3.47 \(\int \frac{(c+d x)^2}{a+b \left (F^{g (e+f x)}\right )^n} \, dx\)

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

[Out]

(c + d*x)^3/(3*a*d) - ((c + d*x)^2*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(a*f*g*n*
Log[F]) - (2*d*(c + d*x)*PolyLog[2, -((b*(F^(g*(e + f*x)))^n)/a)])/(a*f^2*g^2*n^
2*Log[F]^2) + (2*d^2*PolyLog[3, -((b*(F^(g*(e + f*x)))^n)/a)])/(a*f^3*g^3*n^3*Lo
g[F]^3)

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Rubi [A]  time = 0.464419, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2 \[ -\frac{2 d (c+d x) \text{PolyLog}\left (2,-\frac{b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a f^2 g^2 n^2 \log ^2(F)}+\frac{2 d^2 \text{PolyLog}\left (3,-\frac{b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a f^3 g^3 n^3 \log ^3(F)}-\frac{(c+d x)^2 \log \left (\frac{b \left (F^{g (e+f x)}\right )^n}{a}+1\right )}{a f g n \log (F)}+\frac{(c+d x)^3}{3 a d} \]

Antiderivative was successfully verified.

[In]  Int[(c + d*x)^2/(a + b*(F^(g*(e + f*x)))^n),x]

[Out]

(c + d*x)^3/(3*a*d) - ((c + d*x)^2*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(a*f*g*n*
Log[F]) - (2*d*(c + d*x)*PolyLog[2, -((b*(F^(g*(e + f*x)))^n)/a)])/(a*f^2*g^2*n^
2*Log[F]^2) + (2*d^2*PolyLog[3, -((b*(F^(g*(e + f*x)))^n)/a)])/(a*f^3*g^3*n^3*Lo
g[F]^3)

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Rubi in Sympy [A]  time = 52.758, size = 114, normalized size = 0.79 \[ \frac{2 d^{2} \operatorname{Li}_{3}\left (- \frac{a \left (F^{g \left (e + f x\right )}\right )^{- n}}{b}\right )}{a f^{3} g^{3} n^{3} \log{\left (F \right )}^{3}} + \frac{2 d \left (c + d x\right ) \operatorname{Li}_{2}\left (- \frac{a \left (F^{g \left (e + f x\right )}\right )^{- n}}{b}\right )}{a f^{2} g^{2} n^{2} \log{\left (F \right )}^{2}} - \frac{\left (c + d x\right )^{2} \log{\left (\frac{a \left (F^{g \left (e + f x\right )}\right )^{- n}}{b} + 1 \right )}}{a f g n \log{\left (F \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((d*x+c)**2/(a+b*(F**(g*(f*x+e)))**n),x)

[Out]

2*d**2*polylog(3, -a*(F**(g*(e + f*x)))**(-n)/b)/(a*f**3*g**3*n**3*log(F)**3) +
2*d*(c + d*x)*polylog(2, -a*(F**(g*(e + f*x)))**(-n)/b)/(a*f**2*g**2*n**2*log(F)
**2) - (c + d*x)**2*log(a*(F**(g*(e + f*x)))**(-n)/b + 1)/(a*f*g*n*log(F))

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Mathematica [A]  time = 2.66148, size = 0, normalized size = 0. \[ \int \frac{(c+d x)^2}{a+b \left (F^{g (e+f x)}\right )^n} \, dx \]

Verification is Not applicable to the result.

[In]  Integrate[(c + d*x)^2/(a + b*(F^(g*(e + f*x)))^n),x]

[Out]

Integrate[(c + d*x)^2/(a + b*(F^(g*(e + f*x)))^n), x]

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Maple [B]  time = 0.089, size = 1341, normalized size = 9.3 \[ \text{result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((d*x+c)^2/(a+b*(F^(g*(f*x+e)))^n),x)

