3.101.21 \(\int \frac {-64 x+32 e^{2 x} x+(-64 x+e^{2 x} (32 x+64 x^2)) \log (x)+(-128 x+64 e^{2 x} x) \log (x) \log ((2 x-e^{2 x} x) \log (x))}{(-2+e^{2 x}) \log (x)} \, dx\)

Optimal. Leaf size=19 \[ 32 x^2 \log \left (\left (2-e^{2 x}\right ) x \log (x)\right ) \]

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Rubi [A]  time = 1.29, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 32, number of rules used = 14, integrand size = 79, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.177, Rules used = {6742, 2184, 2190, 2531, 2282, 6589, 6688, 14, 43, 2309, 2178, 30, 2555, 12} \begin {gather*} 32 x^2 \log \left (\left (2-e^{2 x}\right ) x \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-64*x + 32*E^(2*x)*x + (-64*x + E^(2*x)*(32*x + 64*x^2))*Log[x] + (-128*x + 64*E^(2*x)*x)*Log[x]*Log[(2*x
- E^(2*x)*x)*Log[x]])/((-2 + E^(2*x))*Log[x]),x]

[Out]

32*x^2*Log[(2 - E^(2*x))*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 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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 2184

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

Rule 2190

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

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2309

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

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

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 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

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 {128 x^2}{-2+e^{2 x}}+\frac {32 x \left (1+\log (x)+2 x \log (x)+2 \log (x) \log \left (-\left (\left (-2+e^{2 x}\right ) x \log (x)\right )\right )\right )}{\log (x)}\right ) \, dx\\ &=32 \int \frac {x \left (1+\log (x)+2 x \log (x)+2 \log (x) \log \left (-\left (\left (-2+e^{2 x}\right ) x \log (x)\right )\right )\right )}{\log (x)} \, dx+128 \int \frac {x^2}{-2+e^{2 x}} \, dx\\ &=-\frac {64 x^3}{3}+32 \int \left (\frac {x (1+\log (x)+2 x \log (x))}{\log (x)}+2 x \log \left (-\left (\left (-2+e^{2 x}\right ) x \log (x)\right )\right )\right ) \, dx+64 \int \frac {e^{2 x} x^2}{-2+e^{2 x}} \, dx\\ &=-\frac {64 x^3}{3}+32 x^2 \log \left (1-\frac {e^{2 x}}{2}\right )+32 \int \frac {x (1+\log (x)+2 x \log (x))}{\log (x)} \, dx-64 \int x \log \left (1-\frac {e^{2 x}}{2}\right ) \, dx+64 \int x \log \left (-\left (\left (-2+e^{2 x}\right ) x \log (x)\right )\right ) \, dx\\ &=-\frac {64 x^3}{3}+32 x^2 \log \left (1-\frac {e^{2 x}}{2}\right )+32 x^2 \log \left (\left (2-e^{2 x}\right ) x \log (x)\right )+32 x \text {Li}_2\left (\frac {e^{2 x}}{2}\right )+32 \int x \left (1+2 x+\frac {1}{\log (x)}\right ) \, dx-32 \int \text {Li}_2\left (\frac {e^{2 x}}{2}\right ) \, dx-64 \int \frac {x \left (-2+e^{2 x}+\left (-2+e^{2 x} (1+2 x)\right ) \log (x)\right )}{2 \left (-2+e^{2 x}\right ) \log (x)} \, dx\\ &=-\frac {64 x^3}{3}+32 x^2 \log \left (1-\frac {e^{2 x}}{2}\right )+32 x^2 \log \left (\left (2-e^{2 x}\right ) x \log (x)\right )+32 x \text {Li}_2\left (\frac {e^{2 x}}{2}\right )-16 \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {x}{2}\right )}{x} \, dx,x,e^{2 x}\right )+32 \int \left (x (1+2 x)+\frac {x}{\log (x)}\right ) \, dx-32 \int \frac {x \left (-2+e^{2 x}+\left (-2+e^{2 x} (1+2 x)\right ) \log (x)\right )}{\left (-2+e^{2 x}\right ) \log (x)} \, dx\\ &=-\frac {64 x^3}{3}+32 x^2 \log \left (1-\frac {e^{2 x}}{2}\right )+32 x^2 \log \left (\left (2-e^{2 x}\right ) x \log (x)\right )+32 x \text {Li}_2\left (\frac {e^{2 x}}{2}\right )-16 \text {Li}_3\left (\frac {e^{2 x}}{2}\right )+32 \int x (1+2 x) \, dx+32 \int \frac {x}{\log (x)} \, dx-32 \int \left (\frac {4 x^2}{-2+e^{2 x}}+\frac {x (1+\log (x)+2 x \log (x))}{\log (x)}\right ) \, dx\\ &=-\frac {64 x^3}{3}+32 x^2 \log \left (1-\frac {e^{2 x}}{2}\right )+32 x^2 \log \left (\left (2-e^{2 x}\right ) x \log (x)\right )+32 x \text {Li}_2\left (\frac {e^{2 x}}{2}\right )-16 \text {Li}_3\left (\frac {e^{2 x}}{2}\right )+32 \int \left (x+2 x^2\right ) \, dx-32 \int \frac {x (1+\log (x)+2 x \log (x))}{\log (x)} \, dx+32 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )-128 \int \frac {x^2}{-2+e^{2 x}} \, dx\\ &=16 x^2+\frac {64 x^3}{3}+32 \text {Ei}(2 \log (x))+32 x^2 \log \left (1-\frac {e^{2 x}}{2}\right )+32 x^2 \log \left (\left (2-e^{2 x}\right ) x \log (x)\right )+32 x \text {Li}_2\left (\frac {e^{2 x}}{2}\right )-16 \text {Li}_3\left (\frac {e^{2 x}}{2}\right )-32 \int x \left (1+2 x+\frac {1}{\log (x)}\right ) \, dx-64 \int \frac {e^{2 x} x^2}{-2+e^{2 x}} \, dx\\ &=16 x^2+\frac {64 x^3}{3}+32 \text {Ei}(2 \log (x))+32 x^2 \log \left (\left (2-e^{2 x}\right ) x \log (x)\right )+32 x \text {Li}_2\left (\frac {e^{2 x}}{2}\right )-16 \text {Li}_3\left (\frac {e^{2 x}}{2}\right )-32 \int \left (x (1+2 x)+\frac {x}{\log (x)}\right ) \, dx+64 \int x \log \left (1-\frac {e^{2 x}}{2}\right ) \, dx\\ &=16 x^2+\frac {64 x^3}{3}+32 \text {Ei}(2 \log (x))+32 x^2 \log \left (\left (2-e^{2 x}\right ) x \log (x)\right )-16 \text {Li}_3\left (\frac {e^{2 x}}{2}\right )-32 \int x (1+2 x) \, dx-32 \int \frac {x}{\log (x)} \, dx+32 \int \text {Li}_2\left (\frac {e^{2 x}}{2}\right ) \, dx\\ &=16 x^2+\frac {64 x^3}{3}+32 \text {Ei}(2 \log (x))+32 x^2 \log \left (\left (2-e^{2 x}\right ) x \log (x)\right )-16 \text {Li}_3\left (\frac {e^{2 x}}{2}\right )+16 \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {x}{2}\right )}{x} \, dx,x,e^{2 x}\right )-32 \int \left (x+2 x^2\right ) \, dx-32 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=32 x^2 \log \left (\left (2-e^{2 x}\right ) x \log (x)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 18, normalized size = 0.95 \begin {gather*} 32 x^2 \log \left (-\left (\left (-2+e^{2 x}\right ) x \log (x)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-64*x + 32*E^(2*x)*x + (-64*x + E^(2*x)*(32*x + 64*x^2))*Log[x] + (-128*x + 64*E^(2*x)*x)*Log[x]*Lo
g[(2*x - E^(2*x)*x)*Log[x]])/((-2 + E^(2*x))*Log[x]),x]

