\(\int \frac {e^{4 x}}{a+b e^{2 x}} \, dx\) [22]

Optimal result

Integrand size = 17, antiderivative size = 31 \[ \int \frac {e^{4 x}}{a+b e^{2 x}} \, dx=\frac {e^{2 x}}{2 b}-\frac {a \log \left (a+b e^{2 x}\right )}{2 b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2280, 45} \[ \int \frac {e^{4 x}}{a+b e^{2 x}} \, dx=\frac {e^{2 x}}{2 b}-\frac {a \log \left (a+b e^{2 x}\right )}{2 b^2} \]

[In]

Int[E^(4*x)/(a + b*E^(2*x)),x]

[Out]

E^(2*x)/(2*b) - (a*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*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x}{a+b x} \, dx,x,e^{2 x}\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {1}{b}-\frac {a}{b (a+b x)}\right ) \, dx,x,e^{2 x}\right ) \\ & = \frac {e^{2 x}}{2 b}-\frac {a \log \left (a+b e^{2 x}\right )}{2 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {e^{4 x}}{a+b e^{2 x}} \, dx=\frac {b e^{2 x}-a \log \left (a+b e^{2 x}\right )}{2 b^2} \]

[In]

Integrate[E^(4*x)/(a + b*E^(2*x)),x]

[Out]

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

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84

method	result	size
default	\(\frac {{\mathrm e}^{2 x}}{2 b}-\frac {a \ln \left (a +b \,{\mathrm e}^{2 x}\right )}{2 b^{2}}\)	\(26\)
norman	\(\frac {{\mathrm e}^{2 x}}{2 b}-\frac {a \ln \left (a +b \,{\mathrm e}^{2 x}\right )}{2 b^{2}}\)	\(26\)
risch	\(\frac {{\mathrm e}^{2 x}}{2 b}-\frac {a \ln \left ({\mathrm e}^{2 x}+\frac {a}{b}\right )}{2 b^{2}}\)	\(28\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {e^{4 x}}{a+b e^{2 x}} \, dx=\frac {b e^{\left (2 \, x\right )} - a \log \left (b e^{\left (2 \, x\right )} + a\right )}{2 \, b^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {e^{4 x}}{a+b e^{2 x}} \, dx=- \frac {a \log {\left (\frac {a}{b} + e^{2 x} \right )}}{2 b^{2}} + \begin {cases} \frac {e^{2 x}}{2 b} & \text {for}\: b \neq 0 \\\frac {x}{b} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

-a*log(a/b + exp(2*x))/(2*b**2) + Piecewise((exp(2*x)/(2*b), Ne(b, 0)), (x/b, True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {e^{4 x}}{a+b e^{2 x}} \, dx=\frac {e^{\left (2 \, x\right )}}{2 \, b} - \frac {a \log \left (b e^{\left (2 \, x\right )} + a\right )}{2 \, b^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {e^{4 x}}{a+b e^{2 x}} \, dx=\frac {e^{\left (2 \, x\right )}}{2 \, b} - \frac {a \log \left ({\left | b e^{\left (2 \, x\right )} + a \right |}\right )}{2 \, b^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {e^{4 x}}{a+b e^{2 x}} \, dx=-\frac {a\,\ln \left (a+b\,{\mathrm {e}}^{2\,x}\right )-b\,{\mathrm {e}}^{2\,x}}{2\,b^2} \]

[In]

int(exp(4*x)/(a + b*exp(2*x)),x)

[Out]

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