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

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

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 45, normalized size of antiderivative = 1.00, 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} \[ -\frac {(c x)^m \, _2F_1\left (1,\frac {m-1}{2};\frac {m+1}{2};-\frac {c x^2}{b}\right )}{b (1-m) x} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^m/(b*x^2 + c*x^4),x]

[Out]

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 42, normalized size = 0.93 \[ \frac {(c x)^m \, _2F_1\left (1,\frac {m-1}{2};\frac {m+1}{2};-\frac {c x^2}{b}\right )}{b (m-1) x} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*x)^m/(b*x^2 + c*x^4),x]

[Out]

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

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fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (c x\right )^{m}}{c x^{4} + b x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c*x)^m/(c*x^4 + b*x^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x\right )^{m}}{c x^{4} + b x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x)^m/(c*x^4 + b*x^2), x)

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x \right )^{m}}{c \,x^{4}+b \,x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x\right )^{m}}{c x^{4} + b x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x)^m/(c*x^4 + b*x^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (c\,x\right )}^m}{c\,x^4+b\,x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^m/(b*x^2 + c*x^4),x)

[Out]

int((c*x)^m/(b*x^2 + c*x^4), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x\right )^{m}}{x^{2} \left (b + c x^{2}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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