3.663 \(\int \frac{x^{-1+m} (a m+b (-1+m) x^2)}{(a+b x^2)^{3/2}} \, dx\)

Optimal. Leaf size=15 \[ \frac{x^m}{\sqrt{a+b x^2}} \]

[Out]

x^m/Sqrt[a + b*x^2]

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Rubi [A]  time = 0.0118946, antiderivative size = 15, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.034, Rules used = {449} \[ \frac{x^m}{\sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x^m/Sqrt[a + b*x^2]

Rule 449

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

Rubi steps

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

Mathematica [C]  time = 0.106674, size = 103, normalized size = 6.87 \[ \frac{x^m \sqrt{\frac{b x^2}{a}+1} \left (b (m-1) x^2 \, _2F_1\left (\frac{3}{2},\frac{m+2}{2};\frac{m+4}{2};-\frac{b x^2}{a}\right )+a (m+2) \, _2F_1\left (\frac{3}{2},\frac{m}{2};\frac{m+2}{2};-\frac{b x^2}{a}\right )\right )}{a (m+2) \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^m*Sqrt[1 + (b*x^2)/a]*(a*(2 + m)*Hypergeometric2F1[3/2, m/2, (2 + m)/2, -((b*x^2)/a)] + b*(-1 + m)*x^2*Hype
rgeometric2F1[3/2, (2 + m)/2, (4 + m)/2, -((b*x^2)/a)]))/(a*(2 + m)*Sqrt[a + b*x^2])

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Maple [A]  time = 0.006, size = 14, normalized size = 0.9 \begin{align*}{{x}^{m}{\frac{1}{\sqrt{b{x}^{2}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(-1+m)*(a*m+b*(-1+m)*x^2)/(b*x^2+a)^(3/2),x)

[Out]

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

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Maxima [A]  time = 2.42674, size = 18, normalized size = 1.2 \begin{align*} \frac{x^{m}}{\sqrt{b x^{2} + a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+m)*(a*m+b*(-1+m)*x^2)/(b*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

x^m/sqrt(b*x^2 + a)

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Fricas [A]  time = 1.57488, size = 39, normalized size = 2.6 \begin{align*} \frac{x x^{m - 1}}{\sqrt{b x^{2} + a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+m)*(a*m+b*(-1+m)*x^2)/(b*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

x*x^(m - 1)/sqrt(b*x^2 + a)

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Sympy [C]  time = 111.034, size = 97, normalized size = 6.47 \begin{align*} \frac{m x^{m} \Gamma \left (\frac{m}{2}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, \frac{m}{2} \\ \frac{m}{2} + 1 \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt{a} \Gamma \left (\frac{m}{2} + 1\right )} + \frac{b x^{2} x^{m} \left (m - 1\right ) \Gamma \left (\frac{m}{2} + 1\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, \frac{m}{2} + 1 \\ \frac{m}{2} + 2 \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac{3}{2}} \Gamma \left (\frac{m}{2} + 2\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(-1+m)*(a*m+b*(-1+m)*x**2)/(b*x**2+a)**(3/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b{\left (m - 1\right )} x^{2} + a m\right )} x^{m - 1}}{{\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^(-1+m)*(a*m+b*(-1+m)*x^2)/(b*x^2+a)^(3/2),x, algorithm="giac")

[Out]

integrate((b*(m - 1)*x^2 + a*m)*x^(m - 1)/(b*x^2 + a)^(3/2), x)