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

Optimal. Leaf size=191 \[ -\frac{d \text{PolyLog}\left (2,-\frac{b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a^2 f^2 g^2 n^2 \log ^2(F)}-\frac{(c+d x) \log \left (\frac{b \left (F^{g (e+f x)}\right )^n}{a}+1\right )}{a^2 f g n \log (F)}+\frac{d \log \left (a+b \left (F^{g (e+f x)}\right )^n\right )}{a^2 f^2 g^2 n^2 \log ^2(F)}+\frac{(c+d x)^2}{2 a^2 d}-\frac{d x}{a^2 f g n \log (F)}+\frac{c+d x}{a f g n \log (F) \left (a+b \left (F^{g (e+f x)}\right )^n\right )} \]

[Out]

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

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Rubi [A]  time = 0.340436, antiderivative size = 191, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 11, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.478, Rules used = {2185, 2184, 2190, 2279, 2391, 2191, 2282, 266, 36, 29, 31} \[ -\frac{d \text{PolyLog}\left (2,-\frac{b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a^2 f^2 g^2 n^2 \log ^2(F)}-\frac{(c+d x) \log \left (\frac{b \left (F^{g (e+f x)}\right )^n}{a}+1\right )}{a^2 f g n \log (F)}+\frac{d \log \left (a+b \left (F^{g (e+f x)}\right )^n\right )}{a^2 f^2 g^2 n^2 \log ^2(F)}+\frac{(c+d x)^2}{2 a^2 d}-\frac{d x}{a^2 f g n \log (F)}+\frac{c+d x}{a f g n \log (F) \left (a+b \left (F^{g (e+f x)}\right )^n\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2185

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

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]

Rule 2191

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

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{c+d x}{\left (a+b \left (F^{g (e+f x)}\right )^n\right )^2} \, dx &=\frac{\int \frac{c+d x}{a+b \left (F^{g (e+f x)}\right )^n} \, dx}{a}-\frac{b \int \frac{\left (F^{g (e+f x)}\right )^n (c+d x)}{\left (a+b \left (F^{g (e+f x)}\right )^n\right )^2} \, dx}{a}\\ &=\frac{(c+d x)^2}{2 a^2 d}+\frac{c+d x}{a f \left (a+b \left (F^{g (e+f x)}\right )^n\right ) g n \log (F)}-\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^2}-\frac{d \int \frac{1}{a+b \left (F^{g (e+f x)}\right )^n} \, dx}{a f g n \log (F)}\\ &=\frac{(c+d x)^2}{2 a^2 d}+\frac{c+d x}{a f \left (a+b \left (F^{g (e+f x)}\right )^n\right ) g n \log (F)}-\frac{(c+d x) \log \left (1+\frac{b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a^2 f g n \log (F)}-\frac{d \operatorname{Subst}\left (\int \frac{1}{x \left (a+b x^n\right )} \, dx,x,F^{g (e+f x)}\right )}{a f^2 g^2 n \log ^2(F)}+\frac{d \int \log \left (1+\frac{b \left (F^{g (e+f x)}\right )^n}{a}\right ) \, dx}{a^2 f g n \log (F)}\\ &=\frac{(c+d x)^2}{2 a^2 d}+\frac{c+d x}{a f \left (a+b \left (F^{g (e+f x)}\right )^n\right ) g n \log (F)}-\frac{(c+d x) \log \left (1+\frac{b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a^2 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^2 f^2 g^2 n^2 \log ^2(F)}-\frac{d \operatorname{Subst}\left (\int \frac{1}{x (a+b 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^2 d}+\frac{c+d x}{a f \left (a+b \left (F^{g (e+f x)}\right )^n\right ) g n \log (F)}-\frac{(c+d x) \log \left (1+\frac{b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a^2 f g n \log (F)}-\frac{d \text{Li}_2\left (-\frac{b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a^2 f^2 g^2 n^2 \log ^2(F)}-\frac{d \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,\left (F^{g (e+f x)}\right )^n\right )}{a^2 f^2 g^2 n^2 \log ^2(F)}+\frac{(b d) \operatorname{Subst}\left (\int \frac{1}{a+b x} \, dx,x,\left (F^{g (e+f x)}\right )^n\right )}{a^2 f^2 g^2 n^2 \log ^2(F)}\\ &=\frac{(c+d x)^2}{2 a^2 d}-\frac{d x}{a^2 f g n \log (F)}+\frac{c+d x}{a f \left (a+b \left (F^{g (e+f x)}\right )^n\right ) g n \log (F)}+\frac{d \log \left (a+b \left (F^{g (e+f x)}\right )^n\right )}{a^2 f^2 g^2 n^2 \log ^2(F)}-\frac{(c+d x) \log \left (1+\frac{b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a^2 f g n \log (F)}-\frac{d \text{Li}_2\left (-\frac{b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a^2 f^2 g^2 n^2 \log ^2(F)}\\ \end{align*}

Mathematica [F]  time = 0.998373, size = 0, normalized size = 0. \[ \int \frac{c+d x}{\left (a+b \left (F^{g (e+f x)}\right )^n\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [B]  time = 0.089, size = 631, normalized size = 3.3 \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)^2,x)

[Out]

