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

Optimal. Leaf size=73 \[ -\frac {(b B-2 A c) \log \left (b+c x^2\right )}{2 b^3}+\frac {\log (x) (b B-2 A c)}{b^3}+\frac {b B-A c}{2 b^2 \left (b+c x^2\right )}-\frac {A}{2 b^2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1584, 446, 77} \[ \frac {b B-A c}{2 b^2 \left (b+c x^2\right )}-\frac {(b B-2 A c) \log \left (b+c x^2\right )}{2 b^3}+\frac {\log (x) (b B-2 A c)}{b^3}-\frac {A}{2 b^2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.06, size = 64, normalized size = 0.88 \[ \frac {\frac {b (b B-A c)}{b+c x^2}+(2 A c-b B) \log \left (b+c x^2\right )+2 \log (x) (b B-2 A c)-\frac {A b}{x^2}}{2 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.80, size = 117, normalized size = 1.60 \[ -\frac {A b^{2} - {\left (B b^{2} - 2 \, A b c\right )} x^{2} + {\left ({\left (B b c - 2 \, A c^{2}\right )} x^{4} + {\left (B b^{2} - 2 \, A b c\right )} x^{2}\right )} \log \left (c x^{2} + b\right ) - 2 \, {\left ({\left (B b c - 2 \, A c^{2}\right )} x^{4} + {\left (B b^{2} - 2 \, A b c\right )} x^{2}\right )} \log \relax (x)}{2 \, {\left (b^{3} c x^{4} + b^{4} x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 80, normalized size = 1.10 \[ \frac {{\left (B b - 2 \, A c\right )} \log \left ({\left | x \right |}\right )}{b^{3}} + \frac {B b x^{2} - 2 \, A c x^{2} - A b}{2 \, {\left (c x^{4} + b x^{2}\right )} b^{2}} - \frac {{\left (B b c - 2 \, A c^{2}\right )} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, b^{3} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 86, normalized size = 1.18 \[ -\frac {A c}{2 \left (c \,x^{2}+b \right ) b^{2}}-\frac {2 A c \ln \relax (x )}{b^{3}}+\frac {A c \ln \left (c \,x^{2}+b \right )}{b^{3}}+\frac {B}{2 \left (c \,x^{2}+b \right ) b}+\frac {B \ln \relax (x )}{b^{2}}-\frac {B \ln \left (c \,x^{2}+b \right )}{2 b^{2}}-\frac {A}{2 b^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.33, size = 76, normalized size = 1.04 \[ \frac {{\left (B b - 2 \, A c\right )} x^{2} - A b}{2 \, {\left (b^{2} c x^{4} + b^{3} x^{2}\right )}} - \frac {{\left (B b - 2 \, A c\right )} \log \left (c x^{2} + b\right )}{2 \, b^{3}} + \frac {{\left (B b - 2 \, A c\right )} \log \left (x^{2}\right )}{2 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.15, size = 78, normalized size = 1.07 \[ \frac {\ln \left (c\,x^2+b\right )\,\left (2\,A\,c-B\,b\right )}{2\,b^3}-\frac {\frac {A}{2\,b}+\frac {x^2\,\left (2\,A\,c-B\,b\right )}{2\,b^2}}{c\,x^4+b\,x^2}-\frac {\ln \relax (x)\,\left (2\,A\,c-B\,b\right )}{b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.88, size = 70, normalized size = 0.96 \[ \frac {- A b + x^{2} \left (- 2 A c + B b\right )}{2 b^{3} x^{2} + 2 b^{2} c x^{4}} + \frac {\left (- 2 A c + B b\right ) \log {\relax (x )}}{b^{3}} - \frac {\left (- 2 A c + B b\right ) \log {\left (\frac {b}{c} + x^{2} \right )}}{2 b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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