3.381 \(\int \frac{1}{13+\frac{2}{x}+15 x} \, dx\)

Optimal. Leaf size=21 \[ \frac{2}{21} \log (3 x+2)-\frac{1}{35} \log (5 x+1) \]

[Out]

(2*Log[2 + 3*x])/21 - Log[1 + 5*x]/35

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Rubi [A]  time = 0.0073871, antiderivative size = 21, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1350, 632, 31} \[ \frac{2}{21} \log (3 x+2)-\frac{1}{35} \log (5 x+1) \]

Antiderivative was successfully verified.

[In]

Int[(13 + 2/x + 15*x)^(-1),x]

[Out]

(2*Log[2 + 3*x])/21 - Log[1 + 5*x]/35

Rule 1350

Int[((a_) + (c_.)*(x_)^(n_.) + (b_.)*(x_)^(mn_))^(p_.), x_Symbol] :> Int[(b + a*x^n + c*x^(2*n))^p/x^(n*p), x]
/; FreeQ[{a, b, c, n}, x] && EqQ[mn, -n] && IntegerQ[p] && PosQ[n]

Rule 632

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && NiceSqrtQ[b^2 - 4*a*
c]

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}{13+\frac{2}{x}+15 x} \, dx &=\int \frac{x}{2+13 x+15 x^2} \, dx\\ &=-\left (\frac{3}{7} \int \frac{1}{3+15 x} \, dx\right )+\frac{10}{7} \int \frac{1}{10+15 x} \, dx\\ &=\frac{2}{21} \log (2+3 x)-\frac{1}{35} \log (1+5 x)\\ \end{align*}

Mathematica [A]  time = 0.0029758, size = 21, normalized size = 1. \[ \frac{2}{21} \log (3 x+2)-\frac{1}{35} \log (5 x+1) \]

Antiderivative was successfully verified.

[In]

Integrate[(13 + 2/x + 15*x)^(-1),x]

[Out]

(2*Log[2 + 3*x])/21 - Log[1 + 5*x]/35

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Maple [A]  time = 0.006, size = 18, normalized size = 0.9 \begin{align*}{\frac{2\,\ln \left ( 2+3\,x \right ) }{21}}-{\frac{\ln \left ( 1+5\,x \right ) }{35}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(13+2/x+15*x),x)

[Out]

2/21*ln(2+3*x)-1/35*ln(1+5*x)

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Maxima [A]  time = 0.998803, size = 23, normalized size = 1.1 \begin{align*} -\frac{1}{35} \, \log \left (5 \, x + 1\right ) + \frac{2}{21} \, \log \left (3 \, x + 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(13+2/x+15*x),x, algorithm="maxima")

[Out]

-1/35*log(5*x + 1) + 2/21*log(3*x + 2)

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Fricas [A]  time = 1.25802, size = 54, normalized size = 2.57 \begin{align*} -\frac{1}{35} \, \log \left (5 \, x + 1\right ) + \frac{2}{21} \, \log \left (3 \, x + 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(13+2/x+15*x),x, algorithm="fricas")

[Out]

-1/35*log(5*x + 1) + 2/21*log(3*x + 2)

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Sympy [A]  time = 0.098802, size = 17, normalized size = 0.81 \begin{align*} - \frac{\log{\left (x + \frac{1}{5} \right )}}{35} + \frac{2 \log{\left (x + \frac{2}{3} \right )}}{21} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(13+2/x+15*x),x)

[Out]

-log(x + 1/5)/35 + 2*log(x + 2/3)/21

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Giac [A]  time = 1.15851, size = 26, normalized size = 1.24 \begin{align*} -\frac{1}{35} \, \log \left ({\left | 5 \, x + 1 \right |}\right ) + \frac{2}{21} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(13+2/x+15*x),x, algorithm="giac")

[Out]

-1/35*log(abs(5*x + 1)) + 2/21*log(abs(3*x + 2))