\(\int \frac {x^2}{2+f^{-c-d x}+f^{c+d x}} \, dx\) [457]

Optimal result

Integrand size = 25, antiderivative size = 75 \[ \int \frac {x^2}{2+f^{-c-d x}+f^{c+d x}} \, dx=\frac {x^2}{d \log (f)}-\frac {x^2}{d \left (1+f^{c+d x}\right ) \log (f)}-\frac {2 x \log \left (1+f^{c+d x}\right )}{d^2 \log ^2(f)}-\frac {2 \operatorname {PolyLog}\left (2,-f^{c+d x}\right )}{d^3 \log ^3(f)} \] Output:

x^2/d/ln(f)-x^2/d/(1+f^(d*x+c))/ln(f)-2*x*ln(1+f^(d*x+c))/d^2/ln(f)^2-2*po 
lylog(2,-f^(d*x+c))/d^3/ln(f)^3
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.84 \[ \int \frac {x^2}{2+f^{-c-d x}+f^{c+d x}} \, dx=\frac {d x \log (f) \left (\frac {d f^{c+d x} x \log (f)}{1+f^{c+d x}}-2 \log \left (1+f^{c+d x}\right )\right )-2 \operatorname {PolyLog}\left (2,-f^{c+d x}\right )}{d^3 \log ^3(f)} \] Input:

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

Output:

(d*x*Log[f]*((d*f^(c + d*x)*x*Log[f])/(1 + f^(c + d*x)) - 2*Log[1 + f^(c + 
 d*x)]) - 2*PolyLog[2, -f^(c + d*x)])/(d^3*Log[f]^3)
 

Rubi [A] (verified)

Time = 1.21 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {2697, 7239, 2621, 2615, 2620, 2715, 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.

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

\(\Big \downarrow \) 2697

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

\(\Big \downarrow \) 7239

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

\(\Big \downarrow \) 2621

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

\(\Big \downarrow \) 2615

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

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

Output:

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

Defintions of rubi rules used

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 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 2697

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

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 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 = 0.09 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.79

method	result	size
risch	\(\frac {x^{2}}{d \ln \left (f \right ) \left (f^{-d x -c}+1\right )}-\frac {x^{2}}{d \ln \left (f \right )}-\frac {2 c x}{d^{2} \ln \left (f \right )}-\frac {c^{2}}{d^{3} \ln \left (f \right )}-\frac {2 \ln \left (f^{-d x} f^{-c}+1\right ) x}{\ln \left (f \right )^{2} d^{2}}+\frac {2 \operatorname {polylog}\left (2, -f^{-d x} f^{-c}\right )}{\ln \left (f \right )^{3} d^{3}}-\frac {2 c \ln \left (f^{-d x} f^{-c}\right )}{\ln \left (f \right )^{2} d^{3}}\)	\(134\)

Input:

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

Output:

1/d/ln(f)*x^2/(f^(-d*x-c)+1)-x^2/d/ln(f)-2/d^2/ln(f)*c*x-1/d^3/ln(f)*c^2-2 
/ln(f)^2/d^2*ln(f^(-d*x)*f^(-c)+1)*x+2/ln(f)^3/d^3*polylog(2,-f^(-d*x)*f^( 
-c))-2/ln(f)^2/d^3*c*ln(f^(-d*x)*f^(-c))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.52 \[ \int \frac {x^2}{2+f^{-c-d x}+f^{c+d x}} \, dx=-\frac {c^{2} \log \left (f\right )^{2} - {\left (d^{2} x^{2} - c^{2}\right )} f^{d x + c} \log \left (f\right )^{2} + 2 \, {\left (f^{d x + c} + 1\right )} {\rm Li}_2\left (-f^{d x + c}\right ) + 2 \, {\left (d f^{d x + c} x \log \left (f\right ) + d x \log \left (f\right )\right )} \log \left (f^{d x + c} + 1\right )}{d^{3} f^{d x + c} \log \left (f\right )^{3} + d^{3} \log \left (f\right )^{3}} \] Input:

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

Output:

-(c^2*log(f)^2 - (d^2*x^2 - c^2)*f^(d*x + c)*log(f)^2 + 2*(f^(d*x + c) + 1 
)*dilog(-f^(d*x + c)) + 2*(d*f^(d*x + c)*x*log(f) + d*x*log(f))*log(f^(d*x 
 + c) + 1))/(d^3*f^(d*x + c)*log(f)^3 + d^3*log(f)^3)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.99 \[ \int \frac {x^2}{2+f^{-c-d x}+f^{c+d x}} \, dx=-\frac {x^{2}}{d f^{d x} f^{c} \log \left (f\right ) + d \log \left (f\right )} + \frac {x^{2}}{d \log \left (f\right )} - \frac {2 \, {\left (d x \log \left (f^{d x} f^{c} + 1\right ) \log \left (f\right ) + {\rm Li}_2\left (-f^{d x} f^{c}\right )\right )}}{d^{3} \log \left (f\right )^{3}} \] Input:

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

Output:

-x^2/(d*f^(d*x)*f^c*log(f) + d*log(f)) + x^2/(d*log(f)) - 2*(d*x*log(f^(d* 
x)*f^c + 1)*log(f) + dilog(-f^(d*x)*f^c))/(d^3*log(f)^3)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^2}{2+f^{-c-d x}+f^{c+d x}} \, dx=\frac {2 f^{d x +c} \left (\int \frac {x}{f^{2 d x +2 c}+2 f^{d x +c}+1}d x \right ) \mathrm {log}\left (f \right )^{2} d^{2}-2 f^{d x +c} \mathrm {log}\left (f^{d x +c}+1\right )+2 f^{d x +c} \mathrm {log}\left (f \right ) d x +2 \left (\int \frac {x}{f^{2 d x +2 c}+2 f^{d x +c}+1}d x \right ) \mathrm {log}\left (f \right )^{2} d^{2}-2 \,\mathrm {log}\left (f^{d x +c}+1\right )-\mathrm {log}\left (f \right )^{2} d^{2} x^{2}}{\mathrm {log}\left (f \right )^{3} d^{3} \left (f^{d x +c}+1\right )} \] Input:

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

Output:

(2*f**(c + d*x)*int(x/(f**(2*c + 2*d*x) + 2*f**(c + d*x) + 1),x)*log(f)**2 
*d**2 - 2*f**(c + d*x)*log(f**(c + d*x) + 1) + 2*f**(c + d*x)*log(f)*d*x + 
 2*int(x/(f**(2*c + 2*d*x) + 2*f**(c + d*x) + 1),x)*log(f)**2*d**2 - 2*log 
(f**(c + d*x) + 1) - log(f)**2*d**2*x**2)/(log(f)**3*d**3*(f**(c + d*x) + 
1))