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

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

[Out]

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

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Rubi [A]  time = 0.0307726, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {1142, 1584, 364} \[ -\frac{(c x)^m \, _2F_1\left (2,\frac{m-3}{2};\frac{m-1}{2};-\frac{c x^2}{b}\right )}{b^2 (3-m) x^3} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^m/(b*x^2 + c*x^4)^2,x]

[Out]

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

Rule 1142

Int[(u_)^(m_.)*((a_.) + (b_.)*(v_)^2 + (c_.)*(v_)^4)^(p_.), x_Symbol] :> Dist[u^m/(Coefficient[v, x, 1]*v^m),
Subst[Int[x^m*(a + b*x^2 + c*x^(2*2))^p, x], x, v], x] /; FreeQ[{a, b, c, m, p}, x] && LinearPairQ[u, v, x]

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 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)^m}{\left (b x^2+c x^4\right )^2} \, dx &=\left (x^{-m} (c x)^m\right ) \operatorname{Subst}\left (\int \frac{x^m}{\left (b x^2+c x^4\right )^2} \, dx,x,x\right )\\ &=\left (x^{-m} (c x)^m\right ) \operatorname{Subst}\left (\int \frac{x^{-4+m}}{\left (b+c x^2\right )^2} \, dx,x,x\right )\\ &=-\frac{(c x)^m \, _2F_1\left (2,\frac{1}{2} (-3+m);\frac{1}{2} (-1+m);-\frac{c x^2}{b}\right )}{b^2 (3-m) x^3}\\ \end{align*}

Mathematica [A]  time = 0.0115341, size = 44, normalized size = 0.98 \[ \frac{(c x)^m \, _2F_1\left (2,\frac{m-3}{2};\frac{m-3}{2}+1;-\frac{c x^2}{b}\right )}{b^2 (m-3) x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.334, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( cx \right ) ^{m}}{ \left ( c{x}^{4}+b{x}^{2} \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x)^m/(c*x^4 + b*x^2)^2, 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}}{c^{2} x^{8} + 2 \, b c x^{6} + b^{2} x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c*x)^m/(c^2*x^8 + 2*b*c*x^6 + b^2*x^4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c*x)**m/(x**4*(b + c*x**2)**2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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