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

Optimal. Leaf size=98 \[ -\frac{d \text{PolyLog}\left (2,-\frac{b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a f^2 g^2 n^2 \log ^2(F)}-\frac{(c+d x) \log \left (\frac{b \left (F^{g (e+f x)}\right )^n}{a}+1\right )}{a f g n \log (F)}+\frac{(c+d x)^2}{2 a d} \]

[Out]

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

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Rubi [A]  time = 0.156513, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2184, 2190, 2279, 2391} \[ -\frac{d \text{PolyLog}\left (2,-\frac{b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a f^2 g^2 n^2 \log ^2(F)}-\frac{(c+d x) \log \left (\frac{b \left (F^{g (e+f x)}\right )^n}{a}+1\right )}{a f g n \log (F)}+\frac{(c+d x)^2}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \frac{c+d x}{a+b \left (F^{g (e+f x)}\right )^n} \, dx &=\frac{(c+d x)^2}{2 a d}-\frac{b \int \frac{\left (F^{g (e+f x)}\right )^n (c+d x)}{a+b \left (F^{g (e+f x)}\right )^n} \, dx}{a}\\ &=\frac{(c+d x)^2}{2 a d}-\frac{(c+d x) \log \left (1+\frac{b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a f g n \log (F)}+\frac{d \int \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)^2}{2 a d}-\frac{(c+d x) \log \left (1+\frac{b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a f g n \log (F)}+\frac{d \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{b x}{a}\right )}{x} \, dx,x,\left (F^{g (e+f x)}\right )^n\right )}{a f^2 g^2 n^2 \log ^2(F)}\\ &=\frac{(c+d x)^2}{2 a d}-\frac{(c+d x) \log \left (1+\frac{b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a f g n \log (F)}-\frac{d \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)}\\ \end{align*}

Mathematica [A]  time = 0.0105853, size = 74, normalized size = 0.76 \[ \frac{d \text{PolyLog}\left (2,-\frac{a \left (F^{g (e+f x)}\right )^{-n}}{b}\right )-f g n \log (F) (c+d x) \log \left (\frac{a \left (F^{g (e+f x)}\right )^{-n}}{b}+1\right )}{a f^2 g^2 n^2 \log ^2(F)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.06, size = 719, normalized size = 7.3 \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)/(a+b*(F^(g*(f*x+e)))^n),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -c{\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 )} + d \int \frac{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)/(a+b*(F^(g*(f*x+e)))^n),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.68967, size = 344, normalized size = 3.51 \begin{align*} \frac{2 \,{\left (d e - c f\right )} g n \log \left (F^{f g n x + e g n} b + a\right ) \log \left (F\right ) +{\left (d f^{2} g^{2} n^{2} x^{2} + 2 \, c f^{2} g^{2} n^{2} x\right )} \log \left (F\right )^{2} - 2 \,{\left (d f g n x + d e g n\right )} \log \left (F\right ) \log \left (\frac{F^{f g n x + e g n} b + a}{a}\right ) - 2 \, d{\rm Li}_2\left (-\frac{F^{f g n x + e g n} b + a}{a} + 1\right )}{2 \, a f^{2} g^{2} n^{2} \log \left (F\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(d*e - c*f)*g*n*log(F^(f*g*n*x + e*g*n)*b + a)*log(F) + (d*f^2*g^2*n^2*x^2 + 2*c*f^2*g^2*n^2*x)*log(F)^
2 - 2*(d*f*g*n*x + d*e*g*n)*log(F)*log((F^(f*g*n*x + e*g*n)*b + a)/a) - 2*d*dilog(-(F^(f*g*n*x + e*g*n)*b + a)
/a + 1))/(a*f^2*g^2*n^2*log(F)^2)

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

[Out]

Integral((c + d*x)/(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{d x + c}{{\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)/(a+b*(F^(g*(f*x+e)))^n),x, algorithm="giac")

[Out]

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