3.19.36 \(\int \frac {(-32 x^2+64 x^4) \log (2 x)-12 e^{3 \log ^2(\log ^2(2 x))} \log (\log ^2(2 x))}{x \log (2 x)} \, dx\)

Optimal. Leaf size=26 \[ -2-e^{3 \log ^2\left (\log ^2(2 x)\right )}+\left (-2+4 x^2\right )^2 \]

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Rubi [A]  time = 0.49, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.102, Rules used = {6742, 14, 2276, 2204, 2209} \begin {gather*} 16 x^4-16 x^2-e^{3 \log ^2\left (\log ^2(2 x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-32*x^2 + 64*x^4)*Log[2*x] - 12*E^(3*Log[Log[2*x]^2]^2)*Log[Log[2*x]^2])/(x*Log[2*x]),x]

[Out]

-E^(3*Log[Log[2*x]^2]^2) - 16*x^2 + 16*x^4

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 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2276

Int[(F_)^(((a_.) + Log[(c_.)*(x_)^(n_.)]^2*(b_.))*(d_.))*((e_.)*(x_))^(m_.), x_Symbol] :> Dist[(e*x)^(m + 1)/(
e*n*(c*x^n)^((m + 1)/n)), Subst[Int[E^(a*d*Log[F] + ((m + 1)*x)/n + b*d*Log[F]*x^2), x], x, Log[c*x^n]], x] /;
FreeQ[{F, a, b, c, d, e, m, n}, x]

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 \left (32 x \left (-1+2 x^2\right )-\frac {12 e^{3 \log ^2\left (\log ^2(2 x)\right )} \log \left (\log ^2(2 x)\right )}{x \log (2 x)}\right ) \, dx\\ &=-\left (12 \int \frac {e^{3 \log ^2\left (\log ^2(2 x)\right )} \log \left (\log ^2(2 x)\right )}{x \log (2 x)} \, dx\right )+32 \int x \left (-1+2 x^2\right ) \, dx\\ &=-\left (12 \operatorname {Subst}\left (\int \frac {e^{3 \log ^2\left (x^2\right )} \log \left (x^2\right )}{x} \, dx,x,\log (2 x)\right )\right )+32 \int \left (-x+2 x^3\right ) \, dx\\ &=-16 x^2+16 x^4-6 \operatorname {Subst}\left (\int e^{3 x^2} x \, dx,x,\log \left (\log ^2(2 x)\right )\right )\\ &=-e^{3 \log ^2\left (\log ^2(2 x)\right )}-16 x^2+16 x^4\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 26, normalized size = 1.00 \begin {gather*} -e^{3 \log ^2\left (\log ^2(2 x)\right )}-16 x^2+16 x^4 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-32*x^2 + 64*x^4)*Log[2*x] - 12*E^(3*Log[Log[2*x]^2]^2)*Log[Log[2*x]^2])/(x*Log[2*x]),x]

[Out]

-E^(3*Log[Log[2*x]^2]^2) - 16*x^2 + 16*x^4

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fricas [A]  time = 0.92, size = 25, normalized size = 0.96 \begin {gather*} 16 \, x^{4} - 16 \, x^{2} - e^{\left (3 \, \log \left (\log \left (2 \, x\right )^{2}\right )^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*log(log(2*x)^2)*exp(3*log(log(2*x)^2)^2)+(64*x^4-32*x^2)*log(2*x))/x/log(2*x),x, algorithm="fri
cas")

[Out]

16*x^4 - 16*x^2 - e^(3*log(log(2*x)^2)^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*log(log(2*x)^2)*exp(3*log(log(2*x)^2)^2)+(64*x^4-32*x^2)*log(2*x))/x/log(2*x),x, algorithm="gia
c")

[Out]

undef

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maple [A]  time = 0.64, size = 26, normalized size = 1.00

method	result	size
default	\(16 x^{4}-16 x^{2}-{\mathrm e}^{3 \ln \left (\ln \left (2 x \right )^{2}\right )^{2}}\)	\(26\)
risch	\(16 x^{4}-16 x^{2}-{\mathrm e}^{\frac {3 \left (i \pi \mathrm {csgn}\left (i \ln \left (2 x \right )^{2}\right )^{3}-2 i \pi \mathrm {csgn}\left (i \ln \left (2 x \right )^{2}\right )^{2} \mathrm {csgn}\left (i \ln \left (2 x \right )\right )+i \pi \,\mathrm {csgn}\left (i \ln \left (2 x \right )^{2}\right ) \mathrm {csgn}\left (i \ln \left (2 x \right )\right )^{2}-4 \ln \left (\ln \left (2 x \right )\right )\right )^{2}}{4}}\)	\(91\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-12*ln(ln(2*x)^2)*exp(3*ln(ln(2*x)^2)^2)+(64*x^4-32*x^2)*ln(2*x))/x/ln(2*x),x,method=_RETURNVERBOSE)

[Out]

16*x^4-16*x^2-exp(3*ln(ln(2*x)^2)^2)

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maxima [A]  time = 0.50, size = 25, normalized size = 0.96 \begin {gather*} 16 \, x^{4} - 16 \, x^{2} - e^{\left (3 \, \log \left (\log \left (2 \, x\right )^{2}\right )^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*log(log(2*x)^2)*exp(3*log(log(2*x)^2)^2)+(64*x^4-32*x^2)*log(2*x))/x/log(2*x),x, algorithm="max
ima")

[Out]

16*x^4 - 16*x^2 - e^(3*log(log(2*x)^2)^2)

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mupad [B]  time = 1.18, size = 34, normalized size = 1.31 \begin {gather*} 16\,x^4-16\,x^2-{\mathrm {e}}^{3\,{\ln \left ({\ln \relax (x)}^2+2\,\ln \relax (2)\,\ln \relax (x)+{\ln \relax (2)}^2\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(2*x)*(32*x^2 - 64*x^4) + 12*log(log(2*x)^2)*exp(3*log(log(2*x)^2)^2))/(x*log(2*x)),x)

[Out]

16*x^4 - 16*x^2 - exp(3*log(log(x)^2 + 2*log(2)*log(x) + log(2)^2)^2)

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sympy [A]  time = 0.49, size = 22, normalized size = 0.85 \begin {gather*} 16 x^{4} - 16 x^{2} - e^{3 \log {\left (\log {\left (2 x \right )}^{2} \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*ln(ln(2*x)**2)*exp(3*ln(ln(2*x)**2)**2)+(64*x**4-32*x**2)*ln(2*x))/x/ln(2*x),x)

[Out]

16*x**4 - 16*x**2 - exp(3*log(log(2*x)**2)**2)

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