3.63 \(\int \frac{x}{\left (b f^{-x}+a f^x\right )^3} \, dx\)

Optimal. Leaf size=196 \[ -\frac{i \text{PolyLog}\left (2,-\frac{i \sqrt{a} f^x}{\sqrt{b}}\right )}{16 a^{3/2} b^{3/2} \log ^2(f)}+\frac{i \text{PolyLog}\left (2,\frac{i \sqrt{a} f^x}{\sqrt{b}}\right )}{16 a^{3/2} b^{3/2} \log ^2(f)}+\frac{x \tan ^{-1}\left (\frac{\sqrt{a} f^x}{\sqrt{b}}\right )}{8 a^{3/2} b^{3/2} \log (f)}+\frac{f^x}{8 a b \log ^2(f) \left (a f^{2 x}+b\right )}+\frac{x f^x}{8 a b \log (f) \left (a f^{2 x}+b\right )}-\frac{x f^x}{4 a \log (f) \left (a f^{2 x}+b\right )^2} \]

[Out]

f^x/(8*a*b*(b + a*f^(2*x))*Log[f]^2) - (f^x*x)/(4*a*(b + a*f^(2*x))^2*Log[f]) +
(f^x*x)/(8*a*b*(b + a*f^(2*x))*Log[f]) + (x*ArcTan[(Sqrt[a]*f^x)/Sqrt[b]])/(8*a^
(3/2)*b^(3/2)*Log[f]) - ((I/16)*PolyLog[2, ((-I)*Sqrt[a]*f^x)/Sqrt[b]])/(a^(3/2)
*b^(3/2)*Log[f]^2) + ((I/16)*PolyLog[2, (I*Sqrt[a]*f^x)/Sqrt[b]])/(a^(3/2)*b^(3/
2)*Log[f]^2)

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Rubi [A]  time = 0.835626, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 22, number of rules used = 9, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.529 \[ -\frac{i \text{PolyLog}\left (2,-\frac{i \sqrt{a} f^x}{\sqrt{b}}\right )}{16 a^{3/2} b^{3/2} \log ^2(f)}+\frac{i \text{PolyLog}\left (2,\frac{i \sqrt{a} f^x}{\sqrt{b}}\right )}{16 a^{3/2} b^{3/2} \log ^2(f)}+\frac{x \tan ^{-1}\left (\frac{\sqrt{a} f^x}{\sqrt{b}}\right )}{8 a^{3/2} b^{3/2} \log (f)}+\frac{f^x}{8 a b \log ^2(f) \left (a f^{2 x}+b\right )}+\frac{x f^x}{8 a b \log (f) \left (a f^{2 x}+b\right )}-\frac{x f^x}{4 a \log (f) \left (a f^{2 x}+b\right )^2} \]

Antiderivative was successfully verified.

[In]  Int[x/(b/f^x + a*f^x)^3,x]

[Out]

f^x/(8*a*b*(b + a*f^(2*x))*Log[f]^2) - (f^x*x)/(4*a*(b + a*f^(2*x))^2*Log[f]) +
(f^x*x)/(8*a*b*(b + a*f^(2*x))*Log[f]) + (x*ArcTan[(Sqrt[a]*f^x)/Sqrt[b]])/(8*a^
(3/2)*b^(3/2)*Log[f]) - ((I/16)*PolyLog[2, ((-I)*Sqrt[a]*f^x)/Sqrt[b]])/(a^(3/2)
*b^(3/2)*Log[f]^2) + ((I/16)*PolyLog[2, (I*Sqrt[a]*f^x)/Sqrt[b]])/(a^(3/2)*b^(3/
2)*Log[f]^2)

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Rubi in Sympy [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{f^{- 3 x} x}{\left (a + b f^{- 2 x}\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(x/(b/(f**x)+a*f**x)**3,x)

[Out]

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

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Mathematica [A]  time = 0.294583, size = 209, normalized size = 1.07 \[ \frac{-\frac{i \text{PolyLog}\left (2,-\frac{i \sqrt{a} f^x}{\sqrt{b}}\right )}{b^{3/2}}+\frac{i \text{PolyLog}\left (2,\frac{i \sqrt{a} f^x}{\sqrt{b}}\right )}{b^{3/2}}+\frac{i x \log (f) \log \left (1-\frac{i \sqrt{a} f^x}{\sqrt{b}}\right )}{b^{3/2}}-\frac{i x \log (f) \log \left (1+\frac{i \sqrt{a} f^x}{\sqrt{b}}\right )}{b^{3/2}}+\frac{2 \sqrt{a} f^x}{a b f^{2 x}+b^2}+\frac{2 \sqrt{a} x f^x \log (f)}{a b f^{2 x}+b^2}-\frac{4 \sqrt{a} x f^x \log (f)}{\left (a f^{2 x}+b\right )^2}}{16 a^{3/2} \log ^2(f)} \]

Antiderivative was successfully verified.

