\(\int \frac {x^2}{a+b f^{c+d x}+c f^{2 c+2 d x}} \, dx\) [526]

Optimal result

Integrand size = 29, antiderivative size = 484 \[ \int \frac {x^2}{a+b f^{c+d x}+c f^{2 c+2 d x}} \, dx=-\frac {2 c x^3}{3 \left (b^2-4 a c-b \sqrt {b^2-4 a c}\right )}-\frac {2 c x^3}{3 \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right )}-\frac {2 c x^2 \log \left (1+\frac {2 c f^{c+d x}}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right ) d \log (f)}+\frac {2 c x^2 \log \left (1+\frac {2 c f^{c+d x}}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right ) d \log (f)}-\frac {4 c x \operatorname {PolyLog}\left (2,-\frac {2 c f^{c+d x}}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right ) d^2 \log ^2(f)}+\frac {4 c x \operatorname {PolyLog}\left (2,-\frac {2 c f^{c+d x}}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right ) d^2 \log ^2(f)}+\frac {4 c \operatorname {PolyLog}\left (3,-\frac {2 c f^{c+d x}}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right ) d^3 \log ^3(f)}-\frac {4 c \operatorname {PolyLog}\left (3,-\frac {2 c f^{c+d x}}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right ) d^3 \log ^3(f)} \]

[Out]

-2*c*x^2*ln(1+2*c*f^(d*x+c)/(b-(-4*a*c+b^2)^(1/2)))/d/ln(f)/(b-(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2)-4*c*x*po
lylog(2,-2*c*f^(d*x+c)/(b-(-4*a*c+b^2)^(1/2)))/d^2/ln(f)^2/(b-(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2)+4*c*polyl
og(3,-2*c*f^(d*x+c)/(b-(-4*a*c+b^2)^(1/2)))/d^3/ln(f)^3/(b-(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2)+2*c*x^2*ln(1
+2*c*f^(d*x+c)/(b+(-4*a*c+b^2)^(1/2)))/d/ln(f)/(-4*a*c+b^2)^(1/2)/(b+(-4*a*c+b^2)^(1/2))+4*c*x*polylog(2,-2*c*
f^(d*x+c)/(b+(-4*a*c+b^2)^(1/2)))/d^2/ln(f)^2/(-4*a*c+b^2)^(1/2)/(b+(-4*a*c+b^2)^(1/2))-4*c*polylog(3,-2*c*f^(
d*x+c)/(b+(-4*a*c+b^2)^(1/2)))/d^3/ln(f)^3/(-4*a*c+b^2)^(1/2)/(b+(-4*a*c+b^2)^(1/2))-2/3*c*x^3/(b^2-4*a*c-b*(-
4*a*c+b^2)^(1/2))-2/3*c*x^3/(b^2-4*a*c+b*(-4*a*c+b^2)^(1/2))

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 484, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2295, 2215, 2221, 2611, 2320, 6724} \[ \int \frac {x^2}{a+b f^{c+d x}+c f^{2 c+2 d x}} \, dx=\frac {4 c \operatorname {PolyLog}\left (3,-\frac {2 c f^{c+d x}}{b-\sqrt {b^2-4 a c}}\right )}{d^3 \log ^3(f) \sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right )}-\frac {4 c \operatorname {PolyLog}\left (3,-\frac {2 c f^{c+d x}}{b+\sqrt {b^2-4 a c}}\right )}{d^3 \log ^3(f) \sqrt {b^2-4 a c} \left (\sqrt {b^2-4 a c}+b\right )}-\frac {4 c x \operatorname {PolyLog}\left (2,-\frac {2 c f^{c+d x}}{b-\sqrt {b^2-4 a c}}\right )}{d^2 \log ^2(f) \sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right )}+\frac {4 c x \operatorname {PolyLog}\left (2,-\frac {2 c f^{c+d x}}{b+\sqrt {b^2-4 a c}}\right )}{d^2 \log ^2(f) \sqrt {b^2-4 a c} \left (\sqrt {b^2-4 a c}+b\right )}-\frac {2 c x^2 \log \left (\frac {2 c f^{c+d x}}{b-\sqrt {b^2-4 a c}}+1\right )}{d \log (f) \sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right )}+\frac {2 c x^2 \log \left (\frac {2 c f^{c+d x}}{\sqrt {b^2-4 a c}+b}+1\right )}{d \log (f) \sqrt {b^2-4 a c} \left (\sqrt {b^2-4 a c}+b\right )}-\frac {2 c x^3}{3 \left (-b \sqrt {b^2-4 a c}-4 a c+b^2\right )}-\frac {2 c x^3}{3 \left (b \sqrt {b^2-4 a c}-4 a c+b^2\right )} \]

