3.64 \(\int \frac {x^2}{(b f^{-x}+a f^x)^3} \, dx\)

Optimal. Leaf size=316 \[ \frac {i \text {Li}_3\left (-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log ^3(f)}-\frac {i \text {Li}_3\left (\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log ^3(f)}-\frac {i x \text {Li}_2\left (-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log ^2(f)}+\frac {i x \text {Li}_2\left (\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log ^2(f)}+\frac {x^2 \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log (f)}-\frac {\tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{4 a^{3/2} b^{3/2} \log ^3(f)}+\frac {x^2 f^x}{8 a b \log (f) \left (a f^{2 x}+b\right )}-\frac {x^2 f^x}{4 a \log (f) \left (a f^{2 x}+b\right )^2}+\frac {x f^x}{4 a b \log ^2(f) \left (a f^{2 x}+b\right )} \]

[Out]

-1/4*arctan(f^x*a^(1/2)/b^(1/2))/a^(3/2)/b^(3/2)/ln(f)^3+1/4*f^x*x/a/b/(b+a*f^(2*x))/ln(f)^2-1/4*f^x*x^2/a/(b+
a*f^(2*x))^2/ln(f)+1/8*f^x*x^2/a/b/(b+a*f^(2*x))/ln(f)+1/8*x^2*arctan(f^x*a^(1/2)/b^(1/2))/a^(3/2)/b^(3/2)/ln(
f)-1/8*I*x*polylog(2,-I*f^x*a^(1/2)/b^(1/2))/a^(3/2)/b^(3/2)/ln(f)^2+1/8*I*x*polylog(2,I*f^x*a^(1/2)/b^(1/2))/
a^(3/2)/b^(3/2)/ln(f)^2+1/8*I*polylog(3,-I*f^x*a^(1/2)/b^(1/2))/a^(3/2)/b^(3/2)/ln(f)^3-1/8*I*polylog(3,I*f^x*
a^(1/2)/b^(1/2))/a^(3/2)/b^(3/2)/ln(f)^3

________________________________________________________________________________________

Rubi [A]  time = 1.16, antiderivative size = 316, normalized size of antiderivative = 1.00, number of steps used = 43, number of rules used = 14, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.737, Rules used = {2283, 2254, 2249, 199, 205, 2245, 14, 2282, 4848, 2391, 12, 5143, 2531, 6589} \[ -\frac {i x \text {PolyLog}\left (2,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log ^2(f)}+\frac {i x \text {PolyLog}\left (2,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log ^2(f)}+\frac {i \text {PolyLog}\left (3,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log ^3(f)}-\frac {i \text {PolyLog}\left (3,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log ^3(f)}+\frac {x^2 \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log (f)}-\frac {\tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{4 a^{3/2} b^{3/2} \log ^3(f)}+\frac {x^2 f^x}{8 a b \log (f) \left (a f^{2 x}+b\right )}-\frac {x^2 f^x}{4 a \log (f) \left (a f^{2 x}+b\right )^2}+\frac {x f^x}{4 a b \log ^2(f) \left (a f^{2 x}+b\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(b/f^x + a*f^x)^3,x]

[Out]

-ArcTan[(Sqrt[a]*f^x)/Sqrt[b]]/(4*a^(3/2)*b^(3/2)*Log[f]^3) + (f^x*x)/(4*a*b*(b + a*f^(2*x))*Log[f]^2) - (f^x*
x^2)/(4*a*(b + a*f^(2*x))^2*Log[f]) + (f^x*x^2)/(8*a*b*(b + a*f^(2*x))*Log[f]) + (x^2*ArcTan[(Sqrt[a]*f^x)/Sqr
t[b]])/(8*a^(3/2)*b^(3/2)*Log[f]) - ((I/8)*x*PolyLog[2, ((-I)*Sqrt[a]*f^x)/Sqrt[b]])/(a^(3/2)*b^(3/2)*Log[f]^2
) + ((I/8)*x*PolyLog[2, (I*Sqrt[a]*f^x)/Sqrt[b]])/(a^(3/2)*b^(3/2)*Log[f]^2) + ((I/8)*PolyLog[3, ((-I)*Sqrt[a]
*f^x)/Sqrt[b]])/(a^(3/2)*b^(3/2)*Log[f]^3) - ((I/8)*PolyLog[3, (I*Sqrt[a]*f^x)/Sqrt[b]])/(a^(3/2)*b^(3/2)*Log[
f]^3)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2245

