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

Optimal. Leaf size=105 \[ -\frac{b^3 (b B-A c)}{2 c^5 \left (b+c x^2\right )}-\frac{b^2 (4 b B-3 A c) \log \left (b+c x^2\right )}{2 c^5}-\frac{x^4 (2 b B-A c)}{4 c^3}+\frac{b x^2 (3 b B-2 A c)}{2 c^4}+\frac{B x^6}{6 c^2} \]

[Out]

(b*(3*b*B - 2*A*c)*x^2)/(2*c^4) - ((2*b*B - A*c)*x^4)/(4*c^3) + (B*x^6)/(6*c^2) - (b^3*(b*B - A*c))/(2*c^5*(b
+ c*x^2)) - (b^2*(4*b*B - 3*A*c)*Log[b + c*x^2])/(2*c^5)

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Rubi [A]  time = 0.135018, antiderivative size = 105, 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} \[ -\frac{b^3 (b B-A c)}{2 c^5 \left (b+c x^2\right )}-\frac{b^2 (4 b B-3 A c) \log \left (b+c x^2\right )}{2 c^5}-\frac{x^4 (2 b B-A c)}{4 c^3}+\frac{b x^2 (3 b B-2 A c)}{2 c^4}+\frac{B x^6}{6 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(3*b*B - 2*A*c)*x^2)/(2*c^4) - ((2*b*B - A*c)*x^4)/(4*c^3) + (B*x^6)/(6*c^2) - (b^3*(b*B - A*c))/(2*c^5*(b
+ c*x^2)) - (b^2*(4*b*B - 3*A*c)*Log[b + c*x^2])/(2*c^5)

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 )^2} \, dx &=\int \frac{x^7 \left (A+B x^2\right )}{\left (b+c x^2\right )^2} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^3 (A+B x)}{(b+c x)^2} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{b (3 b B-2 A c)}{c^4}+\frac{(-2 b B+A c) x}{c^3}+\frac{B x^2}{c^2}+\frac{b^3 (b B-A c)}{c^4 (b+c x)^2}-\frac{b^2 (4 b B-3 A c)}{c^4 (b+c x)}\right ) \, dx,x,x^2\right )\\ &=\frac{b (3 b B-2 A c) x^2}{2 c^4}-\frac{(2 b B-A c) x^4}{4 c^3}+\frac{B x^6}{6 c^2}-\frac{b^3 (b B-A c)}{2 c^5 \left (b+c x^2\right )}-\frac{b^2 (4 b B-3 A c) \log \left (b+c x^2\right )}{2 c^5}\\ \end{align*}

Mathematica [A]  time = 0.0659772, size = 93, normalized size = 0.89 \[ \frac{\frac{6 b^3 (A c-b B)}{b+c x^2}+6 b^2 (3 A c-4 b B) \log \left (b+c x^2\right )+3 c^2 x^4 (A c-2 b B)+6 b c x^2 (3 b B-2 A c)+2 B c^3 x^6}{12 c^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*b*c*(3*b*B - 2*A*c)*x^2 + 3*c^2*(-2*b*B + A*c)*x^4 + 2*B*c^3*x^6 + (6*b^3*(-(b*B) + A*c))/(b + c*x^2) + 6*b
^2*(-4*b*B + 3*A*c)*Log[b + c*x^2])/(12*c^5)

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Maple [A]  time = 0.01, size = 122, normalized size = 1.2 \begin{align*}{\frac{B{x}^{6}}{6\,{c}^{2}}}+{\frac{A{x}^{4}}{4\,{c}^{2}}}-{\frac{B{x}^{4}b}{2\,{c}^{3}}}-{\frac{Ab{x}^{2}}{{c}^{3}}}+{\frac{3\,B{x}^{2}{b}^{2}}{2\,{c}^{4}}}+{\frac{3\,{b}^{2}\ln \left ( c{x}^{2}+b \right ) A}{2\,{c}^{4}}}-2\,{\frac{{b}^{3}\ln \left ( c{x}^{2}+b \right ) B}{{c}^{5}}}+{\frac{{b}^{3}A}{2\,{c}^{4} \left ( c{x}^{2}+b \right ) }}-{\frac{B{b}^{4}}{2\,{c}^{5} \left ( c{x}^{2}+b \right ) }} \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)^2,x)

[Out]

