Integrand size = 18, antiderivative size = 159 \[ \int \frac {x}{a+b e^{-x}+c e^x} \, dx=\frac {x \log \left (1+\frac {2 c e^x}{a-\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c}}-\frac {x \log \left (1+\frac {2 c e^x}{a+\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c}}+\frac {\operatorname {PolyLog}\left (2,-\frac {2 c e^x}{a-\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c}}-\frac {\operatorname {PolyLog}\left (2,-\frac {2 c e^x}{a+\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c}} \]
x*ln(1+2*c*exp(x)/(a-(a^2-4*b*c)^(1/2)))/(a^2-4*b*c)^(1/2)-x*ln(1+2*c*exp( x)/(a+(a^2-4*b*c)^(1/2)))/(a^2-4*b*c)^(1/2)+polylog(2,-2*c*exp(x)/(a-(a^2- 4*b*c)^(1/2)))/(a^2-4*b*c)^(1/2)-polylog(2,-2*c*exp(x)/(a+(a^2-4*b*c)^(1/2 )))/(a^2-4*b*c)^(1/2)
Time = 0.07 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.77 \[ \int \frac {x}{a+b e^{-x}+c e^x} \, dx=\frac {x \left (\log \left (1+\frac {2 c e^x}{a-\sqrt {a^2-4 b c}}\right )-\log \left (1+\frac {2 c e^x}{a+\sqrt {a^2-4 b c}}\right )\right )+\operatorname {PolyLog}\left (2,\frac {2 c e^x}{-a+\sqrt {a^2-4 b c}}\right )-\operatorname {PolyLog}\left (2,-\frac {2 c e^x}{a+\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c}} \]
(x*(Log[1 + (2*c*E^x)/(a - Sqrt[a^2 - 4*b*c])] - Log[1 + (2*c*E^x)/(a + Sq rt[a^2 - 4*b*c])]) + PolyLog[2, (2*c*E^x)/(-a + Sqrt[a^2 - 4*b*c])] - Poly Log[2, (-2*c*E^x)/(a + Sqrt[a^2 - 4*b*c])])/Sqrt[a^2 - 4*b*c]
Time = 0.62 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {2697, 2694, 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.
| \(\displaystyle \int \frac {x}{a+b e^{-x}+c e^x} \, dx\) |
| \(\Big \downarrow \) 2697 |
| \(\displaystyle \int \frac {e^x x}{a e^x+b+c e^{2 x}}dx\) |
| \(\Big \downarrow \) 2694 |
| \(\displaystyle \frac {2 c \int \frac {e^x x}{a+2 c e^x-\sqrt {a^2-4 b c}}dx}{\sqrt {a^2-4 b c}}-\frac {2 c \int \frac {e^x x}{a+2 c e^x+\sqrt {a^2-4 b c}}dx}{\sqrt {a^2-4 b c}}\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle \frac {2 c \left (\frac {x \log \left (\frac {2 c e^x}{a-\sqrt {a^2-4 b c}}+1\right )}{2 c}-\frac {\int \log \left (\frac {2 e^x c}{a-\sqrt {a^2-4 b c}}+1\right )dx}{2 c}\right )}{\sqrt {a^2-4 b c}}-\frac {2 c \left (\frac {x \log \left (\frac {2 c e^x}{\sqrt {a^2-4 b c}+a}+1\right )}{2 c}-\frac {\int \log \left (\frac {2 e^x c}{a+\sqrt {a^2-4 b c}}+1\right )dx}{2 c}\right )}{\sqrt {a^2-4 b c}}\) |
| \(\Big \downarrow \) 2715 |
| \(\displaystyle \frac {2 c \left (\frac {x \log \left (\frac {2 c e^x}{a-\sqrt {a^2-4 b c}}+1\right )}{2 c}-\frac {\int e^{-x} \log \left (\frac {2 e^x c}{a-\sqrt {a^2-4 b c}}+1\right )de^x}{2 c}\right )}{\sqrt {a^2-4 b c}}-\frac {2 c \left (\frac {x \log \left (\frac {2 c e^x}{\sqrt {a^2-4 b c}+a}+1\right )}{2 c}-\frac {\int e^{-x} \log \left (\frac {2 e^x c}{a+\sqrt {a^2-4 b c}}+1\right )de^x}{2 c}\right )}{\sqrt {a^2-4 b c}}\) |
| \(\Big \downarrow \) 2838 |
| \(\displaystyle \frac {2 c \left (\frac {\operatorname {PolyLog}\left (2,-\frac {2 c e^x}{a-\sqrt {a^2-4 b c}}\right )}{2 c}+\frac {x \log \left (\frac {2 c e^x}{a-\sqrt {a^2-4 b c}}+1\right )}{2 c}\right )}{\sqrt {a^2-4 b c}}-\frac {2 c \left (\frac {\operatorname {PolyLog}\left (2,-\frac {2 c e^x}{a+\sqrt {a^2-4 b c}}\right )}{2 c}+\frac {x \log \left (\frac {2 c e^x}{\sqrt {a^2-4 b c}+a}+1\right )}{2 c}\right )}{\sqrt {a^2-4 b c}}\) |
(2*c*((x*Log[1 + (2*c*E^x)/(a - Sqrt[a^2 - 4*b*c])])/(2*c) + PolyLog[2, (- 2*c*E^x)/(a - Sqrt[a^2 - 4*b*c])]/(2*c)))/Sqrt[a^2 - 4*b*c] - (2*c*((x*Log [1 + (2*c*E^x)/(a + Sqrt[a^2 - 4*b*c])])/(2*c) + PolyLog[2, (-2*c*E^x)/(a + Sqrt[a^2 - 4*b*c])]/(2*c)))/Sqrt[a^2 - 4*b*c]
3.6.38.3.1 Defintions of rubi rules used
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]
