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

Optimal. Leaf size=51 \[ -\frac{A \log \left (b+c x^2\right )}{2 b^2}+\frac{A \log (x)}{b^2}-\frac{b B-A c}{2 b c \left (b+c x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.0547747, antiderivative size = 51, 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} \[ -\frac{A \log \left (b+c x^2\right )}{2 b^2}+\frac{A \log (x)}{b^2}-\frac{b B-A c}{2 b c \left (b+c x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0285103, size = 46, normalized size = 0.9 \[ \frac{\frac{b (A c-b B)}{c \left (b+c x^2\right )}-A \log \left (b+c x^2\right )+2 A \log (x)}{2 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.012, size = 53, normalized size = 1. \begin{align*}{\frac{A\ln \left ( x \right ) }{{b}^{2}}}-{\frac{A\ln \left ( c{x}^{2}+b \right ) }{2\,{b}^{2}}}+{\frac{A}{2\,b \left ( c{x}^{2}+b \right ) }}-{\frac{B}{2\,c \left ( c{x}^{2}+b \right ) }} \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)^2,x)

[Out]

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

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Maxima [A]  time = 1.12098, size = 69, normalized size = 1.35 \begin{align*} -\frac{B b - A c}{2 \,{\left (b c^{2} x^{2} + b^{2} c\right )}} - \frac{A \log \left (c x^{2} + b\right )}{2 \, b^{2}} + \frac{A \log \left (x^{2}\right )}{2 \, b^{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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.791719, size = 151, normalized size = 2.96 \begin{align*} -\frac{B b^{2} - A b c +{\left (A c^{2} x^{2} + A b c\right )} \log \left (c x^{2} + b\right ) - 2 \,{\left (A c^{2} x^{2} + A b c\right )} \log \left (x\right )}{2 \,{\left (b^{2} c^{2} x^{2} + b^{3} c\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)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.574344, size = 46, normalized size = 0.9 \begin{align*} \frac{A \log{\left (x \right )}}{b^{2}} - \frac{A \log{\left (\frac{b}{c} + x^{2} \right )}}{2 b^{2}} - \frac{- A c + B b}{2 b^{2} c + 2 b c^{2} 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)**2,x)

[Out]

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

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Giac [A]  time = 1.59705, size = 70, normalized size = 1.37 \begin{align*} -\frac{A \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, b^{2}} + \frac{A \log \left ({\left | x \right |}\right )}{b^{2}} - \frac{B b^{2} - A b c}{2 \,{\left (c x^{2} + b\right )} b^{2} c} \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)^2,x, algorithm="giac")

[Out]

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