3.349 \(\int \frac{(c x)^{-1+m}}{a+b x^2} \, dx\)

Optimal. Leaf size=38 \[ \frac{(c x)^m \, _2F_1\left (1,\frac{m}{2};\frac{m+2}{2};-\frac{b x^2}{a}\right )}{a c m} \]

[Out]

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

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Rubi [A]  time = 0.0085645, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.059, Rules used = {364} \[ \frac{(c x)^m \, _2F_1\left (1,\frac{m}{2};\frac{m+2}{2};-\frac{b x^2}{a}\right )}{a c m} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^(-1 + m)/(a + b*x^2),x]

[Out]

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

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{(c x)^{-1+m}}{a+b x^2} \, dx &=\frac{(c x)^m \, _2F_1\left (1,\frac{m}{2};\frac{2+m}{2};-\frac{b x^2}{a}\right )}{a c m}\\ \end{align*}

Mathematica [A]  time = 0.0073654, size = 38, normalized size = 1. \[ \frac{x (c x)^{m-1} \, _2F_1\left (1,\frac{m}{2};\frac{m}{2}+1;-\frac{b x^2}{a}\right )}{a m} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.028, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( cx \right ) ^{-1+m}}{b{x}^{2}+a}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c x\right )^{m - 1}}{b x^{2} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\left (c x\right )^{m - 1}}{b x^{2} + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c*x)^(m - 1)/(b*x^2 + a), x)

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Sympy [C]  time = 18.7562, size = 39, normalized size = 1.03 \begin{align*} \frac{c^{m} m x^{m} \Phi \left (\frac{b x^{2} e^{i \pi }}{a}, 1, \frac{m}{2}\right ) \Gamma \left (\frac{m}{2}\right )}{4 a c \Gamma \left (\frac{m}{2} + 1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c**m*m*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2)*gamma(m/2)/(4*a*c*gamma(m/2 + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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