3.6.24 \(\int \frac {x}{a+b f^{c+d x}+c f^{2 c+2 d x}} \, dx\) [524]

3.6.24.1 Optimal result

Integrand size = 27, antiderivative size = 338 \[ \int \frac {x}{a+b f^{c+d x}+c f^{2 c+2 d x}} \, dx=-\frac {c x^2}{b^2-4 a c-b \sqrt {b^2-4 a c}}-\frac {c x^2}{b^2-4 a c+b \sqrt {b^2-4 a c}}-\frac {2 c x \log \left (1+\frac {2 c f^{c+d x}}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right ) d \log (f)}+\frac {2 c x \log \left (1+\frac {2 c f^{c+d x}}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right ) d \log (f)}-\frac {2 c \operatorname {PolyLog}\left (2,-\frac {2 c f^{c+d x}}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right ) d^2 \log ^2(f)}+\frac {2 c \operatorname {PolyLog}\left (2,-\frac {2 c f^{c+d x}}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right ) d^2 \log ^2(f)} \]

output

-2*c*x*ln(1+2*c*f^(d*x+c)/(b-(-4*a*c+b^2)^(1/2)))/d/ln(f)/(b-(-4*a*c+b^2)^ 
(1/2))/(-4*a*c+b^2)^(1/2)-2*c*polylog(2,-2*c*f^(d*x+c)/(b-(-4*a*c+b^2)^(1/ 
2)))/d^2/ln(f)^2/(b-(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2)+2*c*x*ln(1+2*c* 
f^(d*x+c)/(b+(-4*a*c+b^2)^(1/2)))/d/ln(f)/(-4*a*c+b^2)^(1/2)/(b+(-4*a*c+b^ 
2)^(1/2))+2*c*polylog(2,-2*c*f^(d*x+c)/(b+(-4*a*c+b^2)^(1/2)))/d^2/ln(f)^2 
/(-4*a*c+b^2)^(1/2)/(b+(-4*a*c+b^2)^(1/2))-c*x^2/(b^2-4*a*c-b*(-4*a*c+b^2) 
^(1/2))-c*x^2/(b^2-4*a*c+b*(-4*a*c+b^2)^(1/2))
 

3.6.24.2 Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 236, normalized size of antiderivative = 0.70 \[ \int \frac {x}{a+b f^{c+d x}+c f^{2 c+2 d x}} \, dx=\frac {-d x \log (f) \left (\left (b+\sqrt {b^2-4 a c}\right ) \log \left (1+\frac {\left (b-\sqrt {b^2-4 a c}\right ) f^{-c-d x}}{2 c}\right )+\left (-b+\sqrt {b^2-4 a c}\right ) \log \left (1+\frac {\left (b+\sqrt {b^2-4 a c}\right ) f^{-c-d x}}{2 c}\right )\right )+\left (b+\sqrt {b^2-4 a c}\right ) \operatorname {PolyLog}\left (2,\frac {\left (-b+\sqrt {b^2-4 a c}\right ) f^{-c-d x}}{2 c}\right )+\left (-b+\sqrt {b^2-4 a c}\right ) \operatorname {PolyLog}\left (2,-\frac {\left (b+\sqrt {b^2-4 a c}\right ) f^{-c-d x}}{2 c}\right )}{2 a \sqrt {b^2-4 a c} d^2 \log ^2(f)} \]

input

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

output

(-(d*x*Log[f]*((b + Sqrt[b^2 - 4*a*c])*Log[1 + ((b - Sqrt[b^2 - 4*a*c])*f^ 
(-c - d*x))/(2*c)] + (-b + Sqrt[b^2 - 4*a*c])*Log[1 + ((b + Sqrt[b^2 - 4*a 
*c])*f^(-c - d*x))/(2*c)])) + (b + Sqrt[b^2 - 4*a*c])*PolyLog[2, ((-b + Sq 
rt[b^2 - 4*a*c])*f^(-c - d*x))/(2*c)] + (-b + Sqrt[b^2 - 4*a*c])*PolyLog[2 
, -1/2*((b + Sqrt[b^2 - 4*a*c])*f^(-c - d*x))/c])/(2*a*Sqrt[b^2 - 4*a*c]*d 
^2*Log[f]^2)
 

3.6.24.3 Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 299, normalized size of antiderivative = 0.88, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2693, 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.

