3.62.71 \(\int \frac {24 x+4 e^{\frac {-3+x^2}{x}} x-8 x^2+(-2 x^2+e^{\frac {-3+x^2}{x}} (3+x^2)) \log (\frac {x^4}{16}) \log (\log (\frac {x^4}{16}))}{x^2 \log (\frac {x^4}{16})} \, dx\)

Optimal. Leaf size=24 \[ \left (6+e^{-\frac {3}{x}+x}-2 x\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right ) \]

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Rubi [B]  time = 1.16, antiderivative size = 97, normalized size of antiderivative = 4.04, number of steps used = 14, number of rules used = 9, integrand size = 78, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {6742, 6688, 2353, 2300, 2178, 2302, 29, 2520, 2288} \begin {gather*} -2 x \log \left (\log \left (\frac {x^4}{16}\right )\right )+6 \log \left (\log \left (\frac {x^4}{16}\right )\right )+\frac {e^{x-\frac {3}{x}} \left (3 \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )+x^2 \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )\right )}{\left (\frac {3}{x^2}+1\right ) x^2 \log \left (\frac {x^4}{16}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(24*x + 4*E^((-3 + x^2)/x)*x - 8*x^2 + (-2*x^2 + E^((-3 + x^2)/x)*(3 + x^2))*Log[x^4/16]*Log[Log[x^4/16]])
/(x^2*Log[x^4/16]),x]

[Out]

6*Log[Log[x^4/16]] - 2*x*Log[Log[x^4/16]] + (E^(-3/x + x)*(3*Log[x^4/16]*Log[Log[x^4/16]] + x^2*Log[x^4/16]*Lo
g[Log[x^4/16]]))/((1 + 3/x^2)*x^2*Log[x^4/16])

Rule 29

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

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 2300

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[x/(n*(c*x^n)^(1/n)), Subst[Int[E^(x/n)*(a +
b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, 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 2353

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

Rule 2520

Int[Log[Log[(d_.)*(x_)^(n_.)]^(p_.)*(c_.)], x_Symbol] :> Simp[x*Log[c*Log[d*x^n]^p], x] - Dist[n*p, Int[1/Log[
d*x^n], x], x] /; FreeQ[{c, d, n, p}, x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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 (-\frac {2 \left (-12+4 x+x \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )\right )}{x \log \left (\frac {x^4}{16}\right )}+\frac {e^{-\frac {3}{x}+x} \left (4 x+3 \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )+x^2 \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )\right )}{x^2 \log \left (\frac {x^4}{16}\right )}\right ) \, dx\\ &=-\left (2 \int \frac {-12+4 x+x \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )}{x \log \left (\frac {x^4}{16}\right )} \, dx\right )+\int \frac {e^{-\frac {3}{x}+x} \left (4 x+3 \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )+x^2 \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )\right )}{x^2 \log \left (\frac {x^4}{16}\right )} \, dx\\ &=\frac {e^{-\frac {3}{x}+x} \left (3 \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )+x^2 \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )\right )}{\left (1+\frac {3}{x^2}\right ) x^2 \log \left (\frac {x^4}{16}\right )}-2 \int \left (\frac {4 (-3+x)}{x \log \left (\frac {x^4}{16}\right )}+\log \left (\log \left (\frac {x^4}{16}\right )\right )\right ) \, dx\\ &=\frac {e^{-\frac {3}{x}+x} \left (3 \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )+x^2 \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )\right )}{\left (1+\frac {3}{x^2}\right ) x^2 \log \left (\frac {x^4}{16}\right )}-2 \int \log \left (\log \left (\frac {x^4}{16}\right )\right ) \, dx-8 \int \frac {-3+x}{x \log \left (\frac {x^4}{16}\right )} \, dx\\ &=-2 x \log \left (\log \left (\frac {x^4}{16}\right )\right )+\frac {e^{-\frac {3}{x}+x} \left (3 \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )+x^2 \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )\right )}{\left (1+\frac {3}{x^2}\right ) x^2 \log \left (\frac {x^4}{16}\right )}-8 \int \left (\frac {1}{\log \left (\frac {x^4}{16}\right )}-\frac {3}{x \log \left (\frac {x^4}{16}\right )}\right ) \, dx+8 \int \frac {1}{\log \left (\frac {x^4}{16}\right )} \, dx\\ &=-2 x \log \left (\log \left (\frac {x^4}{16}\right )\right )+\frac {e^{-\frac {3}{x}+x} \left (3 \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )+x^2 \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )\right )}{\left (1+\frac {3}{x^2}\right ) x^2 \log \left (\frac {x^4}{16}\right )}-8 \int \frac {1}{\log \left (\frac {x^4}{16}\right )} \, dx+24 \int \frac {1}{x \log \left (\frac {x^4}{16}\right )} \, dx+\frac {(4 x) \operatorname {Subst}\left (\int \frac {e^{x/4}}{x} \, dx,x,\log \left (\frac {x^4}{16}\right )\right )}{\sqrt [4]{x^4}}\\ &=\frac {4 x \text {Ei}\left (\frac {1}{4} \log \left (\frac {x^4}{16}\right )\right )}{\sqrt [4]{x^4}}-2 x \log \left (\log \left (\frac {x^4}{16}\right )\right )+\frac {e^{-\frac {3}{x}+x} \left (3 \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )+x^2 \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )\right )}{\left (1+\frac {3}{x^2}\right ) x^2 \log \left (\frac {x^4}{16}\right )}+6 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (\frac {x^4}{16}\right )\right )-\frac {(4 x) \operatorname {Subst}\left (\int \frac {e^{x/4}}{x} \, dx,x,\log \left (\frac {x^4}{16}\right )\right )}{\sqrt [4]{x^4}}\\ &=6 \log \left (\log \left (\frac {x^4}{16}\right )\right )-2 x \log \left (\log \left (\frac {x^4}{16}\right )\right )+\frac {e^{-\frac {3}{x}+x} \left (3 \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )+x^2 \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )\right )}{\left (1+\frac {3}{x^2}\right ) x^2 \log \left (\frac {x^4}{16}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.44, size = 24, normalized size = 1.00 \begin {gather*} \left (6+e^{-\frac {3}{x}+x}-2 x\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(24*x + 4*E^((-3 + x^2)/x)*x - 8*x^2 + (-2*x^2 + E^((-3 + x^2)/x)*(3 + x^2))*Log[x^4/16]*Log[Log[x^4
/16]])/(x^2*Log[x^4/16]),x]

