3.1.27 \(\int \frac {e^{2 x}}{A+B e^{4 x}} \, dx\) [27]

3.1.27.1 Optimal result

Integrand size = 17, antiderivative size = 31 \[ \int \frac {e^{2 x}}{A+B e^{4 x}} \, dx=\frac {\arctan \left (\frac {\sqrt {B} e^{2 x}}{\sqrt {A}}\right )}{2 \sqrt {A} \sqrt {B}} \]

output

1/2*arctan(exp(2*x)*B^(1/2)/A^(1/2))/A^(1/2)/B^(1/2)
 

3.1.27.2 Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {e^{2 x}}{A+B e^{4 x}} \, dx=\frac {\arctan \left (\frac {\sqrt {B} e^{2 x}}{\sqrt {A}}\right )}{2 \sqrt {A} \sqrt {B}} \]

input

Integrate[E^(2*x)/(A + B*E^(4*x)),x]
 

output

ArcTan[(Sqrt[B]*E^(2*x))/Sqrt[A]]/(2*Sqrt[A]*Sqrt[B])
 

3.1.27.3 Rubi [A] (verified)

Time = 0.18 (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 = {2679, 218}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {e^{2 x}}{A+B e^{4 x}} \, dx\)

\(\Big \downarrow \) 2679

\(\displaystyle \frac {1}{2} \int \frac {1}{A+B e^{4 x}}de^{2 x}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\arctan \left (\frac {\sqrt {B} e^{2 x}}{\sqrt {A}}\right )}{2 \sqrt {A} \sqrt {B}}\)

input

Int[E^(2*x)/(A + B*E^(4*x)),x]
 

output

ArcTan[(Sqrt[B]*E^(2*x))/Sqrt[A]]/(2*Sqrt[A]*Sqrt[B])
 

3.1.27.3.1 Defintions of rubi rules used

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 2679

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_.)*(G_)^((h_.)*((f_ 
.) + (g_.)*(x_))), x_Symbol] :> With[{m = FullSimplify[d*e*(Log[F]/(g*h*Log 
[G]))]}, Simp[Denominator[m]/(g*h*Log[G])   Subst[Int[x^(Denominator[m] - 1 
)*(a + b*F^(c*e - d*e*(f/g))*x^Numerator[m])^p, x], x, G^(h*((f + g*x)/Deno 
minator[m]))], x] /; LtQ[m, -1] || GtQ[m, 1]] /; FreeQ[{F, G, a, b, c, d, e 
, f, g, h, p}, x]
 

3.1.27.4 Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65

method	result	size
default	\(\frac {\arctan \left (\frac {B \,{\mathrm e}^{2 x}}{\sqrt {A B}}\right )}{2 \sqrt {A B}}\)	\(20\)
risch	\(-\frac {\ln \left ({\mathrm e}^{2 x}-\frac {A}{\sqrt {-A B}}\right )}{4 \sqrt {-A B}}+\frac {\ln \left ({\mathrm e}^{2 x}+\frac {A}{\sqrt {-A B}}\right )}{4 \sqrt {-A B}}\)	\(47\)

input

int(exp(2*x)/(A+B*exp(4*x)),x,method=_RETURNVERBOSE)
 

output

1/2/(A*B)^(1/2)*arctan(B*exp(x)^2/(A*B)^(1/2))
 

3.1.27.5 Fricas [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 76, normalized size of antiderivative = 2.45 \[ \int \frac {e^{2 x}}{A+B e^{4 x}} \, dx=\left [-\frac {\sqrt {-A B} \log \left (\frac {B e^{\left (4 \, x\right )} - 2 \, \sqrt {-A B} e^{\left (2 \, x\right )} - A}{B e^{\left (4 \, x\right )} + A}\right )}{4 \, A B}, -\frac {\sqrt {A B} \arctan \left (\frac {\sqrt {A B} e^{\left (-2 \, x\right )}}{B}\right )}{2 \, A B}\right ] \]

input

integrate(exp(2*x)/(A+B*exp(4*x)),x, algorithm="fricas")
 

output

[-1/4*sqrt(-A*B)*log((B*e^(4*x) - 2*sqrt(-A*B)*e^(2*x) - A)/(B*e^(4*x) + A 
))/(A*B), -1/2*sqrt(A*B)*arctan(sqrt(A*B)*e^(-2*x)/B)/(A*B)]
 

3.1.27.6 Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71 \[ \int \frac {e^{2 x}}{A+B e^{4 x}} \, dx=\operatorname {RootSum} {\left (16 z^{2} A B + 1, \left ( i \mapsto i \log {\left (4 i A + e^{2 x} \right )} \right )\right )} \]

input

integrate(exp(2*x)/(A+B*exp(4*x)),x)
 

output

RootSum(16*_z**2*A*B + 1, Lambda(_i, _i*log(4*_i*A + exp(2*x))))
 

3.1.27.7 Maxima [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.61 \[ \int \frac {e^{2 x}}{A+B e^{4 x}} \, dx=\frac {\arctan \left (\frac {B e^{\left (2 \, x\right )}}{\sqrt {A B}}\right )}{2 \, \sqrt {A B}} \]

input

integrate(exp(2*x)/(A+B*exp(4*x)),x, algorithm="maxima")
 

output

1/2*arctan(B*e^(2*x)/sqrt(A*B))/sqrt(A*B)
 

3.1.27.8 Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.61 \[ \int \frac {e^{2 x}}{A+B e^{4 x}} \, dx=\frac {\arctan \left (\frac {B e^{\left (2 \, x\right )}}{\sqrt {A B}}\right )}{2 \, \sqrt {A B}} \]

input

integrate(exp(2*x)/(A+B*exp(4*x)),x, algorithm="giac")
 

output

1/2*arctan(B*e^(2*x)/sqrt(A*B))/sqrt(A*B)
 

3.1.27.9 Mupad [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.61 \[ \int \frac {e^{2 x}}{A+B e^{4 x}} \, dx=\frac {\mathrm {atan}\left (\frac {B\,{\mathrm {e}}^{2\,x}}{\sqrt {A\,B}}\right )}{2\,\sqrt {A\,B}} \]

input

int(exp(2*x)/(A + B*exp(4*x)),x)
 

output

atan((B*exp(2*x))/(A*B)^(1/2))/(2*(A*B)^(1/2))
 

3.1.27.10 Reduce [B] (verification not implemented)

Time = 0.00 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.29 \[ \int \frac {e^{2 x}}{A+B e^{4 x}} \, dx=-\frac {\sqrt {b}\, \sqrt {a}\, \left (\mathit {atan} \left (\frac {b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}-2 e^{x} \sqrt {b}}{b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right )+\mathit {atan} \left (\frac {b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+2 e^{x} \sqrt {b}}{b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right )\right )}{2 a b} \]

input

int(e**(2*x)/(e**(4*x)*b + a),x)
 

output

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