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

Optimal. Leaf size=333 \[ \frac {i \text {Li}_2\left (-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^3(f)}-\frac {i \text {Li}_2\left (\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^3(f)}+\frac {i \text {Li}_3\left (-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^3(f)}-\frac {i \text {Li}_3\left (\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^3(f)}-\frac {i x \text {Li}_2\left (-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^2(f)}+\frac {i x \text {Li}_2\left (\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^2(f)}+\frac {x^2 \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log (f)}-\frac {x \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{a^{3/2} \sqrt {b} \log ^2(f)}+\frac {x^2 f^x}{2 a \log (f) \left (a+b f^{2 x}\right )} \]

[Out]

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

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Rubi [A]  time = 0.36, antiderivative size = 333, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {2249, 199, 205, 2245, 14, 12, 2282, 4848, 2391, 5143, 2531, 6589} \[ -\frac {i x \text {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^2(f)}+\frac {i \text {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^3(f)}+\frac {i x \text {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^2(f)}-\frac {i \text {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^3(f)}+\frac {i \text {PolyLog}\left (3,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^3(f)}-\frac {i \text {PolyLog}\left (3,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^3(f)}+\frac {x^2 \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log (f)}-\frac {x \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{a^{3/2} \sqrt {b} \log ^2(f)}+\frac {x^2 f^x}{2 a \log (f) \left (a+b f^{2 x}\right )} \]

Antiderivative was successfully verified.

[In]

Int[(f^x*x^2)/(a + b*f^(2*x))^2,x]

[Out]

