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

Optimal. Leaf size=16 \[ \frac{1}{2} \log \left (b x^2+c x^4\right ) \]

[Out]

Log[b*x^2 + c*x^4]/2

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Rubi [A]  time = 0.0239123, antiderivative size = 15, normalized size of antiderivative = 0.94, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {1584, 446, 72} \[ \frac{1}{2} \log \left (b+c x^2\right )+\log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[x] + Log[b + c*x^2]/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 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

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

Mathematica [A]  time = 0.0061463, size = 15, normalized size = 0.94 \[ \frac{1}{2} \log \left (b+c x^2\right )+\log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[x] + Log[b + c*x^2]/2

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Maple [A]  time = 0.004, size = 14, normalized size = 0.9 \begin{align*} \ln \left ( x \right ) +{\frac{\ln \left ( c{x}^{2}+b \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.996828, size = 23, normalized size = 1.44 \begin{align*} \frac{1}{2} \, \log \left (c x^{2} + b\right ) + \frac{1}{2} \, \log \left (x^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(c*x^2 + b) + 1/2*log(x^2)

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Fricas [A]  time = 1.02086, size = 39, normalized size = 2.44 \begin{align*} \frac{1}{2} \, \log \left (c x^{2} + b\right ) + \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(c*x^2 + b) + log(x)

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Sympy [A]  time = 0.324457, size = 12, normalized size = 0.75 \begin{align*} \log{\left (x \right )} + \frac{\log{\left (\frac{b}{c} + x^{2} \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x) + log(b/c + x**2)/2

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Giac [A]  time = 1.09658, size = 20, normalized size = 1.25 \begin{align*} \frac{1}{2} \, \log \left ({\left | c x^{2} + b \right |}\right ) + \log \left ({\left | x \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(abs(c*x^2 + b)) + log(abs(x))