3.46 \(\int \frac{f^x x^3}{a+b f^{2 x}} \, dx\)

Optimal. Leaf size=268 \[ -\frac{3 i x^2 \text{PolyLog}\left (2,-\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{2 \sqrt{a} \sqrt{b} \log ^2(f)}+\frac{3 i x^2 \text{PolyLog}\left (2,\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{2 \sqrt{a} \sqrt{b} \log ^2(f)}+\frac{3 i x \text{PolyLog}\left (3,-\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \log ^3(f)}-\frac{3 i x \text{PolyLog}\left (3,\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \log ^3(f)}-\frac{3 i \text{PolyLog}\left (4,-\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \log ^4(f)}+\frac{3 i \text{PolyLog}\left (4,\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \log ^4(f)}+\frac{x^3 \tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \log (f)} \]

[Out]

(x^3*ArcTan[(Sqrt[b]*f^x)/Sqrt[a]])/(Sqrt[a]*Sqrt[b]*Log[f]) - (((3*I)/2)*x^2*PolyLog[2, ((-I)*Sqrt[b]*f^x)/Sq
rt[a]])/(Sqrt[a]*Sqrt[b]*Log[f]^2) + (((3*I)/2)*x^2*PolyLog[2, (I*Sqrt[b]*f^x)/Sqrt[a]])/(Sqrt[a]*Sqrt[b]*Log[
f]^2) + ((3*I)*x*PolyLog[3, ((-I)*Sqrt[b]*f^x)/Sqrt[a]])/(Sqrt[a]*Sqrt[b]*Log[f]^3) - ((3*I)*x*PolyLog[3, (I*S
qrt[b]*f^x)/Sqrt[a]])/(Sqrt[a]*Sqrt[b]*Log[f]^3) - ((3*I)*PolyLog[4, ((-I)*Sqrt[b]*f^x)/Sqrt[a]])/(Sqrt[a]*Sqr
t[b]*Log[f]^4) + ((3*I)*PolyLog[4, (I*Sqrt[b]*f^x)/Sqrt[a]])/(Sqrt[a]*Sqrt[b]*Log[f]^4)

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Rubi [A]  time = 0.22996, antiderivative size = 268, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {2249, 205, 2245, 12, 5143, 2531, 6609, 2282, 6589} \[ -\frac{3 i x^2 \text{PolyLog}\left (2,-\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{2 \sqrt{a} \sqrt{b} \log ^2(f)}+\frac{3 i x^2 \text{PolyLog}\left (2,\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{2 \sqrt{a} \sqrt{b} \log ^2(f)}+\frac{3 i x \text{PolyLog}\left (3,-\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \log ^3(f)}-\frac{3 i x \text{PolyLog}\left (3,\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \log ^3(f)}-\frac{3 i \text{PolyLog}\left (4,-\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \log ^4(f)}+\frac{3 i \text{PolyLog}\left (4,\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \log ^4(f)}+\frac{x^3 \tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \log (f)} \]

Antiderivative was successfully verified.

[In]

Int[(f^x*x^3)/(a + b*f^(2*x)),x]

[Out]

(x^3*ArcTan[(Sqrt[b]*f^x)/Sqrt[a]])/(Sqrt[a]*Sqrt[b]*Log[f]) - (((3*I)/2)*x^2*PolyLog[2, ((-I)*Sqrt[b]*f^x)/Sq
rt[a]])/(Sqrt[a]*Sqrt[b]*Log[f]^2) + (((3*I)/2)*x^2*PolyLog[2, (I*Sqrt[b]*f^x)/Sqrt[a]])/(Sqrt[a]*Sqrt[b]*Log[
f]^2) + ((3*I)*x*PolyLog[3, ((-I)*Sqrt[b]*f^x)/Sqrt[a]])/(Sqrt[a]*Sqrt[b]*Log[f]^3) - ((3*I)*x*PolyLog[3, (I*S
qrt[b]*f^x)/Sqrt[a]])/(Sqrt[a]*Sqrt[b]*Log[f]^3) - ((3*I)*PolyLog[4, ((-I)*Sqrt[b]*f^x)/Sqrt[a]])/(Sqrt[a]*Sqr
t[b]*Log[f]^4) + ((3*I)*PolyLog[4, (I*Sqrt[b]*f^x)/Sqrt[a]])/(Sqrt[a]*Sqrt[b]*Log[f]^4)

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 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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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 6609

