3.2750 \(\int \frac{x^m}{(a+b x^{2+2 m})^{3/2}} \, dx\)

Optimal. Leaf size=29 \[ \frac{x^{m+1}}{a (m+1) \sqrt{a+b x^{2 (m+1)}}} \]

[Out]

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

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Rubi [A]  time = 0.0081341, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.053, Rules used = {264} \[ \frac{x^{m+1}}{a (m+1) \sqrt{a+b x^{2 (m+1)}}} \]

Antiderivative was successfully verified.

[In]

Int[x^m/(a + b*x^(2 + 2*m))^(3/2),x]

[Out]

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

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{x^m}{\left (a+b x^{2+2 m}\right )^{3/2}} \, dx &=\frac{x^{1+m}}{a (1+m) \sqrt{a+b x^{2 (1+m)}}}\\ \end{align*}

Mathematica [A]  time = 0.0126102, size = 29, normalized size = 1. \[ \frac{x^{m+1}}{a (m+1) \sqrt{a+b x^{2 m+2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m/(a + b*x^(2 + 2*m))^(3/2),x]

[Out]

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

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Maple [F]  time = 0.047, size = 0, normalized size = 0. \begin{align*} \int{{x}^{m} \left ( a+b{x}^{2+2\,m} \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^m/(a+b*x^(2+2*m))^(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 \, m + 2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.38451, size = 99, normalized size = 3.41 \begin{align*} \frac{\sqrt{b x^{2} x^{2 \, m} + a} x x^{m}}{{\left (a b m + a b\right )} x^{2} x^{2 \, m} + a^{2} m + a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 136.26, size = 121, normalized size = 4.17 \begin{align*} \frac{\sqrt{\pi } x x^{m}{{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, \frac{1}{2} \\ \frac{m}{2 \left (m + 1\right )} + 1 + \frac{1}{2 \left (m + 1\right )} \end{matrix}\middle |{\frac{b x^{2} x^{2 m} e^{i \pi }}{a}} \right )}}{2 a a^{\frac{m}{2 \left (m + 1\right )}} a^{\frac{1}{2 \left (m + 1\right )}} m \Gamma \left (\frac{m}{2 \left (m + 1\right )} + 1 + \frac{1}{2 \left (m + 1\right )}\right ) + 2 a a^{\frac{m}{2 \left (m + 1\right )}} a^{\frac{1}{2 \left (m + 1\right )}} \Gamma \left (\frac{m}{2 \left (m + 1\right )} + 1 + \frac{1}{2 \left (m + 1\right )}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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