3.654 \(\int \frac {x^m}{(a+b x^2)^{3/2}} \, dx\)

Optimal. Leaf size=48 \[ \frac {x^{m+1} \, _2F_1\left (1,\frac {m}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )}{a (m+1) \sqrt {a+b x^2}} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 66, normalized size of antiderivative = 1.38, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {365, 364} \[ \frac {x^{m+1} \sqrt {\frac {b x^2}{a}+1} \, _2F_1\left (\frac {3}{2},\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )}{a (m+1) \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^m/(a + b*x^2)^(3/2),x]

[Out]

(x^(1 + m)*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[3/2, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(a*(1 + 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^m}{\left (a+b x^2\right )^{3/2}} \, dx &=\frac {\sqrt {1+\frac {b x^2}{a}} \int \frac {x^m}{\left (1+\frac {b x^2}{a}\right )^{3/2}} \, dx}{a \sqrt {a+b x^2}}\\ &=\frac {x^{1+m} \sqrt {1+\frac {b x^2}{a}} \, _2F_1\left (\frac {3}{2},\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{a (1+m) \sqrt {a+b x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 68, normalized size = 1.42 \[ \frac {x^{m+1} \sqrt {\frac {b x^2}{a}+1} \, _2F_1\left (\frac {3}{2},\frac {m+1}{2};\frac {m+1}{2}+1;-\frac {b x^2}{a}\right )}{a (m+1) \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m/(a + b*x^2)^(3/2),x]

[Out]

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

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fricas [F]  time = 0.94, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x^{2} + a} x^{m}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(b*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*x^2 + a)*x^m/(b^2*x^4 + 2*a*b*x^2 + a^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{m}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(b*x^2+a)^(3/2),x, algorithm="giac")

[Out]

integrate(x^m/(b*x^2 + a)^(3/2), x)

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maple [F]  time = 0.26, size = 0, normalized size = 0.00 \[ \int \frac {x^{m}}{\left (b \,x^{2}+a \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/(b*x^2+a)^(3/2),x)

[Out]

int(x^m/(b*x^2+a)^(3/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{m}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(b*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^m/(b*x^2 + a)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x^m}{{\left (b\,x^2+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/(a + b*x^2)^(3/2),x)

[Out]

int(x^m/(a + b*x^2)^(3/2), x)

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sympy [C]  time = 1.39, size = 53, normalized size = 1.10 \[ \frac {x x^{m} \Gamma \left (\frac {m}{2} + \frac {1}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {m}{2} + \frac {1}{2} \\ \frac {m}{2} + \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m/(b*x**2+a)**(3/2),x)

[Out]

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

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