3.50.62 \(\int \frac {-5-5 x^2+(-5-15 x^2) \log (x)+(2+6 x^2) \log ^2(x)}{(-5 x-5 x^3) \log (x)+(2 x+2 x^3) \log ^2(x)} \, dx\)

Optimal. Leaf size=22 \[ \log \left (\frac {5 x \left (1+x^2\right )}{e^6 \left (-2+\frac {5}{\log (x)}\right )}\right ) \]

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Rubi [A]  time = 0.96, antiderivative size = 21, normalized size of antiderivative = 0.95, number of steps used = 10, number of rules used = 6, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.102, Rules used = {6741, 6725, 446, 72, 2302, 29} \begin {gather*} \log \left (x^2+1\right )+\log (x)-\log (5-2 \log (x))+\log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5 - 5*x^2 + (-5 - 15*x^2)*Log[x] + (2 + 6*x^2)*Log[x]^2)/((-5*x - 5*x^3)*Log[x] + (2*x + 2*x^3)*Log[x]^2
),x]

[Out]

Log[x] + Log[1 + x^2] - Log[5 - 2*Log[x]] + Log[Log[x]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

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]]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5+5 x^2-\left (-5-15 x^2\right ) \log (x)-\left (2+6 x^2\right ) \log ^2(x)}{x \left (1+x^2\right ) (5-2 \log (x)) \log (x)} \, dx\\ &=\int \left (\frac {1+3 x^2}{x \left (1+x^2\right )}+\frac {1}{x \log (x)}-\frac {2}{x (-5+2 \log (x))}\right ) \, dx\\ &=-\left (2 \int \frac {1}{x (-5+2 \log (x))} \, dx\right )+\int \frac {1+3 x^2}{x \left (1+x^2\right )} \, dx+\int \frac {1}{x \log (x)} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1+3 x}{x (1+x)} \, dx,x,x^2\right )+\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )-\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,-5+2 \log (x)\right )\\ &=-\log (5-2 \log (x))+\log (\log (x))+\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {1}{x}+\frac {2}{1+x}\right ) \, dx,x,x^2\right )\\ &=\log (x)+\log \left (1+x^2\right )-\log (5-2 \log (x))+\log (\log (x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 21, normalized size = 0.95 \begin {gather*} \log (x)+\log \left (1+x^2\right )-\log (5-2 \log (x))+\log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5 - 5*x^2 + (-5 - 15*x^2)*Log[x] + (2 + 6*x^2)*Log[x]^2)/((-5*x - 5*x^3)*Log[x] + (2*x + 2*x^3)*Lo
g[x]^2),x]

[Out]

Log[x] + Log[1 + x^2] - Log[5 - 2*Log[x]] + Log[Log[x]]

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fricas [A]  time = 0.78, size = 19, normalized size = 0.86 \begin {gather*} \log \left (x^{3} + x\right ) - \log \left (2 \, \log \relax (x) - 5\right ) + \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^2+2)*log(x)^2+(-15*x^2-5)*log(x)-5*x^2-5)/((2*x^3+2*x)*log(x)^2+(-5*x^3-5*x)*log(x)),x, algori
thm="fricas")

[Out]

log(x^3 + x) - log(2*log(x) - 5) + log(log(x))

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giac [A]  time = 0.16, size = 21, normalized size = 0.95 \begin {gather*} \log \left (x^{2} + 1\right ) + \log \relax (x) - \log \left (2 \, \log \relax (x) - 5\right ) + \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^2+2)*log(x)^2+(-15*x^2-5)*log(x)-5*x^2-5)/((2*x^3+2*x)*log(x)^2+(-5*x^3-5*x)*log(x)),x, algori
thm="giac")

[Out]

log(x^2 + 1) + log(x) - log(2*log(x) - 5) + log(log(x))

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maple [A]  time = 0.04, size = 18, normalized size = 0.82

method	result	size
risch	\(\ln \left (x^{3}+x \right )+\ln \left (\ln \relax (x )\right )-\ln \left (\ln \relax (x )-\frac {5}{2}\right )\)	\(18\)
norman	\(\ln \relax (x )-\ln \left (2 \ln \relax (x )-5\right )+\ln \left (\ln \relax (x )\right )+\ln \left (x^{2}+1\right )\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((6*x^2+2)*ln(x)^2+(-15*x^2-5)*ln(x)-5*x^2-5)/((2*x^3+2*x)*ln(x)^2+(-5*x^3-5*x)*ln(x)),x,method=_RETURNVER
BOSE)

[Out]

ln(x^3+x)+ln(ln(x))-ln(ln(x)-5/2)

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maxima [A]  time = 0.54, size = 19, normalized size = 0.86 \begin {gather*} \log \left (x^{2} + 1\right ) + \log \relax (x) - \log \left (\log \relax (x) - \frac {5}{2}\right ) + \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^2+2)*log(x)^2+(-15*x^2-5)*log(x)-5*x^2-5)/((2*x^3+2*x)*log(x)^2+(-5*x^3-5*x)*log(x)),x, algori
thm="maxima")

[Out]

log(x^2 + 1) + log(x) - log(log(x) - 5/2) + log(log(x))

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mupad [B]  time = 3.35, size = 21, normalized size = 0.95 \begin {gather*} \ln \relax (x)-\ln \left (2\,\ln \relax (x)-5\right )+\ln \left (\ln \relax (x)\,\left (x^2+1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*x^2 - log(x)^2*(6*x^2 + 2) + log(x)*(15*x^2 + 5) + 5)/(log(x)^2*(2*x + 2*x^3) - log(x)*(5*x + 5*x^3)),
x)

[Out]

log(x) - log(2*log(x) - 5) + log(log(x)*(x^2 + 1))

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sympy [A]  time = 0.19, size = 19, normalized size = 0.86 \begin {gather*} \log {\left (x^{3} + x \right )} - \log {\left (\log {\relax (x )} - \frac {5}{2} \right )} + \log {\left (\log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x**2+2)*ln(x)**2+(-15*x**2-5)*ln(x)-5*x**2-5)/((2*x**3+2*x)*ln(x)**2+(-5*x**3-5*x)*ln(x)),x)

[Out]

log(x**3 + x) - log(log(x) - 5/2) + log(log(x))

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