3.6.38 \(\int \frac {x}{a+b e^{-x}+c e^x} \, dx\) [538]

Optimal. Leaf size=159 \[ \frac {x \log \left (1+\frac {2 c e^x}{a-\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c}}-\frac {x \log \left (1+\frac {2 c e^x}{a+\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c}}+\frac {\text {Li}_2\left (-\frac {2 c e^x}{a-\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c}}-\frac {\text {Li}_2\left (-\frac {2 c e^x}{a+\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c}} \]

[Out]

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)

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Rubi [A]
time = 0.21, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {2299, 2296, 2221, 2317, 2438} \begin {gather*} \frac {\text {PolyLog}\left (2,-\frac {2 c e^x}{a-\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c}}-\frac {\text {PolyLog}\left (2,-\frac {2 c e^x}{\sqrt {a^2-4 b c}+a}\right )}{\sqrt {a^2-4 b c}}+\frac {x \log \left (\frac {2 c e^x}{a-\sqrt {a^2-4 b c}}+1\right )}{\sqrt {a^2-4 b c}}-\frac {x \log \left (\frac {2 c e^x}{\sqrt {a^2-4 b c}+a}+1\right )}{\sqrt {a^2-4 b c}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/(a + b/E^x + c*E^x),x]

[Out]

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

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

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Mathematica [A]
time = 0.08, size = 123, normalized size = 0.77 \begin {gather*} \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 )+\text {Li}_2\left (\frac {2 c e^x}{-a+\sqrt {a^2-4 b c}}\right )-\text {Li}_2\left (-\frac {2 c e^x}{a+\sqrt {a^2-4 b c}}\right )}{\sqrt {a^2-4 b c}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b/E^x + c*E^x),x]

[Out]

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

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Maple [A]
time = 0.02, size = 171, normalized size = 1.08

method	result	size
default	\(\frac {x \left (\ln \left (\frac {-2 c \,{\mathrm e}^{x}+\sqrt {a^{2}-4 c b}-a}{-a +\sqrt {a^{2}-4 c b}}\right )-\ln \left (\frac {2 c \,{\mathrm e}^{x}+\sqrt {a^{2}-4 c b}+a}{a +\sqrt {a^{2}-4 c b}}\right )\right )}{\sqrt {a^{2}-4 c b}}+\frac {\dilog \left (\frac {-2 c \,{\mathrm e}^{x}+\sqrt {a^{2}-4 c b}-a}{-a +\sqrt {a^{2}-4 c b}}\right )-\dilog \left (\frac {2 c \,{\mathrm e}^{x}+\sqrt {a^{2}-4 c b}+a}{a +\sqrt {a^{2}-4 c b}}\right )}{\sqrt {a^{2}-4 c b}}\)	\(171\)
risch	\(\frac {x \left (\ln \left (\frac {-2 c \,{\mathrm e}^{x}+\sqrt {a^{2}-4 c b}-a}{-a +\sqrt {a^{2}-4 c b}}\right )-\ln \left (\frac {2 c \,{\mathrm e}^{x}+\sqrt {a^{2}-4 c b}+a}{a +\sqrt {a^{2}-4 c b}}\right )\right )}{\sqrt {a^{2}-4 c b}}+\frac {\dilog \left (\frac {-2 c \,{\mathrm e}^{x}+\sqrt {a^{2}-4 c b}-a}{-a +\sqrt {a^{2}-4 c b}}\right )}{\sqrt {a^{2}-4 c b}}-\frac {\dilog \left (\frac {2 c \,{\mathrm e}^{x}+\sqrt {a^{2}-4 c b}+a}{a +\sqrt {a^{2}-4 c b}}\right )}{\sqrt {a^{2}-4 c b}}\)	\(180\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a+b/exp(x)+c*exp(x)),x,method=_RETURNVERBOSE)

[Out]

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*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)+(dilog((-2*c*exp(x)+(a^2-4*b*c)^(1/2)-a)/(-a+(a^2-4*b*c)^(1/2)))-dilog((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)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b/exp(x)+c*exp(x)),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

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Fricas [A]
time = 0.36, size = 214, normalized size = 1.35 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b/exp(x)+c*exp(x)),x, algorithm="fricas")

[Out]

(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)*dilog(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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x e^{x}}{a e^{x} + b + c e^{2 x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b/exp(x)+c*exp(x)),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b/exp(x)+c*exp(x)),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x}{a+c\,{\mathrm {e}}^x+b\,{\mathrm {e}}^{-x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + c*exp(x) + b*exp(-x)),x)

[Out]

int(x/(a + c*exp(x) + b*exp(-x)), x)

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