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

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

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Rubi [A]  time = 0.06, antiderivative size = 39, normalized size of antiderivative = 1.00, 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 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]

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 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 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 {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*}

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Mathematica [C]  time = 0.04, 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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e^x+e^{5 x}}{-1+e^x-e^{2 x}+e^{3 x}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

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fricas [A]  time = 1.29, size = 28, normalized size = 0.72 \[ -\arctan \left (e^{x}\right ) + \frac {1}{2} \, e^{\left (2 \, x\right )} + e^{x} - \frac {1}{2} \, \log \left (e^{\left (2 \, x\right )} + 1\right ) + \log \left (e^{x} - 1\right ) \]

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)

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giac [A]  time = 0.63, size = 29, normalized size = 0.74 \[ -\arctan \left (e^{x}\right ) + \frac {1}{2} \, e^{\left (2 \, x\right )} + e^{x} - \frac {1}{2} \, \log \left (e^{\left (2 \, x\right )} + 1\right ) + \log \left ({\left | e^{x} - 1 \right |}\right ) \]

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))

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maple [A]  time = 0.09, size = 29, normalized size = 0.74

method	result	size
default	\(-\frac {\ln \left (1+{\mathrm e}^{2 x}\right )}{2}-\arctan \left ({\mathrm e}^{x}\right )+\ln \left (-1+{\mathrm e}^{x}\right )+{\mathrm e}^{x}+\frac {{\mathrm e}^{2 x}}{2}\)	\(29\)
risch	\(\frac {{\mathrm e}^{2 x}}{2}+{\mathrm e}^{x}+\ln \left (-1+{\mathrm e}^{x}\right )-\frac {\ln \left ({\mathrm e}^{x}-i\right )}{2}+\frac {i \ln \left ({\mathrm e}^{x}-i\right )}{2}-\frac {\ln \left ({\mathrm e}^{x}+i\right )}{2}-\frac {i \ln \left ({\mathrm e}^{x}+i\right )}{2}\)	\(49\)
meijerg	\(\frac {\left (\moverset {\infty }{\munderset {\textit {\_k1} =0}{\sum }}\frac {1-{\mathrm e}^{-x \left (3+\textit {\_k1} \right ) \left (1-\frac {1}{3+\textit {\_k1}}\right )}}{\left (3+\textit {\_k1} \right ) \left (1-\frac {1}{3+\textit {\_k1}}\right )}\right )}{2}-\frac {\left (\moverset {\infty }{\munderset {\textit {\_k1} =0}{\sum }}\frac {1-{\mathrm e}^{x \left (2-\textit {\_k1} \right )}}{2-\textit {\_k1}}\right )}{2}\)	\(75\)

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,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 1.16, size = 28, normalized size = 0.72 \[ -\arctan \left (e^{x}\right ) + \frac {1}{2} \, e^{\left (2 \, x\right )} + e^{x} - \frac {1}{2} \, \log \left (e^{\left (2 \, x\right )} + 1\right ) + \log \left (e^{x} - 1\right ) \]

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)

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mupad [B]  time = 0.09, size = 28, normalized size = 0.72 \[ \frac {{\mathrm {e}}^{2\,x}}{2}-\frac {\ln \left ({\mathrm {e}}^{2\,x}+1\right )}{2}-\mathrm {atan}\left ({\mathrm {e}}^x\right )+\ln \left ({\mathrm {e}}^x-1\right )+{\mathrm {e}}^x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(2*x)/2 - log(exp(2*x) + 1)/2 - atan(exp(x)) + log(exp(x) - 1) + exp(x)

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sympy [A]  time = 0.21, size = 48, normalized size = 1.23 \[ \frac {e^{2 x}}{2} + e^{x} + \log {\left (e^{x} - 1 \right )} + \operatorname {RootSum} {\left (2 z^{2} + 2 z + 1, \left (i \mapsto i \log {\left (\frac {4 i^{2}}{5} - \frac {6 i}{5} + e^{x} - \frac {3}{5} \right )} \right )\right )} \]

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)))

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