3.76 \(\int \frac{x^{11} (A+B x^2)}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=89 \[ \frac{b^2 (b B-A c)}{4 c^4 \left (b+c x^2\right )^2}-\frac{b (3 b B-2 A c)}{2 c^4 \left (b+c x^2\right )}-\frac{(3 b B-A c) \log \left (b+c x^2\right )}{2 c^4}+\frac{B x^2}{2 c^3} \]

[Out]

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

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Rubi [A]  time = 0.101731, antiderivative size = 89, 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} \[ \frac{b^2 (b B-A c)}{4 c^4 \left (b+c x^2\right )^2}-\frac{b (3 b B-2 A c)}{2 c^4 \left (b+c x^2\right )}-\frac{(3 b B-A c) \log \left (b+c x^2\right )}{2 c^4}+\frac{B x^2}{2 c^3} \]

Antiderivative was successfully verified.

[In]

Int[(x^11*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

(B*x^2)/(2*c^3) + (b^2*(b*B - A*c))/(4*c^4*(b + c*x^2)^2) - (b*(3*b*B - 2*A*c))/(2*c^4*(b + c*x^2)) - ((3*b*B
- A*c)*Log[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^{11} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac{x^5 \left (A+B x^2\right )}{\left (b+c x^2\right )^3} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^2 (A+B x)}{(b+c x)^3} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{B}{c^3}-\frac{b^2 (b B-A c)}{c^3 (b+c x)^3}+\frac{b (3 b B-2 A c)}{c^3 (b+c x)^2}+\frac{-3 b B+A c}{c^3 (b+c x)}\right ) \, dx,x,x^2\right )\\ &=\frac{B x^2}{2 c^3}+\frac{b^2 (b B-A c)}{4 c^4 \left (b+c x^2\right )^2}-\frac{b (3 b B-2 A c)}{2 c^4 \left (b+c x^2\right )}-\frac{(3 b B-A c) \log \left (b+c x^2\right )}{2 c^4}\\ \end{align*}

Mathematica [A]  time = 0.0369192, size = 92, normalized size = 1.03 \[ \frac{2 A b c-3 b^2 B}{2 c^4 \left (b+c x^2\right )}+\frac{b^3 B-A b^2 c}{4 c^4 \left (b+c x^2\right )^2}+\frac{(A c-3 b B) \log \left (b+c x^2\right )}{2 c^4}+\frac{B x^2}{2 c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^11*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

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

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Maple [A]  time = 0.009, size = 109, normalized size = 1.2 \begin{align*}{\frac{B{x}^{2}}{2\,{c}^{3}}}+{\frac{\ln \left ( c{x}^{2}+b \right ) A}{2\,{c}^{3}}}-{\frac{3\,\ln \left ( c{x}^{2}+b \right ) Bb}{2\,{c}^{4}}}+{\frac{Ab}{{c}^{3} \left ( c{x}^{2}+b \right ) }}-{\frac{3\,B{b}^{2}}{2\,{c}^{4} \left ( c{x}^{2}+b \right ) }}-{\frac{{b}^{2}A}{4\,{c}^{3} \left ( c{x}^{2}+b \right ) ^{2}}}+{\frac{B{b}^{3}}{4\,{c}^{4} \left ( c{x}^{2}+b \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11*(B*x^2+A)/(c*x^4+b*x^2)^3,x)

[Out]

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

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Maxima [A]  time = 1.14462, size = 127, normalized size = 1.43 \begin{align*} -\frac{5 \, B b^{3} - 3 \, A b^{2} c + 2 \,{\left (3 \, B b^{2} c - 2 \, A b c^{2}\right )} x^{2}}{4 \,{\left (c^{6} x^{4} + 2 \, b c^{5} x^{2} + b^{2} c^{4}\right )}} + \frac{B x^{2}}{2 \, c^{3}} - \frac{{\left (3 \, B b - A 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^11*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.04805, size = 289, normalized size = 3.25 \begin{align*} \frac{2 \, B c^{3} x^{6} + 4 \, B b c^{2} x^{4} - 5 \, B b^{3} + 3 \, A b^{2} c - 4 \,{\left (B b^{2} c - A b c^{2}\right )} x^{2} - 2 \,{\left ({\left (3 \, B b c^{2} - A c^{3}\right )} x^{4} + 3 \, B b^{3} - A b^{2} c + 2 \,{\left (3 \, B b^{2} c - A b c^{2}\right )} x^{2}\right )} \log \left (c x^{2} + b\right )}{4 \,{\left (c^{6} x^{4} + 2 \, b c^{5} x^{2} + b^{2} c^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.46166, size = 94, normalized size = 1.06 \begin{align*} \frac{B x^{2}}{2 c^{3}} - \frac{- 3 A b^{2} c + 5 B b^{3} + x^{2} \left (- 4 A b c^{2} + 6 B b^{2} c\right )}{4 b^{2} c^{4} + 8 b c^{5} x^{2} + 4 c^{6} x^{4}} - \frac{\left (- A c + 3 B b\right ) \log{\left (b + c x^{2} \right )}}{2 c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**11*(B*x**2+A)/(c*x**4+b*x**2)**3,x)

[Out]

B*x**2/(2*c**3) - (-3*A*b**2*c + 5*B*b**3 + x**2*(-4*A*b*c**2 + 6*B*b**2*c))/(4*b**2*c**4 + 8*b*c**5*x**2 + 4*
c**6*x**4) - (-A*c + 3*B*b)*log(b + c*x**2)/(2*c**4)

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Giac [A]  time = 1.21136, size = 126, normalized size = 1.42 \begin{align*} \frac{B x^{2}}{2 \, c^{3}} - \frac{{\left (3 \, B b - A c\right )} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, c^{4}} + \frac{9 \, B b c^{2} x^{4} - 3 \, A c^{3} x^{4} + 12 \, B b^{2} c x^{2} - 2 \, A b c^{2} x^{2} + 4 \, B b^{3}}{4 \,{\left (c x^{2} + b\right )}^{2} c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

1/2*B*x^2/c^3 - 1/2*(3*B*b - A*c)*log(abs(c*x^2 + b))/c^4 + 1/4*(9*B*b*c^2*x^4 - 3*A*c^3*x^4 + 12*B*b^2*c*x^2
- 2*A*b*c^2*x^2 + 4*B*b^3)/((c*x^2 + b)^2*c^4)