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

Optimal. Leaf size=97 \[ \frac{b B-2 A c}{2 b^3 \left (b+c x^2\right )}+\frac{b B-A c}{4 b^2 \left (b+c x^2\right )^2}-\frac{(b B-3 A c) \log \left (b+c x^2\right )}{2 b^4}+\frac{\log (x) (b B-3 A c)}{b^4}-\frac{A}{2 b^3 x^2} \]

[Out]

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

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Rubi [A]  time = 0.117311, antiderivative size = 97, 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} \[ \frac{b B-2 A c}{2 b^3 \left (b+c x^2\right )}+\frac{b B-A c}{4 b^2 \left (b+c x^2\right )^2}-\frac{(b B-3 A c) \log \left (b+c x^2\right )}{2 b^4}+\frac{\log (x) (b B-3 A c)}{b^4}-\frac{A}{2 b^3 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0595325, size = 86, normalized size = 0.89 \[ \frac{\frac{b^2 (b B-A c)}{\left (b+c x^2\right )^2}+\frac{2 b (b B-2 A c)}{b+c x^2}-2 (b B-3 A c) \log \left (b+c x^2\right )+4 \log (x) (b B-3 A c)-\frac{2 A b}{x^2}}{4 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.015, size = 118, normalized size = 1.2 \begin{align*} -{\frac{A}{2\,{b}^{3}{x}^{2}}}-3\,{\frac{A\ln \left ( x \right ) c}{{b}^{4}}}+{\frac{\ln \left ( x \right ) B}{{b}^{3}}}+{\frac{3\,c\ln \left ( c{x}^{2}+b \right ) A}{2\,{b}^{4}}}-{\frac{\ln \left ( c{x}^{2}+b \right ) B}{2\,{b}^{3}}}-{\frac{Ac}{{b}^{3} \left ( c{x}^{2}+b \right ) }}+{\frac{B}{2\,{b}^{2} \left ( c{x}^{2}+b \right ) }}-{\frac{Ac}{4\,{b}^{2} \left ( c{x}^{2}+b \right ) ^{2}}}+{\frac{B}{4\,b \left ( c{x}^{2}+b \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.0992, size = 147, normalized size = 1.52 \begin{align*} \frac{2 \,{\left (B b c - 3 \, A c^{2}\right )} x^{4} - 2 \, A b^{2} + 3 \,{\left (B b^{2} - 3 \, A b c\right )} x^{2}}{4 \,{\left (b^{3} c^{2} x^{6} + 2 \, b^{4} c x^{4} + b^{5} x^{2}\right )}} - \frac{{\left (B b - 3 \, A c\right )} \log \left (c x^{2} + b\right )}{2 \, b^{4}} + \frac{{\left (B b - 3 \, A c\right )} \log \left (x^{2}\right )}{2 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.05911, size = 412, normalized size = 4.25 \begin{align*} \frac{2 \,{\left (B b^{2} c - 3 \, A b c^{2}\right )} x^{4} - 2 \, A b^{3} + 3 \,{\left (B b^{3} - 3 \, A b^{2} c\right )} x^{2} - 2 \,{\left ({\left (B b c^{2} - 3 \, A c^{3}\right )} x^{6} + 2 \,{\left (B b^{2} c - 3 \, A b c^{2}\right )} x^{4} +{\left (B b^{3} - 3 \, A b^{2} c\right )} x^{2}\right )} \log \left (c x^{2} + b\right ) + 4 \,{\left ({\left (B b c^{2} - 3 \, A c^{3}\right )} x^{6} + 2 \,{\left (B b^{2} c - 3 \, A b c^{2}\right )} x^{4} +{\left (B b^{3} - 3 \, A b^{2} c\right )} x^{2}\right )} \log \left (x\right )}{4 \,{\left (b^{4} c^{2} x^{6} + 2 \, b^{5} c x^{4} + b^{6} x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.36029, size = 107, normalized size = 1.1 \begin{align*} \frac{- 2 A b^{2} + x^{4} \left (- 6 A c^{2} + 2 B b c\right ) + x^{2} \left (- 9 A b c + 3 B b^{2}\right )}{4 b^{5} x^{2} + 8 b^{4} c x^{4} + 4 b^{3} c^{2} x^{6}} + \frac{\left (- 3 A c + B b\right ) \log{\left (x \right )}}{b^{4}} - \frac{\left (- 3 A c + B b\right ) \log{\left (\frac{b}{c} + x^{2} \right )}}{2 b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-2*A*b**2 + x**4*(-6*A*c**2 + 2*B*b*c) + x**2*(-9*A*b*c + 3*B*b**2))/(4*b**5*x**2 + 8*b**4*c*x**4 + 4*b**3*c*
*2*x**6) + (-3*A*c + B*b)*log(x)/b**4 - (-3*A*c + B*b)*log(b/c + x**2)/(2*b**4)

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Giac [A]  time = 1.19155, size = 142, normalized size = 1.46 \begin{align*} \frac{{\left (B b - 3 \, A c\right )} \log \left ({\left | x \right |}\right )}{b^{4}} - \frac{{\left (B b c - 3 \, A c^{2}\right )} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, b^{4} c} + \frac{2 \,{\left (B b^{2} c - 3 \, A b c^{2}\right )} x^{4} - 2 \, A b^{3} + 3 \,{\left (B b^{3} - 3 \, A b^{2} c\right )} x^{2}}{4 \,{\left (c x^{2} + b\right )}^{2} b^{4} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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