\(\displaystyle \int \frac {x}{a+b f^{c+d x}+c f^{2 c+2 d x}} \, dx\)

\(\Big \downarrow \) 2693

\(\displaystyle \frac {2 c \int \frac {x}{2 c f^{c+d x}+b-\sqrt {b^2-4 a c}}dx}{\sqrt {b^2-4 a c}}-\frac {2 c \int \frac {x}{2 c f^{c+d x}+b+\sqrt {b^2-4 a c}}dx}{\sqrt {b^2-4 a c}}\)

\(\Big \downarrow \) 2615

\(\displaystyle \frac {2 c \left (\frac {x^2}{2 \left (b-\sqrt {b^2-4 a c}\right )}-\frac {2 c \int \frac {f^{c+d x} x}{2 c f^{c+d x}+b-\sqrt {b^2-4 a c}}dx}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}-\frac {2 c \left (\frac {x^2}{2 \left (\sqrt {b^2-4 a c}+b\right )}-\frac {2 c \int \frac {f^{c+d x} x}{2 c f^{c+d x}+b+\sqrt {b^2-4 a c}}dx}{\sqrt {b^2-4 a c}+b}\right )}{\sqrt {b^2-4 a c}}\)

\(\Big \downarrow \) 2620

\(\displaystyle \frac {2 c \left (\frac {x^2}{2 \left (b-\sqrt {b^2-4 a c}\right )}-\frac {2 c \left (\frac {x \log \left (\frac {2 c f^{c+d x}}{b-\sqrt {b^2-4 a c}}+1\right )}{2 c d \log (f)}-\frac {\int \log \left (\frac {2 c f^{c+d x}}{b-\sqrt {b^2-4 a c}}+1\right )dx}{2 c d \log (f)}\right )}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}-\frac {2 c \left (\frac {x^2}{2 \left (\sqrt {b^2-4 a c}+b\right )}-\frac {2 c \left (\frac {x \log \left (\frac {2 c f^{c+d x}}{\sqrt {b^2-4 a c}+b}+1\right )}{2 c d \log (f)}-\frac {\int \log \left (\frac {2 c f^{c+d x}}{b+\sqrt {b^2-4 a c}}+1\right )dx}{2 c d \log (f)}\right )}{\sqrt {b^2-4 a c}+b}\right )}{\sqrt {b^2-4 a c}}\)

\(\Big \downarrow \) 2715

\(\displaystyle \frac {2 c \left (\frac {x^2}{2 \left (b-\sqrt {b^2-4 a c}\right )}-\frac {2 c \left (\frac {x \log \left (\frac {2 c f^{c+d x}}{b-\sqrt {b^2-4 a c}}+1\right )}{2 c d \log (f)}-\frac {\int f^{-c-d x} \log \left (\frac {2 c f^{c+d x}}{b-\sqrt {b^2-4 a c}}+1\right )df^{c+d x}}{2 c d^2 \log ^2(f)}\right )}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}-\frac {2 c \left (\frac {x^2}{2 \left (\sqrt {b^2-4 a c}+b\right )}-\frac {2 c \left (\frac {x \log \left (\frac {2 c f^{c+d x}}{\sqrt {b^2-4 a c}+b}+1\right )}{2 c d \log (f)}-\frac {\int f^{-c-d x} \log \left (\frac {2 c f^{c+d x}}{b+\sqrt {b^2-4 a c}}+1\right )df^{c+d x}}{2 c d^2 \log ^2(f)}\right )}{\sqrt {b^2-4 a c}+b}\right )}{\sqrt {b^2-4 a c}}\)