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

[Out]

((2*Sqrt[a]*f^x)/(b^2 + a*b*f^(2*x)) - (4*Sqrt[a]*f^x*x*Log[f])/(b + a*f^(2*x))^
2 + (2*Sqrt[a]*f^x*x*Log[f])/(b^2 + a*b*f^(2*x)) + (I*x*Log[f]*Log[1 - (I*Sqrt[a
]*f^x)/Sqrt[b]])/b^(3/2) - (I*x*Log[f]*Log[1 + (I*Sqrt[a]*f^x)/Sqrt[b]])/b^(3/2)
- (I*PolyLog[2, ((-I)*Sqrt[a]*f^x)/Sqrt[b]])/b^(3/2) + (I*PolyLog[2, (I*Sqrt[a]
*f^x)/Sqrt[b]])/b^(3/2))/(16*a^(3/2)*Log[f]^2)

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Maple [C]  time = 0.053, size = 209, normalized size = 1.1 \[{\frac{ \left ( \left ({f}^{x} \right ) ^{2}\ln \left ( f \right ) ax-\ln \left ( f \right ) bx+a \left ({f}^{x} \right ) ^{2}+b \right ){f}^{x}}{8\, \left ( \ln \left ( f \right ) \right ) ^{2}ab \left ( a \left ({f}^{x} \right ) ^{2}+b \right ) ^{2}}}+{\frac{x}{16\,b\ln \left ( f \right ) a}\ln \left ({1 \left ( -a{f}^{x}+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}} \right ){\frac{1}{\sqrt{-ab}}}}-{\frac{x}{16\,b\ln \left ( f \right ) a}\ln \left ({1 \left ( a{f}^{x}+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}} \right ){\frac{1}{\sqrt{-ab}}}}+{\frac{1}{16\, \left ( \ln \left ( f \right ) \right ) ^{2}ab}{\it dilog} \left ({1 \left ( -a{f}^{x}+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}} \right ){\frac{1}{\sqrt{-ab}}}}-{\frac{1}{16\, \left ( \ln \left ( f \right ) \right ) ^{2}ab}{\it dilog} \left ({1 \left ( a{f}^{x}+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}} \right ){\frac{1}{\sqrt{-ab}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(x/(b/(f^x)+a*f^x)^3,x)

[Out]

