3.15.55 \(\int \frac {-240 x-48 x^2+(240 x+48 x^2) \log (x)+((240 x+96 x^2) \log (x)+(45+18 x) \log ^2(x)) \log (\frac {16 x+3 \log (x)}{\log (x)}) \log (2 \log (\frac {16 x+3 \log (x)}{\log (x)}))}{(16 x \log (x)+3 \log ^2(x)) \log (\frac {16 x+3 \log (x)}{\log (x)})} \, dx\)

Optimal. Leaf size=22 \[ \log (5)+3 x (5+x) \log \left (2 \log \left (3+\frac {16 x}{\log (x)}\right )\right ) \]

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Rubi [A]  time = 3.66, antiderivative size = 35, normalized size of antiderivative = 1.59, number of steps used = 17, number of rules used = 4, integrand size = 107, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {6741, 6742, 2555, 12} \begin {gather*} 3 x^2 \log \left (2 \log \left (\frac {16 x}{\log (x)}+3\right )\right )+15 x \log \left (2 \log \left (\frac {16 x}{\log (x)}+3\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-240*x - 48*x^2 + (240*x + 48*x^2)*Log[x] + ((240*x + 96*x^2)*Log[x] + (45 + 18*x)*Log[x]^2)*Log[(16*x +
3*Log[x])/Log[x]]*Log[2*Log[(16*x + 3*Log[x])/Log[x]]])/((16*x*Log[x] + 3*Log[x]^2)*Log[(16*x + 3*Log[x])/Log[
x]]),x]

[Out]

15*x*Log[2*Log[3 + (16*x)/Log[x]]] + 3*x^2*Log[2*Log[3 + (16*x)/Log[x]]]

Rule 12

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

Rule 2555

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[w*Simplify
[D[u, x]/u], x], x] /; InverseFunctionFreeQ[w, x]] /; ProductQ[u]

