\(\int \frac {x^3}{b f^{-x}+a f^x} \, dx\) [57]

Optimal result

Integrand size = 19, antiderivative size = 268 \[ \int \frac {x^3}{b f^{-x}+a f^x} \, dx=\frac {x^3 \arctan \left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {3 i x^2 \operatorname {PolyLog}\left (2,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{2 \sqrt {a} \sqrt {b} \log ^2(f)}+\frac {3 i x^2 \operatorname {PolyLog}\left (2,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{2 \sqrt {a} \sqrt {b} \log ^2(f)}+\frac {3 i x \operatorname {PolyLog}\left (3,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log ^3(f)}-\frac {3 i x \operatorname {PolyLog}\left (3,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log ^3(f)}-\frac {3 i \operatorname {PolyLog}\left (4,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log ^4(f)}+\frac {3 i \operatorname {PolyLog}\left (4,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log ^4(f)} \]

[Out]

x^3*arctan(f^x*a^(1/2)/b^(1/2))/ln(f)/a^(1/2)/b^(1/2)-3/2*I*x^2*polylog(2,-I*f^x*a^(1/2)/b^(1/2))/ln(f)^2/a^(1
/2)/b^(1/2)+3/2*I*x^2*polylog(2,I*f^x*a^(1/2)/b^(1/2))/ln(f)^2/a^(1/2)/b^(1/2)+3*I*x*polylog(3,-I*f^x*a^(1/2)/
b^(1/2))/ln(f)^3/a^(1/2)/b^(1/2)-3*I*x*polylog(3,I*f^x*a^(1/2)/b^(1/2))/ln(f)^3/a^(1/2)/b^(1/2)-3*I*polylog(4,
-I*f^x*a^(1/2)/b^(1/2))/ln(f)^4/a^(1/2)/b^(1/2)+3*I*polylog(4,I*f^x*a^(1/2)/b^(1/2))/ln(f)^4/a^(1/2)/b^(1/2)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {2320, 211, 2298, 12, 5251, 2611, 6744, 6724} \[ \int \frac {x^3}{b f^{-x}+a f^x} \, dx=\frac {x^3 \arctan \left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {3 i x^2 \operatorname {PolyLog}\left (2,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{2 \sqrt {a} \sqrt {b} \log ^2(f)}+\frac {3 i x^2 \operatorname {PolyLog}\left (2,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{2 \sqrt {a} \sqrt {b} \log ^2(f)}-\frac {3 i \operatorname {PolyLog}\left (4,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log ^4(f)}+\frac {3 i \operatorname {PolyLog}\left (4,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log ^4(f)}+\frac {3 i x \operatorname {PolyLog}\left (3,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log ^3(f)}-\frac {3 i x \operatorname {PolyLog}\left (3,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log ^3(f)} \]

[In]

Int[x^3/(b/f^x + a*f^x),x]

[Out]

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

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 2298

Int[(x_)^(m_.)/((b_.)*(F_)^(v_) + (a_.)*(F_)^((c_.) + (d_.)*(x_))), x_Symbol] :> With[{u = IntHide[1/(a*F^(c +
d*x) + b*F^v), x]}, Simp[x^m*u, x] - Dist[m, Int[x^(m - 1)*u, x], x]] /; FreeQ[{F, a, b, c, d}, x] && EqQ[v,
-(c + d*x)] && GtQ[m, 0]

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]

Rule 6744

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]