\(\Big \downarrow \) 2838

\(\displaystyle \frac {2 c \left (\frac {x^2}{2 \left (b-\sqrt {b^2-4 a c}\right )}-\frac {2 c \left (\frac {\operatorname {PolyLog}\left (2,-\frac {2 c f^{c+d x}}{b-\sqrt {b^2-4 a c}}\right )}{2 c d^2 \log ^2(f)}+\frac {x \log \left (\frac {2 c f^{c+d x}}{b-\sqrt {b^2-4 a c}}+1\right )}{2 c d \log (f)}\right )}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}-\frac {2 c \left (\frac {x^2}{2 \left (\sqrt {b^2-4 a c}+b\right )}-\frac {2 c \left (\frac {\operatorname {PolyLog}\left (2,-\frac {2 c f^{c+d x}}{b+\sqrt {b^2-4 a c}}\right )}{2 c d^2 \log ^2(f)}+\frac {x \log \left (\frac {2 c f^{c+d x}}{\sqrt {b^2-4 a c}+b}+1\right )}{2 c d \log (f)}\right )}{\sqrt {b^2-4 a c}+b}\right )}{\sqrt {b^2-4 a c}}\)

input

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

output

(2*c*(x^2/(2*(b - Sqrt[b^2 - 4*a*c])) - (2*c*((x*Log[1 + (2*c*f^(c + d*x)) 
/(b - Sqrt[b^2 - 4*a*c])])/(2*c*d*Log[f]) + PolyLog[2, (-2*c*f^(c + d*x))/ 
(b - Sqrt[b^2 - 4*a*c])]/(2*c*d^2*Log[f]^2)))/(b - Sqrt[b^2 - 4*a*c])))/Sq 
rt[b^2 - 4*a*c] - (2*c*(x^2/(2*(b + Sqrt[b^2 - 4*a*c])) - (2*c*((x*Log[1 + 
 (2*c*f^(c + d*x))/(b + Sqrt[b^2 - 4*a*c])])/(2*c*d*Log[f]) + PolyLog[2, ( 
-2*c*f^(c + d*x))/(b + Sqrt[b^2 - 4*a*c])]/(2*c*d^2*Log[f]^2)))/(b + Sqrt[ 
b^2 - 4*a*c])))/Sqrt[b^2 - 4*a*c]
 

3.6.24.3.1 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 2693

Int[((f_.) + (g_.)*(x_))^(m_.)/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), 
 x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[2*(c/q)   Int[(f + g*x)^m 
/(b - q + 2*c*F^u), x], x] - Simp[2*(c/q)   Int[(f + g*x)^m/(b + q + 2*c*F^ 
u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] 
 && NeQ[b^2 - 4*a*c, 0] && 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]
 

3.6.24.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(854\) vs. \(2(310)=620\).

Time = 0.09 (sec) , antiderivative size = 855, normalized size of antiderivative = 2.53

