3.78 \(\int \frac{x^9 (A+B x^2)}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=67 \[ -\frac{b (b B-A c)}{4 c^3 \left (b+c x^2\right )^2}+\frac{2 b B-A c}{2 c^3 \left (b+c x^2\right )}+\frac{B \log \left (b+c x^2\right )}{2 c^3} \]

[Out]

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

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Rubi [A]  time = 0.0758642, antiderivative size = 67, 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} \[ -\frac{b (b B-A c)}{4 c^3 \left (b+c x^2\right )^2}+\frac{2 b B-A c}{2 c^3 \left (b+c x^2\right )}+\frac{B \log \left (b+c x^2\right )}{2 c^3} \]

Antiderivative was successfully verified.

[In]

Int[(x^9*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

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

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

Mathematica [A]  time = 0.0237069, size = 64, normalized size = 0.96 \[ \frac{-b c \left (A-4 B x^2\right )-2 A c^2 x^2+3 b^2 B+2 B \left (b+c x^2\right )^2 \log \left (b+c x^2\right )}{4 c^3 \left (b+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^9*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

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

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Maple [A]  time = 0.007, size = 80, normalized size = 1.2 \begin{align*}{\frac{B\ln \left ( c{x}^{2}+b \right ) }{2\,{c}^{3}}}-{\frac{A}{2\,{c}^{2} \left ( c{x}^{2}+b \right ) }}+{\frac{Bb}{{c}^{3} \left ( c{x}^{2}+b \right ) }}+{\frac{Ab}{4\,{c}^{2} \left ( c{x}^{2}+b \right ) ^{2}}}-{\frac{B{b}^{2}}{4\,{c}^{3} \left ( c{x}^{2}+b \right ) ^{2}}} \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)^3,x)

[Out]

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

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Maxima [A]  time = 1.19388, size = 97, normalized size = 1.45 \begin{align*} \frac{3 \, B b^{2} - A b c + 2 \,{\left (2 \, B b c - A c^{2}\right )} x^{2}}{4 \,{\left (c^{5} x^{4} + 2 \, b c^{4} x^{2} + b^{2} c^{3}\right )}} + \frac{B \log \left (c x^{2} + 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)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.31711, size = 184, normalized size = 2.75 \begin{align*} \frac{3 \, B b^{2} - A b c + 2 \,{\left (2 \, B b c - A c^{2}\right )} x^{2} + 2 \,{\left (B c^{2} x^{4} + 2 \, B b c x^{2} + B b^{2}\right )} \log \left (c x^{2} + b\right )}{4 \,{\left (c^{5} x^{4} + 2 \, b c^{4} x^{2} + b^{2} c^{3}\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)^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.07545, size = 70, normalized size = 1.04 \begin{align*} \frac{B \log{\left (b + c x^{2} \right )}}{2 c^{3}} + \frac{- A b c + 3 B b^{2} + x^{2} \left (- 2 A c^{2} + 4 B b c\right )}{4 b^{2} c^{3} + 8 b c^{4} x^{2} + 4 c^{5} x^{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)**3,x)

[Out]

B*log(b + c*x**2)/(2*c**3) + (-A*b*c + 3*B*b**2 + x**2*(-2*A*c**2 + 4*B*b*c))/(4*b**2*c**3 + 8*b*c**4*x**2 + 4
*c**5*x**4)

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Giac [A]  time = 1.15932, size = 74, normalized size = 1.1 \begin{align*} \frac{B \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, c^{3}} - \frac{3 \, B c x^{4} + 2 \, B b x^{2} + 2 \, A c x^{2} + A b}{4 \,{\left (c x^{2} + b\right )}^{2} c^{2}} \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)^3,x, algorithm="giac")

[Out]

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