\(\int \frac {x}{2+f^{-c-d x}+f^{c+d x}} \, dx\) [533]

Optimal result

Integrand size = 23, antiderivative size = 50 \[ \int \frac {x}{2+f^{-c-d x}+f^{c+d x}} \, dx=\frac {x}{d \log (f)}-\frac {x}{d \left (1+f^{c+d x}\right ) \log (f)}-\frac {\log \left (1+f^{c+d x}\right )}{d^2 \log ^2(f)} \]

[Out]

x/d/ln(f)-x/d/(1+f^(d*x+c))/ln(f)-ln(1+f^(d*x+c))/d^2/ln(f)^2

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2299, 6820, 2222, 2320, 36, 29, 31} \[ \int \frac {x}{2+f^{-c-d x}+f^{c+d x}} \, dx=-\frac {\log \left (f^{c+d x}+1\right )}{d^2 \log ^2(f)}-\frac {x}{d \log (f) \left (f^{c+d x}+1\right )}+\frac {x}{d \log (f)} \]

[In]

Int[x/(2 + f^(-c - d*x) + f^(c + d*x)),x]

[Out]

x/(d*Log[f]) - x/(d*(1 + f^(c + d*x))*Log[f]) - Log[1 + f^(c + d*x)]/(d^2*Log[f]^2)

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 2222

Int[((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((a_.) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_.)*
((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*((a + b*(F^(g*(e + f*x)))^n)^(p + 1)/(b*f*g*n*(p + 1
)*Log[F])), x] - Dist[d*(m/(b*f*g*n*(p + 1)*Log[F])), Int[(c + d*x)^(m - 1)*(a + b*(F^(g*(e + f*x)))^n)^(p + 1
), x], x] /; FreeQ[{F, a, b, c, d, e, f, g, m, n, p}, x] && NeQ[p, -1]

Rule 2299

Int[(u_)/((a_) + (b_.)*(F_)^(v_) + (c_.)*(F_)^(w_)), x_Symbol] :> Int[u*(F^v/(c + a*F^v + b*F^(2*v))), x] /; F
reeQ[{F, a, b, c}, x] && EqQ[w, -v] && LinearQ[v, x] && If[RationalQ[Coefficient[v, x, 1]], GtQ[Coefficient[v,
x, 1], 0], LtQ[LeafCount[v], LeafCount[w]]]

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]]]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {f^{c+d x} x}{1+2 f^{c+d x}+f^{2 (c+d x)}} \, dx \\ & = \int \frac {f^{c+d x} x}{\left (1+f^{c+d x}\right )^2} \, dx \\ & = -\frac {x}{d \left (1+f^{c+d x}\right ) \log (f)}+\frac {\int \frac {1}{1+f^{c+d x}} \, dx}{d \log (f)} \\ & = -\frac {x}{d \left (1+f^{c+d x}\right ) \log (f)}+\frac {\text {Subst}\left (\int \frac {1}{x (1+x)} \, dx,x,f^{c+d x}\right )}{d^2 \log ^2(f)} \\ & = -\frac {x}{d \left (1+f^{c+d x}\right ) \log (f)}+\frac {\text {Subst}\left (\int \frac {1}{x} \, dx,x,f^{c+d x}\right )}{d^2 \log ^2(f)}-\frac {\text {Subst}\left (\int \frac {1}{1+x} \, dx,x,f^{c+d x}\right )}{d^2 \log ^2(f)} \\ & = \frac {x}{d \log (f)}-\frac {x}{d \left (1+f^{c+d x}\right ) \log (f)}-\frac {\log \left (1+f^{c+d x}\right )}{d^2 \log ^2(f)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.88 \[ \int \frac {x}{2+f^{-c-d x}+f^{c+d x}} \, dx=\frac {\frac {d f^{c+d x} x \log (f)}{1+f^{c+d x}}-\log \left (1+f^{c+d x}\right )}{d^2 \log ^2(f)} \]

[In]

Integrate[x/(2 + f^(-c - d*x) + f^(c + d*x)),x]

[Out]

((d*f^(c + d*x)*x*Log[f])/(1 + f^(c + d*x)) - Log[1 + f^(c + d*x)])/(d^2*Log[f]^2)