Int[(F_)^((e_.)*((c_.) + (d_.)*(x_)))*((a_.) + (b_.)*(F_)^(v_))^(p_)*(x_)^(m_.), x_Symbol] :> With[{u = IntHid
e[F^(e*(c + d*x))*(a + b*F^v)^p, x]}, Dist[x^m, u, x] - Dist[m, Int[x^(m - 1)*u, x], x]] /; FreeQ[{F, a, b, c,
d, e}, x] && EqQ[v, 2*e*(c + d*x)] && GtQ[m, 0] && ILtQ[p, 0]

Rule 2249

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_.)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Wit
h[{m = FullSimplify[(d*e*Log[F])/(g*h*Log[G])]}, Dist[Denominator[m]/(g*h*Log[G]), Subst[Int[x^(Denominator[m]
- 1)*(a + b*F^(c*e - (d*e*f)/g)*x^Numerator[m])^p, x], x, G^((h*(f + g*x))/Denominator[m])], x] /; LtQ[m, -1]
|| GtQ[m, 1]] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x]

Rule 2254

Int[((a_.) + (b_.)*(F_)^(u_))^(p_.)*((c_.) + (d_.)*(F_)^(v_))^(q_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> W
ith[{w = ExpandIntegrand[(e + f*x)^m, (a + b*F^u)^p*(c + d*F^v)^q, x]}, Int[w, x] /; SumQ[w]] /; FreeQ[{F, a,
b, c, d, e, f, m}, x] && IntegersQ[p, q] && LinearQ[{u, v}, x] && RationalQ[Simplify[u/v]]

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 2283

Int[(u_.)*((a_.)*(F_)^(v_) + (b_.)*(F_)^(w_))^(n_), x_Symbol] :> Int[u*F^(n*v)*(a + b*F^ExpandToSum[w - v, x])
^n, x] /; FreeQ[{F, a, b, n}, x] && ILtQ[n, 0] && LinearQ[{v, w}, x]

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 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 4848

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[(I*b)/2, Int[Log[1 - I*c*x
]/x, x], x] - Dist[(I*b)/2, Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]

