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

Optimal. Leaf size=27 \[ \frac {\log (x)}{2}+\frac {3}{14} \log (3 x+2)-\frac {5}{7} \log (5 x+1) \]

[Out]

1/2*ln(x)+3/14*ln(2+3*x)-5/7*ln(1+5*x)

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Rubi [A]  time = 0.02, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {1386, 705, 29, 632, 31} \[ \frac {\log (x)}{2}+\frac {3}{14} \log (3 x+2)-\frac {5}{7} \log (5 x+1) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(13 + 2/x + 15*x)),x]

[Out]

Log[x]/2 + (3*Log[2 + 3*x])/14 - (5*Log[1 + 5*x])/7

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 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 705

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

Rule 1386

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

Rubi steps

\begin {align*} \int \frac {1}{x^2 \left (13+\frac {2}{x}+15 x\right )} \, dx &=\int \frac {1}{x \left (2+13 x+15 x^2\right )} \, dx\\ &=\frac {1}{2} \int \frac {1}{x} \, dx+\frac {1}{2} \int \frac {-13-15 x}{2+13 x+15 x^2} \, dx\\ &=\frac {\log (x)}{2}+\frac {45}{14} \int \frac {1}{10+15 x} \, dx-\frac {75}{7} \int \frac {1}{3+15 x} \, dx\\ &=\frac {\log (x)}{2}+\frac {3}{14} \log (2+3 x)-\frac {5}{7} \log (1+5 x)\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 27, normalized size = 1.00 \[ \frac {\log (x)}{2}+\frac {3}{14} \log (3 x+2)-\frac {5}{7} \log (5 x+1) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(13 + 2/x + 15*x)),x]

[Out]

Log[x]/2 + (3*Log[2 + 3*x])/14 - (5*Log[1 + 5*x])/7

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fricas [A]  time = 0.73, size = 21, normalized size = 0.78 \[ -\frac {5}{7} \, \log \left (5 \, x + 1\right ) + \frac {3}{14} \, \log \left (3 \, x + 2\right ) + \frac {1}{2} \, \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/7*log(5*x + 1) + 3/14*log(3*x + 2) + 1/2*log(x)

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giac [A]  time = 0.37, size = 24, normalized size = 0.89 \[ -\frac {5}{7} \, \log \left ({\left | 5 \, x + 1 \right |}\right ) + \frac {3}{14} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) + \frac {1}{2} \, \log \left ({\left | x \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/7*log(abs(5*x + 1)) + 3/14*log(abs(3*x + 2)) + 1/2*log(abs(x))

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maple [A]  time = 0.01, size = 22, normalized size = 0.81 \[ \frac {\ln \relax (x )}{2}-\frac {5 \ln \left (5 x +1\right )}{7}+\frac {3 \ln \left (3 x +2\right )}{14} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*ln(x)+3/14*ln(3*x+2)-5/7*ln(5*x+1)

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maxima [A]  time = 0.92, size = 21, normalized size = 0.78 \[ -\frac {5}{7} \, \log \left (5 \, x + 1\right ) + \frac {3}{14} \, \log \left (3 \, x + 2\right ) + \frac {1}{2} \, \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/7*log(5*x + 1) + 3/14*log(3*x + 2) + 1/2*log(x)

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mupad [B]  time = 0.09, size = 17, normalized size = 0.63 \[ \frac {3\,\ln \left (x+\frac {2}{3}\right )}{14}-\frac {5\,\ln \left (x+\frac {1}{5}\right )}{7}+\frac {\ln \relax (x)}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(15*x + 2/x + 13)),x)

[Out]

(3*log(x + 2/3))/14 - (5*log(x + 1/5))/7 + log(x)/2

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sympy [A]  time = 0.15, size = 24, normalized size = 0.89 \[ \frac {\log {\relax (x )}}{2} - \frac {5 \log {\left (x + \frac {1}{5} \right )}}{7} + \frac {3 \log {\left (x + \frac {2}{3} \right )}}{14} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x)/2 - 5*log(x + 1/5)/7 + 3*log(x + 2/3)/14

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