3.272 \(\int \frac{x^m (A+B x^2)}{b x^2+c x^4} \, dx\)

Optimal. Leaf size=71 \[ \frac{x^{m-1} (b B-A c) \, _2F_1\left (1,\frac{m-1}{2};\frac{m+1}{2};-\frac{c x^2}{b}\right )}{b c (1-m)}-\frac{B x^{m-1}}{c (1-m)} \]

[Out]

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

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Rubi [A]  time = 0.0470955, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {1584, 459, 364} \[ \frac{x^{m-1} (b B-A c) \, _2F_1\left (1,\frac{m-1}{2};\frac{m+1}{2};-\frac{c x^2}{b}\right )}{b c (1-m)}-\frac{B x^{m-1}}{c (1-m)} \]

Antiderivative was successfully verified.

[In]

Int[(x^m*(A + B*x^2))/(b*x^2 + c*x^4),x]

[Out]

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

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 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 )}{b x^2+c x^4} \, dx &=\int \frac{x^{-2+m} \left (A+B x^2\right )}{b+c x^2} \, dx\\ &=-\frac{B x^{-1+m}}{c (1-m)}-\frac{(b B (-1+m)-A c (-1+m)) \int \frac{x^{-2+m}}{b+c x^2} \, dx}{c (-1+m)}\\ &=-\frac{B x^{-1+m}}{c (1-m)}+\frac{(b B-A c) x^{-1+m} \, _2F_1\left (1,\frac{1}{2} (-1+m);\frac{1+m}{2};-\frac{c x^2}{b}\right )}{b c (1-m)}\\ \end{align*}

Mathematica [A]  time = 0.0615062, size = 55, normalized size = 0.77 \[ \frac{x^{m-1} \left ((A c-b B) \, _2F_1\left (1,\frac{m-1}{2};\frac{m+1}{2};-\frac{c x^2}{b}\right )+b B\right )}{b c (m-1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^m*(A + B*x^2))/(b*x^2 + c*x^4),x]

[Out]

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

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Maple [F]  time = 0.234, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( B{x}^{2}+A \right ){x}^{m}}{c{x}^{4}+b{x}^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(B*x^2+A)/(c*x^4+b*x^2),x)

[Out]

int(x^m*(B*x^2+A)/(c*x^4+b*x^2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{2} + A\right )} x^{m}}{c x^{4} + b x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*x^m/(c*x^4 + b*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="fricas")

[Out]

integral((B*x^2 + A)*x^m/(c*x^4 + b*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*(B*x**2+A)/(c*x**4+b*x**2),x)

[Out]

Integral(x**m*(A + B*x**2)/(x**2*(b + c*x**2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{2} + A\right )} x^{m}}{c x^{4} + b x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)*x^m/(c*x^4 + b*x^2), x)