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

Optimal. Leaf size=70 \[ -\frac{b B-A c}{2 b^2 x^2}+\frac{c (b B-A c) \log \left (b+c x^2\right )}{2 b^3}-\frac{c \log (x) (b B-A c)}{b^3}-\frac{A}{4 b x^4} \]

[Out]

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

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Rubi [A]  time = 0.0722856, antiderivative size = 70, 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} \[ -\frac{b B-A c}{2 b^2 x^2}+\frac{c (b B-A c) \log \left (b+c x^2\right )}{2 b^3}-\frac{c \log (x) (b B-A c)}{b^3}-\frac{A}{4 b x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.028756, size = 70, normalized size = 1. \[ \frac{-b \left (A b-2 A c x^2+2 b B x^2\right )+4 c x^4 \log (x) (A c-b B)+2 c x^4 (b B-A c) \log \left (b+c x^2\right )}{4 b^3 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 81, normalized size = 1.2 \begin{align*} -{\frac{A}{4\,b{x}^{4}}}+{\frac{Ac}{2\,{b}^{2}{x}^{2}}}-{\frac{B}{2\,b{x}^{2}}}+{\frac{A\ln \left ( x \right ){c}^{2}}{{b}^{3}}}-{\frac{Bc\ln \left ( x \right ) }{{b}^{2}}}-{\frac{{c}^{2}\ln \left ( c{x}^{2}+b \right ) A}{2\,{b}^{3}}}+{\frac{c\ln \left ( c{x}^{2}+b \right ) B}{2\,{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 2.59029, size = 95, normalized size = 1.36 \begin{align*} \frac{{\left (B b c - A c^{2}\right )} \log \left (c x^{2} + b\right )}{2 \, b^{3}} - \frac{{\left (B b c - A c^{2}\right )} \log \left (x^{2}\right )}{2 \, b^{3}} - \frac{2 \,{\left (B b - A c\right )} x^{2} + A b}{4 \, b^{2} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.792531, size = 158, normalized size = 2.26 \begin{align*} \frac{2 \,{\left (B b c - A c^{2}\right )} x^{4} \log \left (c x^{2} + b\right ) - 4 \,{\left (B b c - A c^{2}\right )} x^{4} \log \left (x\right ) - A b^{2} - 2 \,{\left (B b^{2} - A b c\right )} x^{2}}{4 \, b^{3} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.997032, size = 61, normalized size = 0.87 \begin{align*} - \frac{A b + x^{2} \left (- 2 A c + 2 B b\right )}{4 b^{2} x^{4}} - \frac{c \left (- A c + B b\right ) \log{\left (x \right )}}{b^{3}} + \frac{c \left (- A c + B b\right ) \log{\left (\frac{b}{c} + x^{2} \right )}}{2 b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14446, size = 135, normalized size = 1.93 \begin{align*} -\frac{{\left (B b c - A c^{2}\right )} \log \left (x^{2}\right )}{2 \, b^{3}} + \frac{{\left (B b c^{2} - A c^{3}\right )} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, b^{3} c} + \frac{3 \, B b c x^{4} - 3 \, A c^{2} x^{4} - 2 \, B b^{2} x^{2} + 2 \, A b c x^{2} - A b^{2}}{4 \, b^{3} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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