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 \[ -\frac {\text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{2 d^2}-\frac {\text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{2 d^2}-\frac {x \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{2 d}-\frac {x \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{2 d}+\frac {x^2}{2} \]

[Out]

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

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Rubi [A]  time = 0.67, antiderivative size = 150, normalized size of antiderivative = 1.00, 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 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 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 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*}

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Mathematica [B]  time = 0.43, size = 398, normalized size = 2.65 \[ \frac {\left (\sqrt {a^2+b^2}+a\right ) \text {Li}_2\left (\frac {\left (\sqrt {a^2+b^2}-a\right ) e^{-c-d x}}{b}\right )+\left (\sqrt {a^2+b^2}-a\right ) \text {Li}_2\left (-\frac {\left (a+\sqrt {a^2+b^2}\right ) e^{-c-d x}}{b}\right )+a \text {Li}_2\left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}-a}\right )-a \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )-a d x \log \left (\frac {\left (a-\sqrt {a^2+b^2}\right ) e^{-c-d x}}{b}+1\right )-d x \sqrt {a^2+b^2} \log \left (\frac {\left (a-\sqrt {a^2+b^2}\right ) e^{-c-d x}}{b}+1\right )+a d x \log \left (\frac {\left (\sqrt {a^2+b^2}+a\right ) e^{-c-d x}}{b}+1\right )-d x \sqrt {a^2+b^2} \log \left (\frac {\left (\sqrt {a^2+b^2}+a\right ) e^{-c-d x}}{b}+1\right )+a d x \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )-a d x \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{2 d^2 \sqrt {a^2+b^2}} \]

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)

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fricas [A]  time = 0.54, size = 251, normalized size = 1.67 \[ \frac {d^{2} x^{2} + c \log \left (2 \, b e^{\left (d x + c\right )} + 2 \, b \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} + 2 \, a\right ) + c \log \left (2 \, b e^{\left (d x + c\right )} - 2 \, b \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} + 2 \, a\right ) - {\left (d x + c\right )} \log \left (-\frac {b \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} e^{\left (d x + c\right )} + a e^{\left (d x + c\right )} - b}{b}\right ) - {\left (d x + c\right )} \log \left (\frac {b \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} e^{\left (d x + c\right )} - a e^{\left (d x + c\right )} + b}{b}\right ) - {\rm Li}_2\left (\frac {b \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} e^{\left (d x + c\right )} + a e^{\left (d x + c\right )} - b}{b} + 1\right ) - {\rm Li}_2\left (-\frac {b \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} e^{\left (d x + c\right )} - a e^{\left (d x + c\right )} + b}{b} + 1\right )}{2 \, d^{2}} \]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a e e^{\left (d x + c\right )} - b e\right )} x}{b e e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a e e^{\left (d x + c\right )} - b e}\,{d x} \]

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)

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maple [B]  time = 0.05, size = 285, normalized size = 1.90 \[ \frac {x^{2}}{2}-\frac {x \ln \left (\frac {b \,{\mathrm e}^{d x} {\mathrm e}^{2 c}+a \,{\mathrm e}^{c}-\sqrt {a^{2} {\mathrm e}^{2 c}+b^{2} {\mathrm e}^{2 c}}}{a \,{\mathrm e}^{c}-\sqrt {a^{2} {\mathrm e}^{2 c}+b^{2} {\mathrm e}^{2 c}}}\right )}{2 d}-\frac {x \ln \left (\frac {b \,{\mathrm e}^{d x} {\mathrm e}^{2 c}+a \,{\mathrm e}^{c}+\sqrt {a^{2} {\mathrm e}^{2 c}+b^{2} {\mathrm e}^{2 c}}}{a \,{\mathrm e}^{c}+\sqrt {a^{2} {\mathrm e}^{2 c}+b^{2} {\mathrm e}^{2 c}}}\right )}{2 d}-\frac {\dilog \left (\frac {b \,{\mathrm e}^{d x} {\mathrm e}^{2 c}+a \,{\mathrm e}^{c}-\sqrt {a^{2} {\mathrm e}^{2 c}+b^{2} {\mathrm e}^{2 c}}}{a \,{\mathrm e}^{c}-\sqrt {a^{2} {\mathrm e}^{2 c}+b^{2} {\mathrm e}^{2 c}}}\right )}{2 d^{2}}-\frac {\dilog \left (\frac {b \,{\mathrm e}^{d x} {\mathrm e}^{2 c}+a \,{\mathrm e}^{c}+\sqrt {a^{2} {\mathrm e}^{2 c}+b^{2} {\mathrm e}^{2 c}}}{a \,{\mathrm e}^{c}+\sqrt {a^{2} {\mathrm e}^{2 c}+b^{2} {\mathrm e}^{2 c}}}\right )}{2 d^{2}} \]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a e e^{\left (d x + c\right )} - b e\right )} x}{b e e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a e e^{\left (d x + c\right )} - b e}\,{d x} \]

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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int -\frac {x\,\left (b\,e-a\,e\,{\mathrm {e}}^{c+d\,x}\right )}{2\,a\,e\,{\mathrm {e}}^{c+d\,x}-b\,e+b\,e\,{\mathrm {e}}^{2\,c+2\,d\,x}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \left (a e^{c} e^{d x} - b\right )}{2 a e^{c} e^{d x} + b e^{2 c} e^{2 d x} - b}\, dx \]

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)

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