\(\int \frac {c+d x}{a+b (F^{g (e+f x)})^n} \, dx\) [48]

Optimal result

Integrand size = 23, antiderivative size = 98 \[ \int \frac {c+d x}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=\frac {(c+d x)^2}{2 a d}-\frac {(c+d x) \log \left (1+\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a f g n \log (F)}-\frac {d \operatorname {PolyLog}\left (2,-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a f^2 g^2 n^2 \log ^2(F)} \] Output:

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

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.76 \[ \int \frac {c+d x}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=\frac {-f g n (c+d x) \log (F) \log \left (1+\frac {a \left (F^{g (e+f x)}\right )^{-n}}{b}\right )+d \operatorname {PolyLog}\left (2,-\frac {a \left (F^{g (e+f x)}\right )^{-n}}{b}\right )}{a f^2 g^2 n^2 \log ^2(F)} \] Input:

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

Output:

(-(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)
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2615, 2620, 2715, 2838}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 2615

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

\(\Big \downarrow \) 2620

\(\displaystyle \frac {(c+d x)^2}{2 a d}-\frac {b \left (\frac {(c+d x) \log \left (\frac {b \left (F^{g (e+f x)}\right )^n}{a}+1\right )}{b f g n \log (F)}-\frac {d \int \log \left (\frac {b \left (F^{g (e+f x)}\right )^n}{a}+1\right )dx}{b f g n \log (F)}\right )}{a}\)

\(\Big \downarrow \) 2715

\(\displaystyle \frac {(c+d x)^2}{2 a d}-\frac {b \left (\frac {(c+d x) \log \left (\frac {b \left (F^{g (e+f x)}\right )^n}{a}+1\right )}{b f g n \log (F)}-\frac {d \int \left (F^{g (e+f x)}\right )^{-n} \log \left (\frac {b \left (F^{g (e+f x)}\right )^n}{a}+1\right )d\left (F^{g (e+f x)}\right )^n}{b f^2 g^2 n^2 \log ^2(F)}\right )}{a}\)

\(\Big \downarrow \) 2838

\(\displaystyle \frac {(c+d x)^2}{2 a d}-\frac {b \left (\frac {(c+d x) \log \left (\frac {b \left (F^{g (e+f x)}\right )^n}{a}+1\right )}{b f g n \log (F)}+\frac {d \operatorname {PolyLog}\left (2,-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{b f^2 g^2 n^2 \log ^2(F)}\right )}{a}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 2615

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] - Simp[ 
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 2620

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[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 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[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 2838

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(455\) vs. \(2(96)=192\).

Time = 0.08 (sec) , antiderivative size = 456, normalized size of antiderivative = 4.65

method	result	size
risch	\(\frac {c \ln \left (F^{n g f x} F^{-n g f x} \left (F^{g \left (f x +e \right )}\right )^{n}\right )}{\ln \left (F \right ) f g n a}-\frac {c \ln \left (\left (F^{g \left (f x +e \right )}\right )^{n} F^{-n g f x} F^{n g f x} b +a \right )}{\ln \left (F \right ) f g n a}+\frac {d \ln \left (F^{g \left (f x +e \right )}\right )^{2}}{2 \ln \left (F \right )^{2} f^{2} g^{2} a}-\frac {d \ln \left (F^{g \left (f x +e \right )}\right ) \ln \left (1+\frac {b \,F^{n g f x} F^{-n g f x} \left (F^{g \left (f x +e \right )}\right )^{n}}{a}\right )}{\ln \left (F \right )^{2} f^{2} g^{2} n a}-\frac {d \operatorname {polylog}\left (2, -\frac {b \,F^{n g f x} F^{-n g f x} \left (F^{g \left (f x +e \right )}\right )^{n}}{a}\right )}{\ln \left (F \right )^{2} f^{2} g^{2} n^{2} a}+\frac {d \ln \left (F^{n g f x} F^{-n g f x} \left (F^{g \left (f x +e \right )}\right )^{n}\right ) x}{\ln \left (F \right ) f g n a}-\frac {d \ln \left (F^{n g f x} F^{-n g f x} \left (F^{g \left (f x +e \right )}\right )^{n}\right ) \ln \left (F^{g \left (f x +e \right )}\right )}{\ln \left (F \right )^{2} f^{2} g^{2} n a}-\frac {d \ln \left (\left (F^{g \left (f x +e \right )}\right )^{n} F^{-n g f x} F^{n g f x} b +a \right ) x}{\ln \left (F \right ) f g n a}+\frac {d \ln \left (\left (F^{g \left (f x +e \right )}\right )^{n} F^{-n g f x} F^{n g f x} b +a \right ) \ln \left (F^{g \left (f x +e \right )}\right )}{\ln \left (F \right )^{2} f^{2} g^{2} n a}\)	\(456\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.50 \[ \int \frac {c+d x}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=\frac {2 \, {\left (d e - c f\right )} g n \log \left (F^{f g n x + e g n} b + a\right ) \log \left (F\right ) + {\left (d f^{2} g^{2} n^{2} x^{2} + 2 \, c f^{2} g^{2} n^{2} x\right )} \log \left (F\right )^{2} - 2 \, {\left (d f g n x + d e g n\right )} \log \left (F\right ) \log \left (\frac {F^{f g n x + e g n} b + a}{a}\right ) - 2 \, d {\rm Li}_2\left (-\frac {F^{f g n x + e g n} b + a}{a} + 1\right )}{2 \, a f^{2} g^{2} n^{2} \log \left (F\right )^{2}} \] Input:

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

Output:

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)
 

Sympy [F]

\[ \int \frac {c+d x}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=\int \frac {c + d x}{a + b \left (F^{e g + f g x}\right )^{n}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {c+d x}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=\int { \frac {d x + c}{{\left (F^{{\left (f x + e\right )} g}\right )}^{n} b + a} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {c+d x}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=\int { \frac {d x + c}{{\left (F^{{\left (f x + e\right )} g}\right )}^{n} b + a} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {c+d x}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=\int \frac {c+d\,x}{a+b\,{\left (F^{g\,\left (e+f\,x\right )}\right )}^n} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {c+d x}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=\frac {\left (\int \frac {x}{f^{f g n x +e g n} b +a}d x \right ) \mathrm {log}\left (f \right ) a d f g n -\mathrm {log}\left (f^{f g n x +e g n} b +a \right ) c +\mathrm {log}\left (f \right ) c f g n x}{\mathrm {log}\left (f \right ) a f g n} \] Input:

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

Output:

(int(x/(f**(e*g*n + f*g*n*x)*b + a),x)*log(f)*a*d*f*g*n - log(f**(e*g*n + 
f*g*n*x)*b + a)*c + log(f)*c*f*g*n*x)/(log(f)*a*f*g*n)