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

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Rubi [A]  time = 0.0195241, 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 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])

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])

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*}

Mathematica [A]  time = 0.0159536, 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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Maple [F]  time = 0.017, size = 0, normalized size = 0. \begin{align*} \int{{x}^{m} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{m}}{{\left (b x^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b x^{2} + a} x^{m}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}, x\right ) \end{align*}

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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Sympy [C]  time = 1.53825, size = 53, normalized size = 1.1 \begin{align*} \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 )} \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{m}}{{\left (b x^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

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)