3.78.37 \(\int \frac {e^{4+2 e^{x+2 x (i \pi +\log (-1+e^2))+x (i \pi +\log (-1+e^2))^2}+x+2 x (i \pi +\log (-1+e^2))+x (i \pi +\log (-1+e^2))^2} (2+4 (i \pi +\log (-1+e^2))+2 (i \pi +\log (-1+e^2))^2)}{-1+e^{4+2 e^{x+2 x (i \pi +\log (-1+e^2))+x (i \pi +\log (-1+e^2))^2}}} \, dx\)

Optimal. Leaf size=30 \[ \log \left (1-e^{4+2 e^{x \left (1+i \pi +\log \left (-1+e^2\right )\right )^2}}\right ) \]

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Rubi [A]  time = 0.97, antiderivative size = 32, normalized size of antiderivative = 1.07, number of steps used = 4, number of rules used = 4, integrand size = 151, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {12, 2282, 2246, 31} \begin {gather*} \log \left (1-\exp \left (4+2 e^{-x \left (\pi -i \left (1+\log \left (e^2-1\right )\right )\right )^2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(4 + 2*E^(x + 2*x*(I*Pi + Log[-1 + E^2]) + x*(I*Pi + Log[-1 + E^2])^2) + x + 2*x*(I*Pi + Log[-1 + E^2])
+ x*(I*Pi + Log[-1 + E^2])^2)*(2 + 4*(I*Pi + Log[-1 + E^2]) + 2*(I*Pi + Log[-1 + E^2])^2))/(-1 + E^(4 + 2*E^(
x + 2*x*(I*Pi + Log[-1 + E^2]) + x*(I*Pi + Log[-1 + E^2])^2))),x]

[Out]

Log[1 - E^(4 + 2/E^(x*(Pi - I*(1 + Log[-1 + E^2]))^2))]

Rule 12

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2246

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

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]]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left (\left (2 \left (\pi -i \left (1+\log \left (-1+e^2\right )\right )\right )^2\right ) \int \frac {\exp \left (4+2 \exp \left (x+2 x \left (i \pi +\log \left (-1+e^2\right )\right )+x \left (i \pi +\log \left (-1+e^2\right )\right )^2\right )+x+2 x \left (i \pi +\log \left (-1+e^2\right )\right )+x \left (i \pi +\log \left (-1+e^2\right )\right )^2\right )}{-1+\exp \left (4+2 \exp \left (x+2 x \left (i \pi +\log \left (-1+e^2\right )\right )+x \left (i \pi +\log \left (-1+e^2\right )\right )^2\right )\right )} \, dx\right )\\ &=2 \operatorname {Subst}\left (\int \frac {e^{4+2 x}}{-1+e^{4+2 x}} \, dx,x,\exp \left (x \left (1+2 \left (i \pi +\log \left (-1+e^2\right )\right )+\left (i \pi +\log \left (-1+e^2\right )\right )^2\right )\right )\right )\\ &=\operatorname {Subst}\left (\int \frac {1}{-1+x} \, dx,x,\exp \left (4+2 \exp \left (x \left (1+2 \left (i \pi +\log \left (-1+e^2\right )\right )+\left (i \pi +\log \left (-1+e^2\right )\right )^2\right )\right )\right )\right )\\ &=\log \left (1-\exp \left (4+2 \exp \left (x \left (1+2 \left (i \pi +\log \left (-1+e^2\right )\right )+\left (i \pi +\log \left (-1+e^2\right )\right )^2\right )\right )\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 32, normalized size = 1.07 \begin {gather*} \log \left (1-e^{4+2 e^{-x \left (\pi -i \left (1+\log \left (-1+e^2\right )\right )\right )^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(4 + 2*E^(x + 2*x*(I*Pi + Log[-1 + E^2]) + x*(I*Pi + Log[-1 + E^2])^2) + x + 2*x*(I*Pi + Log[-1 +
E^2]) + x*(I*Pi + Log[-1 + E^2])^2)*(2 + 4*(I*Pi + Log[-1 + E^2]) + 2*(I*Pi + Log[-1 + E^2])^2))/(-1 + E^(4 +
2*E^(x + 2*x*(I*Pi + Log[-1 + E^2]) + x*(I*Pi + Log[-1 + E^2])^2))),x]

[Out]

Log[1 - E^(4 + 2/E^(x*(Pi - I*(1 + Log[-1 + E^2]))^2))]