[Out]

d^2/a*x^3+2*d^2*polylog(3,-b*(F^(g*(f*x+e)))^n/a)/a/f^3/g^3/n^3/ln(F)^3+2/ln(F)^
2/f^3/g^2/n*d^2*e*(ln(F^(g*(f*x+e)))-g*(f*x+e)*ln(F))/a*ln((F^(g*(f*x+e)))^n)-2/
ln(F)^2/f^3/g^2/n*d^2*e*(ln(F^(g*(f*x+e)))-g*(f*x+e)*ln(F))/a*ln(a+b*(F^(g*(f*x+
e)))^n)-2/ln(F)/f^2/g/n*c*d*e/a*ln((F^(g*(f*x+e)))^n)+2/ln(F)/f^2/g/n*c*d*e/a*ln
(a+b*(F^(g*(f*x+e)))^n)+2/ln(F)^2/f^3/g^2/n*d^2*e/a*ln(1+b*(F^(g*(f*x+e)))^n/a)*
(ln(F^(g*(f*x+e)))-g*(f*x+e)*ln(F))-2/ln(F)/f/g/n*c*d/a*ln(1+b*(F^(g*(f*x+e)))^n
/a)*x-2/ln(F)/f^2/g/n*c*d/a*ln(1+b*(F^(g*(f*x+e)))^n/a)*e+2/f*c*d/a*x*e-1/f^2*d^
2/a*x*e^2-2/ln(F)/f/g*d^2/a*ln(F^(g*(f*x+e)))*x^2+2/ln(F)^2/f^2/g^2*d^2/a*ln(F^(
g*(f*x+e)))^2*x+1/ln(F)^2/f^2/g^2*c*d/a*ln(F^(g*(f*x+e)))^2-1/ln(F)^2/f^2/g^2*d^
2/a*x*(ln(F^(g*(f*x+e)))-g*(f*x+e)*ln(F))^2+1/ln(F)/f/g/n*c^2/a*ln((F^(g*(f*x+e)
))^n)-1/ln(F)/f/g/n*c^2/a*ln(a+b*(F^(g*(f*x+e)))^n)+2*c*d/a*x^2-2/ln(F)/f^2/g*d^
2/a*x*e*(ln(F^(g*(f*x+e)))-g*(f*x+e)*ln(F))+1/ln(F)^3/f^3/g^3/n*d^2*(ln(F^(g*(f*
x+e)))-g*(f*x+e)*ln(F))^2/a*ln((F^(g*(f*x+e)))^n)+2/ln(F)/f/g*c*d/a*x*(ln(F^(g*(
f*x+e)))-g*(f*x+e)*ln(F))-1/ln(F)/f^3/g/n*d^2*e^2/a*ln(a+b*(F^(g*(f*x+e)))^n)-2/
ln(F)^2/f^2/g^2/n*c*d*(ln(F^(g*(f*x+e)))-g*(f*x+e)*ln(F))/a*ln((F^(g*(f*x+e)))^n
)+2/ln(F)^2/f^2/g^2/n*c*d*(ln(F^(g*(f*x+e)))-g*(f*x+e)*ln(F))/a*ln(a+b*(F^(g*(f*
x+e)))^n)-2/ln(F)^2/f^2/g^2/n*c*d/a*ln(1+b*(F^(g*(f*x+e)))^n/a)*(ln(F^(g*(f*x+e)
))-g*(f*x+e)*ln(F))-2/ln(F)/f/g*c*d/a*ln(F^(g*(f*x+e)))*x+1/ln(F)/f^3/g/n*d^2*e^
2/a*ln(1+b*(F^(g*(f*x+e)))^n/a)+1/ln(F)^3/f^3/g^3/n*d^2*(ln(F^(g*(f*x+e)))-g*(f*
x+e)*ln(F))^2/a*ln(1+b*(F^(g*(f*x+e)))^n/a)-1/ln(F)/f/g/n*d^2/a*ln(1+b*(F^(g*(f*
x+e)))^n/a)*x^2-2/ln(F)^2/f^2/g^2/n^2*d^2/a*polylog(2,-b*(F^(g*(f*x+e)))^n/a)*x-
1/ln(F)^3/f^3/g^3/n*d^2*(ln(F^(g*(f*x+e)))-g*(f*x+e)*ln(F))^2/a*ln(a+b*(F^(g*(f*
x+e)))^n)-2/ln(F)^2/f^2/g^2/n^2*c*d/a*polylog(2,-b*(F^(g*(f*x+e)))^n/a)+1/ln(F)/
f^3/g/n*d^2*e^2/a*ln((F^(g*(f*x+e)))^n)-2/3/ln(F)^3/f^3/g^3*d^2/a*ln(F^(g*(f*x+e
)))^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ -c^{2}{\left (\frac{\log \left ({\left (F^{f g x + e g}\right )}^{n} b + a\right )}{a f g n \log \left (F\right )} - \frac{\log \left ({\left (F^{f g x + e g}\right )}^{n}\right )}{a f g n \log \left (F\right )}\right )} + \int \frac{d^{2} x^{2} + 2 \, c d x}{{\left (F^{f g x}\right )}^{n}{\left (F^{e g}\right )}^{n} b + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((d*x + c)^2/((F^((f*x + e)*g))^n*b + a),x, algorithm="maxima")