[Out]

32*x^2*Log[-((-2 + E^(2*x))*x*Log[x])]

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fricas [A]  time = 0.70, size = 20, normalized size = 1.05 \begin {gather*} 32 \, x^{2} \log \left (-{\left (x e^{\left (2 \, x\right )} - 2 \, x\right )} \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((64*x*exp(x)^2-128*x)*log(x)*log((-x*exp(x)^2+2*x)*log(x))+((64*x^2+32*x)*exp(x)^2-64*x)*log(x)+32*
x*exp(x)^2-64*x)/(exp(x)^2-2)/log(x),x, algorithm="fricas")

[Out]

32*x^2*log(-(x*e^(2*x) - 2*x)*log(x))

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giac [A]  time = 0.24, size = 27, normalized size = 1.42 \begin {gather*} 32 \, x^{2} \log \left (-e^{\left (2 \, x\right )} \log \relax (x) + 2 \, \log \relax (x)\right ) + 32 \, x^{2} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((64*x*exp(x)^2-128*x)*log(x)*log((-x*exp(x)^2+2*x)*log(x))+((64*x^2+32*x)*exp(x)^2-64*x)*log(x)+32*
x*exp(x)^2-64*x)/(exp(x)^2-2)/log(x),x, algorithm="giac")

[Out]

32*x^2*log(-e^(2*x)*log(x) + 2*log(x)) + 32*x^2*log(x)

