3.61 \(\int \frac{x^9 (A+B x^2)}{(b x^2+c x^4)^2} \, dx\)

Optimal. Leaf size=83 \[ \frac{b^2 (b B-A c)}{2 c^4 \left (b+c x^2\right )}-\frac{x^2 (2 b B-A c)}{2 c^3}+\frac{b (3 b B-2 A c) \log \left (b+c x^2\right )}{2 c^4}+\frac{B x^4}{4 c^2} \]

[Out]

-((2*b*B - A*c)*x^2)/(2*c^3) + (B*x^4)/(4*c^2) + (b^2*(b*B - A*c))/(2*c^4*(b + c*x^2)) + (b*(3*b*B - 2*A*c)*Lo
g[b + c*x^2])/(2*c^4)

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Rubi [A]  time = 0.0984574, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {1584, 446, 77} \[ \frac{b^2 (b B-A c)}{2 c^4 \left (b+c x^2\right )}-\frac{x^2 (2 b B-A c)}{2 c^3}+\frac{b (3 b B-2 A c) \log \left (b+c x^2\right )}{2 c^4}+\frac{B x^4}{4 c^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^9*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]

[Out]

-((2*b*B - A*c)*x^2)/(2*c^3) + (B*x^4)/(4*c^2) + (b^2*(b*B - A*c))/(2*c^4*(b + c*x^2)) + (b*(3*b*B - 2*A*c)*Lo
g[b + c*x^2])/(2*c^4)

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 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int \frac{x^9 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx &=\int \frac{x^5 \left (A+B x^2\right )}{\left (b+c x^2\right )^2} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^2 (A+B x)}{(b+c x)^2} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{-2 b B+A c}{c^3}+\frac{B x}{c^2}-\frac{b^2 (b B-A c)}{c^3 (b+c x)^2}+\frac{b (3 b B-2 A c)}{c^3 (b+c x)}\right ) \, dx,x,x^2\right )\\ &=-\frac{(2 b B-A c) x^2}{2 c^3}+\frac{B x^4}{4 c^2}+\frac{b^2 (b B-A c)}{2 c^4 \left (b+c x^2\right )}+\frac{b (3 b B-2 A c) \log \left (b+c x^2\right )}{2 c^4}\\ \end{align*}

Mathematica [A]  time = 0.0524287, size = 72, normalized size = 0.87 \[ \frac{\frac{2 b^2 (b B-A c)}{b+c x^2}+2 c x^2 (A c-2 b B)+2 b (3 b B-2 A c) \log \left (b+c x^2\right )+B c^2 x^4}{4 c^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^9*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]

[Out]

(2*c*(-2*b*B + A*c)*x^2 + B*c^2*x^4 + (2*b^2*(b*B - A*c))/(b + c*x^2) + 2*b*(3*b*B - 2*A*c)*Log[b + c*x^2])/(4
*c^4)

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Maple [A]  time = 0.01, size = 98, normalized size = 1.2 \begin{align*}{\frac{B{x}^{4}}{4\,{c}^{2}}}+{\frac{A{x}^{2}}{2\,{c}^{2}}}-{\frac{B{x}^{2}b}{{c}^{3}}}-{\frac{b\ln \left ( c{x}^{2}+b \right ) A}{{c}^{3}}}+{\frac{3\,{b}^{2}\ln \left ( c{x}^{2}+b \right ) B}{2\,{c}^{4}}}-{\frac{{b}^{2}A}{2\,{c}^{3} \left ( c{x}^{2}+b \right ) }}+{\frac{B{b}^{3}}{2\,{c}^{4} \left ( c{x}^{2}+b \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9*(B*x^2+A)/(c*x^4+b*x^2)^2,x)

[Out]

1/4*B*x^4/c^2+1/2/c^2*A*x^2-1/c^3*B*x^2*b-b/c^3*ln(c*x^2+b)*A+3/2*b^2/c^4*ln(c*x^2+b)*B-1/2*b^2/c^3/(c*x^2+b)*
A+1/2*b^3/c^4/(c*x^2+b)*B