[Out]

-c^2*(log((F^(f*g*x + e*g))^n*b + a)/(a*f*g*n*log(F)) - log((F^(f*g*x + e*g))^n)
/(a*f*g*n*log(F))) + integrate((d^2*x^2 + 2*c*d*x)/((F^(f*g*x))^n*(F^(e*g))^n*b
+ a), x)

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Fricas [A]  time = 0.265419, size = 366, normalized size = 2.52 \[ -\frac{3 \,{\left (d^{2} e^{2} - 2 \, c d e f + c^{2} f^{2}\right )} g^{2} n^{2} \log \left (F^{f g n x + e g n} b + a\right ) \log \left (F\right )^{2} -{\left (d^{2} f^{3} g^{3} n^{3} x^{3} + 3 \, c d f^{3} g^{3} n^{3} x^{2} + 3 \, c^{2} f^{3} g^{3} n^{3} x\right )} \log \left (F\right )^{3} + 3 \,{\left (d^{2} f^{2} g^{2} n^{2} x^{2} + 2 \, c d f^{2} g^{2} n^{2} x -{\left (d^{2} e^{2} - 2 \, c d e f\right )} g^{2} n^{2}\right )} \log \left (F\right )^{2} \log \left (\frac{F^{f g n x + e g n} b + a}{a}\right ) + 6 \,{\left (d^{2} f g n x + c d f g n\right )}{\rm Li}_2\left (-\frac{F^{f g n x + e g n} b + a}{a} + 1\right ) \log \left (F\right ) - 6 \, d^{2}{\rm Li}_{3}(-\frac{F^{f g n x + e g n} b}{a})}{3 \, a f^{3} g^{3} n^{3} \log \left (F\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((d*x + c)^2/((F^((f*x + e)*g))^n*b + a),x, algorithm="fricas")

[Out]

-1/3*(3*(d^2*e^2 - 2*c*d*e*f + c^2*f^2)*g^2*n^2*log(F^(f*g*n*x + e*g*n)*b + a)*l
og(F)^2 - (d^2*f^3*g^3*n^3*x^3 + 3*c*d*f^3*g^3*n^3*x^2 + 3*c^2*f^3*g^3*n^3*x)*lo
g(F)^3 + 3*(d^2*f^2*g^2*n^2*x^2 + 2*c*d*f^2*g^2*n^2*x - (d^2*e^2 - 2*c*d*e*f)*g^
2*n^2)*log(F)^2*log((F^(f*g*n*x + e*g*n)*b + a)/a) + 6*(d^2*f*g*n*x + c*d*f*g*n)
*dilog(-(F^(f*g*n*x + e*g*n)*b + a)/a + 1)*log(F) - 6*d^2*polylog(3, -F^(f*g*n*x
+ e*g*n)*b/a))/(a*f^3*g^3*n^3*log(F)^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((d*x+c)**2/(a+b*(F**(g*(f*x+e)))**n),x)

[Out]

Integral((c + d*x)**2/(a + b*(F**(e*g)*F**(f*g*x))**n), x)

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (d x + c\right )}^{2}}{{\left (F^{{\left (f x + e\right )} g}\right )}^{n} b + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((d*x + c)^2/((F^((f*x + e)*g))^n*b + a),x, algorithm="giac")

[Out]

integrate((d*x + c)^2/((F^((f*x + e)*g))^n*b + a), x)