3.504 \(\int \frac{2^{2 x}}{\sqrt{a-2^{-x} b}} \, dx\)

Optimal. Leaf size=96 \[ \frac{3 b^2 \tanh ^{-1}\left (\frac{\sqrt{a-b 2^{-x}}}{\sqrt{a}}\right )}{4 a^{5/2} \log (2)}+\frac{3 b 2^{x-2} \sqrt{a-b 2^{-x}}}{a^2 \log (2)}+\frac{2^{2 x-1} \sqrt{a-b 2^{-x}}}{a \log (2)} \]

[Out]

(2^(-1 + 2*x)*Sqrt[a - b/2^x])/(a*Log[2]) + (3*2^(-2 + x)*b*Sqrt[a - b/2^x])/(a^2*Log[2]) + (3*b^2*ArcTanh[Sqr
t[a - b/2^x]/Sqrt[a]])/(4*a^(5/2)*Log[2])

________________________________________________________________________________________

Rubi [A]  time = 0.0774554, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2248, 51, 63, 208} \[ \frac{3 b^2 \tanh ^{-1}\left (\frac{\sqrt{a-b 2^{-x}}}{\sqrt{a}}\right )}{4 a^{5/2} \log (2)}+\frac{3 b 2^{x-2} \sqrt{a-b 2^{-x}}}{a^2 \log (2)}+\frac{2^{2 x-1} \sqrt{a-b 2^{-x}}}{a \log (2)} \]

Antiderivative was successfully verified.

[In]

Int[2^(2*x)/Sqrt[a - b/2^x],x]

[Out]

(2^(-1 + 2*x)*Sqrt[a - b/2^x])/(a*Log[2]) + (3*2^(-2 + x)*b*Sqrt[a - b/2^x])/(a^2*Log[2]) + (3*b^2*ArcTanh[Sqr
t[a - b/2^x]/Sqrt[a]])/(4*a^(5/2)*Log[2])

Rule 2248

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{2^{2 x}}{\sqrt{a-2^{-x} b}} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{1}{x^3 \sqrt{a-b x}} \, dx,x,2^{-x}\right )}{\log (2)}\\ &=\frac{2^{-1+2 x} \sqrt{a-2^{-x} b}}{a \log (2)}-\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a-b x}} \, dx,x,2^{-x}\right )}{4 a \log (2)}\\ &=\frac{2^{-1+2 x} \sqrt{a-2^{-x} b}}{a \log (2)}+\frac{3\ 2^{-2+x} b \sqrt{a-2^{-x} b}}{a^2 \log (2)}-\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a-b x}} \, dx,x,2^{-x}\right )}{8 a^2 \log (2)}\\ &=\frac{2^{-1+2 x} \sqrt{a-2^{-x} b}}{a \log (2)}+\frac{3\ 2^{-2+x} b \sqrt{a-2^{-x} b}}{a^2 \log (2)}+\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{\frac{a}{b}-\frac{x^2}{b}} \, dx,x,\sqrt{a-2^{-x} b}\right )}{4 a^2 \log (2)}\\ &=\frac{2^{-1+2 x} \sqrt{a-2^{-x} b}}{a \log (2)}+\frac{3\ 2^{-2+x} b \sqrt{a-2^{-x} b}}{a^2 \log (2)}+\frac{3 b^2 \tanh ^{-1}\left (\frac{\sqrt{a-2^{-x} b}}{\sqrt{a}}\right )}{4 a^{5/2} \log (2)}\\ \end{align*}

Mathematica [A]  time = 0.0127632, size = 115, normalized size = 1.2 \[ \frac{2^{-\frac{x}{2}-2} \left (\sqrt{a} 2^{x/2} \left (a^2 2^{2 x+1}+a b 2^x-3 b^2\right )+3 b^2 \sqrt{a 2^x-b} \tanh ^{-1}\left (\frac{\sqrt{a} 2^{x/2}}{\sqrt{a 2^x-b}}\right )\right )}{a^{5/2} \log (2) \sqrt{a-b 2^{-x}}} \]

Antiderivative was successfully verified.

[In]

Integrate[2^(2*x)/Sqrt[a - b/2^x],x]

[Out]

(2^(-2 - x/2)*(2^(x/2)*Sqrt[a]*(2^(1 + 2*x)*a^2 + 2^x*a*b - 3*b^2) + 3*Sqrt[2^x*a - b]*b^2*ArcTanh[(2^(x/2)*Sq
rt[a])/Sqrt[2^x*a - b]]))/(a^(5/2)*Sqrt[a - b/2^x]*Log[2])

________________________________________________________________________________________

Maple [F]  time = 0.013, size = 0, normalized size = 0. \begin{align*} \int{{2}^{2\,x}{\frac{1}{\sqrt{a-{\frac{b}{{2}^{x}}}}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2^(2*x)/(a-b/(2^x))^(1/2),x)

[Out]

int(2^(2*x)/(a-b/(2^x))^(1/2),x)

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.59552, size = 374, normalized size = 3.9 \begin{align*} \left [\frac{3 \, \sqrt{a} b^{2} \log \left (-2 \cdot 2^{x} a - 2 \cdot 2^{x} \sqrt{a} \sqrt{\frac{2^{x} a - b}{2^{x}}} + b\right ) + 2 \,{\left (2 \cdot 2^{2 \, x} a^{2} + 3 \cdot 2^{x} a b\right )} \sqrt{\frac{2^{x} a - b}{2^{x}}}}{8 \, a^{3} \log \left (2\right )}, -\frac{3 \, \sqrt{-a} b^{2} \arctan \left (\frac{\sqrt{-a} \sqrt{\frac{2^{x} a - b}{2^{x}}}}{a}\right ) -{\left (2 \cdot 2^{2 \, x} a^{2} + 3 \cdot 2^{x} a b\right )} \sqrt{\frac{2^{x} a - b}{2^{x}}}}{4 \, a^{3} \log \left (2\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(3*sqrt(a)*b^2*log(-2*2^x*a - 2*2^x*sqrt(a)*sqrt((2^x*a - b)/2^x) + b) + 2*(2*2^(2*x)*a^2 + 3*2^x*a*b)*sq
rt((2^x*a - b)/2^x))/(a^3*log(2)), -1/4*(3*sqrt(-a)*b^2*arctan(sqrt(-a)*sqrt((2^x*a - b)/2^x)/a) - (2*2^(2*x)*
a^2 + 3*2^x*a*b)*sqrt((2^x*a - b)/2^x))/(a^3*log(2))]

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.37684, size = 136, normalized size = 1.42 \begin{align*} \frac{2 \, \sqrt{2^{2 \, x} a - 2^{x} b}{\left (\frac{2 \cdot 2^{x}}{a} + \frac{3 \, b}{a^{2}}\right )} + \frac{3 \, b^{2} \log \left (\sqrt{{\left | a \right |}}{\left | b \right |}\right )}{a^{\frac{5}{2}}} - \frac{3 \, b^{2} \log \left ({\left | -2 \,{\left (2^{x} \sqrt{a} - \sqrt{2^{2 \, x} a - 2^{x} b}\right )} a + \sqrt{a} b \right |}\right )}{a^{\frac{5}{2}}}}{8 \, \log \left (2\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*sqrt(2^(2*x)*a - 2^x*b)*(2*2^x/a + 3*b/a^2) + 3*b^2*log(sqrt(abs(a))*abs(b))/a^(5/2) - 3*b^2*log(abs(-2
*(2^x*sqrt(a) - sqrt(2^(2*x)*a - 2^x*b))*a + sqrt(a)*b))/a^(5/2))/log(2)