1/8*f^x*((f^x)^2*ln(f)*a*x-ln(f)*b*x+a*(f^x)^2+b)/ln(f)^2/b/a/(a*(f^x)^2+b)^2+1/
16/ln(f)/a/b*x/(-a*b)^(1/2)*ln((-a*f^x+(-a*b)^(1/2))/(-a*b)^(1/2))-1/16/ln(f)/a/
b*x/(-a*b)^(1/2)*ln((a*f^x+(-a*b)^(1/2))/(-a*b)^(1/2))+1/16/ln(f)^2/a/b/(-a*b)^(
1/2)*dilog((-a*f^x+(-a*b)^(1/2))/(-a*b)^(1/2))-1/16/ln(f)^2/a/b/(-a*b)^(1/2)*dil
og((a*f^x+(-a*b)^(1/2))/(-a*b)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.348205, size = 564, normalized size = 2.88 \[ \frac{2 \,{\left (a^{2} x \log \left (f\right ) + a^{2}\right )} f^{3 \, x} - 2 \,{\left (a b x \log \left (f\right ) - a b\right )} f^{x} +{\left (a^{2} f^{4 \, x} \sqrt{-\frac{a}{b}} + 2 \, a b f^{2 \, x} \sqrt{-\frac{a}{b}} + b^{2} \sqrt{-\frac{a}{b}}\right )}{\rm Li}_2\left (-\frac{a f^{x} + b \sqrt{-\frac{a}{b}}}{b \sqrt{-\frac{a}{b}}} + 1\right ) -{\left (a^{2} f^{4 \, x} \sqrt{-\frac{a}{b}} + 2 \, a b f^{2 \, x} \sqrt{-\frac{a}{b}} + b^{2} \sqrt{-\frac{a}{b}}\right )}{\rm Li}_2\left (\frac{a f^{x} - b \sqrt{-\frac{a}{b}}}{b \sqrt{-\frac{a}{b}}} + 1\right ) +{\left (a^{2} f^{4 \, x} x \sqrt{-\frac{a}{b}} \log \left (f\right ) + 2 \, a b f^{2 \, x} x \sqrt{-\frac{a}{b}} \log \left (f\right ) + b^{2} x \sqrt{-\frac{a}{b}} \log \left (f\right )\right )} \log \left (\frac{a f^{x} + b \sqrt{-\frac{a}{b}}}{b \sqrt{-\frac{a}{b}}}\right ) -{\left (a^{2} f^{4 \, x} x \sqrt{-\frac{a}{b}} \log \left (f\right ) + 2 \, a b f^{2 \, x} x \sqrt{-\frac{a}{b}} \log \left (f\right ) + b^{2} x \sqrt{-\frac{a}{b}} \log \left (f\right )\right )} \log \left (-\frac{a f^{x} - b \sqrt{-\frac{a}{b}}}{b \sqrt{-\frac{a}{b}}}\right )}{16 \,{\left (a^{4} b f^{4 \, x} \log \left (f\right )^{2} + 2 \, a^{3} b^{2} f^{2 \, x} \log \left (f\right )^{2} + a^{2} b^{3} \log \left (f\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

1/16*(2*(a^2*x*log(f) + a^2)*f^(3*x) - 2*(a*b*x*log(f) - a*b)*f^x + (a^2*f^(4*x)
*sqrt(-a/b) + 2*a*b*f^(2*x)*sqrt(-a/b) + b^2*sqrt(-a/b))*dilog(-(a*f^x + b*sqrt(
-a/b))/(b*sqrt(-a/b)) + 1) - (a^2*f^(4*x)*sqrt(-a/b) + 2*a*b*f^(2*x)*sqrt(-a/b)
+ b^2*sqrt(-a/b))*dilog((a*f^x - b*sqrt(-a/b))/(b*sqrt(-a/b)) + 1) + (a^2*f^(4*x
)*x*sqrt(-a/b)*log(f) + 2*a*b*f^(2*x)*x*sqrt(-a/b)*log(f) + b^2*x*sqrt(-a/b)*log
(f))*log((a*f^x + b*sqrt(-a/b))/(b*sqrt(-a/b))) - (a^2*f^(4*x)*x*sqrt(-a/b)*log(
f) + 2*a*b*f^(2*x)*x*sqrt(-a/b)*log(f) + b^2*x*sqrt(-a/b)*log(f))*log(-(a*f^x -
b*sqrt(-a/b))/(b*sqrt(-a/b))))/(a^4*b*f^(4*x)*log(f)^2 + 2*a^3*b^2*f^(2*x)*log(f
)^2 + a^2*b^3*log(f)^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \[ \frac{f^{- x} \left (a x \log{\left (f \right )} + a\right ) + f^{- 3 x} \left (- b x \log{\left (f \right )} + b\right )}{8 a^{3} b \log{\left (f \right )}^{2} + 16 a^{2} b^{2} f^{- 2 x} \log{\left (f \right )}^{2} + 8 a b^{3} f^{- 4 x} \log{\left (f \right )}^{2}} + \frac{\int \frac{f^{x} x}{a f^{2 x} + b}\, dx}{8 a b} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

(f**(-x)*(a*x*log(f) + a) + f**(-3*x)*(-b*x*log(f) + b))/(8*a**3*b*log(f)**2 + 1
6*a**2*b**2*f**(-2*x)*log(f)**2 + 8*a*b**3*f**(-4*x)*log(f)**2) + Integral(f**x*
x/(a*f**(2*x) + b), x)/(8*a*b)

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{x}{{\left (a f^{x} + \frac{b}{f^{x}}\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integrate(x/(a*f^x + b/f^x)^3, x)