Rule 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-240 x-48 x^2+\left (240 x+48 x^2\right ) \log (x)+\left (\left (240 x+96 x^2\right ) \log (x)+(45+18 x) \log ^2(x)\right ) \log \left (\frac {16 x+3 \log (x)}{\log (x)}\right ) \log \left (2 \log \left (\frac {16 x+3 \log (x)}{\log (x)}\right )\right )}{\log (x) (16 x+3 \log (x)) \log \left (\frac {16 x+3 \log (x)}{\log (x)}\right )} \, dx\\ &=\int \left (\frac {48 x (5+x) (-1+\log (x))}{\log (x) (16 x+3 \log (x)) \log \left (3+\frac {16 x}{\log (x)}\right )}+3 (5+2 x) \log \left (2 \log \left (3+\frac {16 x}{\log (x)}\right )\right )\right ) \, dx\\ &=3 \int (5+2 x) \log \left (2 \log \left (3+\frac {16 x}{\log (x)}\right )\right ) \, dx+48 \int \frac {x (5+x) (-1+\log (x))}{\log (x) (16 x+3 \log (x)) \log \left (3+\frac {16 x}{\log (x)}\right )} \, dx\\ &=15 x \log \left (2 \log \left (3+\frac {16 x}{\log (x)}\right )\right )+3 x^2 \log \left (2 \log \left (3+\frac {16 x}{\log (x)}\right )\right )-3 \int \frac {16 x (5+x) (-1+\log (x))}{\log (x) (16 x+3 \log (x)) \log \left (3+\frac {16 x}{\log (x)}\right )} \, dx+48 \int \left (\frac {5 x (-1+\log (x))}{\log (x) (16 x+3 \log (x)) \log \left (3+\frac {16 x}{\log (x)}\right )}+\frac {x^2 (-1+\log (x))}{\log (x) (16 x+3 \log (x)) \log \left (3+\frac {16 x}{\log (x)}\right )}\right ) \, dx\\ &=15 x \log \left (2 \log \left (3+\frac {16 x}{\log (x)}\right )\right )+3 x^2 \log \left (2 \log \left (3+\frac {16 x}{\log (x)}\right )\right )+48 \int \frac {x^2 (-1+\log (x))}{\log (x) (16 x+3 \log (x)) \log \left (3+\frac {16 x}{\log (x)}\right )} \, dx-48 \int \frac {x (5+x) (-1+\log (x))}{\log (x) (16 x+3 \log (x)) \log \left (3+\frac {16 x}{\log (x)}\right )} \, dx+240 \int \frac {x (-1+\log (x))}{\log (x) (16 x+3 \log (x)) \log \left (3+\frac {16 x}{\log (x)}\right )} \, dx\\ &=15 x \log \left (2 \log \left (3+\frac {16 x}{\log (x)}\right )\right )+3 x^2 \log \left (2 \log \left (3+\frac {16 x}{\log (x)}\right )\right )+48 \int \left (\frac {x^2}{(16 x+3 \log (x)) \log \left (3+\frac {16 x}{\log (x)}\right )}-\frac {x^2}{\log (x) (16 x+3 \log (x)) \log \left (3+\frac {16 x}{\log (x)}\right )}\right ) \, dx-48 \int \left (\frac {5 x (-1+\log (x))}{\log (x) (16 x+3 \log (x)) \log \left (3+\frac {16 x}{\log (x)}\right )}+\frac {x^2 (-1+\log (x))}{\log (x) (16 x+3 \log (x)) \log \left (3+\frac {16 x}{\log (x)}\right )}\right ) \, dx+240 \int \left (\frac {x}{(16 x+3 \log (x)) \log \left (3+\frac {16 x}{\log (x)}\right )}-\frac {x}{\log (x) (16 x+3 \log (x)) \log \left (3+\frac {16 x}{\log (x)}\right )}\right ) \, dx\\ &=15 x \log \left (2 \log \left (3+\frac {16 x}{\log (x)}\right )\right )+3 x^2 \log \left (2 \log \left (3+\frac {16 x}{\log (x)}\right )\right )+48 \int \frac {x^2}{(16 x+3 \log (x)) \log \left (3+\frac {16 x}{\log (x)}\right )} \, dx-48 \int \frac {x^2}{\log (x) (16 x+3 \log (x)) \log \left (3+\frac {16 x}{\log (x)}\right )} \, dx-48 \int \frac {x^2 (-1+\log (x))}{\log (x) (16 x+3 \log (x)) \log \left (3+\frac {16 x}{\log (x)}\right )} \, dx+240 \int \frac {x}{(16 x+3 \log (x)) \log \left (3+\frac {16 x}{\log (x)}\right )} \, dx-240 \int \frac {x}{\log (x) (16 x+3 \log (x)) \log \left (3+\frac {16 x}{\log (x)}\right )} \, dx-240 \int \frac {x (-1+\log (x))}{\log (x) (16 x+3 \log (x)) \log \left (3+\frac {16 x}{\log (x)}\right )} \, dx\\ &=15 x \log \left (2 \log \left (3+\frac {16 x}{\log (x)}\right )\right )+3 x^2 \log \left (2 \log \left (3+\frac {16 x}{\log (x)}\right )\right )-48 \int \left (\frac {x^2}{(16 x+3 \log (x)) \log \left (3+\frac {16 x}{\log (x)}\right )}-\frac {x^2}{\log (x) (16 x+3 \log (x)) \log \left (3+\frac {16 x}{\log (x)}\right )}\right ) \, dx+48 \int \frac {x^2}{(16 x+3 \log (x)) \log \left (3+\frac {16 x}{\log (x)}\right )} \, dx-48 \int \frac {x^2}{\log (x) (16 x+3 \log (x)) \log \left (3+\frac {16 x}{\log (x)}\right )} \, dx-240 \int \left (\frac {x}{(16 x+3 \log (x)) \log \left (3+\frac {16 x}{\log (x)}\right )}-\frac {x}{\log (x) (16 x+3 \log (x)) \log \left (3+\frac {16 x}{\log (x)}\right )}\right ) \, dx+240 \int \frac {x}{(16 x+3 \log (x)) \log \left (3+\frac {16 x}{\log (x)}\right )} \, dx-240 \int \frac {x}{\log (x) (16 x+3 \log (x)) \log \left (3+\frac {16 x}{\log (x)}\right )} \, dx\\ &=15 x \log \left (2 \log \left (3+\frac {16 x}{\log (x)}\right )\right )+3 x^2 \log \left (2 \log \left (3+\frac {16 x}{\log (x)}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 19, normalized size = 0.86 \begin {gather*} 3 x (5+x) \log \left (2 \log \left (3+\frac {16 x}{\log (x)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-240*x - 48*x^2 + (240*x + 48*x^2)*Log[x] + ((240*x + 96*x^2)*Log[x] + (45 + 18*x)*Log[x]^2)*Log[(1
6*x + 3*Log[x])/Log[x]]*Log[2*Log[(16*x + 3*Log[x])/Log[x]]])/((16*x*Log[x] + 3*Log[x]^2)*Log[(16*x + 3*Log[x]
)/Log[x]]),x]

[Out]

3*x*(5 + x)*Log[2*Log[3 + (16*x)/Log[x]]]

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fricas [A]  time = 0.66, size = 26, normalized size = 1.18 \begin {gather*} 3 \, {\left (x^{2} + 5 \, x\right )} \log \left (2 \, \log \left (\frac {16 \, x + 3 \, \log \relax (x)}{\log \relax (x)}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((18*x+45)*log(x)^2+(96*x^2+240*x)*log(x))*log((3*log(x)+16*x)/log(x))*log(2*log((3*log(x)+16*x)/lo
g(x)))+(48*x^2+240*x)*log(x)-48*x^2-240*x)/(3*log(x)^2+16*x*log(x))/log((3*log(x)+16*x)/log(x)),x, algorithm="
fricas")

[Out]

3*(x^2 + 5*x)*log(2*log((16*x + 3*log(x))/log(x)))

