3.486 \(\int \frac{2^x}{a+2^{-2 x} b} \, dx\)

Optimal. Leaf size=43 \[ \frac{2^x}{a \log (2)}-\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} 2^x}{\sqrt{b}}\right )}{a^{3/2} \log (2)} \]

[Out]

2^x/(a*Log[2]) - (Sqrt[b]*ArcTan[(2^x*Sqrt[a])/Sqrt[b]])/(a^(3/2)*Log[2])

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Rubi [A]  time = 0.0400896, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {2249, 193, 321, 205} \[ \frac{2^x}{a \log (2)}-\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} 2^x}{\sqrt{b}}\right )}{a^{3/2} \log (2)} \]

Antiderivative was successfully verified.

[In]

Int[2^x/(a + b/2^(2*x)),x]

[Out]

2^x/(a*Log[2]) - (Sqrt[b]*ArcTan[(2^x*Sqrt[a])/Sqrt[b]])/(a^(3/2)*Log[2])

Rule 2249

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

Rule 193

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b}, x] && LtQ[n, 0]
&& IntegerQ[p]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 205

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

Rubi steps

\begin{align*} \int \frac{2^x}{a+2^{-2 x} b} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{a+\frac{b}{x^2}} \, dx,x,2^x\right )}{\log (2)}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{b+a x^2} \, dx,x,2^x\right )}{\log (2)}\\ &=\frac{2^x}{a \log (2)}-\frac{b \operatorname{Subst}\left (\int \frac{1}{b+a x^2} \, dx,x,2^x\right )}{a \log (2)}\\ &=\frac{2^x}{a \log (2)}-\frac{\sqrt{b} \tan ^{-1}\left (\frac{2^x \sqrt{a}}{\sqrt{b}}\right )}{a^{3/2} \log (2)}\\ \end{align*}

Mathematica [A]  time = 0.0081684, size = 40, normalized size = 0.93 \[ \frac{\frac{2^x}{a}-\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} 2^x}{\sqrt{b}}\right )}{a^{3/2}}}{\log (2)} \]

Antiderivative was successfully verified.

[In]

Integrate[2^x/(a + b/2^(2*x)),x]

[Out]

(2^x/a - (Sqrt[b]*ArcTan[(2^x*Sqrt[a])/Sqrt[b]])/a^(3/2))/Log[2]

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Maple [B]  time = 0.029, size = 74, normalized size = 1.7 \begin{align*}{\frac{{2}^{x}}{a\ln \left ( 2 \right ) }}+{\frac{1}{2\,{a}^{2}\ln \left ( 2 \right ) }\sqrt{-ab}\ln \left ({2}^{x}-{\frac{1}{a}\sqrt{-ab}} \right ) }-{\frac{1}{2\,{a}^{2}\ln \left ( 2 \right ) }\sqrt{-ab}\ln \left ({2}^{x}+{\frac{1}{a}\sqrt{-ab}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2^x/(a+b/(2^(2*x))),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.55053, size = 212, normalized size = 4.93 \begin{align*} \left [\frac{\sqrt{-\frac{b}{a}} \log \left (-\frac{2 \cdot 2^{x} a \sqrt{-\frac{b}{a}} - 2^{2 \, x} a + b}{2^{2 \, x} a + b}\right ) + 2 \cdot 2^{x}}{2 \, a \log \left (2\right )}, -\frac{\sqrt{\frac{b}{a}} \arctan \left (\frac{2^{x} a \sqrt{\frac{b}{a}}}{b}\right ) - 2^{x}}{a \log \left (2\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(-b/a)*log(-(2*2^x*a*sqrt(-b/a) - 2^(2*x)*a + b)/(2^(2*x)*a + b)) + 2*2^x)/(a*log(2)), -(sqrt(b/a)*a
rctan(2^x*a*sqrt(b/a)/b) - 2^x)/(a*log(2))]

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Sympy [A]  time = 0.201809, size = 39, normalized size = 0.91 \begin{align*} \begin{cases} \frac{2^{x}}{a \log{\left (2 \right )}} & \text{for}\: a \log{\left (2 \right )} \neq 0 \\\frac{x}{a} & \text{otherwise} \end{cases} + \frac{\operatorname{RootSum}{\left (4 z^{2} a^{3} + b, \left ( i \mapsto i \log{\left (2^{x} - 2 i a \right )} \right )\right )}}{\log{\left (2 \right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2**x/(a*log(2)), Ne(a*log(2), 0)), (x/a, True)) + RootSum(4*_z**2*a**3 + b, Lambda(_i, _i*log(2**x
- 2*_i*a)))/log(2)

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Giac [A]  time = 1.22902, size = 51, normalized size = 1.19 \begin{align*} -\frac{b \arctan \left (\frac{2^{x} a}{\sqrt{a b}}\right )}{\sqrt{a b} a \log \left (2\right )} + \frac{2^{x}}{a \log \left (2\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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