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

Optimal. Leaf size=49 \[ -\frac {(b B-A c) \log \left (b+c x^2\right )}{2 b^2}+\frac {\log (x) (b B-A c)}{b^2}-\frac {A}{2 b x^2} \]

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Rubi [A]  time = 0.06, antiderivative size = 49, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\frac {(b B-A c) \log \left (b+c x^2\right )}{2 b^2}+\frac {\log (x) (b B-A c)}{b^2}-\frac {A}{2 b x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.02, size = 49, normalized size = 1.00 \begin {gather*} \frac {(A c-b B) \log \left (b+c x^2\right )}{2 b^2}+\frac {\log (x) (b B-A c)}{b^2}-\frac {A}{2 b x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*A/(b*x^2) + ((b*B - A*c)*Log[x])/b^2 + ((-(b*B) + A*c)*Log[b + c*x^2])/(2*b^2)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A+B x^2}{x \left (b x^2+c x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.40, size = 47, normalized size = 0.96 \begin {gather*} -\frac {{\left (B b - A c\right )} x^{2} \log \left (c x^{2} + b\right ) - 2 \, {\left (B b - A c\right )} x^{2} \log \relax (x) + A b}{2 \, b^{2} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 71, normalized size = 1.45 \begin {gather*} \frac {{\left (B b - A c\right )} \log \left (x^{2}\right )}{2 \, b^{2}} - \frac {{\left (B b c - A c^{2}\right )} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, b^{2} c} - \frac {B b x^{2} - A c x^{2} + A b}{2 \, b^{2} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 56, normalized size = 1.14 \begin {gather*} -\frac {A c \ln \relax (x )}{b^{2}}+\frac {A c \ln \left (c \,x^{2}+b \right )}{2 b^{2}}+\frac {B \ln \relax (x )}{b}-\frac {B \ln \left (c \,x^{2}+b \right )}{2 b}-\frac {A}{2 b \,x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.32, size = 48, normalized size = 0.98 \begin {gather*} -\frac {{\left (B b - A c\right )} \log \left (c x^{2} + b\right )}{2 \, b^{2}} + \frac {{\left (B b - A c\right )} \log \left (x^{2}\right )}{2 \, b^{2}} - \frac {A}{2 \, b x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.14, size = 46, normalized size = 0.94 \begin {gather*} \frac {\ln \left (c\,x^2+b\right )\,\left (A\,c-B\,b\right )}{2\,b^2}-\frac {A}{2\,b\,x^2}-\frac {\ln \relax (x)\,\left (A\,c-B\,b\right )}{b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.76, size = 41, normalized size = 0.84 \begin {gather*} - \frac {A}{2 b x^{2}} + \frac {\left (- A c + B b\right ) \log {\relax (x )}}{b^{2}} - \frac {\left (- A c + B b\right ) \log {\left (\frac {b}{c} + x^{2} \right )}}{2 b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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