\(\int \frac {x}{a+b f^{-c-d x}+c f^{c+d x}} \, dx\) [541]

Optimal result

Integrand size = 27, antiderivative size = 203 \[ \int \frac {x}{a+b f^{-c-d x}+c f^{c+d x}} \, dx=\frac {x \log \left (1+\frac {2 c f^{c+d x}}{a-\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d \log (f)}-\frac {x \log \left (1+\frac {2 c f^{c+d x}}{a+\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d \log (f)}+\frac {\operatorname {PolyLog}\left (2,-\frac {2 c f^{c+d x}}{a-\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d^2 \log ^2(f)}-\frac {\operatorname {PolyLog}\left (2,-\frac {2 c f^{c+d x}}{a+\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d^2 \log ^2(f)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2299, 2296, 2221, 2317, 2438} \[ \int \frac {x}{a+b f^{-c-d x}+c f^{c+d x}} \, dx=\frac {\operatorname {PolyLog}\left (2,-\frac {2 c f^{c+d x}}{a-\sqrt {a^2-4 b c}}\right )}{d^2 \log ^2(f) \sqrt {a^2-4 b c}}-\frac {\operatorname {PolyLog}\left (2,-\frac {2 c f^{c+d x}}{a+\sqrt {a^2-4 b c}}\right )}{d^2 \log ^2(f) \sqrt {a^2-4 b c}}+\frac {x \log \left (\frac {2 c f^{c+d x}}{a-\sqrt {a^2-4 b c}}+1\right )}{d \log (f) \sqrt {a^2-4 b c}}-\frac {x \log \left (\frac {2 c f^{c+d x}}{\sqrt {a^2-4 b c}+a}+1\right )}{d \log (f) \sqrt {a^2-4 b c}} \]

[In]

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

[Out]

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

Rule 2221

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]
- 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 2296

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[2*(c/q), Int[(f + g*x)^m*(F^u/(b - q + 2*c*F^u)), x], x] - Dist[2*(c/q), Int[(f + g
*x)^m*(F^u/(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 2299

Int[(u_)/((a_) + (b_.)*(F_)^(v_) + (c_.)*(F_)^(w_)), x_Symbol] :> Int[u*(F^v/(c + a*F^v + b*F^(2*v))), x] /; F
reeQ[{F, a, b, c}, x] && EqQ[w, -v] && LinearQ[v, x] && If[RationalQ[Coefficient[v, x, 1]], GtQ[Coefficient[v,
x, 1], 0], LtQ[LeafCount[v], LeafCount[w]]]

Rule 2317

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 2438

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*} \text {integral}& = \int \frac {f^{c+d x} x}{b+a f^{c+d x}+c f^{2 (c+d x)}} \, dx \\ & = \frac {(2 c) \int \frac {f^{c+d x} x}{a-\sqrt {a^2-4 b c}+2 c f^{c+d x}} \, dx}{\sqrt {a^2-4 b c}}-\frac {(2 c) \int \frac {f^{c+d x} x}{a+\sqrt {a^2-4 b c}+2 c f^{c+d x}} \, dx}{\sqrt {a^2-4 b c}} \\ & = \frac {x \log \left (1+\frac {2 c f^{c+d x}}{a-\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d \log (f)}-\frac {x \log \left (1+\frac {2 c f^{c+d x}}{a+\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d \log (f)}-\frac {\int \log \left (1+\frac {2 c f^{c+d x}}{a-\sqrt {a^2-4 b c}}\right ) \, dx}{\sqrt {a^2-4 b c} d \log (f)}+\frac {\int \log \left (1+\frac {2 c f^{c+d x}}{a+\sqrt {a^2-4 b c}}\right ) \, dx}{\sqrt {a^2-4 b c} d \log (f)} \\ & = \frac {x \log \left (1+\frac {2 c f^{c+d x}}{a-\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d \log (f)}-\frac {x \log \left (1+\frac {2 c f^{c+d x}}{a+\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d \log (f)}-\frac {\text {Subst}\left (\int \frac {\log \left (1+\frac {2 c x}{a-\sqrt {a^2-4 b c}}\right )}{x} \, dx,x,f^{c+d x}\right )}{\sqrt {a^2-4 b c} d^2 \log ^2(f)}+\frac {\text {Subst}\left (\int \frac {\log \left (1+\frac {2 c x}{a+\sqrt {a^2-4 b c}}\right )}{x} \, dx,x,f^{c+d x}\right )}{\sqrt {a^2-4 b c} d^2 \log ^2(f)} \\ & = \frac {x \log \left (1+\frac {2 c f^{c+d x}}{a-\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d \log (f)}-\frac {x \log \left (1+\frac {2 c f^{c+d x}}{a+\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d \log (f)}+\frac {\text {Li}_2\left (-\frac {2 c f^{c+d x}}{a-\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d^2 \log ^2(f)}-\frac {\text {Li}_2\left (-\frac {2 c f^{c+d x}}{a+\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d^2 \log ^2(f)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.73 \[ \int \frac {x}{a+b f^{-c-d x}+c f^{c+d x}} \, dx=\frac {d x \log (f) \left (\log \left (1+\frac {2 c f^{c+d x}}{a-\sqrt {a^2-4 b c}}\right )-\log \left (1+\frac {2 c f^{c+d x}}{a+\sqrt {a^2-4 b c}}\right )\right )+\operatorname {PolyLog}\left (2,\frac {2 c f^{c+d x}}{-a+\sqrt {a^2-4 b c}}\right )-\operatorname {PolyLog}\left (2,-\frac {2 c f^{c+d x}}{a+\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c} d^2 \log ^2(f)} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(432\) vs. \(2(187)=374\).

