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

Optimal. Leaf size=196 \[ -\frac {i \text {Li}_2\left (-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{16 a^{3/2} b^{3/2} \log ^2(f)}+\frac {i \text {Li}_2\left (\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{16 a^{3/2} b^{3/2} \log ^2(f)}+\frac {x \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log (f)}+\frac {f^x}{8 a b \log ^2(f) \left (a f^{2 x}+b\right )}+\frac {x f^x}{8 a b \log (f) \left (a f^{2 x}+b\right )}-\frac {x f^x}{4 a \log (f) \left (a f^{2 x}+b\right )^2} \]

[Out]

1/8*f^x/a/b/(b+a*f^(2*x))/ln(f)^2-1/4*f^x*x/a/(b+a*f^(2*x))^2/ln(f)+1/8*f^x*x/a/b/(b+a*f^(2*x))/ln(f)+1/8*x*ar
ctan(f^x*a^(1/2)/b^(1/2))/a^(3/2)/b^(3/2)/ln(f)-1/16*I*polylog(2,-I*f^x*a^(1/2)/b^(1/2))/a^(3/2)/b^(3/2)/ln(f)
^2+1/16*I*polylog(2,I*f^x*a^(1/2)/b^(1/2))/a^(3/2)/b^(3/2)/ln(f)^2

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Rubi [A]  time = 0.50, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 22, number of rules used = 9, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {2283, 2254, 2249, 199, 205, 2245, 2282, 4848, 2391} \[ -\frac {i \text {PolyLog}\left (2,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{16 a^{3/2} b^{3/2} \log ^2(f)}+\frac {i \text {PolyLog}\left (2,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{16 a^{3/2} b^{3/2} \log ^2(f)}+\frac {x \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log (f)}+\frac {f^x}{8 a b \log ^2(f) \left (a f^{2 x}+b\right )}+\frac {x f^x}{8 a b \log (f) \left (a f^{2 x}+b\right )}-\frac {x f^x}{4 a \log (f) \left (a f^{2 x}+b\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x/(b/f^x + a*f^x)^3,x]

[Out]

f^x/(8*a*b*(b + a*f^(2*x))*Log[f]^2) - (f^x*x)/(4*a*(b + a*f^(2*x))^2*Log[f]) + (f^x*x)/(8*a*b*(b + a*f^(2*x))
*Log[f]) + (x*ArcTan[(Sqrt[a]*f^x)/Sqrt[b]])/(8*a^(3/2)*b^(3/2)*Log[f]) - ((I/16)*PolyLog[2, ((-I)*Sqrt[a]*f^x
)/Sqrt[b]])/(a^(3/2)*b^(3/2)*Log[f]^2) + ((I/16)*PolyLog[2, (I*Sqrt[a]*f^x)/Sqrt[b]])/(a^(3/2)*b^(3/2)*Log[f]^
2)

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 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]

