\(\int \frac {(c x)^m}{(b x^2+c x^4)^3} \, dx\) [408]

Optimal result

Integrand size = 19, antiderivative size = 45 \[ \int \frac {(c x)^m}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {(c x)^m \operatorname {Hypergeometric2F1}\left (3,\frac {1}{2} (-5+m),\frac {1}{2} (-3+m),-\frac {c x^2}{b}\right )}{b^3 (5-m) x^5} \]

[Out]

-(c*x)^m*hypergeom([3, -5/2+1/2*m],[-3/2+1/2*m],-c*x^2/b)/b^3/(5-m)/x^5

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1156, 1598, 371} \[ \int \frac {(c x)^m}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {(c x)^m \operatorname {Hypergeometric2F1}\left (3,\frac {m-5}{2},\frac {m-3}{2},-\frac {c x^2}{b}\right )}{b^3 (5-m) x^5} \]

[In]

Int[(c*x)^m/(b*x^2 + c*x^4)^3,x]

[Out]

-(((c*x)^m*Hypergeometric2F1[3, (-5 + m)/2, (-3 + m)/2, -((c*x^2)/b)])/(b^3*(5 - m)*x^5))

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 1156

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 1598

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*} \text {integral}& = \left (x^{-m} (c x)^m\right ) \text {Subst}\left (\int \frac {x^m}{\left (b x^2+c x^4\right )^3} \, dx,x,x\right ) \\ & = \left (x^{-m} (c x)^m\right ) \text {Subst}\left (\int \frac {x^{-6+m}}{\left (b+c x^2\right )^3} \, dx,x,x\right ) \\ & = -\frac {(c x)^m \, _2F_1\left (3,\frac {1}{2} (-5+m);\frac {1}{2} (-3+m);-\frac {c x^2}{b}\right )}{b^3 (5-m) x^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.98 \[ \int \frac {(c x)^m}{\left (b x^2+c x^4\right )^3} \, dx=\frac {(c x)^m \operatorname {Hypergeometric2F1}\left (3,\frac {1}{2} (-5+m),1+\frac {1}{2} (-5+m),-\frac {c x^2}{b}\right )}{b^3 (-5+m) x^5} \]

[In]

Integrate[(c*x)^m/(b*x^2 + c*x^4)^3,x]

[Out]

((c*x)^m*Hypergeometric2F1[3, (-5 + m)/2, 1 + (-5 + m)/2, -((c*x^2)/b)])/(b^3*(-5 + m)*x^5)

Maple [F]

\[\int \frac {\left (c x \right )^{m}}{\left (c \,x^{4}+b \,x^{2}\right )^{3}}d x\]

[In]

int((c*x)^m/(c*x^4+b*x^2)^3,x)

[Out]

int((c*x)^m/(c*x^4+b*x^2)^3,x)

Fricas [F]

\[ \int \frac {(c x)^m}{\left (b x^2+c x^4\right )^3} \, dx=\int { \frac {\left (c x\right )^{m}}{{\left (c x^{4} + b x^{2}\right )}^{3}} \,d x } \]

[In]

integrate((c*x)^m/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

integral((c*x)^m/(c^3*x^12 + 3*b*c^2*x^10 + 3*b^2*c*x^8 + b^3*x^6), x)

Sympy [F]

\[ \int \frac {(c x)^m}{\left (b x^2+c x^4\right )^3} \, dx=\int \frac {\left (c x\right )^{m}}{x^{6} \left (b + c x^{2}\right )^{3}}\, dx \]

[In]

integrate((c*x)**m/(c*x**4+b*x**2)**3,x)

[Out]

Integral((c*x)**m/(x**6*(b + c*x**2)**3), x)

Maxima [F]

\[ \int \frac {(c x)^m}{\left (b x^2+c x^4\right )^3} \, dx=\int { \frac {\left (c x\right )^{m}}{{\left (c x^{4} + b x^{2}\right )}^{3}} \,d x } \]

[In]

integrate((c*x)^m/(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

integrate((c*x)^m/(c*x^4 + b*x^2)^3, x)

Giac [F]

\[ \int \frac {(c x)^m}{\left (b x^2+c x^4\right )^3} \, dx=\int { \frac {\left (c x\right )^{m}}{{\left (c x^{4} + b x^{2}\right )}^{3}} \,d x } \]

[In]

integrate((c*x)^m/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

integrate((c*x)^m/(c*x^4 + b*x^2)^3, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(c x)^m}{\left (b x^2+c x^4\right )^3} \, dx=\int \frac {{\left (c\,x\right )}^m}{{\left (c\,x^4+b\,x^2\right )}^3} \,d x \]

[In]

int((c*x)^m/(b*x^2 + c*x^4)^3,x)

[Out]

int((c*x)^m/(b*x^2 + c*x^4)^3, x)