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Maxima [A]  time = 1.10743, size = 111, normalized size = 1.34 \begin{align*} \frac{B b^{3} - A b^{2} c}{2 \,{\left (c^{5} x^{2} + b c^{4}\right )}} + \frac{B c x^{4} - 2 \,{\left (2 \, B b - A c\right )} x^{2}}{4 \, c^{3}} + \frac{{\left (3 \, B b^{2} - 2 \, A b c\right )} \log \left (c x^{2} + b\right )}{2 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="maxima")

[Out]

1/2*(B*b^3 - A*b^2*c)/(c^5*x^2 + b*c^4) + 1/4*(B*c*x^4 - 2*(2*B*b - A*c)*x^2)/c^3 + 1/2*(3*B*b^2 - 2*A*b*c)*lo
g(c*x^2 + b)/c^4

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Fricas [A]  time = 0.753855, size = 251, normalized size = 3.02 \begin{align*} \frac{B c^{3} x^{6} -{\left (3 \, B b c^{2} - 2 \, A c^{3}\right )} x^{4} + 2 \, B b^{3} - 2 \, A b^{2} c - 2 \,{\left (2 \, B b^{2} c - A b c^{2}\right )} x^{2} + 2 \,{\left (3 \, B b^{3} - 2 \, A b^{2} c +{\left (3 \, B b^{2} c - 2 \, A b c^{2}\right )} x^{2}\right )} \log \left (c x^{2} + b\right )}{4 \,{\left (c^{5} x^{2} + b c^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="fricas")

[Out]

1/4*(B*c^3*x^6 - (3*B*b*c^2 - 2*A*c^3)*x^4 + 2*B*b^3 - 2*A*b^2*c - 2*(2*B*b^2*c - A*b*c^2)*x^2 + 2*(3*B*b^3 -
2*A*b^2*c + (3*B*b^2*c - 2*A*b*c^2)*x^2)*log(c*x^2 + b))/(c^5*x^2 + b*c^4)

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Sympy [A]  time = 0.812214, size = 78, normalized size = 0.94 \begin{align*} \frac{B x^{4}}{4 c^{2}} + \frac{b \left (- 2 A c + 3 B b\right ) \log{\left (b + c x^{2} \right )}}{2 c^{4}} + \frac{- A b^{2} c + B b^{3}}{2 b c^{4} + 2 c^{5} x^{2}} - \frac{x^{2} \left (- A c + 2 B b\right )}{2 c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9*(B*x**2+A)/(c*x**4+b*x**2)**2,x)

[Out]

B*x**4/(4*c**2) + b*(-2*A*c + 3*B*b)*log(b + c*x**2)/(2*c**4) + (-A*b**2*c + B*b**3)/(2*b*c**4 + 2*c**5*x**2)
- x**2*(-A*c + 2*B*b)/(2*c**3)

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Giac [A]  time = 1.23714, size = 143, normalized size = 1.72 \begin{align*} \frac{{\left (3 \, B b^{2} - 2 \, A b c\right )} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, c^{4}} + \frac{B c^{2} x^{4} - 4 \, B b c x^{2} + 2 \, A c^{2} x^{2}}{4 \, c^{4}} - \frac{3 \, B b^{2} c x^{2} - 2 \, A b c^{2} x^{2} + 2 \, B b^{3} - A b^{2} c}{2 \,{\left (c x^{2} + b\right )} c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="giac")

[Out]

1/2*(3*B*b^2 - 2*A*b*c)*log(abs(c*x^2 + b))/c^4 + 1/4*(B*c^2*x^4 - 4*B*b*c*x^2 + 2*A*c^2*x^2)/c^4 - 1/2*(3*B*b
^2*c*x^2 - 2*A*b*c^2*x^2 + 2*B*b^3 - A*b^2*c)/((c*x^2 + b)*c^4)