\(\int \frac {f^x x^2}{a+b f^{2 x}} \, dx\) [45]

Optimal result

Integrand size = 18, antiderivative size = 184 \[ \int \frac {f^x x^2}{a+b f^{2 x}} \, dx=\frac {x^2 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {i x \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log ^2(f)}+\frac {i x \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log ^2(f)}+\frac {i \operatorname {PolyLog}\left (3,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log ^3(f)}-\frac {i \operatorname {PolyLog}\left (3,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log ^3(f)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {2281, 211, 2277, 12, 5251, 2611, 2320, 6724} \[ \int \frac {f^x x^2}{a+b f^{2 x}} \, dx=\frac {x^2 \arctan \left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}+\frac {i \operatorname {PolyLog}\left (3,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log ^3(f)}-\frac {i \operatorname {PolyLog}\left (3,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log ^3(f)}-\frac {i x \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log ^2(f)}+\frac {i x \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log ^2(f)} \]

[In]

Int[(f^x*x^2)/(a + b*f^(2*x)),x]

[Out]

(x^2*ArcTan[(Sqrt[b]*f^x)/Sqrt[a]])/(Sqrt[a]*Sqrt[b]*Log[f]) - (I*x*PolyLog[2, ((-I)*Sqrt[b]*f^x)/Sqrt[a]])/(S
qrt[a]*Sqrt[b]*Log[f]^2) + (I*x*PolyLog[2, (I*Sqrt[b]*f^x)/Sqrt[a]])/(Sqrt[a]*Sqrt[b]*Log[f]^2) + (I*PolyLog[3
, ((-I)*Sqrt[b]*f^x)/Sqrt[a]])/(Sqrt[a]*Sqrt[b]*Log[f]^3) - (I*PolyLog[3, (I*Sqrt[b]*f^x)/Sqrt[a]])/(Sqrt[a]*S
qrt[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 211

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

Rule 2277

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 2281

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 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 5251

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 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 {x^2 \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-2 \int \frac {x \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)} \, dx \\ & = \frac {x^2 \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {2 \int x \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right ) \, dx}{\sqrt {a} \sqrt {b} \log (f)} \\ & = \frac {x^2 \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {i \int x \log \left (1-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right ) \, dx}{\sqrt {a} \sqrt {b} \log (f)}+\frac {i \int x \log \left (1+\frac {i \sqrt {b} f^x}{\sqrt {a}}\right ) \, dx}{\sqrt {a} \sqrt {b} \log (f)} \\ & = \frac {x^2 \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {i x \text {Li}_2\left (-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log ^2(f)}+\frac {i x \text {Li}_2\left (\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log ^2(f)}+\frac {i \int \text {Li}_2\left (-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right ) \, dx}{\sqrt {a} \sqrt {b} \log ^2(f)}-\frac {i \int \text {Li}_2\left (\frac {i \sqrt {b} f^x}{\sqrt {a}}\right ) \, dx}{\sqrt {a} \sqrt {b} \log ^2(f)} \\ & = \frac {x^2 \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {i x \text {Li}_2\left (-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log ^2(f)}+\frac {i x \text {Li}_2\left (\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log ^2(f)}+\frac {i \text {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {i \sqrt {b} x}{\sqrt {a}}\right )}{x} \, dx,x,f^x\right )}{\sqrt {a} \sqrt {b} \log ^3(f)}-\frac {i \text {Subst}\left (\int \frac {\text {Li}_2\left (\frac {i \sqrt {b} x}{\sqrt {a}}\right )}{x} \, dx,x,f^x\right )}{\sqrt {a} \sqrt {b} \log ^3(f)} \\ & = \frac {x^2 \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {i x \text {Li}_2\left (-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log ^2(f)}+\frac {i x \text {Li}_2\left (\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log ^2(f)}+\frac {i \text {Li}_3\left (-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log ^3(f)}-\frac {i \text {Li}_3\left (\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log ^3(f)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.91 \[ \int \frac {f^x x^2}{a+b f^{2 x}} \, dx=\frac {i \left (x^2 \log ^2(f) \log \left (1-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )-x^2 \log ^2(f) \log \left (1+\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )-2 x \log (f) \operatorname {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )+2 x \log (f) \operatorname {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )+2 \operatorname {PolyLog}\left (3,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )-2 \operatorname {PolyLog}\left (3,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )\right )}{2 \sqrt {a} \sqrt {b} \log ^3(f)} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.96 \[ \int \frac {f^x x^2}{a+b f^{2 x}} \, dx=-\frac {x^{2} \sqrt {-\frac {b}{a}} \log \left (f^{x} \sqrt {-\frac {b}{a}} + 1\right ) \log \left (f\right )^{2} - x^{2} \sqrt {-\frac {b}{a}} \log \left (-f^{x} \sqrt {-\frac {b}{a}} + 1\right ) \log \left (f\right )^{2} - 2 \, x \sqrt {-\frac {b}{a}} {\rm Li}_2\left (f^{x} \sqrt {-\frac {b}{a}}\right ) \log \left (f\right ) + 2 \, x \sqrt {-\frac {b}{a}} {\rm Li}_2\left (-f^{x} \sqrt {-\frac {b}{a}}\right ) \log \left (f\right ) + 2 \, \sqrt {-\frac {b}{a}} {\rm polylog}\left (3, f^{x} \sqrt {-\frac {b}{a}}\right ) - 2 \, \sqrt {-\frac {b}{a}} {\rm polylog}\left (3, -f^{x} \sqrt {-\frac {b}{a}}\right )}{2 \, b \log \left (f\right )^{3}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

integrate(f^x*x^2/(b*f^(2*x) + a), x)

Giac [F]

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

[In]

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

[Out]

integrate(f^x*x^2/(b*f^(2*x) + a), x)

Mupad [F(-1)]

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

[In]

int((f^x*x^2)/(a + b*f^(2*x)),x)

[Out]

int((f^x*x^2)/(a + b*f^(2*x)), x)