3.578 \(\int \frac{(b e-a e e^{c+d x}) x}{b e-2 a e e^{c+d x}-b e e^{2 (c+d x)}} \, dx\)

Optimal. Leaf size=150 \[ -\frac{\text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{2 d^2}-\frac{\text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{2 d^2}-\frac{x \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )}{2 d}-\frac{x \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{2 d}+\frac{x^2}{2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.67244, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 47, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.106, Rules used = {2265, 2184, 2190, 2279, 2391} \[ -\frac{\text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{2 d^2}-\frac{\text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{2 d^2}-\frac{x \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )}{2 d}-\frac{x \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{2 d}+\frac{x^2}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2265

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

Rule 2184

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 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 2279

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 2391

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

Mathematica [B]  time = 0.413307, size = 398, normalized size = 2.65 \[ \frac{\left (\sqrt{a^2+b^2}+a\right ) \text{PolyLog}\left (2,\frac{\left (\sqrt{a^2+b^2}-a\right ) e^{-c-d x}}{b}\right )+\left (\sqrt{a^2+b^2}-a\right ) \text{PolyLog}\left (2,-\frac{\left (\sqrt{a^2+b^2}+a\right ) e^{-c-d x}}{b}\right )+a \text{PolyLog}\left (2,\frac{b e^{c+d x}}{\sqrt{a^2+b^2}-a}\right )-a \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )-a d x \log \left (\frac{\left (a-\sqrt{a^2+b^2}\right ) e^{-c-d x}}{b}+1\right )-d x \sqrt{a^2+b^2} \log \left (\frac{\left (a-\sqrt{a^2+b^2}\right ) e^{-c-d x}}{b}+1\right )+a d x \log \left (\frac{\left (\sqrt{a^2+b^2}+a\right ) e^{-c-d x}}{b}+1\right )-d x \sqrt{a^2+b^2} \log \left (\frac{\left (\sqrt{a^2+b^2}+a\right ) e^{-c-d x}}{b}+1\right )+a d x \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )-a d x \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{2 d^2 \sqrt{a^2+b^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.037, size = 285, normalized size = 1.9 \begin{align*} -{\frac{x}{2\,d}\ln \left ({ \left ({{\rm e}^{2\,c}}{{\rm e}^{dx}}b+{{\rm e}^{c}}a-\sqrt{ \left ({{\rm e}^{c}} \right ) ^{2}{a}^{2}+{{\rm e}^{2\,c}}{b}^{2}} \right ) \left ({{\rm e}^{c}}a-\sqrt{ \left ({{\rm e}^{c}} \right ) ^{2}{a}^{2}+{{\rm e}^{2\,c}}{b}^{2}} \right ) ^{-1}} \right ) }-{\frac{x}{2\,d}\ln \left ({ \left ({{\rm e}^{2\,c}}{{\rm e}^{dx}}b+{{\rm e}^{c}}a+\sqrt{ \left ({{\rm e}^{c}} \right ) ^{2}{a}^{2}+{{\rm e}^{2\,c}}{b}^{2}} \right ) \left ({{\rm e}^{c}}a+\sqrt{ \left ({{\rm e}^{c}} \right ) ^{2}{a}^{2}+{{\rm e}^{2\,c}}{b}^{2}} \right ) ^{-1}} \right ) }-{\frac{1}{2\,{d}^{2}}{\it dilog} \left ({ \left ({{\rm e}^{2\,c}}{{\rm e}^{dx}}b+{{\rm e}^{c}}a+\sqrt{ \left ({{\rm e}^{c}} \right ) ^{2}{a}^{2}+{{\rm e}^{2\,c}}{b}^{2}} \right ) \left ({{\rm e}^{c}}a+\sqrt{ \left ({{\rm e}^{c}} \right ) ^{2}{a}^{2}+{{\rm e}^{2\,c}}{b}^{2}} \right ) ^{-1}} \right ) }-{\frac{1}{2\,{d}^{2}}{\it dilog} \left ({ \left ({{\rm e}^{2\,c}}{{\rm e}^{dx}}b+{{\rm e}^{c}}a-\sqrt{ \left ({{\rm e}^{c}} \right ) ^{2}{a}^{2}+{{\rm e}^{2\,c}}{b}^{2}} \right ) \left ({{\rm e}^{c}}a-\sqrt{ \left ({{\rm e}^{c}} \right ) ^{2}{a}^{2}+{{\rm e}^{2\,c}}{b}^{2}} \right ) ^{-1}} \right ) }+{\frac{{x}^{2}}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*e-a*e*exp(d*x+c))*x/(b*e-2*a*e*exp(d*x+c)-b*e*exp(2*d*x+2*c)),x)

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a e e^{\left (d x + c\right )} - b e\right )} x}{b e e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a e e^{\left (d x + c\right )} - b e}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*e-a*e*exp(d*x+c))*x/(b*e-2*a*e*exp(d*x+c)-b*e*exp(2*d*x+2*c)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.31385, size = 593, normalized size = 3.95 \begin{align*} \frac{d^{2} x^{2} + c \log \left (2 \, b e^{\left (d x + c\right )} + 2 \, b \sqrt{\frac{a^{2} + b^{2}}{b^{2}}} + 2 \, a\right ) + c \log \left (2 \, b e^{\left (d x + c\right )} - 2 \, b \sqrt{\frac{a^{2} + b^{2}}{b^{2}}} + 2 \, a\right ) -{\left (d x + c\right )} \log \left (-\frac{b \sqrt{\frac{a^{2} + b^{2}}{b^{2}}} e^{\left (d x + c\right )} + a e^{\left (d x + c\right )} - b}{b}\right ) -{\left (d x + c\right )} \log \left (\frac{b \sqrt{\frac{a^{2} + b^{2}}{b^{2}}} e^{\left (d x + c\right )} - a e^{\left (d x + c\right )} + b}{b}\right ) -{\rm Li}_2\left (\frac{b \sqrt{\frac{a^{2} + b^{2}}{b^{2}}} e^{\left (d x + c\right )} + a e^{\left (d x + c\right )} - b}{b} + 1\right ) -{\rm Li}_2\left (-\frac{b \sqrt{\frac{a^{2} + b^{2}}{b^{2}}} e^{\left (d x + c\right )} - a e^{\left (d x + c\right )} + b}{b} + 1\right )}{2 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*e-a*e*exp(d*x+c))*x/(b*e-2*a*e*exp(d*x+c)-b*e*exp(2*d*x+2*c)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \left (a e^{c} e^{d x} - b\right )}{2 a e^{c} e^{d x} + b e^{2 c} e^{2 d x} - b}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*e-a*e*exp(d*x+c))*x/(b*e-2*a*e*exp(d*x+c)-b*e*exp(2*d*x+2*c)),x)

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a e e^{\left (d x + c\right )} - b e\right )} x}{b e e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a e e^{\left (d x + c\right )} - b e}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*e-a*e*exp(d*x+c))*x/(b*e-2*a*e*exp(d*x+c)-b*e*exp(2*d*x+2*c)),x, algorithm="giac")

[Out]

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