3.525 \(\int \frac{x^2}{1+2 f^{c+d x}+f^{2 c+2 d x}} \, dx\)

Optimal. Leaf size=145 \[ \frac{2 \text{PolyLog}\left (2,-f^{c+d x}\right )}{d^3 \log ^3(f)}+\frac{2 \text{PolyLog}\left (3,-f^{c+d x}\right )}{d^3 \log ^3(f)}-\frac{2 x \text{PolyLog}\left (2,-f^{c+d x}\right )}{d^2 \log ^2(f)}+\frac{2 x \log \left (f^{c+d x}+1\right )}{d^2 \log ^2(f)}-\frac{x^2 \log \left (f^{c+d x}+1\right )}{d \log (f)}+\frac{x^2}{d \log (f) \left (f^{c+d x}+1\right )}-\frac{x^2}{d \log (f)}+\frac{x^3}{3} \]

[Out]

x^3/3 - x^2/(d*Log[f]) + x^2/(d*(1 + f^(c + d*x))*Log[f]) + (2*x*Log[1 + f^(c +
d*x)])/(d^2*Log[f]^2) - (x^2*Log[1 + f^(c + d*x)])/(d*Log[f]) + (2*PolyLog[2, -f
^(c + d*x)])/(d^3*Log[f]^3) - (2*x*PolyLog[2, -f^(c + d*x)])/(d^2*Log[f]^2) + (2
*PolyLog[3, -f^(c + d*x)])/(d^3*Log[f]^3)

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Rubi [A]  time = 0.646586, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 10, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.37 \[ \frac{2 \text{PolyLog}\left (2,-f^{c+d x}\right )}{d^3 \log ^3(f)}+\frac{2 \text{PolyLog}\left (3,-f^{c+d x}\right )}{d^3 \log ^3(f)}-\frac{2 x \text{PolyLog}\left (2,-f^{c+d x}\right )}{d^2 \log ^2(f)}+\frac{2 x \log \left (f^{c+d x}+1\right )}{d^2 \log ^2(f)}-\frac{x^2 \log \left (f^{c+d x}+1\right )}{d \log (f)}+\frac{x^2}{d \log (f) \left (f^{c+d x}+1\right )}-\frac{x^2}{d \log (f)}+\frac{x^3}{3} \]

Antiderivative was successfully verified.

[In]  Int[x^2/(1 + 2*f^(c + d*x) + f^(2*c + 2*d*x)),x]

[Out]

x^3/3 - x^2/(d*Log[f]) + x^2/(d*(1 + f^(c + d*x))*Log[f]) + (2*x*Log[1 + f^(c +
d*x)])/(d^2*Log[f]^2) - (x^2*Log[1 + f^(c + d*x)])/(d*Log[f]) + (2*PolyLog[2, -f
^(c + d*x)])/(d^3*Log[f]^3) - (2*x*PolyLog[2, -f^(c + d*x)])/(d^2*Log[f]^2) + (2
*PolyLog[3, -f^(c + d*x)])/(d^3*Log[f]^3)

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Rubi in Sympy [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{x^{2}}{2 f^{c + d x} + f^{2 c + 2 d x} + 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(x**2/(1+2*f**(d*x+c)+f**(2*d*x+2*c)),x)

[Out]

Integral(x**2/(2*f**(c + d*x) + f**(2*c + 2*d*x) + 1), x)

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Mathematica [A]  time = 0.347866, size = 123, normalized size = 0.85 \[ \frac{6 \text{PolyLog}\left (3,-f^{c+d x}\right )+(6-6 d x \log (f)) \text{PolyLog}\left (2,-f^{c+d x}\right )-\frac{3 d^2 x^2 \log ^2(f) \left (f^{c+d x}+\left (f^{c+d x}+1\right ) \log \left (f^{c+d x}+1\right )\right )}{f^{c+d x}+1}+6 d x \log (f) \log \left (f^{c+d x}+1\right )+d^3 x^3 \log ^3(f)}{3 d^3 \log ^3(f)} \]

Antiderivative was successfully verified.

