3.450 \(\int \frac {-2+3 x^6}{x (5+2 x^6)} \, dx\)

Optimal. Leaf size=19 \[ \frac {19}{60} \log \left (2 x^6+5\right )-\frac {2 \log (x)}{5} \]

[Out]

-2/5*ln(x)+19/60*ln(2*x^6+5)

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Rubi [A]  time = 0.02, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {446, 72} \[ \frac {19}{60} \log \left (2 x^6+5\right )-\frac {2 \log (x)}{5} \]

Antiderivative was successfully verified.

[In]

Int[(-2 + 3*x^6)/(x*(5 + 2*x^6)),x]

[Out]

(-2*Log[x])/5 + (19*Log[5 + 2*x^6])/60

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {-2+3 x^6}{x \left (5+2 x^6\right )} \, dx &=\frac {1}{6} \operatorname {Subst}\left (\int \frac {-2+3 x}{x (5+2 x)} \, dx,x,x^6\right )\\ &=\frac {1}{6} \operatorname {Subst}\left (\int \left (-\frac {2}{5 x}+\frac {19}{5 (5+2 x)}\right ) \, dx,x,x^6\right )\\ &=-\frac {2 \log (x)}{5}+\frac {19}{60} \log \left (5+2 x^6\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 19, normalized size = 1.00 \[ \frac {19}{60} \log \left (2 x^6+5\right )-\frac {2 \log (x)}{5} \]

Antiderivative was successfully verified.

[In]

Integrate[(-2 + 3*x^6)/(x*(5 + 2*x^6)),x]

[Out]

(-2*Log[x])/5 + (19*Log[5 + 2*x^6])/60

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fricas [A]  time = 0.48, size = 15, normalized size = 0.79 \[ \frac {19}{60} \, \log \left (2 \, x^{6} + 5\right ) - \frac {2}{5} \, \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^6-2)/x/(2*x^6+5),x, algorithm="fricas")

[Out]

19/60*log(2*x^6 + 5) - 2/5*log(x)

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giac [A]  time = 0.41, size = 17, normalized size = 0.89 \[ \frac {19}{60} \, \log \left (2 \, x^{6} + 5\right ) - \frac {1}{15} \, \log \left (x^{6}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^6-2)/x/(2*x^6+5),x, algorithm="giac")

[Out]

19/60*log(2*x^6 + 5) - 1/15*log(x^6)

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maple [A]  time = 0.01, size = 16, normalized size = 0.84 \[ -\frac {2 \ln \relax (x )}{5}+\frac {19 \ln \left (2 x^{6}+5\right )}{60} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^6-2)/x/(2*x^6+5),x)

[Out]

-2/5*ln(x)+19/60*ln(2*x^6+5)

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maxima [A]  time = 0.47, size = 17, normalized size = 0.89 \[ \frac {19}{60} \, \log \left (2 \, x^{6} + 5\right ) - \frac {1}{15} \, \log \left (x^{6}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^6-2)/x/(2*x^6+5),x, algorithm="maxima")

[Out]

19/60*log(2*x^6 + 5) - 1/15*log(x^6)

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mupad [B]  time = 0.09, size = 13, normalized size = 0.68 \[ \frac {19\,\ln \left (x^6+\frac {5}{2}\right )}{60}-\frac {2\,\ln \relax (x)}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^6 - 2)/(x*(2*x^6 + 5)),x)

[Out]

(19*log(x^6 + 5/2))/60 - (2*log(x))/5

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sympy [A]  time = 0.11, size = 17, normalized size = 0.89 \[ - \frac {2 \log {\relax (x )}}{5} + \frac {19 \log {\left (2 x^{6} + 5 \right )}}{60} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x**6-2)/x/(2*x**6+5),x)

[Out]

-2*log(x)/5 + 19*log(2*x**6 + 5)/60

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