(d*x+c)/a/f/(a+b*(F^(g*(f*x+e)))^n)/g/n/ln(F)+1/2/ln(F)^2/a^2/f^2/g^2*d*ln(F^(g*(f*x+e)))^2-1/ln(F)^2/a^2/f^2/
g^2/n*d*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)))+1/ln(F)/a^2/f/g/n*d*ln
(F^(n*g*f*x)*exp(-n*(ln(F)*f*g*x-ln(F^(g*(f*x+e))))))*x-1/ln(F)^2/a^2/f^2/g^2/n*d*ln(F^(n*g*f*x)*exp(-n*(ln(F)
*f*g*x-ln(F^(g*(f*x+e))))))*ln(F^(g*(f*x+e)))-1/ln(F)/a^2/f/g/n*d*ln(a+b*F^(n*g*f*x)*exp(-n*(ln(F)*f*g*x-ln(F^
(g*(f*x+e))))))*x+1/ln(F)^2/a^2/f^2/g^2/n*d*ln(a+b*F^(n*g*f*x)*exp(-n*(ln(F)*f*g*x-ln(F^(g*(f*x+e))))))*ln(F^(
g*(f*x+e)))-1/ln(F)^2/a^2/f^2/g^2/n^2*d*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)^2/a^2/f^2/g^2/n^2*d*ln(F^(n*g*f*x)*exp(-n*(ln(F)*f*g*x-ln(F^(g*(f*x+e))))))+1/ln(F)^2/a^2/f^2/g^2/n^2*d*
ln(a+b*F^(n*g*f*x)*exp(-n*(ln(F)*f*g*x-ln(F^(g*(f*x+e))))))+1/ln(F)/a^2/f/g/n*c*ln(F^(n*g*f*x)*exp(-n*(ln(F)*f
*g*x-ln(F^(g*(f*x+e))))))-1/ln(F)/a^2/f/g/n*c*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*} d{\left (\frac{x}{{\left (F^{f g x}\right )}^{n}{\left (F^{e g}\right )}^{n} a b f g n \log \left (F\right ) + a^{2} f g n \log \left (F\right )} + \int \frac{f g n x \log \left (F\right ) - 1}{{\left (F^{f g x}\right )}^{n}{\left (F^{e g}\right )}^{n} a b f g n \log \left (F\right ) + a^{2} f g n \log \left (F\right )}\,{d x}\right )} + c{\left (\frac{1}{{\left ({\left (F^{f g x + e g}\right )}^{n} a b n + a^{2} n\right )} f g \log \left (F\right )} + \frac{\log \left (F^{f g x + e g}\right )}{a^{2} f g \log \left (F\right )} - \frac{\log \left (\frac{{\left (F^{f g x + e g}\right )}^{n} b + a}{b}\right )}{a^{2} f g n \log \left (F\right )}\right )} \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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.59507, size = 918, normalized size = 4.81 \begin{align*} -\frac{2 \,{\left (a d e - a c f\right )} g n \log \left (F\right ) -{\left (a d f^{2} g^{2} n^{2} x^{2} + 2 \, a c f^{2} g^{2} n^{2} x -{\left (a d e^{2} - 2 \, a c e f\right )} g^{2} n^{2}\right )} \log \left (F\right )^{2} -{\left ({\left (b d f^{2} g^{2} n^{2} x^{2} + 2 \, b c f^{2} g^{2} n^{2} x -{\left (b d e^{2} - 2 \, b c e f\right )} g^{2} n^{2}\right )} \log \left (F\right )^{2} - 2 \,{\left (b d f g n x + b d e g n\right )} \log \left (F\right )\right )} F^{f g n x + e g n} + 2 \,{\left (F^{f g n x + e g n} b d + a d\right )}{\rm Li}_2\left (-\frac{F^{f g n x + e g n} b + a}{a} + 1\right ) - 2 \,{\left ({\left (a d e - a c f\right )} g n \log \left (F\right ) +{\left ({\left (b d e - b c f\right )} g n \log \left (F\right ) + b d\right )} F^{f g n x + e g n} + a d\right )} \log \left (F^{f g n x + e g n} b + a\right ) + 2 \,{\left ({\left (b d f g n x + b d e g n\right )} F^{f g n x + e g n} \log \left (F\right ) +{\left (a d f g n x + a d e g n\right )} \log \left (F\right )\right )} \log \left (\frac{F^{f g n x + e g n} b + a}{a}\right )}{2 \,{\left (F^{f g n x + e g n} a^{2} b f^{2} g^{2} n^{2} \log \left (F\right )^{2} + a^{3} f^{2} g^{2} n^{2} \log \left (F\right )^{2}\right )}} \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)^2,x, algorithm="fricas")

[Out]

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

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

[Out]

(c + d*x)/(a**2*f*g*n*log(F) + a*b*f*g*n*(F**(g*(e + f*x)))**n*log(F)) + (Integral(-d/(a + b*exp(e*g*n*log(F))
*exp(f*g*n*x*log(F))), x) + Integral(c*f*g*n*log(F)/(a + b*exp(e*g*n*log(F))*exp(f*g*n*x*log(F))), x) + Integr
al(d*f*g*n*x*log(F)/(a + b*exp(e*g*n*log(F))*exp(f*g*n*x*log(F))), x))/(a*f*g*n*log(F))

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

[Out]

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