3.7.56 \(\int \frac {x^{2+m}}{\sqrt {a+b x^2}} \, dx\) [656]

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

[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 = 0.01, 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 = {372, 371} \begin {gather*} \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}} \end {gather*}

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 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^{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 = 0.17, size = 65, normalized size = 1.30 \begin {gather*} \frac {x^{3+m} \sqrt {1+\frac {b x^2}{a}} \, _2F_1\left (\frac {1}{2},\frac {3+m}{2};1+\frac {3+m}{2};-\frac {b x^2}{a}\right )}{(3+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^(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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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 = 1.29, size = 54, normalized size = 1.08 \begin {gather*} \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 )} \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**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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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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