[Out]

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

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fricas [A]  time = 1.21, size = 26, normalized size = 1.08 \begin {gather*} -{\left (2 \, x - e^{\left (\frac {x^{2} - 3}{x}\right )} - 6\right )} \log \left (\log \left (\frac {1}{16} \, x^{4}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2+3)*exp((x^2-3)/x)-2*x^2)*log(1/16*x^4)*log(log(1/16*x^4))+4*x*exp((x^2-3)/x)-8*x^2+24*x)/x^2/
log(1/16*x^4),x, algorithm="fricas")

[Out]

-(2*x - e^((x^2 - 3)/x) - 6)*log(log(1/16*x^4))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left (2 \, x^{2} - {\left (x^{2} + 3\right )} e^{\left (\frac {x^{2} - 3}{x}\right )}\right )} \log \left (\frac {1}{16} \, x^{4}\right ) \log \left (\log \left (\frac {1}{16} \, x^{4}\right )\right ) + 8 \, x^{2} - 4 \, x e^{\left (\frac {x^{2} - 3}{x}\right )} - 24 \, x}{x^{2} \log \left (\frac {1}{16} \, x^{4}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2+3)*exp((x^2-3)/x)-2*x^2)*log(1/16*x^4)*log(log(1/16*x^4))+4*x*exp((x^2-3)/x)-8*x^2+24*x)/x^2/
log(1/16*x^4),x, algorithm="giac")

[Out]

integrate(-((2*x^2 - (x^2 + 3)*e^((x^2 - 3)/x))*log(1/16*x^4)*log(log(1/16*x^4)) + 8*x^2 - 4*x*e^((x^2 - 3)/x)
- 24*x)/(x^2*log(1/16*x^4)), x)

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maple [C]  time = 1.12, size = 336, normalized size = 14.00

method	result	size
risch	\(\left (-2 x +{\mathrm e}^{\frac {x^{2}-3}{x}}\right ) \ln \left (-4 \ln \relax (2)+4 \ln \relax (x )-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (-\mathrm {csgn}\left (i x^{2}\right )+\mathrm {csgn}\left (i x \right )\right )^{2}}{2}-\frac {i \pi \,\mathrm {csgn}\left (i x^{3}\right ) \left (-\mathrm {csgn}\left (i x^{3}\right )+\mathrm {csgn}\left (i x^{2}\right )\right ) \left (-\mathrm {csgn}\left (i x^{3}\right )+\mathrm {csgn}\left (i x \right )\right )}{2}-\frac {i \pi \,\mathrm {csgn}\left (i x^{4}\right ) \left (-\mathrm {csgn}\left (i x^{4}\right )+\mathrm {csgn}\left (i x^{3}\right )\right ) \left (-\mathrm {csgn}\left (i x^{4}\right )+\mathrm {csgn}\left (i x \right )\right )}{2}\right )+6 \ln \left (\ln \relax (x )-\frac {i \left (\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right )^{2}+\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right ) \mathrm {csgn}\left (i x^{4}\right )-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{4}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}-\pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )^{2}+\pi \mathrm {csgn}\left (i x^{4}\right )^{3}+\pi \mathrm {csgn}\left (i x^{3}\right )^{3}-\pi \,\mathrm {csgn}\left (i x^{3}\right ) \mathrm {csgn}\left (i x^{4}\right )^{2}-8 i \ln \relax (2)\right )}{8}\right )\)	\(336\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x^2+3)*exp((x^2-3)/x)-2*x^2)*ln(1/16*x^4)*ln(ln(1/16*x^4))+4*x*exp((x^2-3)/x)-8*x^2+24*x)/x^2/ln(1/16*x
^4),x,method=_RETURNVERBOSE)

