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

Optimal. Leaf size=37 \[ \frac{a}{2 b^2 \left (a+b e^{2 x}\right )}+\frac{\log \left (a+b e^{2 x}\right )}{2 b^2} \]

[Out]

a/(2*b^2*(a + b*E^(2*x))) + Log[a + b*E^(2*x)]/(2*b^2)

________________________________________________________________________________________

Rubi [A]  time = 0.037703, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2248, 43} \[ \frac{a}{2 b^2 \left (a+b e^{2 x}\right )}+\frac{\log \left (a+b e^{2 x}\right )}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Int[E^(4*x)/(a + b*E^(2*x))^2,x]

[Out]

a/(2*b^2*(a + b*E^(2*x))) + Log[a + b*E^(2*x)]/(2*b^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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0255883, size = 31, normalized size = 0.84 \[ \frac{\frac{a}{a+b e^{2 x}}+\log \left (a+b e^{2 x}\right )}{2 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a/(a + b*E^(2*x)) + Log[a + b*E^(2*x)])/(2*b^2)

________________________________________________________________________________________

Maple [A]  time = 0.006, size = 32, normalized size = 0.9 \begin{align*}{\frac{\ln \left ( a+b \left ({{\rm e}^{x}} \right ) ^{2} \right ) }{2\,{b}^{2}}}+{\frac{a}{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))^2,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.04988, size = 46, normalized size = 1.24 \begin{align*} \frac{a}{2 \,{\left (b^{3} e^{\left (2 \, x\right )} + a b^{2}\right )}} + \frac{\log \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))^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.48779, size = 92, normalized size = 2.49 \begin{align*} \frac{{\left (b e^{\left (2 \, x\right )} + a\right )} \log \left (b e^{\left (2 \, x\right )} + a\right ) + a}{2 \,{\left (b^{3} e^{\left (2 \, x\right )} + a 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))^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 0.133682, size = 32, normalized size = 0.86 \begin{align*} \frac{a}{2 a b^{2} + 2 b^{3} e^{2 x}} + \frac{\log{\left (\frac{a}{b} + e^{2 x} \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))**2,x)

[Out]

a/(2*a*b**2 + 2*b**3*exp(2*x)) + log(a/b + exp(2*x))/(2*b**2)

________________________________________________________________________________________

Giac [A]  time = 1.27111, size = 43, normalized size = 1.16 \begin{align*} \frac{\log \left ({\left | b e^{\left (2 \, x\right )} + a \right |}\right )}{2 \, b^{2}} + \frac{a}{2 \,{\left (b e^{\left (2 \, x\right )} + a\right )} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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