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[((e + f*x)^m*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]), x] - Dist[(f*m)/(b*c*p*Log[F]), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, x] && GtQ[m,
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 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^3}{a+b f^{2 x}} \, dx &=\frac{x^3 \tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \log (f)}-3 \int \frac{x^2 \tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \log (f)} \, dx\\ &=\frac{x^3 \tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \log (f)}-\frac{3 \int x^2 \tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right ) \, dx}{\sqrt{a} \sqrt{b} \log (f)}\\ &=\frac{x^3 \tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \log (f)}-\frac{(3 i) \int x^2 \log \left (1-\frac{i \sqrt{b} f^x}{\sqrt{a}}\right ) \, dx}{2 \sqrt{a} \sqrt{b} \log (f)}+\frac{(3 i) \int x^2 \log \left (1+\frac{i \sqrt{b} f^x}{\sqrt{a}}\right ) \, dx}{2 \sqrt{a} \sqrt{b} \log (f)}\\ &=\frac{x^3 \tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \log (f)}-\frac{3 i x^2 \text{Li}_2\left (-\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{2 \sqrt{a} \sqrt{b} \log ^2(f)}+\frac{3 i x^2 \text{Li}_2\left (\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{2 \sqrt{a} \sqrt{b} \log ^2(f)}+\frac{(3 i) \int x \text{Li}_2\left (-\frac{i \sqrt{b} f^x}{\sqrt{a}}\right ) \, dx}{\sqrt{a} \sqrt{b} \log ^2(f)}-\frac{(3 i) \int x \text{Li}_2\left (\frac{i \sqrt{b} f^x}{\sqrt{a}}\right ) \, dx}{\sqrt{a} \sqrt{b} \log ^2(f)}\\ &=\frac{x^3 \tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \log (f)}-\frac{3 i x^2 \text{Li}_2\left (-\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{2 \sqrt{a} \sqrt{b} \log ^2(f)}+\frac{3 i x^2 \text{Li}_2\left (\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{2 \sqrt{a} \sqrt{b} \log ^2(f)}+\frac{3 i x \text{Li}_3\left (-\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \log ^3(f)}-\frac{3 i x \text{Li}_3\left (\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \log ^3(f)}-\frac{(3 i) \int \text{Li}_3\left (-\frac{i \sqrt{b} f^x}{\sqrt{a}}\right ) \, dx}{\sqrt{a} \sqrt{b} \log ^3(f)}+\frac{(3 i) \int \text{Li}_3\left (\frac{i \sqrt{b} f^x}{\sqrt{a}}\right ) \, dx}{\sqrt{a} \sqrt{b} \log ^3(f)}\\ &=\frac{x^3 \tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \log (f)}-\frac{3 i x^2 \text{Li}_2\left (-\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{2 \sqrt{a} \sqrt{b} \log ^2(f)}+\frac{3 i x^2 \text{Li}_2\left (\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{2 \sqrt{a} \sqrt{b} \log ^2(f)}+\frac{3 i x \text{Li}_3\left (-\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \log ^3(f)}-\frac{3 i x \text{Li}_3\left (\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \log ^3(f)}-\frac{(3 i) \operatorname{Subst}\left (\int \frac{\text{Li}_3\left (-\frac{i \sqrt{b} x}{\sqrt{a}}\right )}{x} \, dx,x,f^x\right )}{\sqrt{a} \sqrt{b} \log ^4(f)}+\frac{(3 i) \operatorname{Subst}\left (\int \frac{\text{Li}_3\left (\frac{i \sqrt{b} x}{\sqrt{a}}\right )}{x} \, dx,x,f^x\right )}{\sqrt{a} \sqrt{b} \log ^4(f)}\\ &=\frac{x^3 \tan ^{-1}\left (\frac{\sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \log (f)}-\frac{3 i x^2 \text{Li}_2\left (-\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{2 \sqrt{a} \sqrt{b} \log ^2(f)}+\frac{3 i x^2 \text{Li}_2\left (\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{2 \sqrt{a} \sqrt{b} \log ^2(f)}+\frac{3 i x \text{Li}_3\left (-\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \log ^3(f)}-\frac{3 i x \text{Li}_3\left (\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \log ^3(f)}-\frac{3 i \text{Li}_4\left (-\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \log ^4(f)}+\frac{3 i \text{Li}_4\left (\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \log ^4(f)}\\ \end{align*}

Mathematica [A]  time = 0.0514449, size = 224, normalized size = 0.84 \[ \frac{i \left (-3 x^2 \log ^2(f) \text{PolyLog}\left (2,-\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )+3 x^2 \log ^2(f) \text{PolyLog}\left (2,\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )-6 \text{PolyLog}\left (4,-\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )+6 \text{PolyLog}\left (4,\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )+6 x \log (f) \text{PolyLog}\left (3,-\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )-6 x \log (f) \text{PolyLog}\left (3,\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )+x^3 \log ^3(f) \log \left (1-\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )-x^3 \log ^3(f) \log \left (1+\frac{i \sqrt{b} f^x}{\sqrt{a}}\right )\right )}{2 \sqrt{a} \sqrt{b} \log ^4(f)} \]

Antiderivative was successfully verified.

