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

Optimal. Leaf size=484 \[ \frac {4 c \text {Li}_3\left (-\frac {2 c f^{c+d x}}{b-\sqrt {b^2-4 a c}}\right )}{d^3 \log ^3(f) \sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right )}-\frac {4 c \text {Li}_3\left (-\frac {2 c f^{c+d x}}{b+\sqrt {b^2-4 a c}}\right )}{d^3 \log ^3(f) \sqrt {b^2-4 a c} \left (\sqrt {b^2-4 a c}+b\right )}-\frac {4 c x \text {Li}_2\left (-\frac {2 c f^{c+d x}}{b-\sqrt {b^2-4 a c}}\right )}{d^2 \log ^2(f) \sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right )}+\frac {4 c x \text {Li}_2\left (-\frac {2 c f^{c+d x}}{b+\sqrt {b^2-4 a c}}\right )}{d^2 \log ^2(f) \sqrt {b^2-4 a c} \left (\sqrt {b^2-4 a c}+b\right )}-\frac {2 c x^2 \log \left (\frac {2 c f^{c+d x}}{b-\sqrt {b^2-4 a c}}+1\right )}{d \log (f) \sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right )}+\frac {2 c x^2 \log \left (\frac {2 c f^{c+d x}}{\sqrt {b^2-4 a c}+b}+1\right )}{d \log (f) \sqrt {b^2-4 a c} \left (\sqrt {b^2-4 a c}+b\right )}-\frac {2 c x^3}{3 \left (-b \sqrt {b^2-4 a c}-4 a c+b^2\right )}-\frac {2 c x^3}{3 \left (b \sqrt {b^2-4 a c}-4 a c+b^2\right )} \]

[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]  time = 0.87, antiderivative size = 484, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2263, 2184, 2190, 2531, 2282, 6589} \[ -\frac {4 c x \text {PolyLog}\left (2,-\frac {2 c f^{c+d x}}{b-\sqrt {b^2-4 a c}}\right )}{d^2 \log ^2(f) \sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right )}+\frac {4 c x \text {PolyLog}\left (2,-\frac {2 c f^{c+d x}}{\sqrt {b^2-4 a c}+b}\right )}{d^2 \log ^2(f) \sqrt {b^2-4 a c} \left (\sqrt {b^2-4 a c}+b\right )}+\frac {4 c \text {PolyLog}\left (3,-\frac {2 c f^{c+d x}}{b-\sqrt {b^2-4 a c}}\right )}{d^3 \log ^3(f) \sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right )}-\frac {4 c \text {PolyLog}\left (3,-\frac {2 c f^{c+d x}}{\sqrt {b^2-4 a c}+b}\right )}{d^3 \log ^3(f) \sqrt {b^2-4 a c} \left (\sqrt {b^2-4 a c}+b\right )}-\frac {2 c x^2 \log \left (\frac {2 c f^{c+d x}}{b-\sqrt {b^2-4 a c}}+1\right )}{d \log (f) \sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right )}+\frac {2 c x^2 \log \left (\frac {2 c f^{c+d x}}{\sqrt {b^2-4 a c}+b}+1\right )}{d \log (f) \sqrt {b^2-4 a c} \left (\sqrt {b^2-4 a c}+b\right )}-\frac {2 c x^3}{3 \left (-b \sqrt {b^2-4 a c}-4 a c+b^2\right )}-\frac {2 c x^3}{3 \left (b \sqrt {b^2-4 a c}-4 a c+b^2\right )} \]

Antiderivative was successfully verified.

[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 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 2263

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 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 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 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]

Rubi steps

\begin {align*} \int \frac {x^2}{a+b f^{c+d x}+c f^{2 c+2 d x}} \, dx &=\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) \operatorname {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) \operatorname {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 [F]  time = 2.83, size = 0, normalized size = 0.00 \[ \int \frac {x^2}{a+b f^{c+d x}+c f^{2 c+2 d x}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [C]  time = 0.44, size = 694, normalized size = 1.43 \[ \frac {2 \, {\left (b^{2} - 4 \, a c\right )} d^{3} x^{3} \log \relax (f)^{3} - 6 \, {\left (a b d x \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} \log \relax (f) + {\left (b^{2} - 4 \, a c\right )} d x \log \relax (f)\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 \relax (f) - {\left (b^{2} - 4 \, a c\right )} d x \log \relax (f)\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 \relax (f)^{2} - {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} \log \relax (f)^{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 \relax (f)^{2} + {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} \log \relax (f)^{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 \relax (f)^{2} + {\left ({\left (b^{2} - 4 \, a c\right )} d^{2} x^{2} - b^{2} c^{2} + 4 \, a c^{3}\right )} \log \relax (f)^{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 \relax (f)^{2} - {\left ({\left (b^{2} - 4 \, a c\right )} d^{2} x^{2} - b^{2} c^{2} + 4 \, a c^{3}\right )} \log \relax (f)^{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 \relax (f)^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [F]  time = 0.16, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{b \,f^{d x +c}+c \,f^{2 d x +2 c}+a}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[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 details)Is 4*a*c-b^2 positive or negative?

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{a + b f^{c} f^{d x} + c f^{2 c} f^{2 d x}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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*f**(d*x) + c*f**(2*c)*f**(2*d*x)), x)

________________________________________________________________________________________