-((x*ArcTan[(Sqrt[b]*f^x)/Sqrt[a]])/(a^(3/2)*Sqrt[b]*Log[f]^2)) + (f^x*x^2)/(2*a*(a + b*f^(2*x))*Log[f]) + (x^
2*ArcTan[(Sqrt[b]*f^x)/Sqrt[a]])/(2*a^(3/2)*Sqrt[b]*Log[f]) + ((I/2)*PolyLog[2, ((-I)*Sqrt[b]*f^x)/Sqrt[a]])/(
a^(3/2)*Sqrt[b]*Log[f]^3) - ((I/2)*x*PolyLog[2, ((-I)*Sqrt[b]*f^x)/Sqrt[a]])/(a^(3/2)*Sqrt[b]*Log[f]^2) - ((I/
2)*PolyLog[2, (I*Sqrt[b]*f^x)/Sqrt[a]])/(a^(3/2)*Sqrt[b]*Log[f]^3) + ((I/2)*x*PolyLog[2, (I*Sqrt[b]*f^x)/Sqrt[
a]])/(a^(3/2)*Sqrt[b]*Log[f]^2) + ((I/2)*PolyLog[3, ((-I)*Sqrt[b]*f^x)/Sqrt[a]])/(a^(3/2)*Sqrt[b]*Log[f]^3) -
((I/2)*PolyLog[3, (I*Sqrt[b]*f^x)/Sqrt[a]])/(a^(3/2)*Sqrt[b]*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 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 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 {f^x x^2}{\left (a+b f^{2 x}\right )^2} \, dx &=\frac {f^x x^2}{2 a \left (a+b f^{2 x}\right ) \log (f)}+\frac {x^2 \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log (f)}-2 \int x \left (\frac {f^x}{2 a \left (a+b f^{2 x}\right ) \log (f)}+\frac {\tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log (f)}\right ) \, dx\\ &=\frac {f^x x^2}{2 a \left (a+b f^{2 x}\right ) \log (f)}+\frac {x^2 \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log (f)}-2 \int \left (\frac {f^x x}{2 a \left (a+b f^{2 x}\right ) \log (f)}+\frac {x \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log (f)}\right ) \, dx\\ &=\frac {f^x x^2}{2 a \left (a+b f^{2 x}\right ) \log (f)}+\frac {x^2 \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log (f)}-\frac {\int \frac {f^x x}{a+b f^{2 x}} \, dx}{a \log (f)}-\frac {\int x \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right ) \, dx}{a^{3/2} \sqrt {b} \log (f)}\\ &=-\frac {x \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{a^{3/2} \sqrt {b} \log ^2(f)}+\frac {f^x x^2}{2 a \left (a+b f^{2 x}\right ) \log (f)}+\frac {x^2 \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log (f)}+\frac {\int \frac {\tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)} \, dx}{a \log (f)}-\frac {i \int x \log \left (1-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right ) \, dx}{2 a^{3/2} \sqrt {b} \log (f)}+\frac {i \int x \log \left (1+\frac {i \sqrt {b} f^x}{\sqrt {a}}\right ) \, dx}{2 a^{3/2} \sqrt {b} \log (f)}\\ &=-\frac {x \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{a^{3/2} \sqrt {b} \log ^2(f)}+\frac {f^x x^2}{2 a \left (a+b f^{2 x}\right ) \log (f)}+\frac {x^2 \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log (f)}-\frac {i x \text {Li}_2\left (-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^2(f)}+\frac {i x \text {Li}_2\left (\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^2(f)}+\frac {i \int \text {Li}_2\left (-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right ) \, dx}{2 a^{3/2} \sqrt {b} \log ^2(f)}-\frac {i \int \text {Li}_2\left (\frac {i \sqrt {b} f^x}{\sqrt {a}}\right ) \, dx}{2 a^{3/2} \sqrt {b} \log ^2(f)}+\frac {\int \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right ) \, dx}{a^{3/2} \sqrt {b} \log ^2(f)}\\ &=-\frac {x \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{a^{3/2} \sqrt {b} \log ^2(f)}+\frac {f^x x^2}{2 a \left (a+b f^{2 x}\right ) \log (f)}+\frac {x^2 \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log (f)}-\frac {i x \text {Li}_2\left (-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^2(f)}+\frac {i x \text {Li}_2\left (\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^2(f)}+\frac {i \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {i \sqrt {b} x}{\sqrt {a}}\right )}{x} \, dx,x,f^x\right )}{2 a^{3/2} \sqrt {b} \log ^3(f)}-\frac {i \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {i \sqrt {b} x}{\sqrt {a}}\right )}{x} \, dx,x,f^x\right )}{2 a^{3/2} \sqrt {b} \log ^3(f)}+\frac {\operatorname {Subst}\left (\int \frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{x} \, dx,x,f^x\right )}{a^{3/2} \sqrt {b} \log ^3(f)}\\ &=-\frac {x \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{a^{3/2} \sqrt {b} \log ^2(f)}+\frac {f^x x^2}{2 a \left (a+b f^{2 x}\right ) \log (f)}+\frac {x^2 \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log (f)}-\frac {i x \text {Li}_2\left (-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^2(f)}+\frac {i x \text {Li}_2\left (\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^2(f)}+\frac {i \text {Li}_3\left (-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^3(f)}-\frac {i \text {Li}_3\left (\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^3(f)}+\frac {i \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {i \sqrt {b} x}{\sqrt {a}}\right )}{x} \, dx,x,f^x\right )}{2 a^{3/2} \sqrt {b} \log ^3(f)}-\frac {i \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {i \sqrt {b} x}{\sqrt {a}}\right )}{x} \, dx,x,f^x\right )}{2 a^{3/2} \sqrt {b} \log ^3(f)}\\ &=-\frac {x \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{a^{3/2} \sqrt {b} \log ^2(f)}+\frac {f^x x^2}{2 a \left (a+b f^{2 x}\right ) \log (f)}+\frac {x^2 \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log (f)}+\frac {i \text {Li}_2\left (-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^3(f)}-\frac {i x \text {Li}_2\left (-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^2(f)}-\frac {i \text {Li}_2\left (\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^3(f)}+\frac {i x \text {Li}_2\left (\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^2(f)}+\frac {i \text {Li}_3\left (-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^3(f)}-\frac {i \text {Li}_3\left (\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} \log ^3(f)}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 477, normalized size = 1.43 \[ -\frac {\frac {-\frac {i \text {Li}_2\left (-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \log ^2(f)}-\frac {i x \log \left (1+\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \log (f)}+\frac {i x^2}{2 \sqrt {a}}}{2 \sqrt {b}}+\frac {\frac {i \text {Li}_2\left (\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \log ^2(f)}+\frac {i x \log \left (1-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \log (f)}-\frac {i x^2}{2 \sqrt {a}}}{2 \sqrt {b}}}{a \log (f)}+\frac {\frac {\frac {2 i \text {Li}_3\left (-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \log ^3(f)}-\frac {2 i x \text {Li}_2\left (-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \log ^2(f)}-\frac {i x^2 \log \left (1+\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \log (f)}+\frac {i x^3}{3 \sqrt {a}}}{2 \sqrt {b}}+\frac {-\frac {2 i \text {Li}_3\left (\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \log ^3(f)}+\frac {2 i x \text {Li}_2\left (\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \log ^2(f)}+\frac {i x^2 \log \left (1-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \log (f)}-\frac {i x^3}{3 \sqrt {a}}}{2 \sqrt {b}}}{2 a}+\frac {x^2 f^x}{2 a \log (f) \left (a+b f^{2 x}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(f^x*x^2)/(2*a*(a + b*f^(2*x))*Log[f]) - ((((I/2)*x^2)/Sqrt[a] - (I*x*Log[1 + (I*Sqrt[b]*f^x)/Sqrt[a]])/(Sqrt[
a]*Log[f]) - (I*PolyLog[2, ((-I)*Sqrt[b]*f^x)/Sqrt[a]])/(Sqrt[a]*Log[f]^2))/(2*Sqrt[b]) + (((-1/2*I)*x^2)/Sqrt
[a] + (I*x*Log[1 - (I*Sqrt[b]*f^x)/Sqrt[a]])/(Sqrt[a]*Log[f]) + (I*PolyLog[2, (I*Sqrt[b]*f^x)/Sqrt[a]])/(Sqrt[
a]*Log[f]^2))/(2*Sqrt[b]))/(a*Log[f]) + ((((I/3)*x^3)/Sqrt[a] - (I*x^2*Log[1 + (I*Sqrt[b]*f^x)/Sqrt[a]])/(Sqrt
[a]*Log[f]) - ((2*I)*x*PolyLog[2, ((-I)*Sqrt[b]*f^x)/Sqrt[a]])/(Sqrt[a]*Log[f]^2) + ((2*I)*PolyLog[3, ((-I)*Sq
rt[b]*f^x)/Sqrt[a]])/(Sqrt[a]*Log[f]^3))/(2*Sqrt[b]) + (((-1/3*I)*x^3)/Sqrt[a] + (I*x^2*Log[1 - (I*Sqrt[b]*f^x
)/Sqrt[a]])/(Sqrt[a]*Log[f]) + ((2*I)*x*PolyLog[2, (I*Sqrt[b]*f^x)/Sqrt[a]])/(Sqrt[a]*Log[f]^2) - ((2*I)*PolyL
og[3, (I*Sqrt[b]*f^x)/Sqrt[a]])/(Sqrt[a]*Log[f]^3))/(2*Sqrt[b]))/(2*a)

