3.1.20 \(\int \frac {e^{2 x}}{(a+b e^x)^3} \, dx\) [20]

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 = {2280, 37} \begin {gather*} \frac {e^{2 x}}{2 a \left (a+b e^x\right )^2} \end {gather*}

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 2280

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 &=\text {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 = 0.03, size = 26, normalized size = 1.24 \begin {gather*} \frac {-a-2 b e^x}{2 b^2 \left (a+b e^x\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(2*x)/(a + b*E^x)^3,x]

[Out]

(-a - 2*b*E^x)/(2*b^2*(a + b*E^x)^2)

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Maple [A]
time = 0.02, size = 29, normalized size = 1.38

method	result	size
risch	\(-\frac {2 b \,{\mathrm e}^{x}+a}{2 b^{2} \left (a +b \,{\mathrm e}^{x}\right )^{2}}\)	\(21\)
norman	\(\frac {-\frac {{\mathrm e}^{x}}{b}-\frac {a}{2 b^{2}}}{\left (a +b \,{\mathrm e}^{x}\right )^{2}}\)	\(24\)
default	\(-\frac {1}{b^{2} \left (a +b \,{\mathrm e}^{x}\right )}+\frac {a}{2 b^{2} \left (a +b \,{\mathrm e}^{x}\right )^{2}}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (17) = 34\).
time = 0.31, size = 61, normalized size = 2.90 \begin {gather*} -\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 )}} \end {gather*}

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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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (17) = 34\).
time = 0.36, size = 35, normalized size = 1.67 \begin {gather*} -\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 )}} \end {gather*}

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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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (15) = 30\).
time = 0.05, size = 37, normalized size = 1.76 \begin {gather*} \frac {- a - 2 b e^{x}}{2 a^{2} b^{2} + 4 a b^{3} e^{x} + 2 b^{4} e^{2 x}} \end {gather*}

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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Giac [A]
time = 3.67, size = 20, normalized size = 0.95 \begin {gather*} -\frac {2 \, b e^{x} + a}{2 \, {\left (b e^{x} + a\right )}^{2} b^{2}} \end {gather*}

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

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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