Rubi steps \begin{align*} \text {integral}& = \frac {x^3 \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-3 \int \frac {x^2 \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log (f)} \, dx \\ & = \frac {x^3 \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {3 \int x^2 \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right ) \, dx}{\sqrt {a} \sqrt {b} \log (f)} \\ & = \frac {x^3 \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {(3 i) \int x^2 \log \left (1-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right ) \, dx}{2 \sqrt {a} \sqrt {b} \log (f)}+\frac {(3 i) \int x^2 \log \left (1+\frac {i \sqrt {a} f^x}{\sqrt {b}}\right ) \, dx}{2 \sqrt {a} \sqrt {b} \log (f)} \\ & = \frac {x^3 \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {3 i x^2 \text {Li}_2\left (-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{2 \sqrt {a} \sqrt {b} \log ^2(f)}+\frac {3 i x^2 \text {Li}_2\left (\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{2 \sqrt {a} \sqrt {b} \log ^2(f)}+\frac {(3 i) \int x \text {Li}_2\left (-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right ) \, dx}{\sqrt {a} \sqrt {b} \log ^2(f)}-\frac {(3 i) \int x \text {Li}_2\left (\frac {i \sqrt {a} f^x}{\sqrt {b}}\right ) \, dx}{\sqrt {a} \sqrt {b} \log ^2(f)} \\ & = \frac {x^3 \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {3 i x^2 \text {Li}_2\left (-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{2 \sqrt {a} \sqrt {b} \log ^2(f)}+\frac {3 i x^2 \text {Li}_2\left (\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{2 \sqrt {a} \sqrt {b} \log ^2(f)}+\frac {3 i x \text {Li}_3\left (-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log ^3(f)}-\frac {3 i x \text {Li}_3\left (\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log ^3(f)}-\frac {(3 i) \int \text {Li}_3\left (-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right ) \, dx}{\sqrt {a} \sqrt {b} \log ^3(f)}+\frac {(3 i) \int \text {Li}_3\left (\frac {i \sqrt {a} f^x}{\sqrt {b}}\right ) \, dx}{\sqrt {a} \sqrt {b} \log ^3(f)} \\ & = \frac {x^3 \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {3 i x^2 \text {Li}_2\left (-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{2 \sqrt {a} \sqrt {b} \log ^2(f)}+\frac {3 i x^2 \text {Li}_2\left (\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{2 \sqrt {a} \sqrt {b} \log ^2(f)}+\frac {3 i x \text {Li}_3\left (-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log ^3(f)}-\frac {3 i x \text {Li}_3\left (\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log ^3(f)}-\frac {(3 i) \text {Subst}\left (\int \frac {\text {Li}_3\left (-\frac {i \sqrt {a} x}{\sqrt {b}}\right )}{x} \, dx,x,f^x\right )}{\sqrt {a} \sqrt {b} \log ^4(f)}+\frac {(3 i) \text {Subst}\left (\int \frac {\text {Li}_3\left (\frac {i \sqrt {a} x}{\sqrt {b}}\right )}{x} \, dx,x,f^x\right )}{\sqrt {a} \sqrt {b} \log ^4(f)} \\ & = \frac {x^3 \tan ^{-1}\left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log (f)}-\frac {3 i x^2 \text {Li}_2\left (-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{2 \sqrt {a} \sqrt {b} \log ^2(f)}+\frac {3 i x^2 \text {Li}_2\left (\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{2 \sqrt {a} \sqrt {b} \log ^2(f)}+\frac {3 i x \text {Li}_3\left (-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log ^3(f)}-\frac {3 i x \text {Li}_3\left (\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log ^3(f)}-\frac {3 i \text {Li}_4\left (-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log ^4(f)}+\frac {3 i \text {Li}_4\left (\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \log ^4(f)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.84 \[ \int \frac {x^3}{b f^{-x}+a f^x} \, dx=\frac {i \left (x^3 \log ^3(f) \log \left (1-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )-x^3 \log ^3(f) \log \left (1+\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )-3 x^2 \log ^2(f) \operatorname {PolyLog}\left (2,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )+3 x^2 \log ^2(f) \operatorname {PolyLog}\left (2,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )+6 x \log (f) \operatorname {PolyLog}\left (3,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )-6 x \log (f) \operatorname {PolyLog}\left (3,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )-6 \operatorname {PolyLog}\left (4,-\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )+6 \operatorname {PolyLog}\left (4,\frac {i \sqrt {a} f^x}{\sqrt {b}}\right )\right )}{2 \sqrt {a} \sqrt {b} \log ^4(f)} \]

[In]

Integrate[x^3/(b/f^x + a*f^x),x]

[Out]

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

Maple [F]

\[\int \frac {x^{3}}{b \,f^{-x}+a \,f^{x}}d x\]

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.89 \[ \int \frac {x^3}{b f^{-x}+a f^x} \, dx=-\frac {x^{3} \sqrt {-\frac {a}{b}} \log \left (f^{x} \sqrt {-\frac {a}{b}} + 1\right ) \log \left (f\right )^{3} - x^{3} \sqrt {-\frac {a}{b}} \log \left (-f^{x} \sqrt {-\frac {a}{b}} + 1\right ) \log \left (f\right )^{3} - 3 \, x^{2} \sqrt {-\frac {a}{b}} {\rm Li}_2\left (f^{x} \sqrt {-\frac {a}{b}}\right ) \log \left (f\right )^{2} + 3 \, x^{2} \sqrt {-\frac {a}{b}} {\rm Li}_2\left (-f^{x} \sqrt {-\frac {a}{b}}\right ) \log \left (f\right )^{2} + 6 \, x \sqrt {-\frac {a}{b}} \log \left (f\right ) {\rm polylog}\left (3, f^{x} \sqrt {-\frac {a}{b}}\right ) - 6 \, x \sqrt {-\frac {a}{b}} \log \left (f\right ) {\rm polylog}\left (3, -f^{x} \sqrt {-\frac {a}{b}}\right ) - 6 \, \sqrt {-\frac {a}{b}} {\rm polylog}\left (4, f^{x} \sqrt {-\frac {a}{b}}\right ) + 6 \, \sqrt {-\frac {a}{b}} {\rm polylog}\left (4, -f^{x} \sqrt {-\frac {a}{b}}\right )}{2 \, a \log \left (f\right )^{4}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {x^3}{b f^{-x}+a f^x} \, dx=\int \frac {f^{x} x^{3}}{a f^{2 x} + b}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {x^3}{b f^{-x}+a f^x} \, dx=\int { \frac {x^{3}}{a f^{x} + \frac {b}{f^{x}}} \,d x } \]

[In]

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

[Out]

integrate(x^3/(a*f^x + b/f^x), x)

Giac [F]

\[ \int \frac {x^3}{b f^{-x}+a f^x} \, dx=\int { \frac {x^{3}}{a f^{x} + \frac {b}{f^{x}}} \,d x } \]

[In]

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

[Out]

integrate(x^3/(a*f^x + b/f^x), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3}{b f^{-x}+a f^x} \, dx=\int \frac {x^3}{\frac {b}{f^x}+a\,f^x} \,d x \]

[In]

int(x^3/(b/f^x + a*f^x),x)

[Out]

int(x^3/(b/f^x + a*f^x), x)