[In]

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

[Out]

(-2*c*x^3)/(3*(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c])) - (2*c*x^3)/(3*(b^2 - 4*a*c + b*Sqrt[b^2 - 4*a*c])) - (2*c*
x^2*Log[1 + (2*c*f^(c + d*x))/(b - Sqrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a*c]*(b - Sqrt[b^2 - 4*a*c])*d*Log[f]) +
(2*c*x^2*Log[1 + (2*c*f^(c + d*x))/(b + Sqrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c])*d*Log
[f]) - (4*c*x*PolyLog[2, (-2*c*f^(c + d*x))/(b - Sqrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a*c]*(b - Sqrt[b^2 - 4*a*c
])*d^2*Log[f]^2) + (4*c*x*PolyLog[2, (-2*c*f^(c + d*x))/(b + Sqrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a*c]*(b + Sqrt
[b^2 - 4*a*c])*d^2*Log[f]^2) + (4*c*PolyLog[3, (-2*c*f^(c + d*x))/(b - Sqrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a*c]
*(b - Sqrt[b^2 - 4*a*c])*d^3*Log[f]^3) - (4*c*PolyLog[3, (-2*c*f^(c + d*x))/(b + Sqrt[b^2 - 4*a*c])])/(Sqrt[b^
2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c])*d^3*Log[f]^3)

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 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 2295

Int[((f_.) + (g_.)*(x_))^(m_.)/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q = Rt[b^2 - 4*
a*c, 2]}, Dist[2*(c/q), Int[(f + g*x)^m/(b - q + 2*c*F^u), x], x] - Dist[2*(c/q), Int[(f + g*x)^m/(b + q + 2*c
*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[
m, 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 2611

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 6724

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]

