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

Optimal. Leaf size=96 \[ -\frac{b^2 x^2 (b B-A c)}{2 c^4}+\frac{b^3 (b B-A c) \log \left (b+c x^2\right )}{2 c^5}-\frac{x^6 (b B-A c)}{6 c^2}+\frac{b x^4 (b B-A c)}{4 c^3}+\frac{B x^8}{8 c} \]

[Out]

-(b^2*(b*B - A*c)*x^2)/(2*c^4) + (b*(b*B - A*c)*x^4)/(4*c^3) - ((b*B - A*c)*x^6)/(6*c^2) + (B*x^8)/(8*c) + (b^
3*(b*B - A*c)*Log[b + c*x^2])/(2*c^5)

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Rubi [A]  time = 0.126403, antiderivative size = 96, 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} \[ -\frac{b^2 x^2 (b B-A c)}{2 c^4}+\frac{b^3 (b B-A c) \log \left (b+c x^2\right )}{2 c^5}-\frac{x^6 (b B-A c)}{6 c^2}+\frac{b x^4 (b B-A c)}{4 c^3}+\frac{B x^8}{8 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b^2*(b*B - A*c)*x^2)/(2*c^4) + (b*(b*B - A*c)*x^4)/(4*c^3) - ((b*B - A*c)*x^6)/(6*c^2) + (B*x^8)/(8*c) + (b^
3*(b*B - 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^9 \left (A+B x^2\right )}{b x^2+c x^4} \, dx &=\int \frac{x^7 \left (A+B x^2\right )}{b+c x^2} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^3 (A+B x)}{b+c x} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (-\frac{b^2 (b B-A c)}{c^4}+\frac{b (b B-A c) x}{c^3}+\frac{(-b B+A c) x^2}{c^2}+\frac{B x^3}{c}+\frac{b^3 (b B-A c)}{c^4 (b+c x)}\right ) \, dx,x,x^2\right )\\ &=-\frac{b^2 (b B-A c) x^2}{2 c^4}+\frac{b (b B-A c) x^4}{4 c^3}-\frac{(b B-A c) x^6}{6 c^2}+\frac{B x^8}{8 c}+\frac{b^3 (b B-A c) \log \left (b+c x^2\right )}{2 c^5}\\ \end{align*}

Mathematica [A]  time = 0.0366463, size = 92, normalized size = 0.96 \[ \frac{c x^2 \left (6 b^2 c \left (2 A+B x^2\right )-2 b c^2 x^2 \left (3 A+2 B x^2\right )+c^3 x^4 \left (4 A+3 B x^2\right )-12 b^3 B\right )+12 b^3 (b B-A c) \log \left (b+c x^2\right )}{24 c^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*x^2*(-12*b^3*B + 6*b^2*c*(2*A + B*x^2) - 2*b*c^2*x^2*(3*A + 2*B*x^2) + c^3*x^4*(4*A + 3*B*x^2)) + 12*b^3*(b
*B - A*c)*Log[b + c*x^2])/(24*c^5)

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Maple [A]  time = 0.004, size = 110, normalized size = 1.2 \begin{align*}{\frac{B{x}^{8}}{8\,c}}+{\frac{A{x}^{6}}{6\,c}}-{\frac{B{x}^{6}b}{6\,{c}^{2}}}-{\frac{Ab{x}^{4}}{4\,{c}^{2}}}+{\frac{B{x}^{4}{b}^{2}}{4\,{c}^{3}}}+{\frac{A{b}^{2}{x}^{2}}{2\,{c}^{3}}}-{\frac{B{x}^{2}{b}^{3}}{2\,{c}^{4}}}-{\frac{{b}^{3}\ln \left ( c{x}^{2}+b \right ) A}{2\,{c}^{4}}}+{\frac{{b}^{4}\ln \left ( c{x}^{2}+b \right ) B}{2\,{c}^{5}}} \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),x)

[Out]

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

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Maxima [A]  time = 1.01772, size = 131, normalized size = 1.36 \begin{align*} \frac{3 \, B c^{3} x^{8} - 4 \,{\left (B b c^{2} - A c^{3}\right )} x^{6} + 6 \,{\left (B b^{2} c - A b c^{2}\right )} x^{4} - 12 \,{\left (B b^{3} - A b^{2} c\right )} x^{2}}{24 \, c^{4}} + \frac{{\left (B b^{4} - A b^{3} 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^9*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="maxima")

[Out]

1/24*(3*B*c^3*x^8 - 4*(B*b*c^2 - A*c^3)*x^6 + 6*(B*b^2*c - A*b*c^2)*x^4 - 12*(B*b^3 - A*b^2*c)*x^2)/c^4 + 1/2*
(B*b^4 - A*b^3*c)*log(c*x^2 + b)/c^5

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Fricas [A]  time = 0.482255, size = 201, normalized size = 2.09 \begin{align*} \frac{3 \, B c^{4} x^{8} - 4 \,{\left (B b c^{3} - A c^{4}\right )} x^{6} + 6 \,{\left (B b^{2} c^{2} - A b c^{3}\right )} x^{4} - 12 \,{\left (B b^{3} c - A b^{2} c^{2}\right )} x^{2} + 12 \,{\left (B b^{4} - A b^{3} c\right )} \log \left (c x^{2} + b\right )}{24 \, c^{5}} \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),x, algorithm="fricas")

[Out]

1/24*(3*B*c^4*x^8 - 4*(B*b*c^3 - A*c^4)*x^6 + 6*(B*b^2*c^2 - A*b*c^3)*x^4 - 12*(B*b^3*c - A*b^2*c^2)*x^2 + 12*
(B*b^4 - A*b^3*c)*log(c*x^2 + b))/c^5

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Sympy [A]  time = 0.46994, size = 85, normalized size = 0.89 \begin{align*} \frac{B x^{8}}{8 c} + \frac{b^{3} \left (- A c + B b\right ) \log{\left (b + c x^{2} \right )}}{2 c^{5}} - \frac{x^{6} \left (- A c + B b\right )}{6 c^{2}} + \frac{x^{4} \left (- A b c + B b^{2}\right )}{4 c^{3}} - \frac{x^{2} \left (- A b^{2} c + B b^{3}\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),x)

[Out]

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

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Giac [A]  time = 1.23302, size = 136, normalized size = 1.42 \begin{align*} \frac{3 \, B c^{3} x^{8} - 4 \, B b c^{2} x^{6} + 4 \, A c^{3} x^{6} + 6 \, B b^{2} c x^{4} - 6 \, A b c^{2} x^{4} - 12 \, B b^{3} x^{2} + 12 \, A b^{2} c x^{2}}{24 \, c^{4}} + \frac{{\left (B b^{4} - A b^{3} c\right )} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, c^{5}} \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),x, algorithm="giac")

[Out]

1/24*(3*B*c^3*x^8 - 4*B*b*c^2*x^6 + 4*A*c^3*x^6 + 6*B*b^2*c*x^4 - 6*A*b*c^2*x^4 - 12*B*b^3*x^2 + 12*A*b^2*c*x^
2)/c^4 + 1/2*(B*b^4 - A*b^3*c)*log(abs(c*x^2 + b))/c^5