Rubi steps

\begin {align*} \int \frac {x}{\left (b f^{-x}+a f^x\right )^3} \, dx &=\int \frac {f^{3 x} x}{\left (b+a f^{2 x}\right )^3} \, dx\\ &=\int \left (-\frac {b f^x x}{a \left (b+a f^{2 x}\right )^3}+\frac {f^x x}{a \left (b+a f^{2 x}\right )^2}\right ) \, dx\\ &=\frac {\int \frac {f^x x}{\left (b+a f^{2 x}\right )^2} \, dx}{a}-\frac {b \int \frac {f^x x}{\left (b+a f^{2 x}\right )^3} \, dx}{a}\\ &=-\frac {f^x x}{4 a \left (b+a f^{2 x}\right )^2 \log (f)}+\frac {f^x x}{8 a b \left (b+a f^{2 x}\right ) \log (f)}+\frac {x \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}{a}+\frac {b \int \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}{4 a \left (b+a f^{2 x}\right )^2 \log (f)}+\frac {f^x x}{8 a b \left (b+a f^{2 x}\right ) \log (f)}+\frac {x \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}{\left (b+a f^{2 x}\right )^2} \, dx}{4 a \log (f)}+\frac {3 \int \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right ) \, dx}{8 a^{3/2} b^{3/2} \log (f)}-\frac {\int \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right ) \, dx}{2 a^{3/2} b^{3/2} \log (f)}+\frac {3 \int \frac {f^x}{b+a f^{2 x}} \, dx}{8 a b \log (f)}-\frac {\int \frac {f^x}{b+a f^{2 x}} \, dx}{2 a b \log (f)}\\ &=-\frac {f^x x}{4 a \left (b+a f^{2 x}\right )^2 \log (f)}+\frac {f^x x}{8 a b \left (b+a f^{2 x}\right ) \log (f)}+\frac {x \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log (f)}+\frac {\operatorname {Subst}\left (\int \frac {1}{\left (b+a x^2\right )^2} \, dx,x,f^x\right )}{4 a \log ^2(f)}+\frac {3 \operatorname {Subst}\left (\int \frac {\tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{x} \, dx,x,f^x\right )}{8 a^{3/2} b^{3/2} \log ^2(f)}-\frac {\operatorname {Subst}\left (\int \frac {\tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{x} \, dx,x,f^x\right )}{2 a^{3/2} b^{3/2} \log ^2(f)}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{b+a x^2} \, dx,x,f^x\right )}{8 a b \log ^2(f)}-\frac {\operatorname {Subst}\left (\int \frac {1}{b+a x^2} \, dx,x,f^x\right )}{2 a b \log ^2(f)}\\ &=\frac {f^x}{8 a b \left (b+a f^{2 x}\right ) \log ^2(f)}-\frac {\tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log ^2(f)}-\frac {f^x x}{4 a \left (b+a f^{2 x}\right )^2 \log (f)}+\frac {f^x x}{8 a b \left (b+a f^{2 x}\right ) \log (f)}+\frac {x \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log (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 )}{16 a^{3/2} b^{3/2} \log ^2(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 )}{16 a^{3/2} b^{3/2} \log ^2(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 )}{4 a^{3/2} b^{3/2} \log ^2(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 )}{4 a^{3/2} b^{3/2} \log ^2(f)}+\frac {\operatorname {Subst}\left (\int \frac {1}{b+a x^2} \, dx,x,f^x\right )}{8 a b \log ^2(f)}\\ &=\frac {f^x}{8 a b \left (b+a f^{2 x}\right ) \log ^2(f)}-\frac {f^x x}{4 a \left (b+a f^{2 x}\right )^2 \log (f)}+\frac {f^x x}{8 a b \left (b+a f^{2 x}\right ) \log (f)}+\frac {x \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log (f)}-\frac {i \text {Li}_2\left (-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{16 a^{3/2} b^{3/2} \log ^2(f)}+\frac {i \text {Li}_2\left (\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{16 a^{3/2} b^{3/2} \log ^2(f)}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 209, normalized size = 1.07 \[ \frac {-\frac {i \text {Li}_2\left (-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{b^{3/2}}+\frac {i \text {Li}_2\left (\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{b^{3/2}}+\frac {i x \log (f) \log \left (1-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{b^{3/2}}-\frac {i x \log (f) \log \left (1+\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{b^{3/2}}+\frac {2 \sqrt {a} f^x}{a b f^{2 x}+b^2}+\frac {2 \sqrt {a} x f^x \log (f)}{a b f^{2 x}+b^2}-\frac {4 \sqrt {a} x f^x \log (f)}{\left (a f^{2 x}+b\right )^2}}{16 a^{3/2} \log ^2(f)} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(b/f^x + a*f^x)^3,x]

[Out]

((2*Sqrt[a]*f^x)/(b^2 + a*b*f^(2*x)) - (4*Sqrt[a]*f^x*x*Log[f])/(b + a*f^(2*x))^2 + (2*Sqrt[a]*f^x*x*Log[f])/(
b^2 + a*b*f^(2*x)) + (I*x*Log[f]*Log[1 - (I*Sqrt[a]*f^x)/Sqrt[b]])/b^(3/2) - (I*x*Log[f]*Log[1 + (I*Sqrt[a]*f^
x)/Sqrt[b]])/b^(3/2) - (I*PolyLog[2, ((-I)*Sqrt[a]*f^x)/Sqrt[b]])/b^(3/2) + (I*PolyLog[2, (I*Sqrt[a]*f^x)/Sqrt
[b]])/b^(3/2))/(16*a^(3/2)*Log[f]^2)