Time = 0.09 (sec) , antiderivative size = 433, normalized size of antiderivative = 2.13

method	result	size
risch	\(-\frac {\ln \left (\frac {-2 b \,f^{-d x} f^{-c}+\sqrt {a^{2}-4 c b}-a}{-a +\sqrt {a^{2}-4 c b}}\right ) x}{d \ln \left (f \right ) \sqrt {a^{2}-4 c b}}+\frac {\ln \left (\frac {2 b \,f^{-d x} f^{-c}+\sqrt {a^{2}-4 c b}+a}{a +\sqrt {a^{2}-4 c b}}\right ) x}{d \ln \left (f \right ) \sqrt {a^{2}-4 c b}}-\frac {\ln \left (\frac {-2 b \,f^{-d x} f^{-c}+\sqrt {a^{2}-4 c b}-a}{-a +\sqrt {a^{2}-4 c b}}\right ) c}{d^{2} \ln \left (f \right ) \sqrt {a^{2}-4 c b}}+\frac {\ln \left (\frac {2 b \,f^{-d x} f^{-c}+\sqrt {a^{2}-4 c b}+a}{a +\sqrt {a^{2}-4 c b}}\right ) c}{d^{2} \ln \left (f \right ) \sqrt {a^{2}-4 c b}}+\frac {\operatorname {dilog}\left (\frac {-2 b \,f^{-d x} f^{-c}+\sqrt {a^{2}-4 c b}-a}{-a +\sqrt {a^{2}-4 c b}}\right )}{d^{2} \ln \left (f \right )^{2} \sqrt {a^{2}-4 c b}}-\frac {\operatorname {dilog}\left (\frac {2 b \,f^{-d x} f^{-c}+\sqrt {a^{2}-4 c b}+a}{a +\sqrt {a^{2}-4 c b}}\right )}{d^{2} \ln \left (f \right )^{2} \sqrt {a^{2}-4 c b}}+\frac {2 c \arctan \left (\frac {2 b \,f^{-d x} f^{-c}+a}{\sqrt {-a^{2}+4 c b}}\right )}{d^{2} \ln \left (f \right ) \sqrt {-a^{2}+4 c b}}\)	\(433\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.74 \[ \int \frac {x}{a+b f^{-c-d x}+c f^{c+d x}} \, dx=\frac {b c \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} \log \left (2 \, c f^{d x + c} + b \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} + a\right ) \log \left (f\right ) - b c \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} \log \left (2 \, c f^{d x + c} - b \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} + a\right ) \log \left (f\right ) + {\left (b d x + b c\right )} \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} \log \left (f\right ) \log \left (\frac {{\left (b \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} + a\right )} f^{d x + c} + 2 \, b}{2 \, b}\right ) - {\left (b d x + b c\right )} \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} \log \left (f\right ) \log \left (-\frac {{\left (b \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} - a\right )} f^{d x + c} - 2 \, b}{2 \, b}\right ) + b \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} {\rm Li}_2\left (-\frac {{\left (b \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} + a\right )} f^{d x + c} + 2 \, b}{2 \, b} + 1\right ) - b \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} {\rm Li}_2\left (\frac {{\left (b \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} - a\right )} f^{d x + c} - 2 \, b}{2 \, b} + 1\right )}{{\left (a^{2} - 4 \, b c\right )} d^{2} \log \left (f\right )^{2}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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