\(\int \frac {2^x}{\sqrt {a-4^{-x} b}} \, dx\) [495]

Optimal result

Integrand size = 18, antiderivative size = 25 \[ \int \frac {2^x}{\sqrt {a-4^{-x} b}} \, dx=\frac {2^x \sqrt {a-2^{-2 x} b}}{a \log (2)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2281, 197} \[ \int \frac {2^x}{\sqrt {a-4^{-x} b}} \, dx=\frac {2^x \sqrt {a-b 2^{-2 x}}}{a \log (2)} \]

[In]

Int[2^x/Sqrt[a - b/4^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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52 \[ \int \frac {2^x}{\sqrt {a-4^{-x} b}} \, dx=\frac {2^{-x} \left (2^{2 x} a-b\right )}{a \sqrt {a-2^{-2 x} b} \log (2)} \]

[In]

Integrate[2^x/Sqrt[a - b/4^x],x]

[Out]

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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.76

method	result	size
risch	\(\frac {\left (a 2^{2 x}-b \right ) 2^{-x}}{\sqrt {\left (a 2^{2 x}-b \right ) 2^{-2 x}}\, a \ln \left (2\right )}\)	\(44\)

[In]

int(2^x/(a-b/(4^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)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {2^x}{\sqrt {a-4^{-x} b}} \, dx=\frac {2^{x} \sqrt {\frac {2^{2 \, x} a - b}{2^{2 \, x}}}}{a \log \left (2\right )} \]

[In]

integrate(2^x/(a-b/(4^x))^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {2^x}{\sqrt {a-4^{-x} b}} \, dx=\int \frac {2^{x}}{\sqrt {a - 4^{- x} b}}\, dx \]

[In]

integrate(2**x/(a-b/(4**x))**(1/2),x)

[Out]

Integral(2**x/sqrt(a - b/4**x), x)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {2^x}{\sqrt {a-4^{-x} b}} \, dx=\frac {\sqrt {2^{2 \, x} a - b}}{a \log \left (2\right )} \]

[In]

integrate(2^x/(a-b/(4^x))^(1/2),x, algorithm="maxima")

[Out]

sqrt(2^(2*x)*a - b)/(a*log(2))

Giac [F]

\[ \int \frac {2^x}{\sqrt {a-4^{-x} b}} \, dx=\int { \frac {2^{x}}{\sqrt {a - \frac {b}{4^{x}}}} \,d x } \]

[In]

integrate(2^x/(a-b/(4^x))^(1/2),x, algorithm="giac")

[Out]

integrate(2^x/sqrt(a - b/4^x), x)

Mupad [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {2^x}{\sqrt {a-4^{-x} b}} \, dx=\frac {2^x\,\sqrt {a-\frac {b}{2^{2\,x}}}}{a\,\ln \left (2\right )} \]

[In]

int(2^x/(a - b/4^x)^(1/2),x)

[Out]

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