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.28

method	result	size
norman	\(-\frac {x \,{\mathrm e}^{\left (-d x -c \right ) \ln \left (f \right )}}{d \ln \left (f \right ) \left ({\mathrm e}^{\left (-d x -c \right ) \ln \left (f \right )}+1\right )}-\frac {\ln \left ({\mathrm e}^{\left (-d x -c \right ) \ln \left (f \right )}+1\right )}{d^{2} \ln \left (f \right )^{2}}\)	\(64\)
risch	\(-\frac {x}{d \ln \left (f \right )}-\frac {c}{d^{2} \ln \left (f \right )}+\frac {x}{d \ln \left (f \right ) \left (f^{-d x -c}+1\right )}-\frac {\ln \left (f^{-d x -c}+1\right )}{d^{2} \ln \left (f \right )^{2}}\)	\(67\)

[In]

int(x/(2+f^(-d*x-c)+f^(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-x/d/ln(f)*exp((-d*x-c)*ln(f))/(exp((-d*x-c)*ln(f))+1)-1/d^2/ln(f)^2*ln(exp((-d*x-c)*ln(f))+1)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.22 \[ \int \frac {x}{2+f^{-c-d x}+f^{c+d x}} \, dx=\frac {d f^{d x + c} x \log \left (f\right ) - {\left (f^{d x + c} + 1\right )} \log \left (f^{d x + c} + 1\right )}{d^{2} f^{d x + c} \log \left (f\right )^{2} + d^{2} \log \left (f\right )^{2}} \]

[In]

integrate(x/(2+f^(-d*x-c)+f^(d*x+c)),x, algorithm="fricas")

[Out]

(d*f^(d*x + c)*x*log(f) - (f^(d*x + c) + 1)*log(f^(d*x + c) + 1))/(d^2*f^(d*x + c)*log(f)^2 + d^2*log(f)^2)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.84 \[ \int \frac {x}{2+f^{-c-d x}+f^{c+d x}} \, dx=- \frac {x}{d f^{c + d x} \log {\left (f \right )} + d \log {\left (f \right )}} + \frac {x}{d \log {\left (f \right )}} - \frac {\log {\left (f^{c + d x} + 1 \right )}}{d^{2} \log {\left (f \right )}^{2}} \]

[In]

integrate(x/(2+f**(-d*x-c)+f**(d*x+c)),x)

[Out]

-x/(d*f**(c + d*x)*log(f) + d*log(f)) + x/(d*log(f)) - log(f**(c + d*x) + 1)/(d**2*log(f)**2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.14 \[ \int \frac {x}{2+f^{-c-d x}+f^{c+d x}} \, dx=\frac {f^{d x} f^{c} x}{d f^{d x} f^{c} \log \left (f\right ) + d \log \left (f\right )} - \frac {\log \left (\frac {f^{d x} f^{c} + 1}{f^{c}}\right )}{d^{2} \log \left (f\right )^{2}} \]

[In]

integrate(x/(2+f^(-d*x-c)+f^(d*x+c)),x, algorithm="maxima")

[Out]

f^(d*x)*f^c*x/(d*f^(d*x)*f^c*log(f) + d*log(f)) - log((f^(d*x)*f^c + 1)/f^c)/(d^2*log(f)^2)

Giac [F]

\[ \int \frac {x}{2+f^{-c-d x}+f^{c+d x}} \, dx=\int { \frac {x}{f^{d x + c} + f^{-d x - c} + 2} \,d x } \]

[In]

integrate(x/(2+f^(-d*x-c)+f^(d*x+c)),x, algorithm="giac")

[Out]

integrate(x/(f^(d*x + c) + f^(-d*x - c) + 2), x)

Mupad [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.04 \[ \int \frac {x}{2+f^{-c-d x}+f^{c+d x}} \, dx=\frac {f^{d\,x}\,f^c\,x}{d\,\ln \left (f\right )\,\left (f^{d\,x}\,f^c+1\right )}-\frac {\ln \left (f^{d\,x}\,f^c+1\right )}{d^2\,{\ln \left (f\right )}^2} \]

[In]

int(x/(1/f^(c + d*x) + f^(c + d*x) + 2),x)

[Out]

(f^(d*x)*f^c*x)/(d*log(f)*(f^(d*x)*f^c + 1)) - log(f^(d*x)*f^c + 1)/(d^2*log(f)^2)