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

Optimal. Leaf size=145 \[ -\frac{2 x \text{PolyLog}\left (2,-f^{c+d x}\right )}{d^2 \log ^2(f)}+\frac{2 \text{PolyLog}\left (2,-f^{c+d x}\right )}{d^3 \log ^3(f)}+\frac{2 \text{PolyLog}\left (3,-f^{c+d x}\right )}{d^3 \log ^3(f)}+\frac{2 x \log \left (f^{c+d x}+1\right )}{d^2 \log ^2(f)}-\frac{x^2 \log \left (f^{c+d x}+1\right )}{d \log (f)}+\frac{x^2}{d \log (f) \left (f^{c+d x}+1\right )}-\frac{x^2}{d \log (f)}+\frac{x^3}{3} \]

[Out]

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

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Rubi [A]  time = 0.420712, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 10, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.37, Rules used = {6688, 2185, 2184, 2190, 2531, 2282, 6589, 2191, 2279, 2391} \[ -\frac{2 x \text{PolyLog}\left (2,-f^{c+d x}\right )}{d^2 \log ^2(f)}+\frac{2 \text{PolyLog}\left (2,-f^{c+d x}\right )}{d^3 \log ^3(f)}+\frac{2 \text{PolyLog}\left (3,-f^{c+d x}\right )}{d^3 \log ^3(f)}+\frac{2 x \log \left (f^{c+d x}+1\right )}{d^2 \log ^2(f)}-\frac{x^2 \log \left (f^{c+d x}+1\right )}{d \log (f)}+\frac{x^2}{d \log (f) \left (f^{c+d x}+1\right )}-\frac{x^2}{d \log (f)}+\frac{x^3}{3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6688

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

Rule 2185

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 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 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 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]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

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 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]

Rubi steps

\begin{align*} \int \frac{x^2}{1+2 f^{c+d x}+f^{2 c+2 d x}} \, dx &=\int \frac{x^2}{\left (1+f^{c+d x}\right )^2} \, dx\\ &=-\int \frac{f^{c+d x} x^2}{\left (1+f^{c+d x}\right )^2} \, dx+\int \frac{x^2}{1+f^{c+d x}} \, dx\\ &=\frac{x^3}{3}+\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)}-\int \frac{f^{c+d x} x^2}{1+f^{c+d x}} \, dx\\ &=\frac{x^3}{3}-\frac{x^2}{d \log (f)}+\frac{x^2}{d \left (1+f^{c+d x}\right ) \log (f)}-\frac{x^2 \log \left (1+f^{c+d x}\right )}{d \log (f)}+\frac{2 \int \frac{f^{c+d x} x}{1+f^{c+d x}} \, dx}{d \log (f)}+\frac{2 \int x \log \left (1+f^{c+d x}\right ) \, dx}{d \log (f)}\\ &=\frac{x^3}{3}-\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{x^2 \log \left (1+f^{c+d x}\right )}{d \log (f)}-\frac{2 x \text{Li}_2\left (-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{2 \int \text{Li}_2\left (-f^{c+d x}\right ) \, dx}{d^2 \log ^2(f)}\\ &=\frac{x^3}{3}-\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{x^2 \log \left (1+f^{c+d x}\right )}{d \log (f)}-\frac{2 x \text{Li}_2\left (-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{2 \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,f^{c+d x}\right )}{d^3 \log ^3(f)}\\ &=\frac{x^3}{3}-\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{x^2 \log \left (1+f^{c+d x}\right )}{d \log (f)}+\frac{2 \text{Li}_2\left (-f^{c+d x}\right )}{d^3 \log ^3(f)}-\frac{2 x \text{Li}_2\left (-f^{c+d x}\right )}{d^2 \log ^2(f)}+\frac{2 \text{Li}_3\left (-f^{c+d x}\right )}{d^3 \log ^3(f)}\\ \end{align*}

Mathematica [A]  time = 0.230603, size = 123, normalized size = 0.85 \[ \frac{6 \text{PolyLog}\left (3,-f^{c+d x}\right )+(6-6 d x \log (f)) \text{PolyLog}\left (2,-f^{c+d x}\right )-\frac{3 d^2 x^2 \log ^2(f) \left (f^{c+d x}+\left (f^{c+d x}+1\right ) \log \left (f^{c+d x}+1\right )\right )}{f^{c+d x}+1}+6 d x \log (f) \log \left (f^{c+d x}+1\right )+d^3 x^3 \log ^3(f)}{3 d^3 \log ^3(f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^3*x^3*Log[f]^3 + 6*d*x*Log[f]*Log[1 + f^(c + d*x)] - (3*d^2*x^2*Log[f]^2*(f^(c + d*x) + (1 + f^(c + d*x))*L
og[1 + f^(c + d*x)]))/(1 + f^(c + d*x)) + (6 - 6*d*x*Log[f])*PolyLog[2, -f^(c + d*x)] + 6*PolyLog[3, -f^(c + d
*x)])/(3*d^3*Log[f]^3)

