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

Optimal. Leaf size=98 \[ \frac {x^{m-3} (b B (3-m)-A c (5-m)) \, _2F_1\left (1,\frac {m-3}{2};\frac {m-1}{2};-\frac {c x^2}{b}\right )}{2 b^2 c (3-m)}-\frac {x^{m-3} (b B-A c)}{2 b c \left (b+c x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 92, normalized size of antiderivative = 0.94, 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, 457, 364} \[ \frac {x^{m-3} \left (\frac {b B}{c}-\frac {A (5-m)}{3-m}\right ) \, _2F_1\left (1,\frac {m-3}{2};\frac {m-1}{2};-\frac {c x^2}{b}\right )}{2 b^2}-\frac {x^{m-3} (b B-A c)}{2 b c \left (b+c x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

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

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Mathematica [A]  time = 0.10, size = 80, normalized size = 0.82 \[ \frac {x^{m-3} \left ((A c-b B) \, _2F_1\left (2,\frac {m-3}{2};\frac {m-1}{2};-\frac {c x^2}{b}\right )+b B \, _2F_1\left (1,\frac {m-3}{2};\frac {m-1}{2};-\frac {c x^2}{b}\right )\right )}{b^2 c (m-3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.89, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B x^{2} + A\right )} x^{m}}{c^{2} x^{8} + 2 \, b c x^{6} + b^{2} x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.53, size = 0, normalized size = 0.00 \[ \int \frac {\left (B \,x^{2}+A \right ) x^{m}}{\left (c \,x^{4}+b \,x^{2}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^m\,\left (B\,x^2+A\right )}{{\left (c\,x^4+b\,x^2\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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