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maple [C]  time = 0.17, size = 292, normalized size = 15.37

method	result	size
risch	\(32 x^{2} \ln \left ({\mathrm e}^{2 x}-2\right )+32 x^{2} \ln \left (\ln \relax (x )\right )+32 i \pi \,x^{2}-32 i \pi \,x^{2} \mathrm {csgn}\left (i x \ln \relax (x ) \left ({\mathrm e}^{2 x}-2\right )\right )^{2}+16 i \pi \,x^{2} \mathrm {csgn}\left (i \ln \relax (x ) \left ({\mathrm e}^{2 x}-2\right )\right ) \mathrm {csgn}\left (i x \ln \relax (x ) \left ({\mathrm e}^{2 x}-2\right )\right )^{2}+16 i \pi \,x^{2} \mathrm {csgn}\left (i x \ln \relax (x ) \left ({\mathrm e}^{2 x}-2\right )\right )^{3}+16 i \pi \,x^{2} \mathrm {csgn}\left (i \left ({\mathrm e}^{2 x}-2\right )\right ) \mathrm {csgn}\left (i \ln \relax (x ) \left ({\mathrm e}^{2 x}-2\right )\right )^{2}-16 i \pi \,x^{2} \mathrm {csgn}\left (i \ln \relax (x ) \left ({\mathrm e}^{2 x}-2\right )\right )^{3}+16 i \pi \,x^{2} \mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i \ln \relax (x ) \left ({\mathrm e}^{2 x}-2\right )\right )^{2}-16 i \pi \,x^{2} \mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{2 x}-2\right )\right ) \mathrm {csgn}\left (i \ln \relax (x ) \left ({\mathrm e}^{2 x}-2\right )\right )+16 i \pi \,x^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \ln \relax (x ) \left ({\mathrm e}^{2 x}-2\right )\right )^{2}-16 i \pi \,x^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \ln \relax (x ) \left ({\mathrm e}^{2 x}-2\right )\right ) \mathrm {csgn}\left (i x \ln \relax (x ) \left ({\mathrm e}^{2 x}-2\right )\right )+32 x^{2} \ln \relax (x )\)	\(292\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((64*x*exp(x)^2-128*x)*ln(x)*ln((-x*exp(x)^2+2*x)*ln(x))+((64*x^2+32*x)*exp(x)^2-64*x)*ln(x)+32*x*exp(x)^2
-64*x)/(exp(x)^2-2)/ln(x),x,method=_RETURNVERBOSE)

[Out]

32*x^2*ln(exp(2*x)-2)+32*x^2*ln(ln(x))+32*I*Pi*x^2-32*I*Pi*x^2*csgn(I*x*ln(x)*(exp(2*x)-2))^2+16*I*Pi*x^2*csgn
(I*ln(x)*(exp(2*x)-2))*csgn(I*x*ln(x)*(exp(2*x)-2))^2+16*I*Pi*x^2*csgn(I*x*ln(x)*(exp(2*x)-2))^3+16*I*Pi*x^2*c
sgn(I*(exp(2*x)-2))*csgn(I*ln(x)*(exp(2*x)-2))^2-16*I*Pi*x^2*csgn(I*ln(x)*(exp(2*x)-2))^3+16*I*Pi*x^2*csgn(I*l
n(x))*csgn(I*ln(x)*(exp(2*x)-2))^2-16*I*Pi*x^2*csgn(I*ln(x))*csgn(I*(exp(2*x)-2))*csgn(I*ln(x)*(exp(2*x)-2))+1
6*I*Pi*x^2*csgn(I*x)*csgn(I*x*ln(x)*(exp(2*x)-2))^2-16*I*Pi*x^2*csgn(I*x)*csgn(I*ln(x)*(exp(2*x)-2))*csgn(I*x*
ln(x)*(exp(2*x)-2))+32*x^2*ln(x)

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maxima [A]  time = 0.42, size = 30, normalized size = 1.58 \begin {gather*} 32 \, x^{2} \log \relax (x) + 32 \, x^{2} \log \left (-e^{\left (2 \, x\right )} + 2\right ) + 32 \, x^{2} \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((64*x*exp(x)^2-128*x)*log(x)*log((-x*exp(x)^2+2*x)*log(x))+((64*x^2+32*x)*exp(x)^2-64*x)*log(x)+32*
x*exp(x)^2-64*x)/(exp(x)^2-2)/log(x),x, algorithm="maxima")

[Out]

32*x^2*log(x) + 32*x^2*log(-e^(2*x) + 2) + 32*x^2*log(log(x))

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mupad [B]  time = 6.92, size = 20, normalized size = 1.05 \begin {gather*} 32\,x^2\,\ln \left (\ln \relax (x)\,\left (2\,x-x\,{\mathrm {e}}^{2\,x}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(64*x - 32*x*exp(2*x) + log(x)*(64*x - exp(2*x)*(32*x + 64*x^2)) + log(x)*log(log(x)*(2*x - x*exp(2*x)))*
(128*x - 64*x*exp(2*x)))/(log(x)*(exp(2*x) - 2)),x)

[Out]

32*x^2*log(log(x)*(2*x - x*exp(2*x)))

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sympy [A]  time = 0.76, size = 19, normalized size = 1.00 \begin {gather*} 32 x^{2} \log {\left (\left (- x e^{2 x} + 2 x\right ) \log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((64*x*exp(x)**2-128*x)*ln(x)*ln((-x*exp(x)**2+2*x)*ln(x))+((64*x**2+32*x)*exp(x)**2-64*x)*ln(x)+32*
x*exp(x)**2-64*x)/(exp(x)**2-2)/ln(x),x)

[Out]

32*x**2*log((-x*exp(2*x) + 2*x)*log(x))

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