3.25 \(\int \frac {1}{b x+d x^3} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 22, normalized size of antiderivative = 1.00, 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} \[ \frac {\log (x)}{b}-\frac {\log \left (b+d x^2\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[(b*x + d*x^3)^(-1),x]

[Out]

Log[x]/b - Log[b + d*x^2]/(2*b)

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]

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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 22, normalized size = 1.00 \[ \frac {\log (x)}{b}-\frac {\log \left (b+d x^2\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x + d*x^3)^(-1),x]

[Out]

Log[x]/b - Log[b + d*x^2]/(2*b)

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fricas [A]  time = 0.78, size = 18, normalized size = 0.82 \[ -\frac {\log \left (d x^{2} + b\right ) - 2 \, \log \relax (x)}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(log(d*x^2 + b) - 2*log(x))/b

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giac [A]  time = 0.30, size = 24, normalized size = 1.09 \[ \frac {\log \left (x^{2}\right )}{2 \, b} - \frac {\log \left ({\left | d x^{2} + b \right |}\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 21, normalized size = 0.95 \[ \frac {\ln \relax (x )}{b}-\frac {\ln \left (d \,x^{2}+b \right )}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.79, size = 20, normalized size = 0.91 \[ -\frac {\log \left (d x^{2} + b\right )}{2 \, b} + \frac {\log \relax (x)}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*log(d*x^2 + b)/b + log(x)/b

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mupad [B]  time = 2.13, size = 18, normalized size = 0.82 \[ -\frac {\ln \left (d\,x^2+b\right )-2\,\ln \relax (x)}{2\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x + d*x^3),x)

[Out]

-(log(b + d*x^2) - 2*log(x))/(2*b)

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sympy [A]  time = 0.20, size = 15, normalized size = 0.68 \[ \frac {\log {\relax (x )}}{b} - \frac {\log {\left (\frac {b}{d} + x^{2} \right )}}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x)/b - log(b/d + x**2)/(2*b)

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