3.58 \(\int \frac{A (c x)^m}{a+b x^2} \, dx\)

Optimal. Leaf size=45 \[ \frac{A (c x)^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{a c (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.0166907, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {12, 364} \[ \frac{A (c x)^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{a c (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(A*(c*x)^m)/(a + b*x^2),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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{A (c x)^m}{a+b x^2} \, dx &=A \int \frac{(c x)^m}{a+b x^2} \, dx\\ &=\frac{A (c x)^{1+m} \, _2F_1\left (1,\frac{1+m}{2};\frac{3+m}{2};-\frac{b x^2}{a}\right )}{a c (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.0085597, size = 43, normalized size = 0.96 \[ \frac{A x (c x)^m \, _2F_1\left (1,\frac{m+1}{2};\frac{m+1}{2}+1;-\frac{b x^2}{a}\right )}{a (m+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(A*(c*x)^m)/(a + b*x^2),x]

[Out]

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

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Maple [F]  time = 0.03, size = 0, normalized size = 0. \begin{align*} \int{\frac{A \left ( cx \right ) ^{m}}{b{x}^{2}+a}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*integrate((c*x)^m/(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} A}{b x^{2} + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c*x)^m*A/(b*x^2 + a), x)

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Sympy [C]  time = 1.51525, size = 97, normalized size = 2.16 \begin{align*} A \left (\frac{c^{m} m x x^{m} \Phi \left (\frac{b x^{2} e^{i \pi }}{a}, 1, \frac{m}{2} + \frac{1}{2}\right ) \Gamma \left (\frac{m}{2} + \frac{1}{2}\right )}{4 a \Gamma \left (\frac{m}{2} + \frac{3}{2}\right )} + \frac{c^{m} x x^{m} \Phi \left (\frac{b x^{2} e^{i \pi }}{a}, 1, \frac{m}{2} + \frac{1}{2}\right ) \Gamma \left (\frac{m}{2} + \frac{1}{2}\right )}{4 a \Gamma \left (\frac{m}{2} + \frac{3}{2}\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*(c**m*m*x*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(4*a*gamma(m/2 + 3/2)) + c*
*m*x*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(4*a*gamma(m/2 + 3/2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x)^m*A/(b*x^2 + a), x)