method	result	size
risch	\(\frac {x^{2}}{2 a}+\frac {c x}{d a}+\frac {c^{2}}{2 d^{2} a}-\frac {\ln \left (\frac {-2 c \,f^{d x} f^{c}+\sqrt {-4 c a +b^{2}}-b}{-b +\sqrt {-4 c a +b^{2}}}\right ) x}{2 d \ln \left (f \right ) a}-\frac {\ln \left (\frac {-2 c \,f^{d x} f^{c}+\sqrt {-4 c a +b^{2}}-b}{-b +\sqrt {-4 c a +b^{2}}}\right ) c}{2 d^{2} \ln \left (f \right ) a}-\frac {\ln \left (\frac {-2 c \,f^{d x} f^{c}+\sqrt {-4 c a +b^{2}}-b}{-b +\sqrt {-4 c a +b^{2}}}\right ) b x}{2 d \ln \left (f \right ) a \sqrt {-4 c a +b^{2}}}-\frac {\ln \left (\frac {-2 c \,f^{d x} f^{c}+\sqrt {-4 c a +b^{2}}-b}{-b +\sqrt {-4 c a +b^{2}}}\right ) b c}{2 d^{2} \ln \left (f \right ) a \sqrt {-4 c a +b^{2}}}-\frac {\ln \left (\frac {2 c \,f^{d x} f^{c}+\sqrt {-4 c a +b^{2}}+b}{b +\sqrt {-4 c a +b^{2}}}\right ) x}{2 d \ln \left (f \right ) a}-\frac {\ln \left (\frac {2 c \,f^{d x} f^{c}+\sqrt {-4 c a +b^{2}}+b}{b +\sqrt {-4 c a +b^{2}}}\right ) c}{2 d^{2} \ln \left (f \right ) a}+\frac {\ln \left (\frac {2 c \,f^{d x} f^{c}+\sqrt {-4 c a +b^{2}}+b}{b +\sqrt {-4 c a +b^{2}}}\right ) b x}{2 d \ln \left (f \right ) a \sqrt {-4 c a +b^{2}}}+\frac {\ln \left (\frac {2 c \,f^{d x} f^{c}+\sqrt {-4 c a +b^{2}}+b}{b +\sqrt {-4 c a +b^{2}}}\right ) b c}{2 d^{2} \ln \left (f \right ) a \sqrt {-4 c a +b^{2}}}-\frac {\operatorname {dilog}\left (\frac {-2 c \,f^{d x} f^{c}+\sqrt {-4 c a +b^{2}}-b}{-b +\sqrt {-4 c a +b^{2}}}\right )}{2 d^{2} \ln \left (f \right )^{2} a}-\frac {\operatorname {dilog}\left (\frac {-2 c \,f^{d x} f^{c}+\sqrt {-4 c a +b^{2}}-b}{-b +\sqrt {-4 c a +b^{2}}}\right ) b}{2 d^{2} \ln \left (f \right )^{2} a \sqrt {-4 c a +b^{2}}}-\frac {\operatorname {dilog}\left (\frac {2 c \,f^{d x} f^{c}+\sqrt {-4 c a +b^{2}}+b}{b +\sqrt {-4 c a +b^{2}}}\right )}{2 d^{2} \ln \left (f \right )^{2} a}+\frac {\operatorname {dilog}\left (\frac {2 c \,f^{d x} f^{c}+\sqrt {-4 c a +b^{2}}+b}{b +\sqrt {-4 c a +b^{2}}}\right ) b}{2 d^{2} \ln \left (f \right )^{2} a \sqrt {-4 c a +b^{2}}}-\frac {c \ln \left (f^{d x} f^{c}\right )}{d^{2} \ln \left (f \right ) a}+\frac {c \ln \left (a +b \,f^{d x} f^{c}+c \,f^{2 d x} f^{2 c}\right )}{2 d^{2} \ln \left (f \right ) a}+\frac {c b \arctan \left (\frac {2 c \,f^{d x} f^{c}+b}{\sqrt {4 c a -b^{2}}}\right )}{d^{2} \ln \left (f \right ) a \sqrt {4 c a -b^{2}}}\)	\(855\)

input

int(x/(a+b*f^(d*x+c)+c*f^(2*d*x+2*c)),x,method=_RETURNVERBOSE)
 

