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

Optimal. Leaf size=40 \[ -\frac {\log \left (f^{c+d x}+1\right )}{d \log (f)}+\frac {1}{d \log (f) \left (f^{c+d x}+1\right )}+x \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2282, 44} \[ -\frac {\log \left (f^{c+d x}+1\right )}{d \log (f)}+\frac {1}{d \log (f) \left (f^{c+d x}+1\right )}+x \]

Antiderivative was successfully verified.

[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 44

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] && L
tQ[m + n + 2, 0])

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 {1}{1+2 f^{c+d x}+f^{2 c+2 d x}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{x (1+x)^2} \, dx,x,f^{c+d x}\right )}{d \log (f)}\\ &=\frac {\operatorname {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]  time = 0.04, size = 37, normalized size = 0.92 \[ \frac {\frac {1}{f^{c+d x}+1}-\log \left (f^{c+d x}+1\right )+d x \log (f)}{d \log (f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 59, normalized size = 1.48 \[ \frac {d f^{d x + c} x \log \relax (f) + d x \log \relax (f) - {\left (f^{d x + c} + 1\right )} \log \left (f^{d x + c} + 1\right ) + 1}{d f^{d x + c} \log \relax (f) + d \log \relax (f)} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Undef/Unsigned Inf encountered in limitU
ndef/Unsigned Inf encountered in limitUndef/Unsigned Inf encountered in limit1/d/ln(f)*ln(abs(f)^(d*x)*abs(f)^
c)-1/d/ln(f)*ln(abs(f^(d*x)*f^c+1))+1/ln(f)/d/(f^(d*x)*f^c+1)

________________________________________________________________________________________

maple [A]  time = 0.03, size = 68, normalized size = 1.70 \[ -\frac {\ln \left ({\mathrm e}^{\left (d x +c \right ) \ln \relax (f )}+1\right )}{d \ln \relax (f )}+\frac {x \,{\mathrm e}^{\left (d x +c \right ) \ln \relax (f )}+x -\frac {{\mathrm e}^{\left (d x +c \right ) \ln \relax (f )}}{d \ln \relax (f )}}{{\mathrm e}^{\left (d x +c \right ) \ln \relax (f )}+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.97, size = 48, normalized size = 1.20 \[ \frac {d x + c}{d} - \frac {\log \left (f^{d x + c} + 1\right )}{d \log \relax (f)} + \frac {1}{d {\left (f^{d x + c} + 1\right )} \log \relax (f)} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

mupad [B]  time = 3.52, size = 50, normalized size = 1.25 \[ \frac {1}{d\,\ln \relax (f)\,\left (f^{d\,x}\,f^c+1\right )}-\frac {\ln \left (f^{d\,x}\,f^c+1\right )-d\,x\,\ln \relax (f)}{d\,\ln \relax (f)} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

sympy [A]  time = 0.13, size = 34, normalized size = 0.85 \[ x + \frac {1}{d f^{c + d x} \log {\relax (f )} + d \log {\relax (f )}} - \frac {\log {\left (f^{c + d x} + 1 \right )}}{d \log {\relax (f )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________