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

Optimal. Leaf size=75 \[ -\frac {2 \text {Li}_2\left (-f^{c+d x}\right )}{d^3 \log ^3(f)}-\frac {2 x \log \left (f^{c+d x}+1\right )}{d^2 \log ^2(f)}-\frac {x^2}{d \log (f) \left (f^{c+d x}+1\right )}+\frac {x^2}{d \log (f)} \]

[Out]

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

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Rubi [A]  time = 0.49, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {2267, 6688, 2191, 2184, 2190, 2279, 2391} \[ -\frac {2 \text {PolyLog}\left (2,-f^{c+d x}\right )}{d^3 \log ^3(f)}-\frac {2 x \log \left (f^{c+d x}+1\right )}{d^2 \log ^2(f)}-\frac {x^2}{d \log (f) \left (f^{c+d x}+1\right )}+\frac {x^2}{d \log (f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2184

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 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 2191

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 2267

Int[(u_)/((a_) + (b_.)*(F_)^(v_) + (c_.)*(F_)^(w_)), x_Symbol] :> Int[(u*F^v)/(c + a*F^v + b*F^(2*v)), x] /; F
reeQ[{F, a, b, c}, x] && EqQ[w, -v] && LinearQ[v, x] && If[RationalQ[Coefficient[v, x, 1]], GtQ[Coefficient[v,
x, 1], 0], LtQ[LeafCount[v], LeafCount[w]]]

Rule 2279

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 2391

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 6688

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

Rubi steps

\begin {align*} \int \frac {x^2}{2+f^{-c-d x}+f^{c+d x}} \, dx &=\int \frac {f^{c+d x} x^2}{1+2 f^{c+d x}+f^{2 (c+d x)}} \, dx\\ &=\int \frac {f^{c+d x} x^2}{\left (1+f^{c+d x}\right )^2} \, dx\\ &=-\frac {x^2}{d \left (1+f^{c+d x}\right ) \log (f)}+\frac {2 \int \frac {x}{1+f^{c+d x}} \, dx}{d \log (f)}\\ &=\frac {x^2}{d \log (f)}-\frac {x^2}{d \left (1+f^{c+d x}\right ) \log (f)}-\frac {2 \int \frac {f^{c+d x} x}{1+f^{c+d x}} \, dx}{d \log (f)}\\ &=\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 \int \log \left (1+f^{c+d x}\right ) \, dx}{d^2 \log ^2(f)}\\ &=\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 {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,f^{c+d x}\right )}{d^3 \log ^3(f)}\\ &=\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 \text {Li}_2\left (-f^{c+d x}\right )}{d^3 \log ^3(f)}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 63, normalized size = 0.84 \[ \frac {d x \log (f) \left (\frac {d x \log (f) f^{c+d x}}{f^{c+d x}+1}-2 \log \left (f^{c+d x}+1\right )\right )-2 \text {Li}_2\left (-f^{c+d x}\right )}{d^3 \log ^3(f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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)

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fricas [A]  time = 0.42, size = 114, normalized size = 1.52 \[ -\frac {c^{2} \log \relax (f)^{2} - {\left (d^{2} x^{2} - c^{2}\right )} f^{d x + c} \log \relax (f)^{2} + 2 \, {\left (f^{d x + c} + 1\right )} {\rm Li}_2\left (-f^{d x + c}\right ) + 2 \, {\left (d f^{d x + c} x \log \relax (f) + d x \log \relax (f)\right )} \log \left (f^{d x + c} + 1\right )}{d^{3} f^{d x + c} \log \relax (f)^{3} + d^{3} \log \relax (f)^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{f^{d x + c} + f^{-d x - c} + 2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.08, size = 134, normalized size = 1.79 \[ \frac {x^{2}}{\left (f^{-d x -c}+1\right ) d \ln \relax (f )}-\frac {x^{2}}{d \ln \relax (f )}-\frac {2 c x}{d^{2} \ln \relax (f )}-\frac {c^{2}}{d^{3} \ln \relax (f )}-\frac {2 x \ln \left (f^{-c} f^{-d x}+1\right )}{d^{2} \ln \relax (f )^{2}}-\frac {2 c \ln \left (f^{-c} f^{-d x}\right )}{d^{3} \ln \relax (f )^{2}}+\frac {2 \polylog \left (2, -f^{-c} f^{-d x}\right )}{d^{3} \ln \relax (f )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.27, size = 74, normalized size = 0.99 \[ -\frac {x^{2}}{d f^{d x} f^{c} \log \relax (f) + d \log \relax (f)} + \frac {x^{2}}{d \log \relax (f)} - \frac {2 \, {\left (d x \log \left (f^{d x} f^{c} + 1\right ) \log \relax (f) + {\rm Li}_2\left (-f^{d x} f^{c}\right )\right )}}{d^{3} \log \relax (f)^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2}{\frac {1}{f^{c+d\,x}}+f^{c+d\,x}+2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {x^{2}}{d f^{c + d x} \log {\relax (f )} + d \log {\relax (f )}} + \frac {2 \int \frac {x}{e^{c \log {\relax (f )}} e^{d x \log {\relax (f )}} + 1}\, dx}{d \log {\relax (f )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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))

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