Rubi steps \begin{align*} \text {integral}& = \frac {(2 c) \int \frac {x^2}{b-\sqrt {b^2-4 a c}+2 c f^{c+d x}} \, dx}{\sqrt {b^2-4 a c}}-\frac {(2 c) \int \frac {x^2}{b+\sqrt {b^2-4 a c}+2 c f^{c+d x}} \, dx}{\sqrt {b^2-4 a c}} \\ & = -\frac {2 c x^3}{3 \left (b^2-4 a c-b \sqrt {b^2-4 a c}\right )}-\frac {2 c x^3}{3 \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right )}+\frac {\left (4 c^2\right ) \int \frac {f^{c+d x} x^2}{b-\sqrt {b^2-4 a c}+2 c f^{c+d x}} \, dx}{b^2-4 a c-b \sqrt {b^2-4 a c}}+\frac {\left (4 c^2\right ) \int \frac {f^{c+d x} x^2}{b+\sqrt {b^2-4 a c}+2 c f^{c+d x}} \, dx}{b^2-4 a c+b \sqrt {b^2-4 a c}} \\ & = -\frac {2 c x^3}{3 \left (b^2-4 a c-b \sqrt {b^2-4 a c}\right )}-\frac {2 c x^3}{3 \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right )}+\frac {2 c x^2 \log \left (1+\frac {2 c f^{c+d x}}{b-\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) d \log (f)}+\frac {2 c x^2 \log \left (1+\frac {2 c f^{c+d x}}{b+\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) d \log (f)}-\frac {(4 c) \int x \log \left (1+\frac {2 c f^{c+d x}}{b-\sqrt {b^2-4 a c}}\right ) \, dx}{\left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) d \log (f)}-\frac {(4 c) \int x \log \left (1+\frac {2 c f^{c+d x}}{b+\sqrt {b^2-4 a c}}\right ) \, dx}{\left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) d \log (f)} \\ & = -\frac {2 c x^3}{3 \left (b^2-4 a c-b \sqrt {b^2-4 a c}\right )}-\frac {2 c x^3}{3 \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right )}+\frac {2 c x^2 \log \left (1+\frac {2 c f^{c+d x}}{b-\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) d \log (f)}+\frac {2 c x^2 \log \left (1+\frac {2 c f^{c+d x}}{b+\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) d \log (f)}+\frac {4 c x \text {Li}_2\left (-\frac {2 c f^{c+d x}}{b-\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) d^2 \log ^2(f)}+\frac {4 c x \text {Li}_2\left (-\frac {2 c f^{c+d x}}{b+\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) d^2 \log ^2(f)}-\frac {(4 c) \int \text {Li}_2\left (-\frac {2 c f^{c+d x}}{b-\sqrt {b^2-4 a c}}\right ) \, dx}{\left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) d^2 \log ^2(f)}-\frac {(4 c) \int \text {Li}_2\left (-\frac {2 c f^{c+d x}}{b+\sqrt {b^2-4 a c}}\right ) \, dx}{\left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) d^2 \log ^2(f)} \\ & = -\frac {2 c x^3}{3 \left (b^2-4 a c-b \sqrt {b^2-4 a c}\right )}-\frac {2 c x^3}{3 \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right )}+\frac {2 c x^2 \log \left (1+\frac {2 c f^{c+d x}}{b-\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) d \log (f)}+\frac {2 c x^2 \log \left (1+\frac {2 c f^{c+d x}}{b+\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) d \log (f)}+\frac {4 c x \text {Li}_2\left (-\frac {2 c f^{c+d x}}{b-\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) d^2 \log ^2(f)}+\frac {4 c x \text {Li}_2\left (-\frac {2 c f^{c+d x}}{b+\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) d^2 \log ^2(f)}-\frac {(4 c) \text {Subst}\left (\int \frac {\text {Li}_2\left (\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )}{x} \, dx,x,f^{c+d x}\right )}{\left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) d^3 \log ^3(f)}-\frac {(4 c) \text {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{x} \, dx,x,f^{c+d x}\right )}{\left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) d^3 \log ^3(f)} \\ & = -\frac {2 c x^3}{3 \left (b^2-4 a c-b \sqrt {b^2-4 a c}\right )}-\frac {2 c x^3}{3 \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right )}+\frac {2 c x^2 \log \left (1+\frac {2 c f^{c+d x}}{b-\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) d \log (f)}+\frac {2 c x^2 \log \left (1+\frac {2 c f^{c+d x}}{b+\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) d \log (f)}+\frac {4 c x \text {Li}_2\left (-\frac {2 c f^{c+d x}}{b-\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) d^2 \log ^2(f)}+\frac {4 c x \text {Li}_2\left (-\frac {2 c f^{c+d x}}{b+\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) d^2 \log ^2(f)}-\frac {4 c \text {Li}_3\left (-\frac {2 c f^{c+d x}}{b-\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) d^3 \log ^3(f)}-\frac {4 c \text {Li}_3\left (-\frac {2 c f^{c+d x}}{b+\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) d^3 \log ^3(f)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 335, normalized size of antiderivative = 0.69 \[ \int \frac {x^2}{a+b f^{c+d x}+c f^{2 c+2 d x}} \, dx=\frac {2 c \left (\frac {x^2 \log \left (1+\frac {\left (b-\sqrt {b^2-4 a c}\right ) f^{-c-d x}}{2 c}\right )}{-b+\sqrt {b^2-4 a c}}+\frac {x^2 \log \left (1+\frac {\left (b+\sqrt {b^2-4 a c}\right ) f^{-c-d x}}{2 c}\right )}{b+\sqrt {b^2-4 a c}}-\frac {2 \left (d x \log (f) \operatorname {PolyLog}\left (2,\frac {\left (-b+\sqrt {b^2-4 a c}\right ) f^{-c-d x}}{2 c}\right )+\operatorname {PolyLog}\left (3,\frac {\left (-b+\sqrt {b^2-4 a c}\right ) f^{-c-d x}}{2 c}\right )\right )}{\left (-b+\sqrt {b^2-4 a c}\right ) d^2 \log ^2(f)}-\frac {2 \left (d x \log (f) \operatorname {PolyLog}\left (2,-\frac {\left (b+\sqrt {b^2-4 a c}\right ) f^{-c-d x}}{2 c}\right )+\operatorname {PolyLog}\left (3,-\frac {\left (b+\sqrt {b^2-4 a c}\right ) f^{-c-d x}}{2 c}\right )\right )}{\left (b+\sqrt {b^2-4 a c}\right ) d^2 \log ^2(f)}\right )}{\sqrt {b^2-4 a c} d \log (f)} \]

