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

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

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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} $$\begin{array}{c}\log (1-\exp (4+2e^{-x{(\pi -i(1+\log (e^2-1)))}^2}))\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =-(\left(2{(\pi -i(1+\log (-1+e^2)))}^2\right)\int \frac{\exp (4+2\exp (x+2x(i\pi +\log (-1+e^2))+x{(i\pi +\log (-1+e^2))}^2)+x+2x(i\pi +\log (-1+e^2))+x{(i\pi +\log (-1+e^2))}^2)}{-1+\exp (4+2\exp (x+2x(i\pi +\log (-1+e^2))+x{(i\pi +\log (-1+e^2))}^2))}dx)\\ & =2\mathrm{Subst}(\int \frac{e^{4+2x}}{-1+e^{4+2x}}dx,x,\exp \left(x(1+2(i\pi +\log (-1+e^2))+{(i\pi +\log (-1+e^2))}^2)\right))\\ & =\mathrm{Subst}(\int \frac{1}{-1+x}dx,x,\exp (4+2\exp \left(x(1+2(i\pi +\log (-1+e^2))+{(i\pi +\log (-1+e^2))}^2)\right)))\\ & =\log (1-\exp (4+2\exp \left(x(1+2(i\pi +\log (-1+e^2))+{(i\pi +\log (-1+e^2))}^2)\right)))\end{array}\end{array}$$

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

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 $$\begin{array}{c}-x\log {(-e^2+1)}^2-2x\log (-e^2+1)-x+\log (e^{(x\log {(-e^2+1)}^2+2x\log (-e^2+1)+x+2e^{(x\log {(-e^2+1)}^2+2x\log (-e^2+1)+x)}+4)}-e^{(x\log {(-e^2+1)}^2+2x\log (-e^2+1)+x)})\end{array}$$

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 $$\begin{array}{c}\log (e^{(2e^{(x\log {(-e^2+1)}^2+2x\log (-e^2+1)+x)}+4)}-1)\end{array}$$

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	$\ln ({\mathrm{e}}^{{\mathrm{e}}^{x\ln {(1-{\mathrm{e}}^2)}^2+2x\ln (1-{\mathrm{e}}^2)+x}+2}-1)+\ln ({\mathrm{e}}^{{\mathrm{e}}^{x\ln {(1-{\mathrm{e}}^2)}^2+2x\ln (1-{\mathrm{e}}^2)+x}+2}+1)$	$62$
norman	$\ln ({\mathrm{e}}^{{\mathrm{e}}^{x\ln {(1-{\mathrm{e}}^2)}^2+2x\ln (1-{\mathrm{e}}^2)+x}+2}-1)+\ln ({\mathrm{e}}^{{\mathrm{e}}^{x\ln {(1-{\mathrm{e}}^2)}^2+2x\ln (1-{\mathrm{e}}^2)+x}+2}+1)$	$62$
default	$\frac{(2\ln {(1-{\mathrm{e}}^2)}^2+4\ln (1-{\mathrm{e}}^2)+2)(\frac{\ln ({\mathrm{e}}^{{\mathrm{e}}^{x\ln {(1-{\mathrm{e}}^2)}^2+2x\ln (1-{\mathrm{e}}^2)+x}+2}-1)}{2}+\frac{\ln ({\mathrm{e}}^{{\mathrm{e}}^{x\ln {(1-{\mathrm{e}}^2)}^2+2x\ln (1-{\mathrm{e}}^2)+x}+2}+1)}{2})}{\ln {(1-{\mathrm{e}}^2)}^2+2\ln (1-{\mathrm{e}}^2)+1}$	$111$
risch	$\frac{(2\ln {(1-{\mathrm{e}}^2)}^2+4\ln (1-{\mathrm{e}}^2)+2){(1-{\mathrm{e}}^2)}^{2x}{\mathrm{e}}^{x(\ln {(1-{\mathrm{e}}^2)}^2+1)}}{\ln {(1-{\mathrm{e}}^2)}^2+2\ln (1-{\mathrm{e}}^2)+1}-\frac{2\ln {(1-{\mathrm{e}}^2)}^2{(1-{\mathrm{e}}^2)}^{2x}{\mathrm{e}}^{x(\ln {(1-{\mathrm{e}}^2)}^2+1)}}{\ln {(1-{\mathrm{e}}^2)}^2+2\ln (1-{\mathrm{e}}^2)+1}-\frac{4\ln (1-{\mathrm{e}}^2){(1-{\mathrm{e}}^2)}^{2x}{\mathrm{e}}^{x(\ln {(1-{\mathrm{e}}^2)}^2+1)}}{\ln {(1-{\mathrm{e}}^2)}^2+2\ln (1-{\mathrm{e}}^2)+1}-\frac{4\ln {(1-{\mathrm{e}}^2)}^2}{\ln {(1-{\mathrm{e}}^2)}^2+2\ln (1-{\mathrm{e}}^2)+1}-\frac{2{(1-{\mathrm{e}}^2)}^{2x}{\mathrm{e}}^{x(\ln {(1-{\mathrm{e}}^2)}^2+1)}}{\ln {(1-{\mathrm{e}}^2)}^2+2\ln (1-{\mathrm{e}}^2)+1}-\frac{8\ln (1-{\mathrm{e}}^2)}{\ln {(1-{\mathrm{e}}^2)}^2+2\ln (1-{\mathrm{e}}^2)+1}-\frac{4}{\ln {(1-{\mathrm{e}}^2)}^2+2\ln (1-{\mathrm{e}}^2)+1}+\frac{\ln ({\mathrm{e}}^{2{(1-{\mathrm{e}}^2)}^{2x}{\mathrm{e}}^{x(\ln {(1-{\mathrm{e}}^2)}^2+1)}+4}-1)\ln {(1-{\mathrm{e}}^2)}^2}{\ln {(1-{\mathrm{e}}^2)}^2+2\ln (1-{\mathrm{e}}^2)+1}+\frac{2\ln ({\mathrm{e}}^{2{(1-{\mathrm{e}}^2)}^{2x}{\mathrm{e}}^{x(\ln {(1-{\mathrm{e}}^2)}^2+1)}+4}-1)\ln (1-{\mathrm{e}}^2)}{\ln {(1-{\mathrm{e}}^2)}^2+2\ln (1-{\mathrm{e}}^2)+1}+\frac{\ln ({\mathrm{e}}^{2{(1-{\mathrm{e}}^2)}^{2x}{\mathrm{e}}^{x(\ln {(1-{\mathrm{e}}^2)}^2+1)}+4}-1)}{\ln {(1-{\mathrm{e}}^2)}^2+2\ln (1-{\mathrm{e}}^2)+1}$	$501$

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 $$\begin{array}{c}\log (e^{(2e^{(x\log {(-e^2+1)}^2+2x\log (-e^2+1)+x)}+4)}-1)\end{array}$$

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 $$\begin{array}{c}\ln ({\mathrm{e}}^{2{\mathrm{e}}^{x{\ln (1-{\mathrm{e}}^2)}^2}{\mathrm{e}}^x{(1-{\mathrm{e}}^2)}^{2x}+4}-1)\end{array}$$

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 $$\begin{array}{c}\frac{(2+4\log (-1+e^2)+4i\pi +2{(\log (-1+e^2)+i\pi )}^2)\log (e^{2e^{x+2x(\log (-1+e^2)+i\pi )+x{(\log (-1+e^2)+i\pi )}^2}+4}-1)}{2{(1+\log (-1+e^2)+i\pi )}^2}\end{array}$$

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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