[In]  Integrate[x^2/(1 + 2*f^(c + d*x) + f^(2*c + 2*d*x)),x]

[Out]

(d^3*x^3*Log[f]^3 + 6*d*x*Log[f]*Log[1 + f^(c + d*x)] - (3*d^2*x^2*Log[f]^2*(f^(
c + d*x) + (1 + f^(c + d*x))*Log[1 + f^(c + d*x)]))/(1 + f^(c + d*x)) + (6 - 6*d
*x*Log[f])*PolyLog[2, -f^(c + d*x)] + 6*PolyLog[3, -f^(c + d*x)])/(3*d^3*Log[f]^
3)

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Maple [A]  time = 0.039, size = 221, normalized size = 1.5 \[{\frac{{x}^{2}}{d \left ( 1+{f}^{dx+c} \right ) \ln \left ( f \right ) }}+{\frac{{x}^{3}}{3}}-{\frac{{c}^{2}x}{{d}^{2}}}-{\frac{2\,{c}^{3}}{3\,{d}^{3}}}-{\frac{{x}^{2}\ln \left ( 1+{f}^{dx+c} \right ) }{d\ln \left ( f \right ) }}-2\,{\frac{x{\it polylog} \left ( 2,-{f}^{dx+c} \right ) }{ \left ( \ln \left ( f \right ) \right ) ^{2}{d}^{2}}}+2\,{\frac{{\it polylog} \left ( 3,-{f}^{dx+c} \right ) }{{d}^{3} \left ( \ln \left ( f \right ) \right ) ^{3}}}+{\frac{{c}^{2}\ln \left ({f}^{dx+c} \right ) }{{d}^{3}\ln \left ( f \right ) }}-{\frac{{x}^{2}}{d\ln \left ( f \right ) }}-2\,{\frac{cx}{{d}^{2}\ln \left ( f \right ) }}-{\frac{{c}^{2}}{{d}^{3}\ln \left ( f \right ) }}+2\,{\frac{x\ln \left ( 1+{f}^{dx+c} \right ) }{ \left ( \ln \left ( f \right ) \right ) ^{2}{d}^{2}}}+2\,{\frac{{\it polylog} \left ( 2,-{f}^{dx+c} \right ) }{{d}^{3} \left ( \ln \left ( f \right ) \right ) ^{3}}}+2\,{\frac{c\ln \left ({f}^{dx+c} \right ) }{{d}^{3} \left ( \ln \left ( f \right ) \right ) ^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(x^2/(1+2*f^(d*x+c)+f^(2*d*x+2*c)),x)

[Out]

x^2/d/(1+f^(d*x+c))/ln(f)+1/3*x^3-1/d^2*c^2*x-2/3/d^3*c^3-x^2*ln(1+f^(d*x+c))/d/
ln(f)-2*x*polylog(2,-f^(d*x+c))/d^2/ln(f)^2+2*polylog(3,-f^(d*x+c))/d^3/ln(f)^3+
1/d^3/ln(f)*c^2*ln(f^(d*x+c))-x^2/d/ln(f)-2/d^2/ln(f)*c*x-1/d^3/ln(f)*c^2+2*x*ln
(1+f^(d*x+c))/d^2/ln(f)^2+2*polylog(2,-f^(d*x+c))/d^3/ln(f)^3+2/d^3/ln(f)^2*c*ln
(f^(d*x+c))

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Maxima [A]  time = 0.78877, size = 211, normalized size = 1.46 \[ \frac{x^{2}}{d f^{d x} f^{c} \log \left (f\right ) + d \log \left (f\right )} - \frac{\log \left (f^{d x} f^{c} + 1\right ) \log \left (f^{d x}\right )^{2} + 2 \,{\rm Li}_2\left (-f^{d x} f^{c}\right ) \log \left (f^{d x}\right ) - 2 \,{\rm Li}_{3}(-f^{d x} f^{c})}{d^{3} \log \left (f\right )^{3}} + \frac{\log \left (f^{d x}\right )^{3} - 3 \, \log \left (f^{d x}\right )^{2}}{3 \, d^{3} \log \left (f\right )^{3}} + \frac{2 \,{\left (\log \left (f^{d x} f^{c} + 1\right ) \log \left (f^{d x}\right ) +{\rm Li}_2\left (-f^{d x} f^{c}\right )\right )}}{d^{3} \log \left (f\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x^2/(f^(2*d*x + 2*c) + 2*f^(d*x + c) + 1),x, algorithm="maxima")

