3.731 \(\int 2^{\sec (x)} \sec (x) \tan (x) \, dx\)

Optimal. Leaf size=9 \[ \frac {2^{\sec (x)}}{\log (2)} \]

[Out]

2^sec(x)/ln(2)

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Rubi [A]  time = 0.02, antiderivative size = 9, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {4339, 2209} \[ \frac {2^{\sec (x)}}{\log (2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

2^Sec[x]/Log[2]

Rule 2209

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 4339

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*} \int 2^{\sec (x)} \sec (x) \tan (x) \, dx &=-\operatorname {Subst}\left (\int \frac {2^{\frac {1}{x}}}{x^2} \, dx,x,\cos (x)\right )\\ &=\frac {2^{\sec (x)}}{\log (2)}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 9, normalized size = 1.00 \[ \frac {2^{\sec (x)}}{\log (2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

2^Sec[x]/Log[2]

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fricas [A]  time = 0.47, size = 11, normalized size = 1.22 \[ \frac {2^{\left (\frac {1}{\cos \relax (x)}\right )}}{\log \relax (2)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 11, normalized size = 1.22 \[ \frac {2^{\left (\frac {1}{\cos \relax (x)}\right )}}{\log \relax (2)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 10, normalized size = 1.11 \[ \frac {2^{\sec \relax (x )}}{\ln \relax (2)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2^sec(x)*sec(x)*tan(x),x)

[Out]

2^sec(x)/ln(2)

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maxima [A]  time = 0.31, size = 9, normalized size = 1.00 \[ \frac {2^{\sec \relax (x)}}{\log \relax (2)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2^sec(x)/log(2)

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mupad [B]  time = 3.12, size = 11, normalized size = 1.22 \[ \frac {2^{\frac {1}{\cos \relax (x)}}}{\ln \relax (2)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.65, size = 7, normalized size = 0.78 \[ \frac {2^{\sec {\relax (x )}}}{\log {\relax (2 )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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