[In]

Integrate[(f^x*x^3)/(a + b*f^(2*x)),x]

[Out]

((I/2)*(x^3*Log[f]^3*Log[1 - (I*Sqrt[b]*f^x)/Sqrt[a]] - x^3*Log[f]^3*Log[1 + (I*Sqrt[b]*f^x)/Sqrt[a]] - 3*x^2*
Log[f]^2*PolyLog[2, ((-I)*Sqrt[b]*f^x)/Sqrt[a]] + 3*x^2*Log[f]^2*PolyLog[2, (I*Sqrt[b]*f^x)/Sqrt[a]] + 6*x*Log
[f]*PolyLog[3, ((-I)*Sqrt[b]*f^x)/Sqrt[a]] - 6*x*Log[f]*PolyLog[3, (I*Sqrt[b]*f^x)/Sqrt[a]] - 6*PolyLog[4, ((-
I)*Sqrt[b]*f^x)/Sqrt[a]] + 6*PolyLog[4, (I*Sqrt[b]*f^x)/Sqrt[a]]))/(Sqrt[a]*Sqrt[b]*Log[f]^4)

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Maple [F]  time = 0.036, size = 0, normalized size = 0. \begin{align*} \int{\frac{{f}^{x}{x}^{3}}{a+b{f}^{2\,x}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^x*x^3/(a+b*f^(2*x)),x)

[Out]

int(f^x*x^3/(a+b*f^(2*x)),x)

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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^3/(a+b*f^(2*x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [C]  time = 1.58735, size = 552, normalized size = 2.06 \begin{align*} -\frac{x^{3} \sqrt{-\frac{b}{a}} \log \left (f^{x} \sqrt{-\frac{b}{a}} + 1\right ) \log \left (f\right )^{3} - x^{3} \sqrt{-\frac{b}{a}} \log \left (-f^{x} \sqrt{-\frac{b}{a}} + 1\right ) \log \left (f\right )^{3} - 3 \, x^{2} \sqrt{-\frac{b}{a}}{\rm Li}_2\left (f^{x} \sqrt{-\frac{b}{a}}\right ) \log \left (f\right )^{2} + 3 \, x^{2} \sqrt{-\frac{b}{a}}{\rm Li}_2\left (-f^{x} \sqrt{-\frac{b}{a}}\right ) \log \left (f\right )^{2} + 6 \, x \sqrt{-\frac{b}{a}} \log \left (f\right ){\rm polylog}\left (3, f^{x} \sqrt{-\frac{b}{a}}\right ) - 6 \, x \sqrt{-\frac{b}{a}} \log \left (f\right ){\rm polylog}\left (3, -f^{x} \sqrt{-\frac{b}{a}}\right ) - 6 \, \sqrt{-\frac{b}{a}}{\rm polylog}\left (4, f^{x} \sqrt{-\frac{b}{a}}\right ) + 6 \, \sqrt{-\frac{b}{a}}{\rm polylog}\left (4, -f^{x} \sqrt{-\frac{b}{a}}\right )}{2 \, b \log \left (f\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^x*x^3/(a+b*f^(2*x)),x, algorithm="fricas")

[Out]

-1/2*(x^3*sqrt(-b/a)*log(f^x*sqrt(-b/a) + 1)*log(f)^3 - x^3*sqrt(-b/a)*log(-f^x*sqrt(-b/a) + 1)*log(f)^3 - 3*x
^2*sqrt(-b/a)*dilog(f^x*sqrt(-b/a))*log(f)^2 + 3*x^2*sqrt(-b/a)*dilog(-f^x*sqrt(-b/a))*log(f)^2 + 6*x*sqrt(-b/
a)*log(f)*polylog(3, f^x*sqrt(-b/a)) - 6*x*sqrt(-b/a)*log(f)*polylog(3, -f^x*sqrt(-b/a)) - 6*sqrt(-b/a)*polylo
g(4, f^x*sqrt(-b/a)) + 6*sqrt(-b/a)*polylog(4, -f^x*sqrt(-b/a)))/(b*log(f)^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f^{x} x^{3}}{a + b f^{2 x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**x*x**3/(a+b*f**(2*x)),x)

[Out]

Integral(f**x*x**3/(a + b*f**(2*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f^{x} x^{3}}{b f^{2 \, x} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^x*x^3/(a+b*f^(2*x)),x, algorithm="giac")

[Out]

integrate(f^x*x^3/(b*f^(2*x) + a), x)