Rule 5143

Int[ArcTan[(a_.) + (b_.)*(f_)^((c_.) + (d_.)*(x_))]*(x_)^(m_.), x_Symbol] :> Dist[I/2, Int[x^m*Log[1 - I*a - I
*b*f^(c + d*x)], x], x] - Dist[I/2, Int[x^m*Log[1 + I*a + I*b*f^(c + d*x)], x], x] /; FreeQ[{a, b, c, d, f}, x
] && IntegerQ[m] && 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}{\left (b f^{-x}+a f^x\right )^3} \, dx &=\int \frac {f^{3 x} x^2}{\left (b+a f^{2 x}\right )^3} \, dx\\ &=\int \left (-\frac {b f^x x^2}{a \left (b+a f^{2 x}\right )^3}+\frac {f^x x^2}{a \left (b+a f^{2 x}\right )^2}\right ) \, dx\\ &=\frac {\int \frac {f^x x^2}{\left (b+a f^{2 x}\right )^2} \, dx}{a}-\frac {b \int \frac {f^x x^2}{\left (b+a f^{2 x}\right )^3} \, dx}{a}\\ &=-\frac {f^x x^2}{4 a \left (b+a f^{2 x}\right )^2 \log (f)}+\frac {f^x x^2}{8 a b \left (b+a f^{2 x}\right ) \log (f)}+\frac {x^2 \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log (f)}-\frac {2 \int x \left (\frac {f^x}{2 b \left (b+a f^{2 x}\right ) \log (f)}+\frac {\tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{2 \sqrt {a} b^{3/2} \log (f)}\right ) \, dx}{a}+\frac {(2 b) \int x \left (\frac {f^x}{4 b \left (b+a f^{2 x}\right )^2 \log (f)}+\frac {3 f^x}{8 b^2 \left (b+a f^{2 x}\right ) \log (f)}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{8 \sqrt {a} b^{5/2} \log (f)}\right ) \, dx}{a}\\ &=-\frac {f^x x^2}{4 a \left (b+a f^{2 x}\right )^2 \log (f)}+\frac {f^x x^2}{8 a b \left (b+a f^{2 x}\right ) \log (f)}+\frac {x^2 \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log (f)}-\frac {2 \int \left (\frac {f^x x}{2 b \left (b+a f^{2 x}\right ) \log (f)}+\frac {x \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{2 \sqrt {a} b^{3/2} \log (f)}\right ) \, dx}{a}+\frac {(2 b) \int \left (\frac {f^x x}{4 b \left (b+a f^{2 x}\right )^2 \log (f)}+\frac {3 f^x x}{8 b^2 \left (b+a f^{2 x}\right ) \log (f)}+\frac {3 x \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{8 \sqrt {a} b^{5/2} \log (f)}\right ) \, dx}{a}\\ &=-\frac {f^x x^2}{4 a \left (b+a f^{2 x}\right )^2 \log (f)}+\frac {f^x x^2}{8 a b \left (b+a f^{2 x}\right ) \log (f)}+\frac {x^2 \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log (f)}+\frac {\int \frac {f^x x}{\left (b+a f^{2 x}\right )^2} \, dx}{2 a \log (f)}+\frac {3 \int x \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right ) \, dx}{4 a^{3/2} b^{3/2} \log (f)}-\frac {\int x \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right ) \, dx}{a^{3/2} b^{3/2} \log (f)}+\frac {3 \int \frac {f^x x}{b+a f^{2 x}} \, dx}{4 a b \log (f)}-\frac {\int \frac {f^x x}{b+a f^{2 x}} \, dx}{a b \log (f)}\\ &=\frac {f^x x}{4 a b \left (b+a f^{2 x}\right ) \log ^2(f)}-\frac {f^x x^2}{4 a \left (b+a f^{2 x}\right )^2 \log (f)}+\frac {f^x x^2}{8 a b \left (b+a f^{2 x}\right ) \log (f)}+\frac {x^2 \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log (f)}-\frac {\int \left (\frac {f^x}{2 b \left (b+a f^{2 x}\right ) \log (f)}+\frac {\tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{2 \sqrt {a} b^{3/2} \log (f)}\right ) \, dx}{2 a \log (f)}+\frac {(3 i) \int x \log \left (1-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right ) \, dx}{8 a^{3/2} b^{3/2} \log (f)}-\frac {(3 i) \int x \log \left (1+\frac {i \sqrt {a} f^x}{\sqrt {b}}\right ) \, dx}{8 a^{3/2} b^{3/2} \log (f)}-\frac {i \int x \log \left (1-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right ) \, dx}{2 a^{3/2} b^{3/2} \log (f)}+\frac {i \int x \log \left (1+\frac {i \sqrt {a} f^x}{\sqrt {b}}\right ) \, dx}{2 a^{3/2} b^{3/2} \log (f)}-\frac {3 \int \frac {\tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log (f)} \, dx}{4 a b \log (f)}+\frac {\int \frac {\tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log (f)} \, dx}{a b \log (f)}\\ &=\frac {f^x x}{4 a b \left (b+a f^{2 x}\right ) \log ^2(f)}-\frac {f^x x^2}{4 a \left (b+a f^{2 x}\right )^2 \log (f)}+\frac {f^x x^2}{8 a b \left (b+a f^{2 x}\right ) \log (f)}+\frac {x^2 \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log (f)}-\frac {i x \text {Li}_2\left (-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log ^2(f)}+\frac {i x \text {Li}_2\left (\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log ^2(f)}-\frac {(3 i) \int \text {Li}_2\left (-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right ) \, dx}{8 a^{3/2} b^{3/2} \log ^2(f)}+\frac {(3 i) \int \text {Li}_2\left (\frac {i \sqrt {a} f^x}{\sqrt {b}}\right ) \, dx}{8 a^{3/2} b^{3/2} \log ^2(f)}+\frac {i \int \text {Li}_2\left (-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right ) \, dx}{2 