3.5.94 \(\int \frac {2^x}{\sqrt {a+2^{-2 x} b}} \, dx\) [494]

Optimal. Leaf size=24 \[ \frac {2^x \sqrt {a+2^{-2 x} b}}{a \log (2)} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2281, 197} \begin {gather*} \frac {2^x \sqrt {a+b 2^{-2 x}}}{a \log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[2^x/Sqrt[a + b/2^(2*x)],x]

[Out]

(2^x*Sqrt[a + b/2^(2*x)])/(a*Log[2])

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 2281

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]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 35, normalized size = 1.46 \begin {gather*} \frac {2^{-x} \left (2^{2 x} a+b\right )}{a \sqrt {a+2^{-2 x} b} \log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[2^x/Sqrt[a + b/2^(2*x)],x]

[Out]

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

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Maple [A]
time = 0.01, size = 40, normalized size = 1.67

method	result	size
risch	\(\frac {\left (a 2^{2 x}+b \right ) 2^{-x}}{\sqrt {\left (a 2^{2 x}+b \right ) 2^{-2 x}}\, a \ln \left (2\right )}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2^x/(a+b/(2^(2*x)))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 24, normalized size = 1.00 \begin {gather*} \frac {2^{x} \sqrt {a + \frac {b}{2^{2 \, x}}}}{a \log \left (2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2^x/(a+b/(2^(2*x)))^(1/2),x, algorithm="maxima")

[Out]

2^x*sqrt(a + b/2^(2*x))/(a*log(2))

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Fricas [A]
time = 0.42, size = 30, normalized size = 1.25 \begin {gather*} \frac {2^{x} \sqrt {\frac {2^{2 \, x} a + b}{2^{2 \, x}}}}{a \log \left (2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2^x/(a+b/(2^(2*x)))^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2^{x}}{\sqrt {a + 2^{- 2 x} b}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2**x/(a+b/(2**(2*x)))**(1/2),x)

[Out]

Integral(2**x/sqrt(a + b/2**(2*x)), x)

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Giac [A]
time = 2.84, size = 29, normalized size = 1.21 \begin {gather*} \frac {\frac {\sqrt {2^{2 \, x} a + b}}{a} - \frac {\sqrt {b}}{a}}{\log \left (2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2^x/(a+b/(2^(2*x)))^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 3.55, size = 24, normalized size = 1.00 \begin {gather*} \frac {2^x\,\sqrt {a+\frac {b}{2^{2\,x}}}}{a\,\ln \left (2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2^x/(a + b/2^(2*x))^(1/2),x)

[Out]

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

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