[In]

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

[Out]

(2*c*((x^2*Log[1 + ((b - Sqrt[b^2 - 4*a*c])*f^(-c - d*x))/(2*c)])/(-b + Sqrt[b^2 - 4*a*c]) + (x^2*Log[1 + ((b
+ Sqrt[b^2 - 4*a*c])*f^(-c - d*x))/(2*c)])/(b + Sqrt[b^2 - 4*a*c]) - (2*(d*x*Log[f]*PolyLog[2, ((-b + Sqrt[b^2
- 4*a*c])*f^(-c - d*x))/(2*c)] + PolyLog[3, ((-b + Sqrt[b^2 - 4*a*c])*f^(-c - d*x))/(2*c)]))/((-b + Sqrt[b^2
- 4*a*c])*d^2*Log[f]^2) - (2*(d*x*Log[f]*PolyLog[2, -1/2*((b + Sqrt[b^2 - 4*a*c])*f^(-c - d*x))/c] + PolyLog[3
, -1/2*((b + Sqrt[b^2 - 4*a*c])*f^(-c - d*x))/c]))/((b + Sqrt[b^2 - 4*a*c])*d^2*Log[f]^2)))/(Sqrt[b^2 - 4*a*c]
*d*Log[f])

Maple [F]

\[\int \frac {x^{2}}{a +b \,f^{d x +c}+c \,f^{2 d x +2 c}}d x\]

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 694, normalized size of antiderivative = 1.43 \[ \int \frac {x^2}{a+b f^{c+d x}+c f^{2 c+2 d x}} \, dx=\frac {2 \, {\left (b^{2} - 4 \, a c\right )} d^{3} x^{3} \log \left (f\right )^{3} - 6 \, {\left (a b d x \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} \log \left (f\right ) + {\left (b^{2} - 4 \, a c\right )} d x \log \left (f\right )\right )} {\rm Li}_2\left (-\frac {{\left (a \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} + b\right )} f^{d x + c} + 2 \, a}{2 \, a} + 1\right ) + 6 \, {\left (a b d x \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} \log \left (f\right ) - {\left (b^{2} - 4 \, a c\right )} d x \log \left (f\right )\right )} {\rm Li}_2\left (\frac {{\left (a \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} - b\right )} f^{d x + c} - 2 \, a}{2 \, a} + 1\right ) + 3 \, {\left (a b c^{2} \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} \log \left (f\right )^{2} - {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} \log \left (f\right )^{2}\right )} \log \left (2 \, c f^{d x + c} + a \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} + b\right ) - 3 \, {\left (a b c^{2} \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} \log \left (f\right )^{2} + {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} \log \left (f\right )^{2}\right )} \log \left (2 \, c f^{d x + c} - a \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} + b\right ) - 3 \, {\left ({\left (a b d^{2} x^{2} - a b c^{2}\right )} \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} \log \left (f\right )^{2} + {\left ({\left (b^{2} - 4 \, a c\right )} d^{2} x^{2} - b^{2} c^{2} + 4 \, a c^{3}\right )} \log \left (f\right )^{2}\right )} \log \left (\frac {{\left (a \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} + b\right )} f^{d x + c} + 2 \, a}{2 \, a}\right ) + 3 \, {\left ({\left (a b d^{2} x^{2} - a b c^{2}\right )} \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} \log \left (f\right )^{2} - {\left ({\left (b^{2} - 4 \, a c\right )} d^{2} x^{2} - b^{2} c^{2} + 4 \, a c^{3}\right )} \log \left (f\right )^{2}\right )} \log \left (-\frac {{\left (a \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} - b\right )} f^{d x + c} - 2 \, a}{2 \, a}\right ) + 6 \, {\left (a b \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} + b^{2} - 4 \, a c\right )} {\rm polylog}\left (3, -\frac {{\left (a \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} + b\right )} f^{d x + c}}{2 \, a}\right ) - 6 \, {\left (a b \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} - b^{2} + 4 \, a c\right )} {\rm polylog}\left (3, \frac {{\left (a \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} - b\right )} f^{d x + c}}{2 \, a}\right )}{6 \, {\left (a b^{2} - 4 \, a^{2} c\right )} d^{3} \log \left (f\right )^{3}} \]

