3.526 \(\int \frac{e^x+e^{5 x}}{-1+e^x-e^{2 x}+e^{3 x}} \, dx\)

Optimal. Leaf size=39 \[ e^x+\frac{e^{2 x}}{2}+\log \left (1-e^x\right )-\frac{1}{2} \log \left (e^{2 x}+1\right )-\tan ^{-1}\left (e^x\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0615397, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {2282, 2074, 635, 203, 260} \[ e^x+\frac{e^{2 x}}{2}+\log \left (1-e^x\right )-\frac{1}{2} \log \left (e^{2 x}+1\right )-\tan ^{-1}\left (e^x\right ) \]

Antiderivative was successfully verified.

[In]

Int[(E^x + E^(5*x))/(-1 + E^x - E^(2*x) + E^(3*x)),x]

[Out]

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

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 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 260

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

Rubi steps

\begin{align*} \int \frac{e^x+e^{5 x}}{-1+e^x-e^{2 x}+e^{3 x}} \, dx &=\operatorname{Subst}\left (\int \frac{-1-x^4}{1-x+x^2-x^3} \, dx,x,e^x\right )\\ &=\operatorname{Subst}\left (\int \left (1+\frac{1}{-1+x}+x+\frac{-1-x}{1+x^2}\right ) \, dx,x,e^x\right )\\ &=e^x+\frac{e^{2 x}}{2}+\log \left (1-e^x\right )+\operatorname{Subst}\left (\int \frac{-1-x}{1+x^2} \, dx,x,e^x\right )\\ &=e^x+\frac{e^{2 x}}{2}+\log \left (1-e^x\right )-\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,e^x\right )-\operatorname{Subst}\left (\int \frac{x}{1+x^2} \, dx,x,e^x\right )\\ &=e^x+\frac{e^{2 x}}{2}-\tan ^{-1}\left (e^x\right )+\log \left (1-e^x\right )-\frac{1}{2} \log \left (1+e^{2 x}\right )\\ \end{align*}

Mathematica [C]  time = 0.0423789, size = 51, normalized size = 1.31 \[ \frac{1}{2} \left (2 e^x+e^{2 x}+(-1+i) \log \left (-e^x+i\right )+2 \log \left (1-e^x\right )-(1+i) \log \left (e^x+i\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(E^x + E^(5*x))/(-1 + E^x - E^(2*x) + E^(3*x)),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.012, size = 29, normalized size = 0.7 \begin{align*} -{\frac{\ln \left ( \left ({{\rm e}^{x}} \right ) ^{2}+1 \right ) }{2}}-\arctan \left ({{\rm e}^{x}} \right ) +\ln \left ( -1+{{\rm e}^{x}} \right ) +{{\rm e}^{x}}+{\frac{ \left ({{\rm e}^{x}} \right ) ^{2}}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)+exp(5*x))/(-1+exp(x)-exp(2*x)+exp(3*x)),x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.41349, size = 38, normalized size = 0.97 \begin{align*} -\arctan \left (e^{x}\right ) + \frac{1}{2} \, e^{\left (2 \, x\right )} + e^{x} - \frac{1}{2} \, \log \left (e^{\left (2 \, x\right )} + 1\right ) + \log \left (e^{x} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)+exp(5*x))/(-1+exp(x)-exp(2*x)+exp(3*x)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.94914, size = 97, normalized size = 2.49 \begin{align*} -\arctan \left (e^{x}\right ) + \frac{1}{2} \, e^{\left (2 \, x\right )} + e^{x} - \frac{1}{2} \, \log \left (e^{\left (2 \, x\right )} + 1\right ) + \log \left (e^{x} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)+exp(5*x))/(-1+exp(x)-exp(2*x)+exp(3*x)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 0.186641, size = 48, normalized size = 1.23 \begin{align*} \frac{e^{2 x}}{2} + e^{x} + \log{\left (e^{x} - 1 \right )} + \operatorname{RootSum}{\left (2 z^{2} + 2 z + 1, \left ( i \mapsto i \log{\left (\frac{4 i^{2}}{5} - \frac{6 i}{5} + e^{x} - \frac{3}{5} \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)+exp(5*x))/(-1+exp(x)-exp(2*x)+exp(3*x)),x)

[Out]

exp(2*x)/2 + exp(x) + log(exp(x) - 1) + RootSum(2*_z**2 + 2*_z + 1, Lambda(_i, _i*log(4*_i**2/5 - 6*_i/5 + exp
(x) - 3/5)))

________________________________________________________________________________________

Giac [A]  time = 1.13056, size = 39, normalized size = 1. \begin{align*} -\arctan \left (e^{x}\right ) + \frac{1}{2} \, e^{\left (2 \, x\right )} + e^{x} - \frac{1}{2} \, \log \left (e^{\left (2 \, x\right )} + 1\right ) + \log \left ({\left | e^{x} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)+exp(5*x))/(-1+exp(x)-exp(2*x)+exp(3*x)),x, algorithm="giac")

[Out]

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