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fricas [B]  time = 0.72, size = 105, normalized size = 3.50 \begin {gather*} -x \log \left (-e^{2} + 1\right )^{2} - 2 \, x \log \left (-e^{2} + 1\right ) - x + \log \left (e^{\left (x \log \left (-e^{2} + 1\right )^{2} + 2 \, x \log \left (-e^{2} + 1\right ) + x + 2 \, e^{\left (x \log \left (-e^{2} + 1\right )^{2} + 2 \, x \log \left (-e^{2} + 1\right ) + x\right )} + 4\right )} - e^{\left (x \log \left (-e^{2} + 1\right )^{2} + 2 \, x \log \left (-e^{2} + 1\right ) + x\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x*log(-e^2 + 1)^2 - 2*x*log(-e^2 + 1) - x + log(e^(x*log(-e^2 + 1)^2 + 2*x*log(-e^2 + 1) + x + 2*e^(x*log(-e^
2 + 1)^2 + 2*x*log(-e^2 + 1) + x) + 4) - e^(x*log(-e^2 + 1)^2 + 2*x*log(-e^2 + 1) + x))

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giac [A]  time = 0.95, size = 32, normalized size = 1.07 \begin {gather*} \log \left (e^{\left (2 \, e^{\left (x \log \left (-e^{2} + 1\right )^{2} + 2 \, x \log \left (-e^{2} + 1\right ) + x\right )} + 4\right )} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(e^(2*e^(x*log(-e^2 + 1)^2 + 2*x*log(-e^2 + 1) + x) + 4) - 1)

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maple [B]  time = 0.28, size = 62, normalized size = 2.07

method	result	size
derivativedivides	\(\ln \left ({\mathrm e}^{{\mathrm e}^{x \ln \left (1-{\mathrm e}^{2}\right )^{2}+2 x \ln \left (1-{\mathrm e}^{2}\right )+x}+2}-1\right )+\ln \left ({\mathrm e}^{{\mathrm e}^{x \ln \left (1-{\mathrm e}^{2}\right )^{2}+2 x \ln \left (1-{\mathrm e}^{2}\right )+x}+2}+1\right )\)	\(62\)
norman	\(\ln \left ({\mathrm e}^{{\mathrm e}^{x \ln \left (1-{\mathrm e}^{2}\right )^{2}+2 x \ln \left (1-{\mathrm e}^{2}\right )+x}+2}-1\right )+\ln \left ({\mathrm e}^{{\mathrm e}^{x \ln \left (1-{\mathrm e}^{2}\right )^{2}+2 x \ln \left (1-{\mathrm e}^{2}\right )+x}+2}+1\right )\)	\(62\)
default	\(\frac {\left (2 \ln \left (1-{\mathrm e}^{2}\right )^{2}+4 \ln \left (1-{\mathrm e}^{2}\right )+2\right ) \left (\frac {\ln \left ({\mathrm e}^{{\mathrm e}^{x \ln \left (1-{\mathrm e}^{2}\right )^{2}+2 x \ln \left (1-{\mathrm e}^{2}\right )+x}+2}-1\right )}{2}+\frac {\ln \left ({\mathrm e}^{{\mathrm e}^{x \ln \left (1-{\mathrm e}^{2}\right )^{2}+2 x \ln \left (1-{\mathrm e}^{2}\right )+x}+2}+1\right )}{2}\right )}{\ln \left (1-{\mathrm e}^{2}\right )^{2}+2 \ln \left (1-{\mathrm e}^{2}\right )+1}\)	\(111\)
risch	\(\frac {\left (2 \ln \left (1-{\mathrm e}^{2}\right )^{2}+4 \ln \left (1-{\mathrm e}^{2}\right )+2\right ) \left (1-{\mathrm e}^{2}\right )^{2 x} {\mathrm e}^{x \left (\ln \left (1-{\mathrm e}^{2}\right )^{2}+1\right )}}{\ln \left (1-{\mathrm e}^{2}\right )^{2}+2 \ln \left (1-{\mathrm e}^{2}\right )+1}-\frac {2 \ln \left (1-{\mathrm e}^{2}\right )^{2} \left (1-{\mathrm e}^{2}\right )^{2 x} {\mathrm e}^{x \left (\ln \left (1-{\mathrm e}^{2}\right )^{2}+1\right )}}{\ln \left (1-{\mathrm e}^{2}\right )^{2}+2 \ln \left (1-{\mathrm e}^{2}\right )+1}-\frac {4 \ln \left (1-{\mathrm e}^{2}\right ) \left (1-{\mathrm e}^{2}\right )^{2 x} {\mathrm e}^{x \left (\ln \left (1-{\mathrm e}^{2}\right )^{2}+1\right )}}{\ln \left (1-{\mathrm e}^{2}\right )^{2}+2 \ln \left (1-{\mathrm e}^{2}\right )+1}-\frac {4 \ln \left (1-{\mathrm e}^{2}\right )^{2}}{\ln \left (1-{\mathrm e}^{2}\right )^{2}+2 \ln \left (1-{\mathrm e}^{2}\right )+1}-\frac {2 \left (1-{\mathrm e}^{2}\right )^{2 x} {\mathrm e}^{x \left (\ln \left (1-{\mathrm e}^{2}\right )^{2}+1\right )}}{\ln \left (1-{\mathrm e}^{2}\right )^{2}+2 \ln \left (1-{\mathrm e}^{2}\right )+1}-\frac {8 \ln \left (1-{\mathrm e}^{2}\right )}{\ln \left (1-{\mathrm e}^{2}\right )^{2}+2 \ln \left (1-{\mathrm e}^{2}\right )+1}-\frac {4}{\ln \left (1-{\mathrm e}^{2}\right )^{2}+2 \ln \left (1-{\mathrm e}^{2}\right )+1}+\frac {\ln \left ({\mathrm e}^{2 \left (1-{\mathrm e}^{2}\right )^{2 x} {\mathrm e}^{x \left (\ln \left (1-{\mathrm e}^{2}\right )^{2}+1\right )}+4}-1\right ) \ln \left (1-{\mathrm e}^{2}\right )^{2}}{\ln \left (1-{\mathrm e}^{2}\right )^{2}+2 \ln \left (1-{\mathrm e}^{2}\right )+1}+\frac {2 \ln \left ({\mathrm e}^{2 \left (1-{\mathrm e}^{2}\right )^{2 x} {\mathrm e}^{x \left (\ln \left (1-{\mathrm e}^{2}\right )^{2}+1\right )}+4}-1\right ) \ln \left (1-{\mathrm e}^{2}\right )}{\ln \left (1-{\mathrm e}^{2}\right )^{2}+2 \ln \left (1-{\mathrm e}^{2}\right )+1}+\frac {\ln \left ({\mathrm e}^{2 \left (1-{\mathrm e}^{2}\right )^{2 x} {\mathrm e}^{x \left (\ln \left (1-{\mathrm e}^{2}\right )^{2}+1\right )}+4}-1\right )}{\ln \left (1-{\mathrm e}^{2}\right )^{2}+2 \ln \left (1-{\mathrm e}^{2}\right )+1}\)	\(501\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(exp(exp(x*ln(1-exp(2))^2+2*x*ln(1-exp(2))+x)+2)-1)+ln(exp(exp(x*ln(1-exp(2))^2+2*x*ln(1-exp(2))+x)+2)+1)

