\(\int \frac {x}{1+2 f^{c+d x}+f^{2 c+2 d x}} \, dx\) [523]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {6820, 2216, 2215, 2221, 2317, 2438, 2222, 2320, 36, 29, 31} \[ \int \frac {x}{1+2 f^{c+d x}+f^{2 c+2 d x}} \, dx=-\frac {\operatorname {PolyLog}\left (2,-f^{c+d x}\right )}{d^2 \log ^2(f)}+\frac {\log \left (f^{c+d x}+1\right )}{d^2 \log ^2(f)}-\frac {x \log \left (f^{c+d x}+1\right )}{d \log (f)}+\frac {x}{d \log (f) \left (f^{c+d x}+1\right )}-\frac {x}{d \log (f)}+\frac {x^2}{2} \]

[In]

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

[Out]

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

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 2215

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] - Dist[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 2216

Int[((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Dis
t[1/a, Int[(c + d*x)^m*(a + b*(F^(g*(e + f*x)))^n)^(p + 1), x], x] - Dist[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 2221

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]
- Dist[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 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 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[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 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 2438

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 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 {x}{\left (1+f^{c+d x}\right )^2} \, dx \\ & = -\int \frac {f^{c+d x} x}{\left (1+f^{c+d x}\right )^2} \, dx+\int \frac {x}{1+f^{c+d x}} \, dx \\ & = \frac {x^2}{2}+\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)}-\int \frac {f^{c+d x} x}{1+f^{c+d x}} \, dx \\ & = \frac {x^2}{2}+\frac {x}{d \left (1+f^{c+d x}\right ) \log (f)}-\frac {x \log \left (1+f^{c+d x}\right )}{d \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 {\int \log \left (1+f^{c+d x}\right ) \, dx}{d \log (f)} \\ & = \frac {x^2}{2}+\frac {x}{d \left (1+f^{c+d x}\right ) \log (f)}-\frac {x \log \left (1+f^{c+d x}\right )}{d \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 {\text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,f^{c+d x}\right )}{d^2 \log ^2(f)} \\ & = \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 {\text {Li}_2\left (-f^{c+d x}\right )}{d^2 \log ^2(f)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.92 \[ \int \frac {x}{1+2 f^{c+d x}+f^{2 c+2 d x}} \, dx=\frac {1}{2} x \left (x+\frac {2}{d \log (f)+d f^{c+d x} \log (f)}\right )+\frac {\log \left (1+f^{c+d x}\right )}{d^2 \log ^2(f)}-\frac {x \left (1+\log \left (1+f^{c+d x}\right )\right )}{d \log (f)}-\frac {\operatorname {PolyLog}\left (2,-f^{c+d x}\right )}{d^2 \log ^2(f)} \]

[In]

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

[Out]

(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)

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.49

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.49 \[ \int \frac {x}{1+2 f^{c+d x}+f^{2 c+2 d x}} \, dx=\frac {{\left (d^{2} x^{2} - c^{2}\right )} \log \left (f\right )^{2} + {\left ({\left (d^{2} x^{2} - c^{2}\right )} \log \left (f\right )^{2} - 2 \, {\left (d x + c\right )} \log \left (f\right )\right )} f^{d x + c} - 2 \, {\left (f^{d x + c} + 1\right )} {\rm Li}_2\left (-f^{d x + c}\right ) - 2 \, {\left (d x \log \left (f\right ) + {\left (d x \log \left (f\right ) - 1\right )} f^{d x + c} - 1\right )} \log \left (f^{d x + c} + 1\right ) - 2 \, c \log \left (f\right )}{2 \, {\left (d^{2} f^{d x + c} \log \left (f\right )^{2} + d^{2} \log \left (f\right )^{2}\right )}} \]

[In]

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

[Out]

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

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

[In]

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

[Out]

x/(d*f**(c + d*x)*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)

none

Time = 0.21 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.99 \[ \int \frac {x}{1+2 f^{c+d x}+f^{2 c+2 d x}} \, dx=\frac {1}{2} \, x^{2} + \frac {x}{d f^{d x} f^{c} \log \left (f\right ) + d \log \left (f\right )} - \frac {x}{d \log \left (f\right )} - \frac {d x \log \left (f^{d x} f^{c} + 1\right ) \log \left (f\right ) + {\rm Li}_2\left (-f^{d x} f^{c}\right )}{d^{2} \log \left (f\right )^{2}} + \frac {\log \left (f^{d x} f^{c} + 1\right )}{d^{2} \log \left (f\right )^{2}} \]

[In]

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

[Out]

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]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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