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

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Maxima [A]  time = 1.11953, size = 144, normalized size = 1.37 \begin{align*} -\frac{B b^{4} - A b^{3} c}{2 \,{\left (c^{6} x^{2} + b c^{5}\right )}} + \frac{2 \, B c^{2} x^{6} - 3 \,{\left (2 \, B b c - A c^{2}\right )} x^{4} + 6 \,{\left (3 \, B b^{2} - 2 \, A b c\right )} x^{2}}{12 \, c^{4}} - \frac{{\left (4 \, B b^{3} - 3 \, A b^{2} c\right )} \log \left (c x^{2} + b\right )}{2 \, c^{5}} \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)^2,x, algorithm="maxima")

[Out]

-1/2*(B*b^4 - A*b^3*c)/(c^6*x^2 + b*c^5) + 1/12*(2*B*c^2*x^6 - 3*(2*B*b*c - A*c^2)*x^4 + 6*(3*B*b^2 - 2*A*b*c)
*x^2)/c^4 - 1/2*(4*B*b^3 - 3*A*b^2*c)*log(c*x^2 + b)/c^5

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Fricas [A]  time = 0.644268, size = 309, normalized size = 2.94 \begin{align*} \frac{2 \, B c^{4} x^{8} -{\left (4 \, B b c^{3} - 3 \, A c^{4}\right )} x^{6} - 6 \, B b^{4} + 6 \, A b^{3} c + 3 \,{\left (4 \, B b^{2} c^{2} - 3 \, A b c^{3}\right )} x^{4} + 6 \,{\left (3 \, B b^{3} c - 2 \, A b^{2} c^{2}\right )} x^{2} - 6 \,{\left (4 \, B b^{4} - 3 \, A b^{3} c +{\left (4 \, B b^{3} c - 3 \, A b^{2} c^{2}\right )} x^{2}\right )} \log \left (c x^{2} + b\right )}{12 \,{\left (c^{6} x^{2} + b c^{5}\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)^2,x, algorithm="fricas")

[Out]

1/12*(2*B*c^4*x^8 - (4*B*b*c^3 - 3*A*c^4)*x^6 - 6*B*b^4 + 6*A*b^3*c + 3*(4*B*b^2*c^2 - 3*A*b*c^3)*x^4 + 6*(3*B
*b^3*c - 2*A*b^2*c^2)*x^2 - 6*(4*B*b^4 - 3*A*b^3*c + (4*B*b^3*c - 3*A*b^2*c^2)*x^2)*log(c*x^2 + b))/(c^6*x^2 +
b*c^5)

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Sympy [A]  time = 0.87239, size = 102, normalized size = 0.97 \begin{align*} \frac{B x^{6}}{6 c^{2}} - \frac{b^{2} \left (- 3 A c + 4 B b\right ) \log{\left (b + c x^{2} \right )}}{2 c^{5}} - \frac{- A b^{3} c + B b^{4}}{2 b c^{5} + 2 c^{6} x^{2}} - \frac{x^{4} \left (- A c + 2 B b\right )}{4 c^{3}} + \frac{x^{2} \left (- 2 A b c + 3 B b^{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)**2,x)

[Out]

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

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Giac [A]  time = 1.21849, size = 182, normalized size = 1.73 \begin{align*} -\frac{{\left (4 \, B b^{3} - 3 \, A b^{2} c\right )} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, c^{5}} + \frac{2 \, B c^{4} x^{6} - 6 \, B b c^{3} x^{4} + 3 \, A c^{4} x^{4} + 18 \, B b^{2} c^{2} x^{2} - 12 \, A b c^{3} x^{2}}{12 \, c^{6}} + \frac{4 \, B b^{3} c x^{2} - 3 \, A b^{2} c^{2} x^{2} + 3 \, B b^{4} - 2 \, A b^{3} c}{2 \,{\left (c x^{2} + b\right )} c^{5}} \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)^2,x, algorithm="giac")

[Out]

-1/2*(4*B*b^3 - 3*A*b^2*c)*log(abs(c*x^2 + b))/c^5 + 1/12*(2*B*c^4*x^6 - 6*B*b*c^3*x^4 + 3*A*c^4*x^4 + 18*B*b^
2*c^2*x^2 - 12*A*b*c^3*x^2)/c^6 + 1/2*(4*B*b^3*c*x^2 - 3*A*b^2*c^2*x^2 + 3*B*b^4 - 2*A*b^3*c)/((c*x^2 + b)*c^5
)