a^{3/2} b^{3/2} \log ^2(f)}-\frac {i \int \text {Li}_2\left (\frac {i \sqrt {a} f^x}{\sqrt {b}}\right ) \, dx}{2 a^{3/2} b^{3/2} \log ^2(f)}-\frac {\int \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right ) \, dx}{4 a^{3/2} b^{3/2} \log ^2(f)}-\frac {3 \int \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right ) \, dx}{4 a^{3/2} b^{3/2} \log ^2(f)}+\frac {\int \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right ) \, dx}{a^{3/2} b^{3/2} \log ^2(f)}-\frac {\int \frac {f^x}{b+a f^{2 x}} \, dx}{4 a b \log ^2(f)}\\ &=\frac {f^x x}{4 a b \left (b+a f^{2 x}\right ) \log ^2(f)}-\frac {f^x x^2}{4 a \left (b+a f^{2 x}\right )^2 \log (f)}+\frac {f^x x^2}{8 a b \left (b+a f^{2 x}\right ) \log (f)}+\frac {x^2 \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log (f)}-\frac {i x \text {Li}_2\left (-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log ^2(f)}+\frac {i x \text {Li}_2\left (\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log ^2(f)}-\frac {(3 i) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {i \sqrt {a} x}{\sqrt {b}}\right )}{x} \, dx,x,f^x\right )}{8 a^{3/2} b^{3/2} \log ^3(f)}+\frac {(3 i) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {i \sqrt {a} x}{\sqrt {b}}\right )}{x} \, dx,x,f^x\right )}{8 a^{3/2} b^{3/2} \log ^3(f)}+\frac {i \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {i \sqrt {a} x}{\sqrt {b}}\right )}{x} \, dx,x,f^x\right )}{2 a^{3/2} b^{3/2} \log ^3(f)}-\frac {i \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {i \sqrt {a} x}{\sqrt {b}}\right )}{x} \, dx,x,f^x\right )}{2 a^{3/2} b^{3/2} \log ^3(f)}-\frac {\operatorname {Subst}\left (\int \frac {\tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{x} \, dx,x,f^x\right )}{4 a^{3/2} b^{3/2} \log ^3(f)}-\frac {3 \operatorname {Subst}\left (\int \frac {\tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{x} \, dx,x,f^x\right )}{4 a^{3/2} b^{3/2} \log ^3(f)}+\frac {\operatorname {Subst}\left (\int \frac {\tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{x} \, dx,x,f^x\right )}{a^{3/2} b^{3/2} \log ^3(f)}-\frac {\operatorname {Subst}\left (\int \frac {1}{b+a x^2} \, dx,x,f^x\right )}{4 a b \log ^3(f)}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{4 a^{3/2} b^{3/2} \log ^3(f)}+\frac {f^x x}{4 a b \left (b+a f^{2 x}\right ) \log ^2(f)}-\frac {f^x x^2}{4 a \left (b+a f^{2 x}\right )^2 \log (f)}+\frac {f^x x^2}{8 a b \left (b+a f^{2 x}\right ) \log (f)}+\frac {x^2 \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log (f)}-\frac {i x \text {Li}_2\left (-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log ^2(f)}+\frac {i x \text {Li}_2\left (\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log ^2(f)}+\frac {i \text {Li}_3\left (-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log ^3(f)}-\frac {i \text {Li}_3\left (\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log ^3(f)}-\frac {i \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {i \sqrt {a} x}{\sqrt {b}}\right )}{x} \, dx,x,f^x\right )}{8 a^{3/2} b^{3/2} \log ^3(f)}+\frac {i \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {i \sqrt {a} x}{\sqrt {b}}\right )}{x} \, dx,x,f^x\right )}{8 a^{3/2} b^{3/2} \log ^3(f)}-\frac {(3 i) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {i \sqrt {a} x}{\sqrt {b}}\right )}{x} \, dx,x,f^x\right )}{8 a^{3/2} b^{3/2} \log ^3(f)}+\frac {(3 i) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {i \sqrt {a} x}{\sqrt {b}}\right )}{x} \, dx,x,f^x\right )}{8 a^{3/2} b^{3/2} \log ^3(f)}+\frac {i \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {i \sqrt {a} x}{\sqrt {b}}\right )}{x} \, dx,x,f^x\right )}{2 a^{3/2} b^{3/2} \log ^3(f)}-\frac {i \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {i \sqrt {a} x}{\sqrt {b}}\right )}{x} \, dx,x,f^x\right )}{2 a^{3/2} b^{3/2} \log ^3(f)}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{4 a^{3/2} b^{3/2} \log ^3(f)}+\frac {f^x x}{4 a b \left (b+a f^{2 x}\right ) \log ^2(f)}-\frac {f^x x^2}{4 a \left (b+a f^{2 x}\right )^2 \log (f)}+\frac {f^x x^2}{8 a b \left (b+a f^{2 x}\right ) \log (f)}+\frac {x^2 \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log (f)}-\frac {i x \text {Li}_2\left (-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log ^2(f)}+\frac {i x \text {Li}_2\left (\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log ^2(f)}+\frac {i \text {Li}_3\left (-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log ^3(f)}-\frac {i \text {Li}_3\left (\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log ^3(f)}\\ \end {align*}