[Out]

(-2*x+exp((x^2-3)/x))*ln(-4*ln(2)+4*ln(x)-1/2*I*Pi*csgn(I*x^2)*(-csgn(I*x^2)+csgn(I*x))^2-1/2*I*Pi*csgn(I*x^3)
*(-csgn(I*x^3)+csgn(I*x^2))*(-csgn(I*x^3)+csgn(I*x))-1/2*I*Pi*csgn(I*x^4)*(-csgn(I*x^4)+csgn(I*x^3))*(-csgn(I*
x^4)+csgn(I*x)))+6*ln(ln(x)-1/8*I*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x)*csgn(I
*x^2)*csgn(I*x^3)-Pi*csgn(I*x)*csgn(I*x^3)^2+Pi*csgn(I*x)*csgn(I*x^3)*csgn(I*x^4)-Pi*csgn(I*x)*csgn(I*x^4)^2+P
i*csgn(I*x^2)^3-Pi*csgn(I*x^2)*csgn(I*x^3)^2+Pi*csgn(I*x^4)^3+Pi*csgn(I*x^3)^3-Pi*csgn(I*x^3)*csgn(I*x^4)^2-8*
I*ln(2)))

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maxima [A]  time = 0.48, size = 43, normalized size = 1.79 \begin {gather*} -4 \, x \log \relax (2) + 2 \, e^{\left (x - \frac {3}{x}\right )} \log \relax (2) - {\left (2 \, x - e^{\left (x - \frac {3}{x}\right )} - 6\right )} \log \left (-\log \relax (2) + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2+3)*exp((x^2-3)/x)-2*x^2)*log(1/16*x^4)*log(log(1/16*x^4))+4*x*exp((x^2-3)/x)-8*x^2+24*x)/x^2/
log(1/16*x^4),x, algorithm="maxima")

[Out]

-4*x*log(2) + 2*e^(x - 3/x)*log(2) - (2*x - e^(x - 3/x) - 6)*log(-log(2) + log(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {24\,x+4\,x\,{\mathrm {e}}^{\frac {x^2-3}{x}}-8\,x^2+\ln \left (\ln \left (\frac {x^4}{16}\right )\right )\,\ln \left (\frac {x^4}{16}\right )\,\left ({\mathrm {e}}^{\frac {x^2-3}{x}}\,\left (x^2+3\right )-2\,x^2\right )}{x^2\,\ln \left (\frac {x^4}{16}\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((24*x + 4*x*exp((x^2 - 3)/x) - 8*x^2 + log(log(x^4/16))*log(x^4/16)*(exp((x^2 - 3)/x)*(x^2 + 3) - 2*x^2))/
(x^2*log(x^4/16)),x)

[Out]

int((24*x + 4*x*exp((x^2 - 3)/x) - 8*x^2 + log(log(x^4/16))*log(x^4/16)*(exp((x^2 - 3)/x)*(x^2 + 3) - 2*x^2))/
(x^2*log(x^4/16)), x)

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sympy [A]  time = 6.61, size = 37, normalized size = 1.54 \begin {gather*} - 2 x \log {\left (\log {\left (\frac {x^{4}}{16} \right )} \right )} + e^{\frac {x^{2} - 3}{x}} \log {\left (\log {\left (\frac {x^{4}}{16} \right )} \right )} + 6 \log {\left (\log {\left (\frac {x^{4}}{16} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x**2+3)*exp((x**2-3)/x)-2*x**2)*ln(1/16*x**4)*ln(ln(1/16*x**4))+4*x*exp((x**2-3)/x)-8*x**2+24*x)/
x**2/ln(1/16*x**4),x)

[Out]

-2*x*log(log(x**4/16)) + exp((x**2 - 3)/x)*log(log(x**4/16)) + 6*log(log(x**4/16))

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