3.31.84 \(\int \frac {e^{-2 x} (e^{2 x} (-1-x)+(54 x+e^{2 x} x-54 x^3) \log (x)+e^{2 x} (1+x) \log (x) \log (\frac {1+x}{\log (x)}))}{(1+x) \log (x)} \, dx\)

Optimal. Leaf size=22 \[ 27 e^{-2 x} x^2+x \log \left (\frac {1+x}{\log (x)}\right ) \]

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Rubi [A]  time = 0.93, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 8, integrand size = 66, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {6688, 6742, 2196, 2176, 2194, 43, 2298, 2549} \begin {gather*} 27 e^{-2 x} x^2+x \log \left (\frac {x+1}{\log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(2*x)*(-1 - x) + (54*x + E^(2*x)*x - 54*x^3)*Log[x] + E^(2*x)*(1 + x)*Log[x]*Log[(1 + x)/Log[x]])/(E^(2
*x)*(1 + x)*Log[x]),x]

[Out]

(27*x^2)/E^(2*x) + x*Log[(1 + x)/Log[x]]

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 2176

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

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2549

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[x*Simplify[D[u, x]/u], x], x] /; ProductQ[
u]

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 {e^{-2 x} x \left (54+e^{2 x}-54 x^2\right )}{1+x}-\frac {1}{\log (x)}+\log \left (\frac {1+x}{\log (x)}\right )\right ) \, dx\\ &=\int \frac {e^{-2 x} x \left (54+e^{2 x}-54 x^2\right )}{1+x} \, dx-\int \frac {1}{\log (x)} \, dx+\int \log \left (\frac {1+x}{\log (x)}\right ) \, dx\\ &=x \log \left (\frac {1+x}{\log (x)}\right )-\text {li}(x)+\int \left (-54 e^{-2 x} (-1+x) x+\frac {x}{1+x}\right ) \, dx-\int \frac {-1-x+x \log (x)}{(1+x) \log (x)} \, dx\\ &=x \log \left (\frac {1+x}{\log (x)}\right )-\text {li}(x)-54 \int e^{-2 x} (-1+x) x \, dx+\int \frac {x}{1+x} \, dx-\int \left (\frac {x}{1+x}-\frac {1}{\log (x)}\right ) \, dx\\ &=x \log \left (\frac {1+x}{\log (x)}\right )-\text {li}(x)-54 \int \left (-e^{-2 x} x+e^{-2 x} x^2\right ) \, dx+\int \left (1+\frac {1}{-1-x}\right ) \, dx-\int \frac {x}{1+x} \, dx+\int \frac {1}{\log (x)} \, dx\\ &=x-\log (1+x)+x \log \left (\frac {1+x}{\log (x)}\right )+54 \int e^{-2 x} x \, dx-54 \int e^{-2 x} x^2 \, dx-\int \left (1+\frac {1}{-1-x}\right ) \, dx\\ &=-27 e^{-2 x} x+27 e^{-2 x} x^2+x \log \left (\frac {1+x}{\log (x)}\right )+27 \int e^{-2 x} \, dx-54 \int e^{-2 x} x \, dx\\ &=-\frac {27}{2} e^{-2 x}+27 e^{-2 x} x^2+x \log \left (\frac {1+x}{\log (x)}\right )-27 \int e^{-2 x} \, dx\\ &=27 e^{-2 x} x^2+x \log \left (\frac {1+x}{\log (x)}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 22, normalized size = 1.00 \begin {gather*} 27 e^{-2 x} x^2+x \log \left (\frac {1+x}{\log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*x)*(-1 - x) + (54*x + E^(2*x)*x - 54*x^3)*Log[x] + E^(2*x)*(1 + x)*Log[x]*Log[(1 + x)/Log[x]])
/(E^(2*x)*(1 + x)*Log[x]),x]

[Out]

(27*x^2)/E^(2*x) + x*Log[(1 + x)/Log[x]]

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fricas [A]  time = 1.05, size = 26, normalized size = 1.18 \begin {gather*} {\left (x e^{\left (2 \, x\right )} \log \left (\frac {x + 1}{\log \relax (x)}\right ) + 27 \, x^{2}\right )} e^{\left (-2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+1)*exp(x)^2*log(x)*log((x+1)/log(x))+(x*exp(x)^2-54*x^3+54*x)*log(x)+(-x-1)*exp(x)^2)/(x+1)/exp(
x)^2/log(x),x, algorithm="fricas")