output

1/2/a*x^2+1/d/a*c*x+1/2/d^2/a*c^2-1/2/d/ln(f)/a*ln((-2*c*f^(d*x)*f^c+(-4*a 
*c+b^2)^(1/2)-b)/(-b+(-4*a*c+b^2)^(1/2)))*x-1/2/d^2/ln(f)/a*ln((-2*c*f^(d* 
x)*f^c+(-4*a*c+b^2)^(1/2)-b)/(-b+(-4*a*c+b^2)^(1/2)))*c-1/2/d/ln(f)/a/(-4* 
a*c+b^2)^(1/2)*ln((-2*c*f^(d*x)*f^c+(-4*a*c+b^2)^(1/2)-b)/(-b+(-4*a*c+b^2) 
^(1/2)))*b*x-1/2/d^2/ln(f)/a/(-4*a*c+b^2)^(1/2)*ln((-2*c*f^(d*x)*f^c+(-4*a 
*c+b^2)^(1/2)-b)/(-b+(-4*a*c+b^2)^(1/2)))*b*c-1/2/d/ln(f)/a*ln((2*c*f^(d*x 
)*f^c+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))*x-1/2/d^2/ln(f)/a*ln(( 
2*c*f^(d*x)*f^c+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))*c+1/2/d/ln(f 
)/a/(-4*a*c+b^2)^(1/2)*ln((2*c*f^(d*x)*f^c+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a* 
c+b^2)^(1/2)))*b*x+1/2/d^2/ln(f)/a/(-4*a*c+b^2)^(1/2)*ln((2*c*f^(d*x)*f^c+ 
(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))*b*c-1/2/d^2/ln(f)^2/a*dilog( 
(-2*c*f^(d*x)*f^c+(-4*a*c+b^2)^(1/2)-b)/(-b+(-4*a*c+b^2)^(1/2)))-1/2/d^2/l 
n(f)^2/a/(-4*a*c+b^2)^(1/2)*dilog((-2*c*f^(d*x)*f^c+(-4*a*c+b^2)^(1/2)-b)/ 
(-b+(-4*a*c+b^2)^(1/2)))*b-1/2/d^2/ln(f)^2/a*dilog((2*c*f^(d*x)*f^c+(-4*a* 
c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))+1/2/d^2/ln(f)^2/a/(-4*a*c+b^2)^(1/ 
2)*dilog((2*c*f^(d*x)*f^c+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))*b- 
1/d^2/ln(f)*c/a*ln(f^(d*x)*f^c)+1/2/d^2/ln(f)*c/a*ln(a+b*f^(d*x)*f^c+c*(f^ 
(d*x))^2*(f^c)^2)+1/d^2/ln(f)*c/a*b/(4*a*c-b^2)^(1/2)*arctan((2*c*f^(d*x)* 
f^c+b)/(4*a*c-b^2)^(1/2))
 

3.6.24.5 Fricas [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 497, normalized size of antiderivative = 1.47 \[ \int \frac {x}{a+b f^{c+d x}+c f^{2 c+2 d x}} \, dx=\frac {{\left (b^{2} - 4 \, a c\right )} d^{2} x^{2} \log \left (f\right )^{2} - {\left (a b \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} + b^{2} - 4 \, a c\right )} {\rm Li}_2\left (-\frac {{\left (a \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} + b\right )} f^{d x + c} + 2 \, a}{2 \, a} + 1\right ) + {\left (a b \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} - b^{2} + 4 \, a c\right )} {\rm Li}_2\left (\frac {{\left (a \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} - b\right )} f^{d x + c} - 2 \, a}{2 \, a} + 1\right ) - {\left (a b c \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} \log \left (f\right ) - {\left (b^{2} c - 4 \, a c^{2}\right )} \log \left (f\right )\right )} \log \left (2 \, c f^{d x + c} + a \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} + b\right ) + {\left (a b c \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} \log \left (f\right ) + {\left (b^{2} c - 4 \, a c^{2}\right )} \log \left (f\right )\right )} \log \left (2 \, c f^{d x + c} - a \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} + b\right ) - {\left ({\left (a b d x + a b c\right )} \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} \log \left (f\right ) + {\left (b^{2} c - 4 \, a c^{2} + {\left (b^{2} - 4 \, a c\right )} d x\right )} \log \left (f\right )\right )} \log \left (\frac {{\left (a \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} + b\right )} f^{d x + c} + 2 \, a}{2 \, a}\right ) + {\left ({\left (a b d x + a b c\right )} \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} \log \left (f\right ) - {\left (b^{2} c - 4 \, a c^{2} + {\left (b^{2} - 4 \, a c\right )} d x\right )} \log \left (f\right )\right )} \log \left (-\frac {{\left (a \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} - b\right )} f^{d x + c} - 2 \, a}{2 \, a}\right )}{2 \, {\left (a b^{2} - 4 \, a^{2} c\right )} d^{2} \log \left (f\right )^{2}} \]

