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

Optimal. Leaf size=172 \[ -\frac{i \text{PolyLog}\left (2,-\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{4 a^{3/2} \sqrt{b} \log ^2(f)}+\frac{i \text{PolyLog}\left (2,\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{4 a^{3/2} \sqrt{b} \log ^2(f)}-\frac{\tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right )}{2 a^{3/2} \sqrt{b} \log ^2(f)}+\frac{x \tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right )}{2 a^{3/2} \sqrt{b} \log (f)}+\frac{x f^x}{2 a \log (f) \left (a+b f^{2 x}\right )} \]

[Out]

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

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Rubi [A]  time = 0.158377, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {2249, 199, 205, 2245, 2282, 4848, 2391} \[ -\frac{i \text{PolyLog}\left (2,-\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{4 a^{3/2} \sqrt{b} \log ^2(f)}+\frac{i \text{PolyLog}\left (2,\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{4 a^{3/2} \sqrt{b} \log ^2(f)}-\frac{\tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right )}{2 a^{3/2} \sqrt{b} \log ^2(f)}+\frac{x \tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right )}{2 a^{3/2} \sqrt{b} \log (f)}+\frac{x f^x}{2 a \log (f) \left (a+b f^{2 x}\right )} \]

Antiderivative was successfully verified.

[In]

Int[(f^x*x)/(a + b*f^(2*x))^2,x]

[Out]

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

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 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 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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.116771, size = 271, normalized size = 1.58 \[ \frac{\frac{-\frac{i \text{PolyLog}\left (2,-\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \log ^2(f)}-\frac{i x \log \left (1+\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \log (f)}+\frac{i x^2}{2 \sqrt{a}}}{2 \sqrt{b}}+\frac{\frac{i \text{PolyLog}\left (2,\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \log ^2(f)}+\frac{i x \log \left (1-\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \log (f)}-\frac{i x^2}{2 \sqrt{a}}}{2 \sqrt{b}}}{2 a}+\frac{x f^x}{2 a \log (f) \left (a+b f^{2 x}\right )}-\frac{\left (\frac{b f^{2 x}}{a}+1\right ) \tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right )}{2 \sqrt{a} \sqrt{b} \log ^2(f) \left (a+b f^{2 x}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.053, size = 195, normalized size = 1.1 \begin{align*}{\frac{{f}^{x}x}{2\,\ln \left ( f \right ) a \left ( a+b \left ({f}^{x} \right ) ^{2} \right ) }}+{\frac{x}{4\,\ln \left ( f \right ) a}\ln \left ({ \left ( -b{f}^{x}+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}} \right ){\frac{1}{\sqrt{-ab}}}}-{\frac{x}{4\,\ln \left ( f \right ) a}\ln \left ({ \left ( b{f}^{x}+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}} \right ){\frac{1}{\sqrt{-ab}}}}+{\frac{1}{4\, \left ( \ln \left ( f \right ) \right ) ^{2}a}{\it dilog} \left ({ \left ( -b{f}^{x}+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}} \right ){\frac{1}{\sqrt{-ab}}}}-{\frac{1}{4\, \left ( \ln \left ( f \right ) \right ) ^{2}a}{\it dilog} \left ({ \left ( b{f}^{x}+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}} \right ){\frac{1}{\sqrt{-ab}}}}-{\frac{1}{2\, \left ( \ln \left ( f \right ) \right ) ^{2}a}\arctan \left ({b{f}^{x}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/ln(f)/a*f^x*x/(a+b*(f^x)^2)+1/4/ln(f)/a*x/(-a*b)^(1/2)*ln((-b*f^x+(-a*b)^(1/2))/(-a*b)^(1/2))-1/4/ln(f)/a*
x/(-a*b)^(1/2)*ln((b*f^x+(-a*b)^(1/2))/(-a*b)^(1/2))+1/4/ln(f)^2/a/(-a*b)^(1/2)*dilog((-b*f^x+(-a*b)^(1/2))/(-
a*b)^(1/2))-1/4/ln(f)^2/a/(-a*b)^(1/2)*dilog((b*f^x+(-a*b)^(1/2))/(-a*b)^(1/2))-1/2/ln(f)^2/a/(a*b)^(1/2)*arct
an(b*f^x/(a*b)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.53727, size = 670, normalized size = 3.9 \begin{align*} \frac{2 \, b f^{x} x \log \left (f\right ) +{\left (b f^{2 \, x} \sqrt{-\frac{b}{a}} + a \sqrt{-\frac{b}{a}}\right )}{\rm Li}_2\left (f^{x} \sqrt{-\frac{b}{a}}\right ) -{\left (b f^{2 \, x} \sqrt{-\frac{b}{a}} + a \sqrt{-\frac{b}{a}}\right )}{\rm Li}_2\left (-f^{x} \sqrt{-\frac{b}{a}}\right ) -{\left (b f^{2 \, x} \sqrt{-\frac{b}{a}} + a \sqrt{-\frac{b}{a}}\right )} \log \left (2 \, b f^{x} + 2 \, a \sqrt{-\frac{b}{a}}\right ) +{\left (b f^{2 \, x} \sqrt{-\frac{b}{a}} + a \sqrt{-\frac{b}{a}}\right )} \log \left (2 \, b f^{x} - 2 \, a \sqrt{-\frac{b}{a}}\right ) -{\left (b f^{2 \, x} x \sqrt{-\frac{b}{a}} \log \left (f\right ) + a x \sqrt{-\frac{b}{a}} \log \left (f\right )\right )} \log \left (f^{x} \sqrt{-\frac{b}{a}} + 1\right ) +{\left (b f^{2 \, x} x \sqrt{-\frac{b}{a}} \log \left (f\right ) + a x \sqrt{-\frac{b}{a}} \log \left (f\right )\right )} \log \left (-f^{x} \sqrt{-\frac{b}{a}} + 1\right )}{4 \,{\left (a b^{2} f^{2 \, x} \log \left (f\right )^{2} + a^{2} b \log \left (f\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*b*f^x*x*log(f) + (b*f^(2*x)*sqrt(-b/a) + a*sqrt(-b/a))*dilog(f^x*sqrt(-b/a)) - (b*f^(2*x)*sqrt(-b/a) +
a*sqrt(-b/a))*dilog(-f^x*sqrt(-b/a)) - (b*f^(2*x)*sqrt(-b/a) + a*sqrt(-b/a))*log(2*b*f^x + 2*a*sqrt(-b/a)) + (
b*f^(2*x)*sqrt(-b/a) + a*sqrt(-b/a))*log(2*b*f^x - 2*a*sqrt(-b/a)) - (b*f^(2*x)*x*sqrt(-b/a)*log(f) + a*x*sqrt
(-b/a)*log(f))*log(f^x*sqrt(-b/a) + 1) + (b*f^(2*x)*x*sqrt(-b/a)*log(f) + a*x*sqrt(-b/a)*log(f))*log(-f^x*sqrt
(-b/a) + 1))/(a*b^2*f^(2*x)*log(f)^2 + a^2*b*log(f)^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{f^{x} x}{2 a^{2} \log{\left (f \right )} + 2 a b f^{2 x} \log{\left (f \right )}} + \frac{\int - \frac{f^{x}}{a + b f^{2 x}}\, dx + \int \frac{f^{x} x \log{\left (f \right )}}{a + b f^{2 x}}\, dx}{2 a \log{\left (f \right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f^{x} x}{{\left (b f^{2 \, x} + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^x*x/(b*f^(2*x) + a)^2, x)