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

Optimal. Leaf size=191 \[ -\frac {(c+d x) \log \left (\frac {b \left (F^{g (e+f x)}\right )^n}{a}+1\right )}{a^2 f g n \log (F)}-\frac {d \text {Li}_2\left (-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a^2 f^2 g^2 n^2 \log ^2(F)}+\frac {d \log \left (a+b \left (F^{g (e+f x)}\right )^n\right )}{a^2 f^2 g^2 n^2 \log ^2(F)}+\frac {(c+d x)^2}{2 a^2 d}-\frac {d x}{a^2 f g n \log (F)}+\frac {c+d x}{a f g n \log (F) \left (a+b \left (F^{g (e+f x)}\right )^n\right )} \]

[Out]

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

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Rubi [A]  time = 0.34, antiderivative size = 191, normalized size of antiderivative = 1.00, 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 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]

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 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 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 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 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 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 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 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 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}{\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*}

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Mathematica [F]  time = 1.13, size = 0, normalized size = 0.00 \[ \int \frac {c+d x}{\left (a+b \left (F^{g (e+f x)}\right )^n\right )^2} \, dx \]

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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fricas [B]  time = 0.43, size = 400, normalized size = 2.09 \[ -\frac {2 \, {\left (a d e - a c f\right )} g n \log \relax (F) - {\left (a d f^{2} g^{2} n^{2} x^{2} + 2 \, a c f^{2} g^{2} n^{2} x - {\left (a d e^{2} - 2 \, a c e f\right )} g^{2} n^{2}\right )} \log \relax (F)^{2} - {\left ({\left (b d f^{2} g^{2} n^{2} x^{2} + 2 \, b c f^{2} g^{2} n^{2} x - {\left (b d e^{2} - 2 \, b c e f\right )} g^{2} n^{2}\right )} \log \relax (F)^{2} - 2 \, {\left (b d f g n x + b d e g n\right )} \log \relax (F)\right )} F^{f g n x + e g n} + 2 \, {\left (F^{f g n x + e g n} b d + a d\right )} {\rm Li}_2\left (-\frac {F^{f g n x + e g n} b + a}{a} + 1\right ) - 2 \, {\left ({\left (a d e - a c f\right )} g n \log \relax (F) + {\left ({\left (b d e - b c f\right )} g n \log \relax (F) + b d\right )} F^{f g n x + e g n} + a d\right )} \log \left (F^{f g n x + e g n} b + a\right ) + 2 \, {\left ({\left (b d f g n x + b d e g n\right )} F^{f g n x + e g n} \log \relax (F) + {\left (a d f g n x + a d e g n\right )} \log \relax (F)\right )} \log \left (\frac {F^{f g n x + e g n} b + a}{a}\right )}{2 \, {\left (F^{f g n x + e g n} a^{2} b f^{2} g^{2} n^{2} \log \relax (F)^{2} + a^{3} f^{2} g^{2} n^{2} \log \relax (F)^{2}\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {d x + c}{{\left ({\left (F^{{\left (f x + e\right )} g}\right )}^{n} b + a\right )}^{2}}\,{d x} \]

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)

