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

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)*PolyL
og[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*Log[F]^3)

________________________________________________________________________________________

Rubi [A]  time = 0.275688, 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, Rules used = {2184, 2190, 2531, 2282, 6589} \[ -\frac{2 d (c+d x) \text{PolyLog}\left (2,-\frac{b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a f^2 g^2 n^2 \log ^2(F)}+\frac{2 d^2 \text{PolyLog}\left (3,-\frac{b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a f^3 g^3 n^3 \log ^3(F)}-\frac{(c+d x)^2 \log \left (\frac{b \left (F^{g (e+f x)}\right )^n}{a}+1\right )}{a f g n \log (F)}+\frac{(c+d x)^3}{3 a d} \]

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)*PolyL
og[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*Log[F]^3)

Rule 2184

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[(c
+ d*x)^(m + 1)/(a*d*(m + 1)), x] - Dist[b/a, Int[((c + d*x)^m*(F^(g*(e + f*x)))^n)/(a + b*(F^(g*(e + f*x)))^n)
, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2190

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

Mathematica [A]  time = 0.0329647, size = 121, normalized size = 0.83 \[ \frac{2 d f g n \log (F) (c+d x) \text{PolyLog}\left (2,-\frac{a \left (F^{g (e+f x)}\right )^{-n}}{b}\right )+2 d^2 \text{PolyLog}\left (3,-\frac{a \left (F^{g (e+f x)}\right )^{-n}}{b}\right )-f^2 g^2 n^2 \log ^2(F) (c+d x)^2 \log \left (\frac{a \left (F^{g (e+f x)}\right )^{-n}}{b}+1\right )}{a f^3 g^3 n^3 \log ^3(F)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.09, size = 1765, normalized size = 12.2 \begin{align*} \text{result too large to display} \end{align*}

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]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -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} \end{align*}

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, 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 [C]  time = 1.7492, size = 597, normalized size = 4.12 \begin{align*} -\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 polylog}\left (3, -\frac{F^{f g n x + e g n} b}{a}\right )}{3 \, a f^{3} g^{3} n^{3} \log \left (F\right )^{3}} \end{align*}

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, 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)*log(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)*log(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. \begin{align*} \int \frac{\left (c + d x\right )^{2}}{a + b \left (F^{e g} F^{f g x}\right )^{n}}\, dx \end{align*}

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

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, algorithm="giac")

[Out]

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