m/2 + 5/2,), b*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*gamma(m/
2 + 5/2))

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