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Maple [A]  time = 0.061, size = 232, normalized size = 1.6 \begin{align*}{\frac{{x}^{2}}{d \left ( 1+{f}^{dx+c} \right ) \ln \left ( f \right ) }}+{\frac{{x}^{3}}{3}}-{\frac{{c}^{2}x}{{d}^{2}}}-{\frac{2\,{c}^{3}}{3\,{d}^{3}}}-{\frac{\ln \left ({f}^{dx}{f}^{c}+1 \right ){x}^{2}}{d\ln \left ( f \right ) }}-2\,{\frac{{\it polylog} \left ( 2,-{f}^{dx}{f}^{c} \right ) x}{ \left ( \ln \left ( f \right ) \right ) ^{2}{d}^{2}}}+2\,{\frac{{\it polylog} \left ( 3,-{f}^{dx}{f}^{c} \right ) }{ \left ( \ln \left ( f \right ) \right ) ^{3}{d}^{3}}}+{\frac{{c}^{2}\ln \left ({f}^{dx}{f}^{c} \right ) }{\ln \left ( f \right ){d}^{3}}}-{\frac{{x}^{2}}{d\ln \left ( f \right ) }}-2\,{\frac{cx}{\ln \left ( f \right ){d}^{2}}}-{\frac{{c}^{2}}{\ln \left ( f \right ){d}^{3}}}+2\,{\frac{\ln \left ({f}^{dx}{f}^{c}+1 \right ) x}{ \left ( \ln \left ( f \right ) \right ) ^{2}{d}^{2}}}+2\,{\frac{{\it polylog} \left ( 2,-{f}^{dx}{f}^{c} \right ) }{ \left ( \ln \left ( f \right ) \right ) ^{3}{d}^{3}}}+2\,{\frac{c\ln \left ({f}^{dx}{f}^{c} \right ) }{ \left ( \ln \left ( f \right ) \right ) ^{2}{d}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.01225, size = 211, normalized size = 1.46 \begin{align*} \frac{x^{2}}{d f^{d x} f^{c} \log \left (f\right ) + d \log \left (f\right )} - \frac{\log \left (f^{d x} f^{c} + 1\right ) \log \left (f^{d x}\right )^{2} + 2 \,{\rm Li}_2\left (-f^{d x} f^{c}\right ) \log \left (f^{d x}\right ) - 2 \,{\rm Li}_{3}(-f^{d x} f^{c})}{d^{3} \log \left (f\right )^{3}} + \frac{\log \left (f^{d x}\right )^{3} - 3 \, \log \left (f^{d x}\right )^{2}}{3 \, d^{3} \log \left (f\right )^{3}} + \frac{2 \,{\left (\log \left (f^{d x} f^{c} + 1\right ) \log \left (f^{d x}\right ) +{\rm Li}_2\left (-f^{d x} f^{c}\right )\right )}}{d^{3} \log \left (f\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C]  time = 1.36791, size = 522, normalized size = 3.6 \begin{align*} \frac{3 \, c^{2} \log \left (f\right )^{2} +{\left (d^{3} x^{3} + c^{3}\right )} \log \left (f\right )^{3} +{\left ({\left (d^{3} x^{3} + c^{3}\right )} \log \left (f\right )^{3} - 3 \,{\left (d^{2} x^{2} - c^{2}\right )} \log \left (f\right )^{2}\right )} f^{d x + c} - 6 \,{\left (d x \log \left (f\right ) +{\left (d x \log \left (f\right ) - 1\right )} f^{d x + c} - 1\right )}{\rm Li}_2\left (-f^{d x + c}\right ) - 3 \,{\left (d^{2} x^{2} \log \left (f\right )^{2} - 2 \, d x \log \left (f\right ) +{\left (d^{2} x^{2} \log \left (f\right )^{2} - 2 \, d x \log \left (f\right )\right )} f^{d x + c}\right )} \log \left (f^{d x + c} + 1\right ) + 6 \,{\left (f^{d x + c} + 1\right )}{\rm polylog}\left (3, -f^{d x + c}\right )}{3 \,{\left (d^{3} f^{d x + c} \log \left (f\right )^{3} + d^{3} \log \left (f\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*c^2*log(f)^2 + (d^3*x^3 + c^3)*log(f)^3 + ((d^3*x^3 + c^3)*log(f)^3 - 3*(d^2*x^2 - c^2)*log(f)^2)*f^(d*
x + c) - 6*(d*x*log(f) + (d*x*log(f) - 1)*f^(d*x + c) - 1)*dilog(-f^(d*x + c)) - 3*(d^2*x^2*log(f)^2 - 2*d*x*l
og(f) + (d^2*x^2*log(f)^2 - 2*d*x*log(f))*f^(d*x + c))*log(f^(d*x + c) + 1) + 6*(f^(d*x + c) + 1)*polylog(3, -
f^(d*x + c)))/(d^3*f^(d*x + c)*log(f)^3 + d^3*log(f)^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{x^{2}}{d f^{c + d x} \log{\left (f \right )} + d \log{\left (f \right )}} + \frac{\int - \frac{2 x}{e^{c \log{\left (f \right )}} e^{d x \log{\left (f \right )}} + 1}\, dx + \int \frac{d x^{2} \log{\left (f \right )}}{e^{c \log{\left (f \right )}} e^{d x \log{\left (f \right )}} + 1}\, dx}{d \log{\left (f \right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{f^{2 \, d x + 2 \, c} + 2 \, f^{d x + c} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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