3.24 \(\int \frac{e^{4 x}}{(a+b e^{2 x})^3} \, dx\)

Optimal. Leaf size=23 \[ \frac{e^{4 x}}{4 a \left (a+b e^{2 x}\right )^2} \]

[Out]

E^(4*x)/(4*a*(a + b*E^(2*x))^2)

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Rubi [A]  time = 0.0267732, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2248, 37} \[ \frac{e^{4 x}}{4 a \left (a+b e^{2 x}\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[E^(4*x)/(a + b*E^(2*x))^3,x]

[Out]

E^(4*x)/(4*a*(a + b*E^(2*x))^2)

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]

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]

Rubi steps

\begin{align*} \int \frac{e^{4 x}}{\left (a+b e^{2 x}\right )^3} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x}{(a+b x)^3} \, dx,x,e^{2 x}\right )\\ &=\frac{e^{4 x}}{4 a \left (a+b e^{2 x}\right )^2}\\ \end{align*}

Mathematica [A]  time = 0.009315, size = 23, normalized size = 1. \[ \frac{e^{4 x}}{4 a \left (a+b e^{2 x}\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(4*x)/(a + b*E^(2*x))^3,x]

[Out]

E^(4*x)/(4*a*(a + b*E^(2*x))^2)

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Maple [A]  time = 0.006, size = 33, normalized size = 1.4 \begin{align*}{\frac{a}{4\,{b}^{2} \left ( a+b \left ({{\rm e}^{x}} \right ) ^{2} \right ) ^{2}}}-{\frac{1}{2\,{b}^{2} \left ( a+b \left ({{\rm e}^{x}} \right ) ^{2} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(4*x)/(a+b*exp(2*x))^3,x)

[Out]

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

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Maxima [B]  time = 1.02851, size = 90, normalized size = 3.91 \begin{align*} -\frac{b e^{\left (2 \, x\right )}}{2 \,{\left (b^{4} e^{\left (4 \, x\right )} + 2 \, a b^{3} e^{\left (2 \, x\right )} + a^{2} b^{2}\right )}} - \frac{a}{4 \,{\left (b^{4} e^{\left (4 \, x\right )} + 2 \, a b^{3} e^{\left (2 \, x\right )} + a^{2} b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(4*x)/(a+b*exp(2*x))^3,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.45707, size = 89, normalized size = 3.87 \begin{align*} -\frac{2 \, b e^{\left (2 \, x\right )} + a}{4 \,{\left (b^{4} e^{\left (4 \, x\right )} + 2 \, a b^{3} e^{\left (2 \, x\right )} + a^{2} b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(4*x)/(a+b*exp(2*x))^3,x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 0.15944, size = 41, normalized size = 1.78 \begin{align*} \frac{- a - 2 b e^{2 x}}{4 a^{2} b^{2} + 8 a b^{3} e^{2 x} + 4 b^{4} e^{4 x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-a - 2*b*exp(2*x))/(4*a**2*b**2 + 8*a*b**3*exp(2*x) + 4*b**4*exp(4*x))

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Giac [A]  time = 1.25, size = 32, normalized size = 1.39 \begin{align*} -\frac{2 \, b e^{\left (2 \, x\right )} + a}{4 \,{\left (b e^{\left (2 \, x\right )} + a\right )}^{2} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(4*x)/(a+b*exp(2*x))^3,x, algorithm="giac")

[Out]

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