[In]

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

[Out]

1/6*(2*(b^2 - 4*a*c)*d^3*x^3*log(f)^3 - 6*(a*b*d*x*sqrt((b^2 - 4*a*c)/a^2)*log(f) + (b^2 - 4*a*c)*d*x*log(f))*
dilog(-1/2*((a*sqrt((b^2 - 4*a*c)/a^2) + b)*f^(d*x + c) + 2*a)/a + 1) + 6*(a*b*d*x*sqrt((b^2 - 4*a*c)/a^2)*log
(f) - (b^2 - 4*a*c)*d*x*log(f))*dilog(1/2*((a*sqrt((b^2 - 4*a*c)/a^2) - b)*f^(d*x + c) - 2*a)/a + 1) + 3*(a*b*
c^2*sqrt((b^2 - 4*a*c)/a^2)*log(f)^2 - (b^2*c^2 - 4*a*c^3)*log(f)^2)*log(2*c*f^(d*x + c) + a*sqrt((b^2 - 4*a*c
)/a^2) + b) - 3*(a*b*c^2*sqrt((b^2 - 4*a*c)/a^2)*log(f)^2 + (b^2*c^2 - 4*a*c^3)*log(f)^2)*log(2*c*f^(d*x + c)
- a*sqrt((b^2 - 4*a*c)/a^2) + b) - 3*((a*b*d^2*x^2 - a*b*c^2)*sqrt((b^2 - 4*a*c)/a^2)*log(f)^2 + ((b^2 - 4*a*c
)*d^2*x^2 - b^2*c^2 + 4*a*c^3)*log(f)^2)*log(1/2*((a*sqrt((b^2 - 4*a*c)/a^2) + b)*f^(d*x + c) + 2*a)/a) + 3*((
a*b*d^2*x^2 - a*b*c^2)*sqrt((b^2 - 4*a*c)/a^2)*log(f)^2 - ((b^2 - 4*a*c)*d^2*x^2 - b^2*c^2 + 4*a*c^3)*log(f)^2
)*log(-1/2*((a*sqrt((b^2 - 4*a*c)/a^2) - b)*f^(d*x + c) - 2*a)/a) + 6*(a*b*sqrt((b^2 - 4*a*c)/a^2) + b^2 - 4*a
*c)*polylog(3, -1/2*(a*sqrt((b^2 - 4*a*c)/a^2) + b)*f^(d*x + c)/a) - 6*(a*b*sqrt((b^2 - 4*a*c)/a^2) - b^2 + 4*
a*c)*polylog(3, 1/2*(a*sqrt((b^2 - 4*a*c)/a^2) - b)*f^(d*x + c)/a))/((a*b^2 - 4*a^2*c)*d^3*log(f)^3)

Sympy [F]

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

[In]

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

[Out]

Integral(x**2/(a + b*f**(c + d*x) + c*f**(2*c + 2*d*x)), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^2}{a+b f^{c+d x}+c f^{2 c+2 d x}} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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