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

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

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Rubi [A]  time = 0.10, antiderivative size = 83, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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 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]))))

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 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^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*}

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Mathematica [A]  time = 0.06, size = 72, normalized size = 0.87 \begin {gather*} \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} \end {gather*}

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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^9 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.38, size = 121, normalized size = 1.46 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.17, size = 106, normalized size = 1.28 \begin {gather*} \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 {gather*}

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)

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maple [A]  time = 0.05, size = 98, normalized size = 1.18 \begin {gather*} \frac {B \,x^{4}}{4 c^{2}}+\frac {A \,x^{2}}{2 c^{2}}-\frac {B b \,x^{2}}{c^{3}}-\frac {A \,b^{2}}{2 \left (c \,x^{2}+b \right ) c^{3}}-\frac {A b \ln \left (c \,x^{2}+b \right )}{c^{3}}+\frac {B \,b^{3}}{2 \left (c \,x^{2}+b \right ) c^{4}}+\frac {3 B \,b^{2} \ln \left (c \,x^{2}+b \right )}{2 c^{4}} \end {gather*}

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 = 1.32, size = 82, normalized size = 0.99 \begin {gather*} \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 {gather*}

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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mupad [B]  time = 0.07, size = 86, normalized size = 1.04 \begin {gather*} x^2\,\left (\frac {A}{2\,c^2}-\frac {B\,b}{c^3}\right )+\frac {\ln \left (c\,x^2+b\right )\,\left (3\,B\,b^2-2\,A\,b\,c\right )}{2\,c^4}+\frac {B\,x^4}{4\,c^2}+\frac {B\,b^3-A\,b^2\,c}{2\,c\,\left (c^4\,x^2+b\,c^3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.70, size = 78, normalized size = 0.94 \begin {gather*} \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}} + x^{2} \left (\frac {A}{2 c^{2}} - \frac {B b}{c^{3}}\right ) + \frac {- A b^{2} c + B b^{3}}{2 b c^{4} + 2 c^{5} x^{2}} \end {gather*}

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

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