3.26.93 \(\int \frac {x+2 x^2+x^3+(1+2 x+2 x^2+(1+2 x) \log (3)) \log (x)}{(x^2+2 x^3+x^4) \log (x)} \, dx\) [2593]

Optimal. Leaf size=23 \[ \frac {1+2 x+\log (3)}{-x-x^2}+\log (\log (x)) \]

[Out]

(1+2*x+ln(3))/(-x^2-x)+ln(ln(x))

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Rubi [A]
time = 0.22, antiderivative size = 26, normalized size of antiderivative = 1.13, number of steps used = 9, number of rules used = 7, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {1608, 27, 6820, 14, 907, 2339, 29} \begin {gather*} \log (\log (x))-\frac {1+\log (3)}{x}-\frac {1-\log (3)}{x+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x + 2*x^2 + x^3 + (1 + 2*x + 2*x^2 + (1 + 2*x)*Log[3])*Log[x])/((x^2 + 2*x^3 + x^4)*Log[x]),x]

[Out]

-((1 - Log[3])/(1 + x)) - (1 + Log[3])/x + Log[Log[x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 29

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

Rule 907

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 1608

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2339

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 6820

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x+2 x^2+x^3+\left (1+2 x+2 x^2+(1+2 x) \log (3)\right ) \log (x)}{x^2 \left (1+2 x+x^2\right ) \log (x)} \, dx\\ &=\int \frac {x+2 x^2+x^3+\left (1+2 x+2 x^2+(1+2 x) \log (3)\right ) \log (x)}{x^2 (1+x)^2 \log (x)} \, dx\\ &=\int \frac {\frac {1+2 x^2+\log (3)+x (2+\log (9))}{(1+x)^2}+\frac {x}{\log (x)}}{x^2} \, dx\\ &=\int \left (\frac {1+2 x^2+\log (3)+x (2+\log (9))}{x^2 (1+x)^2}+\frac {1}{x \log (x)}\right ) \, dx\\ &=\int \frac {1+2 x^2+\log (3)+x (2+\log (9))}{x^2 (1+x)^2} \, dx+\int \frac {1}{x \log (x)} \, dx\\ &=\int \left (\frac {1-\log (3)}{(1+x)^2}+\frac {1+\log (3)}{x^2}\right ) \, dx+\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )\\ &=-\frac {1-\log (3)}{1+x}-\frac {1+\log (3)}{x}+\log (\log (x))\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.04, size = 28, normalized size = 1.22 \begin {gather*} \frac {-1-\log (3)}{x}+\frac {-1-\log (3)+\log (9)}{1+x}+\log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x + 2*x^2 + x^3 + (1 + 2*x + 2*x^2 + (1 + 2*x)*Log[3])*Log[x])/((x^2 + 2*x^3 + x^4)*Log[x]),x]

[Out]

(-1 - Log[3])/x + (-1 - Log[3] + Log[9])/(1 + x) + Log[Log[x]]

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Maple [A]
time = 0.07, size = 32, normalized size = 1.39

method	result	size
risch	\(-\frac {1+2 x +\ln \left (3\right )}{\left (x +1\right ) x}+\ln \left (\ln \left (x \right )\right )\)	\(22\)
norman	\(\frac {-2 x -1-\ln \left (3\right )}{\left (x +1\right ) x}+\ln \left (\ln \left (x \right )\right )\)	\(23\)
default	\(\ln \left (\ln \left (x \right )\right )-\frac {1}{x}-\frac {1}{x +1}-\frac {\ln \left (3\right )}{x}+\frac {\ln \left (3\right )}{x +1}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x+1)*ln(3)+2*x^2+2*x+1)*ln(x)+x^3+2*x^2+x)/(x^4+2*x^3+x^2)/ln(x),x,method=_RETURNVERBOSE)

[Out]

ln(ln(x))-1/x-1/(x+1)-ln(3)/x+ln(3)/(x+1)

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Maxima [A]
time = 0.51, size = 20, normalized size = 0.87 \begin {gather*} -\frac {2 \, x + \log \left (3\right ) + 1}{x^{2} + x} + \log \left (\log \left (x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((1+2*x)*log(3)+2*x^2+2*x+1)*log(x)+x^3+2*x^2+x)/(x^4+2*x^3+x^2)/log(x),x, algorithm="maxima")

[Out]

-(2*x + log(3) + 1)/(x^2 + x) + log(log(x))

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Fricas [A]
time = 0.35, size = 26, normalized size = 1.13 \begin {gather*} \frac {{\left (x^{2} + x\right )} \log \left (\log \left (x\right )\right ) - 2 \, x - \log \left (3\right ) - 1}{x^{2} + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((1+2*x)*log(3)+2*x^2+2*x+1)*log(x)+x^3+2*x^2+x)/(x^4+2*x^3+x^2)/log(x),x, algorithm="fricas")

[Out]

((x^2 + x)*log(log(x)) - 2*x - log(3) - 1)/(x^2 + x)

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Sympy [A]
time = 0.15, size = 19, normalized size = 0.83 \begin {gather*} \log {\left (\log {\left (x \right )} \right )} + \frac {- 2 x - \log {\left (3 \right )} - 1}{x^{2} + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((1+2*x)*ln(3)+2*x**2+2*x+1)*ln(x)+x**3+2*x**2+x)/(x**4+2*x**3+x**2)/ln(x),x)

[Out]

log(log(x)) + (-2*x - log(3) - 1)/(x**2 + x)

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Giac [A]
time = 0.39, size = 23, normalized size = 1.00 \begin {gather*} -\frac {\log \left (3\right ) + 1}{x} + \frac {\log \left (3\right ) - 1}{x + 1} + \log \left (\log \left (x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((1+2*x)*log(3)+2*x^2+2*x+1)*log(x)+x^3+2*x^2+x)/(x^4+2*x^3+x^2)/log(x),x, algorithm="giac")

[Out]

-(log(3) + 1)/x + (log(3) - 1)/(x + 1) + log(log(x))

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Mupad [B]
time = 1.53, size = 27, normalized size = 1.17 \begin {gather*} \ln \left (\ln \left (x\right )\right )-\frac {x\,\left (\ln \left (3\right )+1\right )-2\,x^3}{x^3+x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + log(x)*(2*x + log(3)*(2*x + 1) + 2*x^2 + 1) + 2*x^2 + x^3)/(log(x)*(x^2 + 2*x^3 + x^4)),x)

[Out]

log(log(x)) - (x*(log(3) + 1) - 2*x^3)/(x^2 + x^3)

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