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fricas [C]  time = 0.43, size = 388, normalized size = 1.17 \[ \frac {2 \, b f^{x} x^{2} \log \relax (f)^{2} + 2 \, {\left ({\left (b x \log \relax (f) - b\right )} f^{2 \, x} \sqrt {-\frac {b}{a}} + {\left (a x \log \relax (f) - a\right )} \sqrt {-\frac {b}{a}}\right )} {\rm Li}_2\left (f^{x} \sqrt {-\frac {b}{a}}\right ) - 2 \, {\left ({\left (b x \log \relax (f) - b\right )} f^{2 \, x} \sqrt {-\frac {b}{a}} + {\left (a x \log \relax (f) - a\right )} \sqrt {-\frac {b}{a}}\right )} {\rm Li}_2\left (-f^{x} \sqrt {-\frac {b}{a}}\right ) - {\left ({\left (b x^{2} \log \relax (f)^{2} - 2 \, b x \log \relax (f)\right )} f^{2 \, x} \sqrt {-\frac {b}{a}} + {\left (a x^{2} \log \relax (f)^{2} - 2 \, a x \log \relax (f)\right )} \sqrt {-\frac {b}{a}}\right )} \log \left (f^{x} \sqrt {-\frac {b}{a}} + 1\right ) + {\left ({\left (b x^{2} \log \relax (f)^{2} - 2 \, b x \log \relax (f)\right )} f^{2 \, x} \sqrt {-\frac {b}{a}} + {\left (a x^{2} \log \relax (f)^{2} - 2 \, a x \log \relax (f)\right )} \sqrt {-\frac {b}{a}}\right )} \log \left (-f^{x} \sqrt {-\frac {b}{a}} + 1\right ) - 2 \, {\left (b f^{2 \, x} \sqrt {-\frac {b}{a}} + a \sqrt {-\frac {b}{a}}\right )} {\rm polylog}\left (3, f^{x} \sqrt {-\frac {b}{a}}\right ) + 2 \, {\left (b f^{2 \, x} \sqrt {-\frac {b}{a}} + a \sqrt {-\frac {b}{a}}\right )} {\rm polylog}\left (3, -f^{x} \sqrt {-\frac {b}{a}}\right )}{4 \, {\left (a b^{2} f^{2 \, x} \log \relax (f)^{3} + a^{2} b \log \relax (f)^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.19, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} f^{x}}{\left (b \,f^{2 x}+a \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f^x*x^2)/(a + b*f^(2*x))^2,x)

[Out]

int((f^x*x^2)/(a + b*f^(2*x))^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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