\(\int 2^{\sec (x)} \sec (x) \tan (x) \, dx\) [731]

Optimal result

Integrand size = 9, antiderivative size = 9 \[ \int 2^{\sec (x)} \sec (x) \tan (x) \, dx=\frac {2^{\sec (x)}}{\log (2)} \]

[Out]

2^sec(x)/ln(2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {4424, 2240} \[ \int 2^{\sec (x)} \sec (x) \tan (x) \, dx=\frac {2^{\sec (x)}}{\log (2)} \]

[In]

Int[2^Sec[x]*Sec[x]*Tan[x],x]

[Out]

2^Sec[x]/Log[2]

Rule 2240

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^n*(
F^(a + b*(c + d*x)^n)/(b*f*n*(c + d*x)^n*Log[F])), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 4424

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFactors[Cos[c*(a + b*x)], x]}, Dist[-(b*
c)^(-1), Subst[Int[SubstFor[1/x, Cos[c*(a + b*x)]/d, u, x], x], x, Cos[c*(a + b*x)]/d], x] /; FunctionOfQ[Cos[
c*(a + b*x)]/d, u, x, True]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Tan] || EqQ[F, tan])

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {2^{\frac {1}{x}}}{x^2} \, dx,x,\cos (x)\right ) \\ & = \frac {2^{\sec (x)}}{\log (2)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00 \[ \int 2^{\sec (x)} \sec (x) \tan (x) \, dx=\frac {2^{\sec (x)}}{\log (2)} \]

[In]

Integrate[2^Sec[x]*Sec[x]*Tan[x],x]

[Out]

2^Sec[x]/Log[2]

Maple [A] (verified)

Time = 21.10 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.11

method	result	size
derivativedivides	\(\frac {2^{\sec \left (x \right )}}{\ln \left (2\right )}\)	\(10\)
default	\(\frac {2^{\sec \left (x \right )}}{\ln \left (2\right )}\)	\(10\)
risch	\(\frac {2^{\sec \left (x \right )}}{\ln \left (2\right )}\)	\(10\)

[In]

int(2^sec(x)*sec(x)*tan(x),x,method=_RETURNVERBOSE)

[Out]

2^sec(x)/ln(2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.22 \[ \int 2^{\sec (x)} \sec (x) \tan (x) \, dx=\frac {2^{\left (\frac {1}{\cos \left (x\right )}\right )}}{\log \left (2\right )} \]

[In]

integrate(2^sec(x)*sec(x)*tan(x),x, algorithm="fricas")

[Out]

2^(1/cos(x))/log(2)

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.78 \[ \int 2^{\sec (x)} \sec (x) \tan (x) \, dx=\frac {2^{\sec {\left (x \right )}}}{\log {\left (2 \right )}} \]

[In]

integrate(2**sec(x)*sec(x)*tan(x),x)

[Out]

2**sec(x)/log(2)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00 \[ \int 2^{\sec (x)} \sec (x) \tan (x) \, dx=\frac {2^{\sec \left (x\right )}}{\log \left (2\right )} \]

[In]

integrate(2^sec(x)*sec(x)*tan(x),x, algorithm="maxima")

[Out]

2^sec(x)/log(2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.22 \[ \int 2^{\sec (x)} \sec (x) \tan (x) \, dx=\frac {2^{\left (\frac {1}{\cos \left (x\right )}\right )}}{\log \left (2\right )} \]

[In]

integrate(2^sec(x)*sec(x)*tan(x),x, algorithm="giac")

[Out]

2^(1/cos(x))/log(2)

Mupad [B] (verification not implemented)

Time = 27.58 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.22 \[ \int 2^{\sec (x)} \sec (x) \tan (x) \, dx=\frac {2^{\frac {1}{\cos \left (x\right )}}}{\ln \left (2\right )} \]

[In]

int((2^(1/cos(x))*tan(x))/cos(x),x)

[Out]

2^(1/cos(x))/log(2)