\(\int \frac {x}{1+2 f^{c+d x}+f^{2 c+2 d x}} \, dx\) [446]

Optimal result

Integrand size = 25, antiderivative size = 96 \[ \int \frac {x}{1+2 f^{c+d x}+f^{2 c+2 d x}} \, dx=\frac {x^2}{2}-\frac {x}{d \log (f)}+\frac {x}{d \left (1+f^{c+d x}\right ) \log (f)}+\frac {\log \left (1+f^{c+d x}\right )}{d^2 \log ^2(f)}-\frac {x \log \left (1+f^{c+d x}\right )}{d \log (f)}-\frac {\operatorname {PolyLog}\left (2,-f^{c+d x}\right )}{d^2 \log ^2(f)} \] Output:

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.92 \[ \int \frac {x}{1+2 f^{c+d x}+f^{2 c+2 d x}} \, dx=\frac {1}{2} x \left (x+\frac {2}{d \log (f)+d f^{c+d x} \log (f)}\right )+\frac {\log \left (1+f^{c+d x}\right )}{d^2 \log ^2(f)}-\frac {x \left (1+\log \left (1+f^{c+d x}\right )\right )}{d \log (f)}-\frac {\operatorname {PolyLog}\left (2,-f^{c+d x}\right )}{d^2 \log ^2(f)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.15 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.02, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {7239, 2616, 2615, 2620, 2621, 2715, 2720, 47, 14, 16, 2838}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {x}{\left (f^{c+d x}+1\right )^2}dx\)

\(\Big \downarrow \) 2616

\(\displaystyle \int \frac {x}{f^{c+d x}+1}dx-\int \frac {f^{c+d x} x}{\left (f^{c+d x}+1\right )^2}dx\)

\(\Big \downarrow \) 2615

\(\displaystyle -\int \frac {f^{c+d x} x}{\left (f^{c+d x}+1\right )^2}dx-\int \frac {f^{c+d x} x}{f^{c+d x}+1}dx+\frac {x^2}{2}\)

\(\Big \downarrow \) 2620

\(\displaystyle -\int \frac {f^{c+d x} x}{\left (f^{c+d x}+1\right )^2}dx+\frac {\int \log \left (f^{c+d x}+1\right )dx}{d \log (f)}-\frac {x \log \left (f^{c+d x}+1\right )}{d \log (f)}+\frac {x^2}{2}\)

\(\Big \downarrow \) 2621

\(\displaystyle -\frac {\int \frac {1}{f^{c+d x}+1}dx}{d \log (f)}+\frac {\int \log \left (f^{c+d x}+1\right )dx}{d \log (f)}-\frac {x \log \left (f^{c+d x}+1\right )}{d \log (f)}+\frac {x}{d \log (f) \left (f^{c+d x}+1\right )}+\frac {x^2}{2}\)

\(\Big \downarrow \) 2715

\(\displaystyle \frac {\int f^{-c-d x} \log \left (f^{c+d x}+1\right )df^{c+d x}}{d^2 \log ^2(f)}-\frac {\int \frac {1}{f^{c+d x}+1}dx}{d \log (f)}-\frac {x \log \left (f^{c+d x}+1\right )}{d \log (f)}+\frac {x}{d \log (f) \left (f^{c+d x}+1\right )}+\frac {x^2}{2}\)

\(\Big \downarrow \) 2720

\(\displaystyle -\frac {\int \frac {f^{-c-d x}}{f^{c+d x}+1}df^{c+d x}}{d^2 \log ^2(f)}+\frac {\int f^{-c-d x} \log \left (f^{c+d x}+1\right )df^{c+d x}}{d^2 \log ^2(f)}-\frac {x \log \left (f^{c+d x}+1\right )}{d \log (f)}+\frac {x}{d \log (f) \left (f^{c+d x}+1\right )}+\frac {x^2}{2}\)

\(\Big \downarrow \) 47

\(\displaystyle -\frac {\int f^{-c-d x}df^{c+d x}-\int \frac {1}{f^{c+d x}+1}df^{c+d x}}{d^2 \log ^2(f)}+\frac {\int f^{-c-d x} \log \left (f^{c+d x}+1\right )df^{c+d x}}{d^2 \log ^2(f)}-\frac {x \log \left (f^{c+d x}+1\right )}{d \log (f)}+\frac {x}{d \log (f) \left (f^{c+d x}+1\right )}+\frac {x^2}{2}\)

\(\Big \downarrow \) 14

\(\displaystyle -\frac {\log \left (f^{c+d x}\right )-\int \frac {1}{f^{c+d x}+1}df^{c+d x}}{d^2 \log ^2(f)}+\frac {\int f^{-c-d x} \log \left (f^{c+d x}+1\right )df^{c+d x}}{d^2 \log ^2(f)}-\frac {x \log \left (f^{c+d x}+1\right )}{d \log (f)}+\frac {x}{d \log (f) \left (f^{c+d x}+1\right )}+\frac {x^2}{2}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {\int f^{-c-d x} \log \left (f^{c+d x}+1\right )df^{c+d x}}{d^2 \log ^2(f)}-\frac {\log \left (f^{c+d x}\right )-\log \left (f^{c+d x}+1\right )}{d^2 \log ^2(f)}-\frac {x \log \left (f^{c+d x}+1\right )}{d \log (f)}+\frac {x}{d \log (f) \left (f^{c+d x}+1\right )}+\frac {x^2}{2}\)

