\(\int \frac {f+g x}{d+e (F^{c (a+b x)})^n} \, dx\) [35]

Optimal result

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

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

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.76 \[ \int \frac {f+g x}{d+e \left (F^{c (a+b x)}\right )^n} \, dx=\frac {-b c n (f+g x) \log (F) \log \left (1+\frac {d \left (F^{c (a+b x)}\right )^{-n}}{e}\right )+g \operatorname {PolyLog}\left (2,-\frac {d \left (F^{c (a+b x)}\right )^{-n}}{e}\right )}{b^2 c^2 d n^2 \log ^2(F)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.72 (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.

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

\(\Big \downarrow \) 2615

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.52 \[ \int \frac {f+g x}{d+e \left (F^{c (a+b x)}\right )^n} \, dx=-\frac {2 \, {\left (b c f - a c g\right )} n \log \left (F^{b c n x + a c n} e + d\right ) \log \left (F\right ) - {\left (b^{2} c^{2} g n^{2} x^{2} + 2 \, b^{2} c^{2} f n^{2} x\right )} \log \left (F\right )^{2} + 2 \, {\left (b c g n x + a c g n\right )} \log \left (F\right ) \log \left (\frac {F^{b c n x + a c n} e + d}{d}\right ) + 2 \, g {\rm Li}_2\left (-\frac {F^{b c n x + a c n} e + d}{d} + 1\right )}{2 \, b^{2} c^{2} d n^{2} \log \left (F\right )^{2}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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