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fricas [B]  time = 0.42, size = 352, normalized size = 1.80 \[ \frac {2 \, {\left (a^{2} x \log \relax (f) + a^{2}\right )} f^{3 \, x} - 2 \, {\left (a b x \log \relax (f) - a b\right )} f^{x} + {\left (a^{2} f^{4 \, x} \sqrt {-\frac {a}{b}} + 2 \, a b f^{2 \, x} \sqrt {-\frac {a}{b}} + b^{2} \sqrt {-\frac {a}{b}}\right )} {\rm Li}_2\left (f^{x} \sqrt {-\frac {a}{b}}\right ) - {\left (a^{2} f^{4 \, x} \sqrt {-\frac {a}{b}} + 2 \, a b f^{2 \, x} \sqrt {-\frac {a}{b}} + b^{2} \sqrt {-\frac {a}{b}}\right )} {\rm Li}_2\left (-f^{x} \sqrt {-\frac {a}{b}}\right ) - {\left (a^{2} f^{4 \, x} x \sqrt {-\frac {a}{b}} \log \relax (f) + 2 \, a b f^{2 \, x} x \sqrt {-\frac {a}{b}} \log \relax (f) + b^{2} x \sqrt {-\frac {a}{b}} \log \relax (f)\right )} \log \left (f^{x} \sqrt {-\frac {a}{b}} + 1\right ) + {\left (a^{2} f^{4 \, x} x \sqrt {-\frac {a}{b}} \log \relax (f) + 2 \, a b f^{2 \, x} x \sqrt {-\frac {a}{b}} \log \relax (f) + b^{2} x \sqrt {-\frac {a}{b}} \log \relax (f)\right )} \log \left (-f^{x} \sqrt {-\frac {a}{b}} + 1\right )}{16 \, {\left (a^{4} b f^{4 \, x} \log \relax (f)^{2} + 2 \, a^{3} b^{2} f^{2 \, x} \log \relax (f)^{2} + a^{2} b^{3} \log \relax (f)^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(2*(a^2*x*log(f) + a^2)*f^(3*x) - 2*(a*b*x*log(f) - a*b)*f^x + (a^2*f^(4*x)*sqrt(-a/b) + 2*a*b*f^(2*x)*sq
rt(-a/b) + b^2*sqrt(-a/b))*dilog(f^x*sqrt(-a/b)) - (a^2*f^(4*x)*sqrt(-a/b) + 2*a*b*f^(2*x)*sqrt(-a/b) + b^2*sq
rt(-a/b))*dilog(-f^x*sqrt(-a/b)) - (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))*log(f^x*sqrt(-a/b) + 1) + (a^2*f^(4*x)*x*sqrt(-a/b)*log(f) + 2*a*b*f^(2*x)*x*sqrt(-a/b)*l
og(f) + b^2*x*sqrt(-a/b)*log(f))*log(-f^x*sqrt(-a/b) + 1))/(a^4*b*f^(4*x)*log(f)^2 + 2*a^3*b^2*f^(2*x)*log(f)^
2 + a^2*b^3*log(f)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/(a*f^x + b/f^x)^3, x)

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maple [A]  time = 0.08, size = 209, normalized size = 1.07 \[ \frac {x \ln \left (\frac {-a \,f^{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )}{16 \sqrt {-a b}\, a b \ln \relax (f )}-\frac {x \ln \left (\frac {a \,f^{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )}{16 \sqrt {-a b}\, a b \ln \relax (f )}+\frac {\left (a x \,f^{2 x} \ln \relax (f )-b x \ln \relax (f )+a \,f^{2 x}+b \right ) f^{x}}{8 \left (a \,f^{2 x}+b \right )^{2} a b \ln \relax (f )^{2}}+\frac {\dilog \left (\frac {-a \,f^{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )}{16 \sqrt {-a b}\, a b \ln \relax (f )^{2}}-\frac {\dilog \left (\frac {a \,f^{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )}{16 \sqrt {-a b}\, a b \ln \relax (f )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*f^x*(ln(f)*(f^x)^2*a*x-ln(f)*b*x+a*(f^x)^2+b)/ln(f)^2/b/a/(a*(f^x)^2+b)^2+1/16/ln(f)/a/b*x/(-a*b)^(1/2)*ln
((-a*f^x+(-a*b)^(1/2))/(-a*b)^(1/2))-1/16/ln(f)/a/b*x/(-a*b)^(1/2)*ln((a*f^x+(-a*b)^(1/2))/(-a*b)^(1/2))+1/16/
ln(f)^2/a/b/(-a*b)^(1/2)*dilog((-a*f^x+(-a*b)^(1/2))/(-a*b)^(1/2))-1/16/ln(f)^2/a/b/(-a*b)^(1/2)*dilog((a*f^x+
(-a*b)^(1/2))/(-a*b)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x}{{\left (\frac {b}{f^x}+a\,f^x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(b/f^x + a*f^x)^3,x)

[Out]

int(x/(b/f^x + a*f^x)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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