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giac [A]  time = 0.64, size = 27, normalized size = 1.23 \begin {gather*} 3 \, {\left (x^{2} + 5 \, x\right )} \log \left (2 \, \log \left (16 \, x + 3 \, \log \relax (x)\right ) - 2 \, \log \left (\log \relax (x)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((18*x+45)*log(x)^2+(96*x^2+240*x)*log(x))*log((3*log(x)+16*x)/log(x))*log(2*log((3*log(x)+16*x)/lo
g(x)))+(48*x^2+240*x)*log(x)-48*x^2-240*x)/(3*log(x)^2+16*x*log(x))/log((3*log(x)+16*x)/log(x)),x, algorithm="
giac")

[Out]

3*(x^2 + 5*x)*log(2*log(16*x + 3*log(x)) - 2*log(log(x)))

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maple [C]  time = 0.24, size = 101, normalized size = 4.59

method	result	size
risch	\(\left (3 x^{2}+15 x \right ) \ln \left (8 \ln \relax (2)-2 \ln \left (\ln \relax (x )\right )+2 \ln \left (\frac {3 \ln \relax (x )}{16}+x \right )-i \pi \,\mathrm {csgn}\left (\frac {i \left (\frac {3 \ln \relax (x )}{16}+x \right )}{\ln \relax (x )}\right ) \left (-\mathrm {csgn}\left (\frac {i \left (\frac {3 \ln \relax (x )}{16}+x \right )}{\ln \relax (x )}\right )+\mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right )\right ) \left (-\mathrm {csgn}\left (\frac {i \left (\frac {3 \ln \relax (x )}{16}+x \right )}{\ln \relax (x )}\right )+\mathrm {csgn}\left (i \left (\frac {3 \ln \relax (x )}{16}+x \right )\right )\right )\right )\)	\(101\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((18*x+45)*ln(x)^2+(96*x^2+240*x)*ln(x))*ln((3*ln(x)+16*x)/ln(x))*ln(2*ln((3*ln(x)+16*x)/ln(x)))+(48*x^2+
240*x)*ln(x)-48*x^2-240*x)/(3*ln(x)^2+16*x*ln(x))/ln((3*ln(x)+16*x)/ln(x)),x,method=_RETURNVERBOSE)

[Out]

(3*x^2+15*x)*ln(8*ln(2)-2*ln(ln(x))+2*ln(3/16*ln(x)+x)-I*Pi*csgn(I/ln(x)*(3/16*ln(x)+x))*(-csgn(I/ln(x)*(3/16*
ln(x)+x))+csgn(I/ln(x)))*(-csgn(I/ln(x)*(3/16*ln(x)+x))+csgn(I*(3/16*ln(x)+x))))

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maxima [A]  time = 0.78, size = 38, normalized size = 1.73 \begin {gather*} 3 \, x^{2} \log \relax (2) + 15 \, x \log \relax (2) + 3 \, {\left (x^{2} + 5 \, x\right )} \log \left (\log \left (16 \, x + 3 \, \log \relax (x)\right ) - \log \left (\log \relax (x)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((18*x+45)*log(x)^2+(96*x^2+240*x)*log(x))*log((3*log(x)+16*x)/log(x))*log(2*log((3*log(x)+16*x)/lo
g(x)))+(48*x^2+240*x)*log(x)-48*x^2-240*x)/(3*log(x)^2+16*x*log(x))/log((3*log(x)+16*x)/log(x)),x, algorithm="
maxima")

[Out]

3*x^2*log(2) + 15*x*log(2) + 3*(x^2 + 5*x)*log(log(16*x + 3*log(x)) - log(log(x)))

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mupad [B]  time = 1.59, size = 28, normalized size = 1.27 \begin {gather*} \left (3\,x^2+15\,x\right )\,\left (\ln \relax (2)+\ln \left (\ln \left (\frac {16\,x+3\,\ln \relax (x)}{\ln \relax (x)}\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(240*x - log(x)*(240*x + 48*x^2) + 48*x^2 - log(2*log((16*x + 3*log(x))/log(x)))*log((16*x + 3*log(x))/lo
g(x))*(log(x)*(240*x + 96*x^2) + log(x)^2*(18*x + 45)))/(log((16*x + 3*log(x))/log(x))*(3*log(x)^2 + 16*x*log(
x))),x)

[Out]

(15*x + 3*x^2)*(log(2) + log(log((16*x + 3*log(x))/log(x))))

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sympy [A]  time = 1.45, size = 24, normalized size = 1.09 \begin {gather*} \left (3 x^{2} + 15 x\right ) \log {\left (2 \log {\left (\frac {16 x + 3 \log {\relax (x )}}{\log {\relax (x )}} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((18*x+45)*ln(x)**2+(96*x**2+240*x)*ln(x))*ln((3*ln(x)+16*x)/ln(x))*ln(2*ln((3*ln(x)+16*x)/ln(x)))+
(48*x**2+240*x)*ln(x)-48*x**2-240*x)/(3*ln(x)**2+16*x*ln(x))/ln((3*ln(x)+16*x)/ln(x)),x)

[Out]

(3*x**2 + 15*x)*log(2*log((16*x + 3*log(x))/log(x)))

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