\(\int \frac {25+e+30 e^{2 x}+5 e^{4 x}}{25+e+10 e^{2 x}+e^{4 x}} \, dx\) [2259]

Optimal result

Integrand size = 35, antiderivative size = 18 \[ \int \frac {25+e+30 e^{2 x}+5 e^{4 x}}{25+e+10 e^{2 x}+e^{4 x}} \, dx=x+\log \left (-e-\left (5+e^{2 x}\right )^2\right ) \]

[Out]

ln(-(exp(x)^2+5)^2-exp(1))+x

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {2320, 1642, 642} \[ \int \frac {25+e+30 e^{2 x}+5 e^{4 x}}{25+e+10 e^{2 x}+e^{4 x}} \, dx=x+\log \left (10 e^{2 x}+e^{4 x}+25+e\right ) \]

[In]

Int[(25 + E + 30*E^(2*x) + 5*E^(4*x))/(25 + E + 10*E^(2*x) + E^(4*x)),x]

[Out]

x + Log[25 + E + 10*E^(2*x) + E^(4*x)]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1642

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 2320

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*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {25+e+30 x+5 x^2}{x \left (25+e+10 x+x^2\right )} \, dx,x,e^{2 x}\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {1}{x}+\frac {4 (5+x)}{25+e+10 x+x^2}\right ) \, dx,x,e^{2 x}\right ) \\ & = x+2 \text {Subst}\left (\int \frac {5+x}{25+e+10 x+x^2} \, dx,x,e^{2 x}\right ) \\ & = x+\log \left (25+e+10 e^{2 x}+e^{4 x}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.50 \[ \int \frac {25+e+30 e^{2 x}+5 e^{4 x}}{25+e+10 e^{2 x}+e^{4 x}} \, dx=\frac {1}{2} \log \left (e^{2 x}\right )+\log \left (25+e+10 e^{2 x}+e^{4 x}\right ) \]

[In]

Integrate[(25 + E + 30*E^(2*x) + 5*E^(4*x))/(25 + E + 10*E^(2*x) + E^(4*x)),x]

[Out]

Log[E^(2*x)]/2 + Log[25 + E + 10*E^(2*x) + E^(4*x)]

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00

method	result	size
norman	\(x +\ln \left ({\mathrm e}^{4 x}+10 \,{\mathrm e}^{2 x}+{\mathrm e}+25\right )\)	\(18\)
risch	\(x +\ln \left ({\mathrm e}^{4 x}+10 \,{\mathrm e}^{2 x}+{\mathrm e}+25\right )\)	\(18\)
parallelrisch	\(x +\ln \left ({\mathrm e}^{4 x}+10 \,{\mathrm e}^{2 x}+{\mathrm e}+25\right )\)	\(18\)
derivativedivides	\(\ln \left (\left ({\mathrm e}^{4 x}+10 \,{\mathrm e}^{2 x}+{\mathrm e}+25\right ) {\mathrm e}^{x}\right )\)	\(19\)
default	\(\ln \left (\left ({\mathrm e}^{4 x}+10 \,{\mathrm e}^{2 x}+{\mathrm e}+25\right ) {\mathrm e}^{x}\right )\)	\(19\)

[In]

int((5*exp(x)^4+30*exp(x)^2+exp(1)+25)/(exp(x)^4+10*exp(x)^2+exp(1)+25),x,method=_RETURNVERBOSE)

[Out]

x+ln(exp(x)^4+10*exp(x)^2+exp(1)+25)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {25+e+30 e^{2 x}+5 e^{4 x}}{25+e+10 e^{2 x}+e^{4 x}} \, dx=x + \log \left (e + e^{\left (4 \, x\right )} + 10 \, e^{\left (2 \, x\right )} + 25\right ) \]

[In]

integrate((5*exp(x)^4+30*exp(x)^2+exp(1)+25)/(exp(x)^4+10*exp(x)^2+exp(1)+25),x, algorithm="fricas")

[Out]

x + log(e + e^(4*x) + 10*e^(2*x) + 25)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {25+e+30 e^{2 x}+5 e^{4 x}}{25+e+10 e^{2 x}+e^{4 x}} \, dx=x + \log {\left (e^{4 x} + 10 e^{2 x} + e + 25 \right )} \]

[In]

integrate((5*exp(x)**4+30*exp(x)**2+exp(1)+25)/(exp(x)**4+10*exp(x)**2+exp(1)+25),x)

[Out]

x + log(exp(4*x) + 10*exp(2*x) + E + 25)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {25+e+30 e^{2 x}+5 e^{4 x}}{25+e+10 e^{2 x}+e^{4 x}} \, dx=x + \log \left (e + e^{\left (4 \, x\right )} + 10 \, e^{\left (2 \, x\right )} + 25\right ) \]

[In]

integrate((5*exp(x)^4+30*exp(x)^2+exp(1)+25)/(exp(x)^4+10*exp(x)^2+exp(1)+25),x, algorithm="maxima")

[Out]

x + log(e + e^(4*x) + 10*e^(2*x) + 25)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {25+e+30 e^{2 x}+5 e^{4 x}}{25+e+10 e^{2 x}+e^{4 x}} \, dx=x + \log \left (e + e^{\left (4 \, x\right )} + 10 \, e^{\left (2 \, x\right )} + 25\right ) \]

[In]

integrate((5*exp(x)^4+30*exp(x)^2+exp(1)+25)/(exp(x)^4+10*exp(x)^2+exp(1)+25),x, algorithm="giac")

[Out]

x + log(e + e^(4*x) + 10*e^(2*x) + 25)

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {25+e+30 e^{2 x}+5 e^{4 x}}{25+e+10 e^{2 x}+e^{4 x}} \, dx=x+\ln \left (10\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+\mathrm {e}+25\right ) \]

[In]

int((30*exp(2*x) + 5*exp(4*x) + exp(1) + 25)/(10*exp(2*x) + exp(4*x) + exp(1) + 25),x)

[Out]

x + log(10*exp(2*x) + exp(4*x) + exp(1) + 25)