input

integrate(x/(a+b*f^(d*x+c)+c*f^(2*d*x+2*c)),x, algorithm="fricas")
 

output

1/2*((b^2 - 4*a*c)*d^2*x^2*log(f)^2 - (a*b*sqrt((b^2 - 4*a*c)/a^2) + b^2 - 
 4*a*c)*dilog(-1/2*((a*sqrt((b^2 - 4*a*c)/a^2) + b)*f^(d*x + c) + 2*a)/a + 
 1) + (a*b*sqrt((b^2 - 4*a*c)/a^2) - b^2 + 4*a*c)*dilog(1/2*((a*sqrt((b^2 
- 4*a*c)/a^2) - b)*f^(d*x + c) - 2*a)/a + 1) - (a*b*c*sqrt((b^2 - 4*a*c)/a 
^2)*log(f) - (b^2*c - 4*a*c^2)*log(f))*log(2*c*f^(d*x + c) + a*sqrt((b^2 - 
 4*a*c)/a^2) + b) + (a*b*c*sqrt((b^2 - 4*a*c)/a^2)*log(f) + (b^2*c - 4*a*c 
^2)*log(f))*log(2*c*f^(d*x + c) - a*sqrt((b^2 - 4*a*c)/a^2) + b) - ((a*b*d 
*x + a*b*c)*sqrt((b^2 - 4*a*c)/a^2)*log(f) + (b^2*c - 4*a*c^2 + (b^2 - 4*a 
*c)*d*x)*log(f))*log(1/2*((a*sqrt((b^2 - 4*a*c)/a^2) + b)*f^(d*x + c) + 2* 
a)/a) + ((a*b*d*x + a*b*c)*sqrt((b^2 - 4*a*c)/a^2)*log(f) - (b^2*c - 4*a*c 
^2 + (b^2 - 4*a*c)*d*x)*log(f))*log(-1/2*((a*sqrt((b^2 - 4*a*c)/a^2) - b)* 
f^(d*x + c) - 2*a)/a))/((a*b^2 - 4*a^2*c)*d^2*log(f)^2)
 

3.6.24.6 Sympy [F]

\[ \int \frac {x}{a+b f^{c+d x}+c f^{2 c+2 d x}} \, dx=\int \frac {x}{a + b f^{c + d x} + c f^{2 c + 2 d x}}\, dx \]

input

integrate(x/(a+b*f**(d*x+c)+c*f**(2*d*x+2*c)),x)
 

output

Integral(x/(a + b*f**(c + d*x) + c*f**(2*c + 2*d*x)), x)
 

3.6.24.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {x}{a+b f^{c+d x}+c f^{2 c+2 d x}} \, dx=\text {Exception raised: ValueError} \]

input

integrate(x/(a+b*f^(d*x+c)+c*f^(2*d*x+2*c)),x, algorithm="maxima")
 

output

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for 
 more deta
 

3.6.24.8 Giac [F]

\[ \int \frac {x}{a+b f^{c+d x}+c f^{2 c+2 d x}} \, dx=\int { \frac {x}{c f^{2 \, d x + 2 \, c} + b f^{d x + c} + a} \,d x } \]

input

integrate(x/(a+b*f^(d*x+c)+c*f^(2*d*x+2*c)),x, algorithm="giac")
 

output

integrate(x/(c*f^(2*d*x + 2*c) + b*f^(d*x + c) + a), x)
 

3.6.24.9 Mupad [F(-1)]

Timed out. \[ \int \frac {x}{a+b f^{c+d x}+c f^{2 c+2 d x}} \, dx=\int \frac {x}{a+b\,f^{c+d\,x}+c\,f^{2\,c+2\,d\,x}} \,d x \]

input

int(x/(a + b*f^(c + d*x) + c*f^(2*c + 2*d*x)),x)
 

output

int(x/(a + b*f^(c + d*x) + c*f^(2*c + 2*d*x)), x)