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

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

[Out]

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

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Rubi [A]  time = 0.108416, 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} \[ -\frac{c (b B-A c)}{2 b^3 \left (b+c x^2\right )}-\frac{b B-2 A c}{2 b^3 x^2}+\frac{c (2 b B-3 A c) \log \left (b+c x^2\right )}{2 b^4}-\frac{c \log (x) (2 b B-3 A c)}{b^4}-\frac{A}{4 b^2 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0968646, size = 85, normalized size = 0.88 \[ -\frac{\frac{A b^2}{x^4}+\frac{2 b c (b B-A c)}{b+c x^2}+\frac{2 b (b B-2 A c)}{x^2}+2 c (3 A c-2 b B) \log \left (b+c x^2\right )-4 c \log (x) (3 A c-2 b B)}{4 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.013, size = 114, normalized size = 1.2 \begin{align*} -{\frac{A}{4\,{x}^{4}{b}^{2}}}+{\frac{Ac}{{b}^{3}{x}^{2}}}-{\frac{B}{2\,{b}^{2}{x}^{2}}}+3\,{\frac{A\ln \left ( x \right ){c}^{2}}{{b}^{4}}}-2\,{\frac{Bc\ln \left ( x \right ) }{{b}^{3}}}-{\frac{3\,{c}^{2}\ln \left ( c{x}^{2}+b \right ) A}{2\,{b}^{4}}}+{\frac{c\ln \left ( c{x}^{2}+b \right ) B}{{b}^{3}}}+{\frac{A{c}^{2}}{2\,{b}^{3} \left ( c{x}^{2}+b \right ) }}-{\frac{Bc}{2\,{b}^{2} \left ( c{x}^{2}+b \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.17452, size = 143, normalized size = 1.47 \begin{align*} -\frac{2 \,{\left (2 \, B b c - 3 \, A c^{2}\right )} x^{4} + A b^{2} +{\left (2 \, B b^{2} - 3 \, A b c\right )} x^{2}}{4 \,{\left (b^{3} c x^{6} + b^{4} x^{4}\right )}} + \frac{{\left (2 \, B b c - 3 \, A c^{2}\right )} \log \left (c x^{2} + b\right )}{2 \, b^{4}} - \frac{{\left (2 \, B b c - 3 \, A c^{2}\right )} \log \left (x^{2}\right )}{2 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.614585, size = 327, normalized size = 3.37 \begin{align*} -\frac{2 \,{\left (2 \, B b^{2} c - 3 \, A b c^{2}\right )} x^{4} + A b^{3} +{\left (2 \, B b^{3} - 3 \, A b^{2} c\right )} x^{2} - 2 \,{\left ({\left (2 \, B b c^{2} - 3 \, A c^{3}\right )} x^{6} +{\left (2 \, B b^{2} c - 3 \, A b c^{2}\right )} x^{4}\right )} \log \left (c x^{2} + b\right ) + 4 \,{\left ({\left (2 \, B b c^{2} - 3 \, A c^{3}\right )} x^{6} +{\left (2 \, B b^{2} c - 3 \, A b c^{2}\right )} x^{4}\right )} \log \left (x\right )}{4 \,{\left (b^{4} c x^{6} + b^{5} x^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.29994, size = 100, normalized size = 1.03 \begin{align*} - \frac{A b^{2} + x^{4} \left (- 6 A c^{2} + 4 B b c\right ) + x^{2} \left (- 3 A b c + 2 B b^{2}\right )}{4 b^{4} x^{4} + 4 b^{3} c x^{6}} - \frac{c \left (- 3 A c + 2 B b\right ) \log{\left (x \right )}}{b^{4}} + \frac{c \left (- 3 A c + 2 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((B*x**2+A)/x/(c*x**4+b*x**2)**2,x)

[Out]

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

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Giac [A]  time = 1.14966, size = 203, normalized size = 2.09 \begin{align*} -\frac{{\left (2 \, B b c - 3 \, A c^{2}\right )} \log \left (x^{2}\right )}{2 \, b^{4}} + \frac{{\left (2 \, B b c^{2} - 3 \, A c^{3}\right )} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, b^{4} c} - \frac{2 \, B b c^{2} x^{2} - 3 \, A c^{3} x^{2} + 3 \, B b^{2} c - 4 \, A b c^{2}}{2 \,{\left (c x^{2} + b\right )} b^{4}} + \frac{6 \, B b c x^{4} - 9 \, A c^{2} x^{4} - 2 \, B b^{2} x^{2} + 4 \, A b c x^{2} - A b^{2}}{4 \, b^{4} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*B*b*c - 3*A*c^2)*log(x^2)/b^4 + 1/2*(2*B*b*c^2 - 3*A*c^3)*log(abs(c*x^2 + b))/(b^4*c) - 1/2*(2*B*b*c^2
*x^2 - 3*A*c^3*x^2 + 3*B*b^2*c - 4*A*b*c^2)/((c*x^2 + b)*b^4) + 1/4*(6*B*b*c*x^4 - 9*A*c^2*x^4 - 2*B*b^2*x^2 +
4*A*b*c*x^2 - A*b^2)/(b^4*x^4)