3.20 \(\int \frac {e^{2 x}}{(a+b e^x)^3} \, dx\)

Optimal. Leaf size=21 \[ \frac {e^{2 x}}{2 a \left (a+b e^x\right )^2} \]

[Out]

1/2*exp(2*x)/a/(a+b*exp(x))^2

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Rubi [A]  time = 0.02, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2248, 37} \[ \frac {e^{2 x}}{2 a \left (a+b e^x\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*x)/(a + b*E^x)^3,x]

[Out]

E^(2*x)/(2*a*(a + b*E^x)^2)

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 2248

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_.)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Wit
h[{m = FullSimplify[(g*h*Log[G])/(d*e*Log[F])]}, Dist[(Denominator[m]*G^(f*h - (c*g*h)/d))/(d*e*Log[F]), Subst
[Int[x^(Numerator[m] - 1)*(a + b*x^Denominator[m])^p, x], x, F^((e*(c + d*x))/Denominator[m])], x] /; LeQ[m, -
1] || GeQ[m, 1]] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x]

Rubi steps

\begin {align*} \int \frac {e^{2 x}}{\left (a+b e^x\right )^3} \, dx &=\operatorname {Subst}\left (\int \frac {x}{(a+b x)^3} \, dx,x,e^x\right )\\ &=\frac {e^{2 x}}{2 a \left (a+b e^x\right )^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 21, normalized size = 1.00 \[ \frac {e^{2 x}}{2 a \left (a+b e^x\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*x)/(a + b*E^x)^3,x]

[Out]

E^(2*x)/(2*a*(a + b*E^x)^2)

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fricas [B]  time = 0.39, size = 35, normalized size = 1.67 \[ -\frac {2 \, b e^{x} + a}{2 \, {\left (b^{4} e^{\left (2 \, x\right )} + 2 \, a b^{3} e^{x} + a^{2} b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*x)/(a+b*exp(x))^3,x, algorithm="fricas")

[Out]

-1/2*(2*b*e^x + a)/(b^4*e^(2*x) + 2*a*b^3*e^x + a^2*b^2)

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giac [A]  time = 0.40, size = 20, normalized size = 0.95 \[ -\frac {2 \, b e^{x} + a}{2 \, {\left (b e^{x} + a\right )}^{2} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*x)/(a+b*exp(x))^3,x, algorithm="giac")

[Out]

-1/2*(2*b*e^x + a)/((b*e^x + a)^2*b^2)

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maple [A]  time = 0.02, size = 29, normalized size = 1.38 \[ \frac {a}{2 \left (b \,{\mathrm e}^{x}+a \right )^{2} b^{2}}-\frac {1}{\left (b \,{\mathrm e}^{x}+a \right ) b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*x)/(b*exp(x)+a)^3,x)

[Out]

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

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maxima [B]  time = 0.48, size = 61, normalized size = 2.90 \[ -\frac {b e^{x}}{b^{4} e^{\left (2 \, x\right )} + 2 \, a b^{3} e^{x} + a^{2} b^{2}} - \frac {a}{2 \, {\left (b^{4} e^{\left (2 \, x\right )} + 2 \, a b^{3} e^{x} + a^{2} b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*x)/(a+b*exp(x))^3,x, algorithm="maxima")

[Out]

-b*e^x/(b^4*e^(2*x) + 2*a*b^3*e^x + a^2*b^2) - 1/2*a/(b^4*e^(2*x) + 2*a*b^3*e^x + a^2*b^2)

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mupad [B]  time = 3.56, size = 29, normalized size = 1.38 \[ \frac {{\mathrm {e}}^{2\,x}}{2\,a\,\left (a^2+2\,{\mathrm {e}}^x\,a\,b+{\mathrm {e}}^{2\,x}\,b^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*x)/(a + b*exp(x))^3,x)

[Out]

exp(2*x)/(2*a*(b^2*exp(2*x) + a^2 + 2*a*b*exp(x)))

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sympy [B]  time = 0.13, size = 37, normalized size = 1.76 \[ \frac {- a - 2 b e^{x}}{2 a^{2} b^{2} + 4 a b^{3} e^{x} + 2 b^{4} e^{2 x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*x)/(a+b*exp(x))**3,x)

[Out]

(-a - 2*b*exp(x))/(2*a**2*b**2 + 4*a*b**3*exp(x) + 2*b**4*exp(2*x))

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