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maple [B]  time = 0.11, size = 591, normalized size = 3.09 \[ \frac {d x \ln \left (F^{f g n x} F^{-f g n x} \left (F^{\left (f x +e \right ) g}\right )^{n}\right )}{a^{2} f g n \ln \relax (F )}-\frac {d x \ln \left (b \,F^{f g n x} F^{-f g n x} \left (F^{\left (f x +e \right ) g}\right )^{n}+a \right )}{a^{2} f g n \ln \relax (F )}+\frac {c \ln \left (F^{f g n x} F^{-f g n x} \left (F^{\left (f x +e \right ) g}\right )^{n}\right )}{a^{2} f g n \ln \relax (F )}-\frac {c \ln \left (b \,F^{f g n x} F^{-f g n x} \left (F^{\left (f x +e \right ) g}\right )^{n}+a \right )}{a^{2} f g n \ln \relax (F )}+\frac {d x +c}{\left (b \left (F^{\left (f x +e \right ) g}\right )^{n}+a \right ) a f g n \ln \relax (F )}+\frac {d \ln \left (F^{\left (f x +e \right ) g}\right )^{2}}{2 a^{2} f^{2} g^{2} \ln \relax (F )^{2}}-\frac {d \ln \left (F^{\left (f x +e \right ) g}\right ) \ln \left (F^{f g n x} F^{-f g n x} \left (F^{\left (f x +e \right ) g}\right )^{n}\right )}{a^{2} f^{2} g^{2} n \ln \relax (F )^{2}}-\frac {d \ln \left (F^{\left (f x +e \right ) g}\right ) \ln \left (\frac {b \,F^{f g n x} F^{-f g n x} \left (F^{\left (f x +e \right ) g}\right )^{n}}{a}+1\right )}{a^{2} f^{2} g^{2} n \ln \relax (F )^{2}}+\frac {d \ln \left (F^{\left (f x +e \right ) g}\right ) \ln \left (b \,F^{f g n x} F^{-f g n x} \left (F^{\left (f x +e \right ) g}\right )^{n}+a \right )}{a^{2} f^{2} g^{2} n \ln \relax (F )^{2}}-\frac {d \polylog \left (2, -\frac {b \,F^{f g n x} F^{-f g n x} \left (F^{\left (f x +e \right ) g}\right )^{n}}{a}\right )}{a^{2} f^{2} g^{2} n^{2} \ln \relax (F )^{2}}-\frac {d \ln \left (F^{f g n x} F^{-f g n x} \left (F^{\left (f x +e \right ) g}\right )^{n}\right )}{a^{2} f^{2} g^{2} n^{2} \ln \relax (F )^{2}}+\frac {d \ln \left (b \,F^{f g n x} F^{-f g n x} \left (F^{\left (f x +e \right ) g}\right )^{n}+a \right )}{a^{2} f^{2} g^{2} n^{2} \ln \relax (F )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ d {\left (\frac {x}{{\left (F^{f g x}\right )}^{n} {\left (F^{e g}\right )}^{n} a b f g n \log \relax (F) + a^{2} f g n \log \relax (F)} + \int \frac {f g n x \log \relax (F) - 1}{{\left (F^{f g x}\right )}^{n} {\left (F^{e g}\right )}^{n} a b f g n \log \relax (F) + a^{2} f g n \log \relax (F)}\,{d x}\right )} + c {\left (\frac {1}{{\left ({\left (F^{f g x + e g}\right )}^{n} a b n + a^{2} n\right )} f g \log \relax (F)} + \frac {\log \left (F^{f g x + e g}\right )}{a^{2} f g \log \relax (F)} - \frac {\log \left (\frac {{\left (F^{f g x + e g}\right )}^{n} b + a}{b}\right )}{a^{2} f g n \log \relax (F)}\right )} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {c+d\,x}{{\left (a+b\,{\left (F^{g\,\left (e+f\,x\right )}\right )}^n\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {c + d x}{a^{2} f g n \log {\relax (F )} + a b f g n \left (F^{g \left (e + f x\right )}\right )^{n} \log {\relax (F )}} + \frac {\int \left (- \frac {d}{a + b e^{e g n \log {\relax (F )}} e^{f g n x \log {\relax (F )}}}\right )\, dx + \int \frac {c f g n \log {\relax (F )}}{a + b e^{e g n \log {\relax (F )}} e^{f g n x \log {\relax (F )}}}\, dx + \int \frac {d f g n x \log {\relax (F )}}{a + b e^{e g n \log {\relax (F )}} e^{f g n x \log {\relax (F )}}}\, dx}{a f g n \log {\relax (F )}} \]

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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