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

Optimal. Leaf size=110 \[ -\frac {i \text {Li}_2\left (-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {b} \log ^2(f)}+\frac {i \text {Li}_2\left (\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {b} \log ^2(f)}+\frac {x \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {2249, 205, 2245, 12, 2282, 4848, 2391} \[ -\frac {i \text {PolyLog}\left (2,-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {b} \log ^2(f)}+\frac {i \text {PolyLog}\left (2,\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {b} \log ^2(f)}+\frac {x \tan ^{-1}\left (\frac {\sqrt {b} f^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (f)} \]

Antiderivative was successfully verified.

[In]

Int[(f^x*x)/(a + b*f^(2*x)),x]

[Out]

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

Rule 12

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

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

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Mathematica [A]  time = 0.07, size = 108, normalized size = 0.98 \[ \frac {i \left (-\text {Li}_2\left (-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )+\text {Li}_2\left (\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )+x \log (f) \left (\log \left (1-\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )-\log \left (1+\frac {i \sqrt {b} f^x}{\sqrt {a}}\right )\right )\right )}{2 \sqrt {a} \sqrt {b} \log ^2(f)} \]

Antiderivative was successfully verified.

[In]

Integrate[(f^x*x)/(a + b*f^(2*x)),x]

[Out]

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

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fricas [A]  time = 0.45, size = 112, normalized size = 1.02 \[ -\frac {x \sqrt {-\frac {b}{a}} \log \left (f^{x} \sqrt {-\frac {b}{a}} + 1\right ) \log \relax (f) - x \sqrt {-\frac {b}{a}} \log \left (-f^{x} \sqrt {-\frac {b}{a}} + 1\right ) \log \relax (f) - \sqrt {-\frac {b}{a}} {\rm Li}_2\left (f^{x} \sqrt {-\frac {b}{a}}\right ) + \sqrt {-\frac {b}{a}} {\rm Li}_2\left (-f^{x} \sqrt {-\frac {b}{a}}\right )}{2 \, b \log \relax (f)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^x*x/(b*f^(2*x) + a), x)

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maple [A]  time = 0.06, size = 134, normalized size = 1.22 \[ \frac {x \ln \left (\frac {-b \,f^{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )}{2 \sqrt {-a b}\, \ln \relax (f )}-\frac {x \ln \left (\frac {b \,f^{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )}{2 \sqrt {-a b}\, \ln \relax (f )}+\frac {\dilog \left (\frac {-b \,f^{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )}{2 \sqrt {-a b}\, \ln \relax (f )^{2}}-\frac {\dilog \left (\frac {b \,f^{x}+\sqrt {-a b}}{\sqrt {-a b}}\right )}{2 \sqrt {-a b}\, \ln \relax (f )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/ln(f)*x/(-a*b)^(1/2)*ln((-b*f^x+(-a*b)^(1/2))/(-a*b)^(1/2))-1/2/ln(f)*x/(-a*b)^(1/2)*ln((b*f^x+(-a*b)^(1/2
))/(-a*b)^(1/2))+1/2/ln(f)^2/(-a*b)^(1/2)*dilog((-b*f^x+(-a*b)^(1/2))/(-a*b)^(1/2))-1/2/ln(f)^2/(-a*b)^(1/2)*d
ilog((b*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 \[ \int \frac {f^{x} x}{b f^{2 \, x} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^x*x/(b*f^(2*x) + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {f^x\,x}{a+b\,f^{2\,x}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f^x*x)/(a + b*f^(2*x)),x)

[Out]

int((f^x*x)/(a + b*f^(2*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(f**x*x/(a + b*f**(2*x)), x)

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