3.16 \(\int \frac{1}{a x+b x^3} \, dx\)

Optimal. Leaf size=22 \[ \frac{\log (x)}{a}-\frac{\log \left (a+b x^2\right )}{2 a} \]

[Out]

Log[x]/a - Log[a + b*x^2]/(2*a)

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Rubi [A]  time = 0.0120126, antiderivative size = 22, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.454, Rules used = {1593, 266, 36, 29, 31} \[ \frac{\log (x)}{a}-\frac{\log \left (a+b x^2\right )}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[(a*x + b*x^3)^(-1),x]

[Out]

Log[x]/a - Log[a + b*x^2]/(2*a)

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.0042559, size = 22, normalized size = 1. \[ \frac{\log (x)}{a}-\frac{\log \left (a+b x^2\right )}{2 a} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*x + b*x^3)^(-1),x]

[Out]

Log[x]/a - Log[a + b*x^2]/(2*a)

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Maple [A]  time = 0.005, size = 21, normalized size = 1. \begin{align*}{\frac{\ln \left ( x \right ) }{a}}-{\frac{\ln \left ( b{x}^{2}+a \right ) }{2\,a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^3+a*x),x)

[Out]

ln(x)/a-1/2*ln(b*x^2+a)/a

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Maxima [A]  time = 1.06132, size = 27, normalized size = 1.23 \begin{align*} -\frac{\log \left (b x^{2} + a\right )}{2 \, a} + \frac{\log \left (x\right )}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^3+a*x),x, algorithm="maxima")

[Out]

-1/2*log(b*x^2 + a)/a + log(x)/a

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Fricas [A]  time = 1.40941, size = 49, normalized size = 2.23 \begin{align*} -\frac{\log \left (b x^{2} + a\right ) - 2 \, \log \left (x\right )}{2 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^3+a*x),x, algorithm="fricas")

[Out]

-1/2*(log(b*x^2 + a) - 2*log(x))/a

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Sympy [A]  time = 0.178339, size = 15, normalized size = 0.68 \begin{align*} \frac{\log{\left (x \right )}}{a} - \frac{\log{\left (\frac{a}{b} + x^{2} \right )}}{2 a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x**3+a*x),x)

[Out]

log(x)/a - log(a/b + x**2)/(2*a)

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Giac [A]  time = 1.18213, size = 32, normalized size = 1.45 \begin{align*} \frac{\log \left (x^{2}\right )}{2 \, a} - \frac{\log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^3+a*x),x, algorithm="giac")

[Out]

1/2*log(x^2)/a - 1/2*log(abs(b*x^2 + a))/a