3.47.87 \(\int \frac {-1+(5 x+16 x^2) \log (x) \log (\log (x))}{16 x \log (x) \log (\log (x))} \, dx\)

Optimal. Leaf size=24 \[ e^4+\frac {1}{16} (x+4 x (1+2 x)-\log (\log (\log (x)))) \]

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Rubi [A]  time = 0.16, antiderivative size = 21, normalized size of antiderivative = 0.88, number of steps used = 6, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {12, 6688, 2302, 29} \begin {gather*} \frac {x^2}{2}+\frac {5 x}{16}-\frac {1}{16} \log (\log (\log (x))) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + (5*x + 16*x^2)*Log[x]*Log[Log[x]])/(16*x*Log[x]*Log[Log[x]]),x]

[Out]

(5*x)/16 + x^2/2 - Log[Log[Log[x]]]/16

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 29

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

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 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{16} \int \frac {-1+\left (5 x+16 x^2\right ) \log (x) \log (\log (x))}{x \log (x) \log (\log (x))} \, dx\\ &=\frac {1}{16} \int \left (5+16 x-\frac {1}{x \log (x) \log (\log (x))}\right ) \, dx\\ &=\frac {5 x}{16}+\frac {x^2}{2}-\frac {1}{16} \int \frac {1}{x \log (x) \log (\log (x))} \, dx\\ &=\frac {5 x}{16}+\frac {x^2}{2}-\frac {1}{16} \operatorname {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,\log (x)\right )\\ &=\frac {5 x}{16}+\frac {x^2}{2}-\frac {1}{16} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (\log (x))\right )\\ &=\frac {5 x}{16}+\frac {x^2}{2}-\frac {1}{16} \log (\log (\log (x)))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 21, normalized size = 0.88 \begin {gather*} \frac {5 x}{16}+\frac {x^2}{2}-\frac {1}{16} \log (\log (\log (x))) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + (5*x + 16*x^2)*Log[x]*Log[Log[x]])/(16*x*Log[x]*Log[Log[x]]),x]

[Out]

(5*x)/16 + x^2/2 - Log[Log[Log[x]]]/16

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fricas [A]  time = 0.68, size = 15, normalized size = 0.62 \begin {gather*} \frac {1}{2} \, x^{2} + \frac {5}{16} \, x - \frac {1}{16} \, \log \left (\log \left (\log \relax (x)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*((16*x^2+5*x)*log(x)*log(log(x))-1)/x/log(x)/log(log(x)),x, algorithm="fricas")

[Out]

1/2*x^2 + 5/16*x - 1/16*log(log(log(x)))

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giac [A]  time = 0.12, size = 15, normalized size = 0.62 \begin {gather*} \frac {1}{2} \, x^{2} + \frac {5}{16} \, x - \frac {1}{16} \, \log \left (\log \left (\log \relax (x)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*((16*x^2+5*x)*log(x)*log(log(x))-1)/x/log(x)/log(log(x)),x, algorithm="giac")

[Out]

1/2*x^2 + 5/16*x - 1/16*log(log(log(x)))

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maple [A]  time = 0.04, size = 16, normalized size = 0.67

method	result	size
default	\(\frac {x^{2}}{2}+\frac {5 x}{16}-\frac {\ln \left (\ln \left (\ln \relax (x )\right )\right )}{16}\)	\(16\)
norman	\(\frac {x^{2}}{2}+\frac {5 x}{16}-\frac {\ln \left (\ln \left (\ln \relax (x )\right )\right )}{16}\)	\(16\)
risch	\(\frac {x^{2}}{2}+\frac {5 x}{16}-\frac {\ln \left (\ln \left (\ln \relax (x )\right )\right )}{16}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/16*((16*x^2+5*x)*ln(x)*ln(ln(x))-1)/x/ln(x)/ln(ln(x)),x,method=_RETURNVERBOSE)

[Out]

1/2*x^2+5/16*x-1/16*ln(ln(ln(x)))

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maxima [A]  time = 0.36, size = 15, normalized size = 0.62 \begin {gather*} \frac {1}{2} \, x^{2} + \frac {5}{16} \, x - \frac {1}{16} \, \log \left (\log \left (\log \relax (x)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*((16*x^2+5*x)*log(x)*log(log(x))-1)/x/log(x)/log(log(x)),x, algorithm="maxima")

[Out]

1/2*x^2 + 5/16*x - 1/16*log(log(log(x)))

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mupad [B]  time = 3.36, size = 15, normalized size = 0.62 \begin {gather*} \frac {5\,x}{16}-\frac {\ln \left (\ln \left (\ln \relax (x)\right )\right )}{16}+\frac {x^2}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((log(log(x))*log(x)*(5*x + 16*x^2))/16 - 1/16)/(x*log(log(x))*log(x)),x)

[Out]

(5*x)/16 - log(log(log(x)))/16 + x^2/2

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sympy [A]  time = 0.25, size = 17, normalized size = 0.71 \begin {gather*} \frac {x^{2}}{2} + \frac {5 x}{16} - \frac {\log {\left (\log {\left (\log {\relax (x )} \right )} \right )}}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*((16*x**2+5*x)*ln(x)*ln(ln(x))-1)/x/ln(x)/ln(ln(x)),x)

[Out]

x**2/2 + 5*x/16 - log(log(log(x)))/16

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