\(\Big \downarrow \) 2838

\(\displaystyle -\frac {\operatorname {PolyLog}\left (2,-f^{c+d x}\right )}{d^2 \log ^2(f)}-\frac {\log \left (f^{c+d x}\right )-\log \left (f^{c+d x}+1\right )}{d^2 \log ^2(f)}-\frac {x \log \left (f^{c+d x}+1\right )}{d \log (f)}+\frac {x}{d \log (f) \left (f^{c+d x}+1\right )}+\frac {x^2}{2}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 14

Int[(a_.)/(x_), x_Symbol] :> Simp[a*Log[x], x] /; FreeQ[a, x]
 

rule 16

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

rule 47

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

rule 2615

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x 
_))))^(n_.)), x_Symbol] :> Simp[(c + d*x)^(m + 1)/(a*d*(m + 1)), x] - Simp[ 
b/a   Int[(c + d*x)^m*((F^(g*(e + f*x)))^n/(a + b*(F^(g*(e + f*x)))^n)), x] 
, x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2616

Int[((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_)*((c_.) + 
(d_.)*(x_))^(m_.), x_Symbol] :> Simp[1/a   Int[(c + d*x)^m*(a + b*(F^(g*(e 
+ f*x)))^n)^(p + 1), x], x] - Simp[b/a   Int[(c + d*x)^m*(F^(g*(e + f*x)))^ 
n*(a + b*(F^(g*(e + f*x)))^n)^p, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n 
}, x] && ILtQ[p, 0] && IGtQ[m, 0]
 

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2621

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] - Simp[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 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{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 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

rule 7239

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.49

method	result	size
risch	\(\frac {x}{d \left (1+f^{d x +c}\right ) \ln \left (f \right )}+\frac {x^{2}}{2}+\frac {c x}{d}+\frac {c^{2}}{2 d^{2}}-\frac {\ln \left (1+f^{d x} f^{c}\right ) x}{d \ln \left (f \right )}-\frac {\operatorname {polylog}\left (2, -f^{d x} f^{c}\right )}{d^{2} \ln \left (f \right )^{2}}+\frac {\ln \left (1+f^{d x} f^{c}\right )}{d^{2} \ln \left (f \right )^{2}}-\frac {\ln \left (f^{d x} f^{c}\right )}{d^{2} \ln \left (f \right )^{2}}-\frac {c \ln \left (f^{d x} f^{c}\right )}{d^{2} \ln \left (f \right )}\)	\(143\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.49 \[ \int \frac {x}{1+2 f^{c+d x}+f^{2 c+2 d x}} \, dx=\frac {{\left (d^{2} x^{2} - c^{2}\right )} \log \left (f\right )^{2} + {\left ({\left (d^{2} x^{2} - c^{2}\right )} \log \left (f\right )^{2} - 2 \, {\left (d x + c\right )} \log \left (f\right )\right )} f^{d x + c} - 2 \, {\left (f^{d x + c} + 1\right )} {\rm Li}_2\left (-f^{d x + c}\right ) - 2 \, {\left (d x \log \left (f\right ) + {\left (d x \log \left (f\right ) - 1\right )} f^{d x + c} - 1\right )} \log \left (f^{d x + c} + 1\right ) - 2 \, c \log \left (f\right )}{2 \, {\left (d^{2} f^{d x + c} \log \left (f\right )^{2} + d^{2} \log \left (f\right )^{2}\right )}} \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {x}{1+2 f^{c+d x}+f^{2 c+2 d x}} \, dx=\frac {x}{d e^{\frac {\left (2 c + 2 d x\right ) \log {\left (f \right )}}{2}} \log {\left (f \right )} + d \log {\left (f \right )}} + \frac {\int \frac {d x \log {\left (f \right )}}{e^{c \log {\left (f \right )}} e^{d x \log {\left (f \right )}} + 1}\, dx + \int \left (- \frac {1}{e^{c \log {\left (f \right )}} e^{d x \log {\left (f \right )}} + 1}\right )\, dx}{d \log {\left (f \right )}} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.99 \[ \int \frac {x}{1+2 f^{c+d x}+f^{2 c+2 d x}} \, dx=\frac {1}{2} \, x^{2} + \frac {x}{d f^{d x} f^{c} \log \left (f\right ) + d \log \left (f\right )} - \frac {x}{d \log \left (f\right )} - \frac {d x \log \left (f^{d x} f^{c} + 1\right ) \log \left (f\right ) + {\rm Li}_2\left (-f^{d x} f^{c}\right )}{d^{2} \log \left (f\right )^{2}} + \frac {\log \left (f^{d x} f^{c} + 1\right )}{d^{2} \log \left (f\right )^{2}} \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {x}{1+2 f^{c+d x}+f^{2 c+2 d x}} \, dx=\int { \frac {x}{f^{2 \, d x + 2 \, c} + 2 \, f^{d x + c} + 1} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x}{1+2 f^{c+d x}+f^{2 c+2 d x}} \, dx=\int \frac {x}{f^{2\,c+2\,d\,x}+2\,f^{c+d\,x}+1} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {x}{1+2 f^{c+d x}+f^{2 c+2 d x}} \, dx=\int \frac {x}{f^{2 d x +2 c}+2 f^{d x +c}+1}d x \] Input:

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

Output:

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