3.1.23 \(\int \frac {e^{4 x}}{(a+b e^{2 x})^2} \, dx\) [23]

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]

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

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Rubi [A]
time = 0.03, antiderivative size = 37, normalized size of antiderivative = 1.00, 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 = {2280, 45} \begin {gather*} \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 {gather*}

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 45

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])

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

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Mathematica [A]
time = 0.04, size = 33, normalized size = 0.89 \begin {gather*} \frac {\frac {a}{a+b e^{2 x}}+\log \left (b \left (a+b e^{2 x}\right )\right )}{2 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.01, size = 32, normalized size = 0.86

method	result	size
default	\(\frac {a}{2 b^{2} \left (a +b \,{\mathrm e}^{2 x}\right )}+\frac {\ln \left (a +b \,{\mathrm e}^{2 x}\right )}{2 b^{2}}\)	\(32\)
norman	\(\frac {a}{2 b^{2} \left (a +b \,{\mathrm e}^{2 x}\right )}+\frac {\ln \left (a +b \,{\mathrm e}^{2 x}\right )}{2 b^{2}}\)	\(32\)
risch	\(\frac {a}{2 b^{2} \left (a +b \,{\mathrm e}^{2 x}\right )}+\frac {\ln \left ({\mathrm e}^{2 x}+\frac {a}{b}\right )}{2 b^{2}}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(4*x)/(a+b*exp(2*x))^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 34, normalized size = 0.92 \begin {gather*} \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 {gather*}

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

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Fricas [A]
time = 0.46, size = 38, normalized size = 1.03 \begin {gather*} \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 {gather*}

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)

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Sympy [A]
time = 0.06, size = 32, normalized size = 0.86 \begin {gather*} \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 {gather*}

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)

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Giac [A]
time = 3.49, size = 32, normalized size = 0.86 \begin {gather*} \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 {gather*}

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)

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Mupad [B]
time = 3.55, size = 34, normalized size = 0.92 \begin {gather*} \frac {\ln \left (a+b\,{\mathrm {e}}^{2\,x}\right )}{2\,b^2}-\frac {{\mathrm {e}}^{2\,x}}{2\,b\,\left (a+b\,{\mathrm {e}}^{2\,x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(4*x)/(a + b*exp(2*x))^2,x)

[Out]

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

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