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Mathematica [A]  time = 0.52, size = 254, normalized size = 0.80 \[ \frac {\frac {3 i \left (2 \text {Li}_3\left (-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )-2 \text {Li}_3\left (\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )-2 x \log (f) \text {Li}_2\left (-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )+2 x \log (f) \text {Li}_2\left (\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )+x^2 \log ^2(f) \log \left (1-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )-x^2 \log ^2(f) \log \left (1+\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )\right )}{b^{3/2}}-\frac {12 \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{b^{3/2}}-\frac {12 \sqrt {a} x^2 f^x \log ^2(f)}{\left (a f^{2 x}+b\right )^2}+\frac {6 \sqrt {a} x f^x \log (f) (x \log (f)+2)}{b \left (a f^{2 x}+b\right )}}{48 a^{3/2} \log ^3(f)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(b/f^x + a*f^x)^3,x]

[Out]

((-12*ArcTan[(Sqrt[a]*f^x)/Sqrt[b]])/b^(3/2) - (12*Sqrt[a]*f^x*x^2*Log[f]^2)/(b + a*f^(2*x))^2 + (6*Sqrt[a]*f^
x*x*Log[f]*(2 + x*Log[f]))/(b*(b + a*f^(2*x))) + ((3*I)*(x^2*Log[f]^2*Log[1 - (I*Sqrt[a]*f^x)/Sqrt[b]] - x^2*L
og[f]^2*Log[1 + (I*Sqrt[a]*f^x)/Sqrt[b]] - 2*x*Log[f]*PolyLog[2, ((-I)*Sqrt[a]*f^x)/Sqrt[b]] + 2*x*Log[f]*Poly
Log[2, (I*Sqrt[a]*f^x)/Sqrt[b]] + 2*PolyLog[3, ((-I)*Sqrt[a]*f^x)/Sqrt[b]] - 2*PolyLog[3, (I*Sqrt[a]*f^x)/Sqrt
[b]]))/b^(3/2))/(48*a^(3/2)*Log[f]^3)

