3.5.42 \(\int \frac {1}{15+\frac {2}{x^2}+\frac {13}{x}} \, dx\) [442]

Optimal. Leaf size=26 \[ \frac {x}{15}-\frac {4}{63} \log (2+3 x)+\frac {1}{175} \log (1+5 x) \]

[Out]

1/15*x-4/63*ln(2+3*x)+1/175*ln(1+5*x)

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Rubi [A]
time = 0.01, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1354, 717, 646, 31} \begin {gather*} \frac {x}{15}-\frac {4}{63} \log (3 x+2)+\frac {1}{175} \log (5 x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(15 + 2/x^2 + 13/x)^(-1),x]

[Out]

x/15 - (4*Log[2 + 3*x])/63 + Log[1 + 5*x]/175

Rule 31

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

Rule 646

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 717

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[e*((d + e*x)^(m - 1)/(c*(
m - 1))), x] + Dist[1/c, Int[(d + e*x)^(m - 2)*(Simp[c*d^2 - a*e^2 + e*(2*c*d - b*e)*x, 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] && GtQ[m, 1]

Rule 1354

Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[x^(2*n*p)*(c + b/x^n + a/x^(2*n))^p,
x] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && LtQ[n, 0] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {1}{15+\frac {2}{x^2}+\frac {13}{x}} \, dx &=\int \frac {x^2}{2+13 x+15 x^2} \, dx\\ &=\frac {x}{15}+\frac {1}{15} \int \frac {-2-13 x}{2+13 x+15 x^2} \, dx\\ &=\frac {x}{15}+\frac {3}{35} \int \frac {1}{3+15 x} \, dx-\frac {20}{21} \int \frac {1}{10+15 x} \, dx\\ &=\frac {x}{15}-\frac {4}{63} \log (2+3 x)+\frac {1}{175} \log (1+5 x)\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 26, normalized size = 1.00 \begin {gather*} \frac {x}{15}-\frac {4}{63} \log (2+3 x)+\frac {1}{175} \log (1+5 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(15 + 2/x^2 + 13/x)^(-1),x]

[Out]

x/15 - (4*Log[2 + 3*x])/63 + Log[1 + 5*x]/175

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Maple [A]
time = 0.02, size = 21, normalized size = 0.81

method	result	size
default	\(\frac {x}{15}-\frac {4 \ln \left (2+3 x \right )}{63}+\frac {\ln \left (1+5 x \right )}{175}\)	\(21\)
norman	\(\frac {x}{15}-\frac {4 \ln \left (2+3 x \right )}{63}+\frac {\ln \left (1+5 x \right )}{175}\)	\(21\)
risch	\(\frac {x}{15}-\frac {4 \ln \left (2+3 x \right )}{63}+\frac {\ln \left (1+5 x \right )}{175}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(15+2/x^2+13/x),x,method=_RETURNVERBOSE)

[Out]

1/15*x-4/63*ln(2+3*x)+1/175*ln(1+5*x)

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Maxima [A]
time = 0.30, size = 20, normalized size = 0.77 \begin {gather*} \frac {1}{15} \, x + \frac {1}{175} \, \log \left (5 \, x + 1\right ) - \frac {4}{63} \, \log \left (3 \, x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*x + 1/175*log(5*x + 1) - 4/63*log(3*x + 2)

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Fricas [A]
time = 0.36, size = 20, normalized size = 0.77 \begin {gather*} \frac {1}{15} \, x + \frac {1}{175} \, \log \left (5 \, x + 1\right ) - \frac {4}{63} \, \log \left (3 \, x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*x + 1/175*log(5*x + 1) - 4/63*log(3*x + 2)

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Sympy [A]
time = 0.04, size = 20, normalized size = 0.77 \begin {gather*} \frac {x}{15} + \frac {\log {\left (x + \frac {1}{5} \right )}}{175} - \frac {4 \log {\left (x + \frac {2}{3} \right )}}{63} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/15 + log(x + 1/5)/175 - 4*log(x + 2/3)/63

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Giac [A]
time = 4.82, size = 22, normalized size = 0.85 \begin {gather*} \frac {1}{15} \, x + \frac {1}{175} \, \log \left ({\left | 5 \, x + 1 \right |}\right ) - \frac {4}{63} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*x + 1/175*log(abs(5*x + 1)) - 4/63*log(abs(3*x + 2))

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Mupad [B]
time = 0.08, size = 16, normalized size = 0.62 \begin {gather*} \frac {x}{15}-\frac {4\,\ln \left (x+\frac {2}{3}\right )}{63}+\frac {\ln \left (x+\frac {1}{5}\right )}{175} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(13/x + 2/x^2 + 15),x)

[Out]

x/15 - (4*log(x + 2/3))/63 + log(x + 1/5)/175

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