[Out]

(x*e^(2*x)*log((x + 1)/log(x)) + 27*x^2)*e^(-2*x)

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giac [A]  time = 0.28, size = 22, normalized size = 1.00 \begin {gather*} 27 \, x^{2} e^{\left (-2 \, x\right )} + x \log \left (x + 1\right ) - x \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+1)*exp(x)^2*log(x)*log((x+1)/log(x))+(x*exp(x)^2-54*x^3+54*x)*log(x)+(-x-1)*exp(x)^2)/(x+1)/exp(
x)^2/log(x),x, algorithm="giac")

[Out]

27*x^2*e^(-2*x) + x*log(x + 1) - x*log(log(x))

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maple [A]  time = 0.58, size = 22, normalized size = 1.00

method	result	size
default	\(27 x^{2} {\mathrm e}^{-2 x}+\ln \left (\frac {x +1}{\ln \relax (x )}\right ) x\)	\(22\)
risch	\(x \ln \left (x +1\right )+\frac {x \left (-i {\mathrm e}^{2 x} \pi \,\mathrm {csgn}\left (i \left (x +1\right )\right ) \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i \left (x +1\right )}{\ln \relax (x )}\right )+i {\mathrm e}^{2 x} \pi \,\mathrm {csgn}\left (i \left (x +1\right )\right ) \mathrm {csgn}\left (\frac {i \left (x +1\right )}{\ln \relax (x )}\right )^{2}+i {\mathrm e}^{2 x} \pi \,\mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i \left (x +1\right )}{\ln \relax (x )}\right )^{2}-i {\mathrm e}^{2 x} \pi \mathrm {csgn}\left (\frac {i \left (x +1\right )}{\ln \relax (x )}\right )^{3}-2 \,{\mathrm e}^{2 x} \ln \left (\ln \relax (x )\right )+54 x \right ) {\mathrm e}^{-2 x}}{2}\)	\(140\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x+1)*exp(x)^2*ln(x)*ln((x+1)/ln(x))+(x*exp(x)^2-54*x^3+54*x)*ln(x)+(-x-1)*exp(x)^2)/(x+1)/exp(x)^2/ln(x)
,x,method=_RETURNVERBOSE)

[Out]

27*x^2/exp(x)^2+ln((x+1)/ln(x))*x

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maxima [A]  time = 0.58, size = 22, normalized size = 1.00 \begin {gather*} 27 \, x^{2} e^{\left (-2 \, x\right )} + x \log \left (x + 1\right ) - x \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+1)*exp(x)^2*log(x)*log((x+1)/log(x))+(x*exp(x)^2-54*x^3+54*x)*log(x)+(-x-1)*exp(x)^2)/(x+1)/exp(
x)^2/log(x),x, algorithm="maxima")

[Out]

27*x^2*e^(-2*x) + x*log(x + 1) - x*log(log(x))

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mupad [B]  time = 1.99, size = 21, normalized size = 0.95 \begin {gather*} 27\,x^2\,{\mathrm {e}}^{-2\,x}+x\,\ln \left (\frac {x+1}{\ln \relax (x)}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-2*x)*(log(x)*(54*x + x*exp(2*x) - 54*x^3) - exp(2*x)*(x + 1) + exp(2*x)*log(x)*log((x + 1)/log(x))*(
x + 1)))/(log(x)*(x + 1)),x)

[Out]

27*x^2*exp(-2*x) + x*log((x + 1)/log(x))

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sympy [A]  time = 0.85, size = 37, normalized size = 1.68 \begin {gather*} 27 x^{2} e^{- 2 x} + \left (x + \frac {1}{6}\right ) \log {\left (\frac {x + 1}{\log {\relax (x )}} \right )} - \frac {\log {\left (6 x + 6 \right )}}{6} + \frac {\log {\left (\log {\relax (x )} \right )}}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+1)*exp(x)**2*ln(x)*ln((x+1)/ln(x))+(x*exp(x)**2-54*x**3+54*x)*ln(x)+(-x-1)*exp(x)**2)/(x+1)/exp(
x)**2/ln(x),x)

[Out]

27*x**2*exp(-2*x) + (x + 1/6)*log((x + 1)/log(x)) - log(6*x + 6)/6 + log(log(x))/6

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