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maxima [A]  time = 0.39, size = 32, normalized size = 1.07 \begin {gather*} \log \left (e^{\left (2 \, e^{\left (x \log \left (-e^{2} + 1\right )^{2} + 2 \, x \log \left (-e^{2} + 1\right ) + x\right )} + 4\right )} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(e^(2*e^(x*log(-e^2 + 1)^2 + 2*x*log(-e^2 + 1) + x) + 4) - 1)

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mupad [B]  time = 6.64, size = 32, normalized size = 1.07 \begin {gather*} \ln \left ({\mathrm {e}}^{2\,{\mathrm {e}}^{x\,{\ln \left (1-{\mathrm {e}}^2\right )}^2}\,{\mathrm {e}}^x\,{\left (1-{\mathrm {e}}^2\right )}^{2\,x}+4}-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x + 2*x*log(1 - exp(2)) + x*log(1 - exp(2))^2)*exp(2*exp(x + 2*x*log(1 - exp(2)) + x*log(1 - exp(2))^
2) + 4)*(4*log(1 - exp(2)) + 2*log(1 - exp(2))^2 + 2))/(exp(2*exp(x + 2*x*log(1 - exp(2)) + x*log(1 - exp(2))^
2) + 4) - 1),x)

[Out]

log(exp(2*exp(x*log(1 - exp(2))^2)*exp(x)*(1 - exp(2))^(2*x) + 4) - 1)

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sympy [B]  time = 123.74, size = 82, normalized size = 2.73 \begin {gather*} \frac {\left (2 + 4 \log {\left (-1 + e^{2} \right )} + 4 i \pi + 2 \left (\log {\left (-1 + e^{2} \right )} + i \pi \right )^{2}\right ) \log {\left (e^{2 e^{x + 2 x \left (\log {\left (-1 + e^{2} \right )} + i \pi \right ) + x \left (\log {\left (-1 + e^{2} \right )} + i \pi \right )^{2}} + 4} - 1 \right )}}{2 \left (1 + \log {\left (-1 + e^{2} \right )} + i \pi \right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*ln(1-exp(2))**2+4*ln(1-exp(2))+2)*exp(x*ln(1-exp(2))**2+2*x*ln(1-exp(2))+x)*exp(exp(x*ln(1-exp(2)
)**2+2*x*ln(1-exp(2))+x)+2)**2/(exp(exp(x*ln(1-exp(2))**2+2*x*ln(1-exp(2))+x)+2)**2-1),x)

[Out]

(2 + 4*log(-1 + exp(2)) + 4*I*pi + 2*(log(-1 + exp(2)) + I*pi)**2)*log(exp(2*exp(x + 2*x*(log(-1 + exp(2)) + I
*pi) + x*(log(-1 + exp(2)) + I*pi)**2) + 4) - 1)/(2*(1 + log(-1 + exp(2)) + I*pi)**2)

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