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fricas [C]  time = 0.44, size = 674, normalized size = 2.13 \[ \frac {2 \, {\left (a^{2} x^{2} \log \relax (f)^{2} + 2 \, a^{2} x \log \relax (f)\right )} f^{3 \, x} - 2 \, {\left (a b x^{2} \log \relax (f)^{2} - 2 \, a b x \log \relax (f)\right )} f^{x} + 2 \, {\left (a^{2} f^{4 \, x} x \sqrt {-\frac {a}{b}} \log \relax (f) + 2 \, a b f^{2 \, x} x \sqrt {-\frac {a}{b}} \log \relax (f) + b^{2} x \sqrt {-\frac {a}{b}} \log \relax (f)\right )} {\rm Li}_2\left (f^{x} \sqrt {-\frac {a}{b}}\right ) - 2 \, {\left (a^{2} f^{4 \, x} x \sqrt {-\frac {a}{b}} \log \relax (f) + 2 \, a b f^{2 \, x} x \sqrt {-\frac {a}{b}} \log \relax (f) + b^{2} x \sqrt {-\frac {a}{b}} \log \relax (f)\right )} {\rm Li}_2\left (-f^{x} \sqrt {-\frac {a}{b}}\right ) - 2 \, {\left (a^{2} f^{4 \, x} \sqrt {-\frac {a}{b}} + 2 \, a b f^{2 \, x} \sqrt {-\frac {a}{b}} + b^{2} \sqrt {-\frac {a}{b}}\right )} \log \left (2 \, a f^{x} + 2 \, b \sqrt {-\frac {a}{b}}\right ) + 2 \, {\left (a^{2} f^{4 \, x} \sqrt {-\frac {a}{b}} + 2 \, a b f^{2 \, x} \sqrt {-\frac {a}{b}} + b^{2} \sqrt {-\frac {a}{b}}\right )} \log \left (2 \, a f^{x} - 2 \, b \sqrt {-\frac {a}{b}}\right ) - {\left (a^{2} f^{4 \, x} x^{2} \sqrt {-\frac {a}{b}} \log \relax (f)^{2} + 2 \, a b f^{2 \, x} x^{2} \sqrt {-\frac {a}{b}} \log \relax (f)^{2} + b^{2} x^{2} \sqrt {-\frac {a}{b}} \log \relax (f)^{2}\right )} \log \left (f^{x} \sqrt {-\frac {a}{b}} + 1\right ) + {\left (a^{2} f^{4 \, x} x^{2} \sqrt {-\frac {a}{b}} \log \relax (f)^{2} + 2 \, a b f^{2 \, x} x^{2} \sqrt {-\frac {a}{b}} \log \relax (f)^{2} + b^{2} x^{2} \sqrt {-\frac {a}{b}} \log \relax (f)^{2}\right )} \log \left (-f^{x} \sqrt {-\frac {a}{b}} + 1\right ) - 2 \, {\left (a^{2} f^{4 \, x} \sqrt {-\frac {a}{b}} + 2 \, a b f^{2 \, x} \sqrt {-\frac {a}{b}} + b^{2} \sqrt {-\frac {a}{b}}\right )} {\rm polylog}\left (3, f^{x} \sqrt {-\frac {a}{b}}\right ) + 2 \, {\left (a^{2} f^{4 \, x} \sqrt {-\frac {a}{b}} + 2 \, a b f^{2 \, x} \sqrt {-\frac {a}{b}} + b^{2} \sqrt {-\frac {a}{b}}\right )} {\rm polylog}\left (3, -f^{x} \sqrt {-\frac {a}{b}}\right )}{16 \, {\left (a^{4} b f^{4 \, x} \log \relax (f)^{3} + 2 \, a^{3} b^{2} f^{2 \, x} \log \relax (f)^{3} + a^{2} b^{3} \log \relax (f)^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b/(f^x)+a*f^x)^3,x, algorithm="fricas")

[Out]

1/16*(2*(a^2*x^2*log(f)^2 + 2*a^2*x*log(f))*f^(3*x) - 2*(a*b*x^2*log(f)^2 - 2*a*b*x*log(f))*f^x + 2*(a^2*f^(4*
x)*x*sqrt(-a/b)*log(f) + 2*a*b*f^(2*x)*x*sqrt(-a/b)*log(f) + b^2*x*sqrt(-a/b)*log(f))*dilog(f^x*sqrt(-a/b)) -
2*(a^2*f^(4*x)*x*sqrt(-a/b)*log(f) + 2*a*b*f^(2*x)*x*sqrt(-a/b)*log(f) + b^2*x*sqrt(-a/b)*log(f))*dilog(-f^x*s
qrt(-a/b)) - 2*(a^2*f^(4*x)*sqrt(-a/b) + 2*a*b*f^(2*x)*sqrt(-a/b) + b^2*sqrt(-a/b))*log(2*a*f^x + 2*b*sqrt(-a/
b)) + 2*(a^2*f^(4*x)*sqrt(-a/b) + 2*a*b*f^(2*x)*sqrt(-a/b) + b^2*sqrt(-a/b))*log(2*a*f^x - 2*b*sqrt(-a/b)) - (
a^2*f^(4*x)*x^2*sqrt(-a/b)*log(f)^2 + 2*a*b*f^(2*x)*x^2*sqrt(-a/b)*log(f)^2 + b^2*x^2*sqrt(-a/b)*log(f)^2)*log
(f^x*sqrt(-a/b) + 1) + (a^2*f^(4*x)*x^2*sqrt(-a/b)*log(f)^2 + 2*a*b*f^(2*x)*x^2*sqrt(-a/b)*log(f)^2 + b^2*x^2*
sqrt(-a/b)*log(f)^2)*log(-f^x*sqrt(-a/b) + 1) - 2*(a^2*f^(4*x)*sqrt(-a/b) + 2*a*b*f^(2*x)*sqrt(-a/b) + b^2*sqr
t(-a/b))*polylog(3, f^x*sqrt(-a/b)) + 2*(a^2*f^(4*x)*sqrt(-a/b) + 2*a*b*f^(2*x)*sqrt(-a/b) + b^2*sqrt(-a/b))*p
olylog(3, -f^x*sqrt(-a/b)))/(a^4*b*f^(4*x)*log(f)^3 + 2*a^3*b^2*f^(2*x)*log(f)^3 + a^2*b^3*log(f)^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{{\left (a f^{x} + \frac {b}{f^{x}}\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b/(f^x)+a*f^x)^3,x, algorithm="giac")

