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]

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

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Rubi [A]  time = 0.01, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 31

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

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 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]

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*}

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Mathematica [A]  time = 0.00, size = 21, normalized size = 1.00 \[ \frac {2}{21} \log (3 x+2)-\frac {1}{35} \log (5 x+1) \]

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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fricas [A]  time = 0.60, size = 17, normalized size = 0.81 \[ -\frac {1}{35} \, \log \left (5 \, x + 1\right ) + \frac {2}{21} \, \log \left (3 \, x + 2\right ) \]

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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giac [A]  time = 0.29, size = 19, normalized size = 0.90 \[ -\frac {1}{35} \, \log \left ({\left | 5 \, x + 1 \right |}\right ) + \frac {2}{21} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \]

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))

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maple [A]  time = 0.00, size = 18, normalized size = 0.86 \[ -\frac {\ln \left (5 x +1\right )}{35}+\frac {2 \ln \left (3 x +2\right )}{21} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.17, size = 17, normalized size = 0.81 \[ -\frac {1}{35} \, \log \left (5 \, x + 1\right ) + \frac {2}{21} \, \log \left (3 \, x + 2\right ) \]

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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mupad [B]  time = 2.11, size = 13, normalized size = 0.62 \[ \frac {2\,\ln \left (x+\frac {2}{3}\right )}{21}-\frac {\ln \left (x+\frac {1}{5}\right )}{35} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(15*x + 2/x + 13),x)

[Out]

(2*log(x + 2/3))/21 - log(x + 1/5)/35

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sympy [A]  time = 0.11, size = 17, normalized size = 0.81 \[ - \frac {\log {\left (x + \frac {1}{5} \right )}}{35} + \frac {2 \log {\left (x + \frac {2}{3} \right )}}{21} \]

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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