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]}, Simp[2*(c/q) Int [(f + g*x)^m*(F^u/(b - q + 2*c*F^u)), x], x] - Simp[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]
Int[(u_)/((a_) + (b_.)*(F_)^(v_) + (c_.)*(F_)^(w_)), x_Symbol] :> Int[u*(F^ v/(c + a*F^v + b*F^(2*v))), x] /; FreeQ[{F, a, b, c}, x] && EqQ[w, -v] && L inearQ[v, x] && If[RationalQ[D[v, x]], GtQ[D[v, x], 0], LtQ[LeafCount[v], L eafCount[w]]]
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]
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]
Time = 0.03 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.08
| method | result | size |
| default | \(\frac {x \left (\ln \left (\frac {-2 c \,{\mathrm e}^{x}+\sqrt {a^{2}-4 c b}-a}{-a +\sqrt {a^{2}-4 c b}}\right )-\ln \left (\frac {2 c \,{\mathrm e}^{x}+\sqrt {a^{2}-4 c b}+a}{a +\sqrt {a^{2}-4 c b}}\right )\right )}{\sqrt {a^{2}-4 c b}}+\frac {\operatorname {dilog}\left (\frac {-2 c \,{\mathrm e}^{x}+\sqrt {a^{2}-4 c b}-a}{-a +\sqrt {a^{2}-4 c b}}\right )-\operatorname {dilog}\left (\frac {2 c \,{\mathrm e}^{x}+\sqrt {a^{2}-4 c b}+a}{a +\sqrt {a^{2}-4 c b}}\right )}{\sqrt {a^{2}-4 c b}}\) | \(171\) |
| risch | \(\frac {x \left (\ln \left (\frac {-2 c \,{\mathrm e}^{x}+\sqrt {a^{2}-4 c b}-a}{-a +\sqrt {a^{2}-4 c b}}\right )-\ln \left (\frac {2 c \,{\mathrm e}^{x}+\sqrt {a^{2}-4 c b}+a}{a +\sqrt {a^{2}-4 c b}}\right )\right )}{\sqrt {a^{2}-4 c b}}+\frac {\operatorname {dilog}\left (\frac {-2 c \,{\mathrm e}^{x}+\sqrt {a^{2}-4 c b}-a}{-a +\sqrt {a^{2}-4 c b}}\right )}{\sqrt {a^{2}-4 c b}}-\frac {\operatorname {dilog}\left (\frac {2 c \,{\mathrm e}^{x}+\sqrt {a^{2}-4 c b}+a}{a +\sqrt {a^{2}-4 c b}}\right )}{\sqrt {a^{2}-4 c b}}\) | \(180\) |
x*(ln((-2*c*exp(x)+(a^2-4*b*c)^(1/2)-a)/(-a+(a^2-4*b*c)^(1/2)))-ln((2*c*ex p(x)+(a^2-4*b*c)^(1/2)+a)/(a+(a^2-4*b*c)^(1/2))))/(a^2-4*b*c)^(1/2)+(dilog ((-2*c*exp(x)+(a^2-4*b*c)^(1/2)-a)/(-a+(a^2-4*b*c)^(1/2)))-dilog((2*c*exp( x)+(a^2-4*b*c)^(1/2)+a)/(a+(a^2-4*b*c)^(1/2))))/(a^2-4*b*c)^(1/2)
Time = 0.34 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.35 \[ \int \frac {x}{a+b e^{-x}+c e^x} \, dx=\frac {b x \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} \log \left (\frac {b \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} e^{x} + a e^{x} + 2 \, b}{2 \, b}\right ) - b x \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} \log \left (-\frac {b \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} e^{x} - a e^{x} - 2 \, b}{2 \, b}\right ) + b \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} {\rm Li}_2\left (-\frac {b \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} e^{x} + a e^{x} + 2 \, b}{2 \, b} + 1\right ) - b \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} {\rm Li}_2\left (\frac {b \sqrt {\frac {a^{2} - 4 \, b c}{b^{2}}} e^{x} - a e^{x} - 2 \, b}{2 \, b} + 1\right )}{a^{2} - 4 \, b c} \]
(b*x*sqrt((a^2 - 4*b*c)/b^2)*log(1/2*(b*sqrt((a^2 - 4*b*c)/b^2)*e^x + a*e^ x + 2*b)/b) - b*x*sqrt((a^2 - 4*b*c)/b^2)*log(-1/2*(b*sqrt((a^2 - 4*b*c)/b ^2)*e^x - a*e^x - 2*b)/b) + b*sqrt((a^2 - 4*b*c)/b^2)*dilog(-1/2*(b*sqrt(( a^2 - 4*b*c)/b^2)*e^x + a*e^x + 2*b)/b + 1) - b*sqrt((a^2 - 4*b*c)/b^2)*di log(1/2*(b*sqrt((a^2 - 4*b*c)/b^2)*e^x - a*e^x - 2*b)/b + 1))/(a^2 - 4*b*c )
\[ \int \frac {x}{a+b e^{-x}+c e^x} \, dx=\int \frac {x e^{x}}{a e^{x} + b + c e^{2 x}}\, dx \]
Exception generated. \[ \int \frac {x}{a+b e^{-x}+c e^x} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a^2-4*b*c>0)', see `assume?` for more deta
\[ \int \frac {x}{a+b e^{-x}+c e^x} \, dx=\int { \frac {x}{b e^{\left (-x\right )} + c e^{x} + a} \,d x } \]
Timed out. \[ \int \frac {x}{a+b e^{-x}+c e^x} \, dx=\int \frac {x}{a+c\,{\mathrm {e}}^x+b\,{\mathrm {e}}^{-x}} \,d x \]