[Out]

x^2/(d*f^(d*x)*f^c*log(f) + d*log(f)) - (log(f^(d*x)*f^c + 1)*log(f^(d*x))^2 + 2
*dilog(-f^(d*x)*f^c)*log(f^(d*x)) - 2*polylog(3, -f^(d*x)*f^c))/(d^3*log(f)^3) +
1/3*(log(f^(d*x))^3 - 3*log(f^(d*x))^2)/(d^3*log(f)^3) + 2*(log(f^(d*x)*f^c + 1
)*log(f^(d*x)) + dilog(-f^(d*x)*f^c))/(d^3*log(f)^3)

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Fricas [A]  time = 0.306643, size = 284, normalized size = 1.96 \[ \frac{3 \, c^{2} \log \left (f\right )^{2} +{\left (d^{3} x^{3} + c^{3}\right )} \log \left (f\right )^{3} +{\left ({\left (d^{3} x^{3} + c^{3}\right )} \log \left (f\right )^{3} - 3 \,{\left (d^{2} x^{2} - c^{2}\right )} \log \left (f\right )^{2}\right )} f^{d x + c} - 6 \,{\left (d x \log \left (f\right ) +{\left (d x \log \left (f\right ) - 1\right )} f^{d x + c} - 1\right )}{\rm Li}_2\left (-f^{d x + c}\right ) - 3 \,{\left (d^{2} x^{2} \log \left (f\right )^{2} - 2 \, d x \log \left (f\right ) +{\left (d^{2} x^{2} \log \left (f\right )^{2} - 2 \, d x \log \left (f\right )\right )} f^{d x + c}\right )} \log \left (f^{d x + c} + 1\right ) + 6 \,{\left (f^{d x + c} + 1\right )}{\rm Li}_{3}(-f^{d x + c})}{3 \,{\left (d^{3} f^{d x + c} \log \left (f\right )^{3} + d^{3} \log \left (f\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x^2/(f^(2*d*x + 2*c) + 2*f^(d*x + c) + 1),x, algorithm="fricas")

[Out]

1/3*(3*c^2*log(f)^2 + (d^3*x^3 + c^3)*log(f)^3 + ((d^3*x^3 + c^3)*log(f)^3 - 3*(
d^2*x^2 - c^2)*log(f)^2)*f^(d*x + c) - 6*(d*x*log(f) + (d*x*log(f) - 1)*f^(d*x +
c) - 1)*dilog(-f^(d*x + c)) - 3*(d^2*x^2*log(f)^2 - 2*d*x*log(f) + (d^2*x^2*log
(f)^2 - 2*d*x*log(f))*f^(d*x + c))*log(f^(d*x + c) + 1) + 6*(f^(d*x + c) + 1)*po
lylog(3, -f^(d*x + c)))/(d^3*f^(d*x + c)*log(f)^3 + d^3*log(f)^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \[ \frac{x^{2}}{d f^{c + d x} \log{\left (f \right )} + d \log{\left (f \right )}} + \frac{\int \left (- \frac{2 x}{e^{c \log{\left (f \right )}} e^{d x \log{\left (f \right )}} + 1}\right )\, dx + \int \frac{d x^{2} \log{\left (f \right )}}{e^{c \log{\left (f \right )}} e^{d x \log{\left (f \right )}} + 1}\, dx}{d \log{\left (f \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x**2/(1+2*f**(d*x+c)+f**(2*d*x+2*c)),x)

[Out]

x**2/(d*f**(c + d*x)*log(f) + d*log(f)) + (Integral(-2*x/(exp(c*log(f))*exp(d*x*
log(f)) + 1), x) + Integral(d*x**2*log(f)/(exp(c*log(f))*exp(d*x*log(f)) + 1), x
))/(d*log(f))

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{x^{2}}{f^{2 \, d x + 2 \, c} + 2 \, f^{d x + c} + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x^2/(f^(2*d*x + 2*c) + 2*f^(d*x + c) + 1),x, algorithm="giac")

[Out]

integrate(x^2/(f^(2*d*x + 2*c) + 2*f^(d*x + c) + 1), x)