\(\int \frac {1}{1+2 f^{c+d x}+f^{2 c+2 d x}} \, dx\) [520]

Optimal result

Integrand size = 23, antiderivative size = 40 \[ \int \frac {1}{1+2 f^{c+d x}+f^{2 c+2 d x}} \, dx=x+\frac {1}{d \left (1+f^{c+d x}\right ) \log (f)}-\frac {\log \left (1+f^{c+d x}\right )}{d \log (f)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2320, 46} \[ \int \frac {1}{1+2 f^{c+d x}+f^{2 c+2 d x}} \, dx=-\frac {\log \left (f^{c+d x}+1\right )}{d \log (f)}+\frac {1}{d \log (f) \left (f^{c+d x}+1\right )}+x \]

[In]

Int[(1 + 2*f^(c + d*x) + f^(2*c + 2*d*x))^(-1),x]

[Out]

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

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.10 \[ \int \frac {1}{1+2 f^{c+d x}+f^{2 c+2 d x}} \, dx=\frac {\frac {1}{1+f^{c+d x}}+\log \left (f^{c+d x}\right )-\log \left (d \left (1+f^{c+d x}\right ) \log (f)\right )}{d \log (f)} \]

[In]

Integrate[(1 + 2*f^(c + d*x) + f^(2*c + 2*d*x))^(-1),x]

[Out]

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

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.15

method	result	size
risch	\(x +\frac {c}{d}+\frac {1}{d \left (1+f^{d x +c}\right ) \ln \left (f \right )}-\frac {\ln \left (1+f^{d x +c}\right )}{d \ln \left (f \right )}\)	\(46\)
norman	\(\frac {x +x \,{\mathrm e}^{\left (d x +c \right ) \ln \left (f \right )}-\frac {{\mathrm e}^{\left (d x +c \right ) \ln \left (f \right )}}{d \ln \left (f \right )}}{{\mathrm e}^{\left (d x +c \right ) \ln \left (f \right )}+1}-\frac {\ln \left ({\mathrm e}^{\left (d x +c \right ) \ln \left (f \right )}+1\right )}{d \ln \left (f \right )}\)	\(68\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.48 \[ \int \frac {1}{1+2 f^{c+d x}+f^{2 c+2 d x}} \, dx=\frac {d f^{d x + c} x \log \left (f\right ) + d x \log \left (f\right ) - {\left (f^{d x + c} + 1\right )} \log \left (f^{d x + c} + 1\right ) + 1}{d f^{d x + c} \log \left (f\right ) + d \log \left (f\right )} \]

[In]

integrate(1/(1+2*f^(d*x+c)+f^(2*d*x+2*c)),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.85 \[ \int \frac {1}{1+2 f^{c+d x}+f^{2 c+2 d x}} \, dx=x + \frac {1}{d f^{c + d x} \log {\left (f \right )} + d \log {\left (f \right )}} - \frac {\log {\left (f^{c + d x} + 1 \right )}}{d \log {\left (f \right )}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.20 \[ \int \frac {1}{1+2 f^{c+d x}+f^{2 c+2 d x}} \, dx=\frac {d x + c}{d} - \frac {\log \left (f^{d x + c} + 1\right )}{d \log \left (f\right )} + \frac {1}{d {\left (f^{d x + c} + 1\right )} \log \left (f\right )} \]

[In]

integrate(1/(1+2*f^(d*x+c)+f^(2*d*x+2*c)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.48 \[ \int \frac {1}{1+2 f^{c+d x}+f^{2 c+2 d x}} \, dx=\frac {\frac {\log \left ({\left | f \right |}^{d x} {\left | f \right |}^{c}\right )}{\log \left (f\right )} - \frac {\log \left ({\left | f^{d x} f^{c} + 1 \right |}\right )}{\log \left (f\right )} + \frac {1}{{\left (f^{d x} f^{c} + 1\right )} \log \left (f\right )}}{d} \]

[In]

integrate(1/(1+2*f^(d*x+c)+f^(2*d*x+2*c)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.25 \[ \int \frac {1}{1+2 f^{c+d x}+f^{2 c+2 d x}} \, dx=\frac {1}{d\,\ln \left (f\right )\,\left (f^{d\,x}\,f^c+1\right )}-\frac {\ln \left (f^{d\,x}\,f^c+1\right )-d\,x\,\ln \left (f\right )}{d\,\ln \left (f\right )} \]

[In]

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

[Out]

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