3.1.3 \(\int \frac {x}{a+b e^{c+d x}} \, dx\) [3]

Optimal. Leaf size=58 \[ \frac {x^2}{2 a}-\frac {x \log \left (1+\frac {b e^{c+d x}}{a}\right )}{a d}-\frac {\text {Li}_2\left (-\frac {b e^{c+d x}}{a}\right )}{a d^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.07, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2215, 2221, 2317, 2438} \begin {gather*} -\frac {\text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a}\right )}{a d^2}-\frac {x \log \left (\frac {b e^{c+d x}}{a}+1\right )}{a d}+\frac {x^2}{2 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/(a + b*E^(c + d*x)),x]

[Out]

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

Rule 2215

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

________________________________________________________________________________________

Mathematica [A]
time = 0.02, size = 53, normalized size = 0.91 \begin {gather*} -\frac {x \log \left (1+\frac {a e^{-c-d x}}{b}\right )}{a d}+\frac {\text {Li}_2\left (-\frac {a e^{-c-d x}}{b}\right )}{a d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*E^(c + d*x)),x]

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.01, size = 90, normalized size = 1.55

method	result	size
derivativedivides	\(\frac {\frac {\left (d x +c \right )^{2}}{2 a}-\frac {\left (d x +c \right ) \ln \left (1+\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a}-\frac {\polylog \left (2, -\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a}-\frac {c \ln \left ({\mathrm e}^{d x +c}\right )}{a}+\frac {c \ln \left (a +b \,{\mathrm e}^{d x +c}\right )}{a}}{d^{2}}\)	\(90\)
default	\(\frac {\frac {\left (d x +c \right )^{2}}{2 a}-\frac {\left (d x +c \right ) \ln \left (1+\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a}-\frac {\polylog \left (2, -\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a}-\frac {c \ln \left ({\mathrm e}^{d x +c}\right )}{a}+\frac {c \ln \left (a +b \,{\mathrm e}^{d x +c}\right )}{a}}{d^{2}}\)	\(90\)
risch	\(\frac {x^{2}}{2 a}+\frac {c x}{d a}+\frac {c^{2}}{2 d^{2} a}-\frac {x \ln \left (1+\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a d}-\frac {\ln \left (1+\frac {b \,{\mathrm e}^{d x +c}}{a}\right ) c}{d^{2} a}-\frac {\polylog \left (2, -\frac {b \,{\mathrm e}^{d x +c}}{a}\right )}{a \,d^{2}}-\frac {c \ln \left ({\mathrm e}^{d x +c}\right )}{d^{2} a}+\frac {c \ln \left (a +b \,{\mathrm e}^{d x +c}\right )}{d^{2} a}\)	\(133\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.29, size = 48, normalized size = 0.83 \begin {gather*} \frac {x^{2}}{2 \, a} - \frac {d x \log \left (\frac {b e^{\left (d x + c\right )}}{a} + 1\right ) + {\rm Li}_2\left (-\frac {b e^{\left (d x + c\right )}}{a}\right )}{a d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.37, size = 72, normalized size = 1.24 \begin {gather*} \frac {d^{2} x^{2} + 2 \, c \log \left (b e^{\left (d x + c\right )} + a\right ) - 2 \, {\left (d x + c\right )} \log \left (\frac {b e^{\left (d x + c\right )} + a}{a}\right ) - 2 \, {\rm Li}_2\left (-\frac {b e^{\left (d x + c\right )} + a}{a} + 1\right )}{2 \, a d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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(d*x+c)),x, algorithm="giac")

[Out]

integrate(x/(b*e^(d*x + c) + a), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x}{a+b\,{\mathrm {e}}^{c+d\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*exp(c + d*x)),x)

[Out]

int(x/(a + b*exp(c + d*x)), x)

________________________________________________________________________________________