[Out]

integrate(x^2/(a*f^x + b/f^x)^3, x)

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maple [F]  time = 0.17, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\left (a \,f^{x}+b \,f^{-x}\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(b/(f^x)+a*f^x)^3,x)

[Out]

int(x^2/(b/(f^x)+a*f^x)^3,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {{\left (a x^{2} \log \relax (f) + 2 \, a x\right )} f^{3 \, x} - {\left (b x^{2} \log \relax (f) - 2 \, b x\right )} f^{x}}{8 \, {\left (a^{3} b f^{4 \, x} \log \relax (f)^{2} + 2 \, a^{2} b^{2} f^{2 \, x} \log \relax (f)^{2} + a b^{3} \log \relax (f)^{2}\right )}} + \int \frac {{\left (x^{2} \log \relax (f)^{2} - 2\right )} f^{x}}{8 \, {\left (a^{2} b f^{2 \, x} \log \relax (f)^{2} + a b^{2} \log \relax (f)^{2}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b/(f^x)+a*f^x)^3,x, algorithm="maxima")

[Out]

1/8*((a*x^2*log(f) + 2*a*x)*f^(3*x) - (b*x^2*log(f) - 2*b*x)*f^x)/(a^3*b*f^(4*x)*log(f)^2 + 2*a^2*b^2*f^(2*x)*
log(f)^2 + a*b^3*log(f)^2) + integrate(1/8*(x^2*log(f)^2 - 2)*f^x/(a^2*b*f^(2*x)*log(f)^2 + a*b^2*log(f)^2), x
)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^2}{{\left (\frac {b}{f^x}+a\,f^x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(b/f^x + a*f^x)^3,x)

[Out]

int(x^2/(b/f^x + a*f^x)^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {f^{3 x} \left (a x^{2} \log {\relax (f )} + 2 a x\right ) + f^{x} \left (- b x^{2} \log {\relax (f )} + 2 b x\right )}{8 a^{3} b f^{4 x} \log {\relax (f )}^{2} + 16 a^{2} b^{2} f^{2 x} \log {\relax (f )}^{2} + 8 a b^{3} \log {\relax (f )}^{2}} + \frac {\int \left (- \frac {2 f^{x}}{a f^{2 x} + b}\right )\, dx + \int \frac {f^{x} x^{2} \log {\relax (f )}^{2}}{a f^{2 x} + b}\, dx}{8 a b \log {\relax (f )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(b/(f**x)+a*f**x)**3,x)

[Out]

(f**(3*x)*(a*x**2*log(f) + 2*a*x) + f**x*(-b*x**2*log(f) + 2*b*x))/(8*a**3*b*f**(4*x)*log(f)**2 + 16*a**2*b**2
*f**(2*x)*log(f)**2 + 8*a*b**3*log(f)**2) + (Integral(-2*f**x/(a*f**(2*x) + b), x) + Integral(f**x*x**2*log(f)
**2/(a*f**(2*x) + b), x))/(8*a*b*log(f)**2)

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