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

Optimal. Leaf size=93 \[ \frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b 2^{-x}}}{\sqrt {a}}\right )}{4 a^{5/2} \log (2)}-\frac {3 b 2^{x-2} \sqrt {a+b 2^{-x}}}{a^2 \log (2)}+\frac {2^{2 x-1} \sqrt {a+b 2^{-x}}}{a \log (2)} \]

[Out]

3/4*b^2*arctanh((a+b/(2^x))^(1/2)/a^(1/2))/a^(5/2)/ln(2)+2^(-1+2*x)*(a+b/(2^x))^(1/2)/a/ln(2)-3*2^(-2+x)*b*(a+
b/(2^x))^(1/2)/a^2/ln(2)

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Rubi [A]  time = 0.07, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, 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 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]

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]

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*}

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Mathematica [A]  time = 0.04, size = 111, normalized size = 1.19 \[ \frac {2^{-\frac {x}{2}-2} \left (\sqrt {a} 2^{x/2} \left (a^2 2^{2 x+1}-a b 2^x-3 b^2\right )+3 b^2 \sqrt {a 2^x+b} \tanh ^{-1}\left (\frac {\sqrt {a} 2^{x/2}}{\sqrt {a 2^x+b}}\right )\right )}{a^{5/2} \log (2) \sqrt {a+b 2^{-x}}} \]

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

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fricas [A]  time = 0.43, size = 166, normalized size = 1.78 \[ \left [\frac {3 \, \sqrt {a} b^{2} \log \left (2 \cdot 2^{x} a + 2 \cdot 2^{x} \sqrt {a} \sqrt {\frac {2^{x} a + b}{2^{x}}} + b\right ) + 2 \, {\left (2 \cdot 2^{2 \, x} a^{2} - 3 \cdot 2^{x} a b\right )} \sqrt {\frac {2^{x} a + b}{2^{x}}}}{8 \, a^{3} \log \relax (2)}, -\frac {3 \, \sqrt {-a} b^{2} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {2^{x} a + b}{2^{x}}}}{a}\right ) - {\left (2 \cdot 2^{2 \, x} a^{2} - 3 \cdot 2^{x} a b\right )} \sqrt {\frac {2^{x} a + b}{2^{x}}}}{4 \, a^{3} \log \relax (2)}\right ] \]

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)*sqr
t((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))]

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giac [A]  time = 0.64, size = 94, normalized size = 1.01 \[ \frac {2 \, \sqrt {2^{2 \, x} a + 2^{x} b} {\left (\frac {2 \cdot 2^{x}}{a} - \frac {3 \, b}{a^{2}}\right )} - \frac {3 \, b^{2} \log \left ({\left | -2 \, {\left (2^{x} \sqrt {a} - \sqrt {2^{2 \, x} a + 2^{x} b}\right )} \sqrt {a} - b \right |}\right )}{a^{\frac {5}{2}}} + \frac {3 \, b^{2} \log \left ({\left | b \right |}\right )}{a^{\frac {5}{2}}}}{8 \, \log \relax (2)} \]

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

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[ \int \frac {2^{2 x}}{\sqrt {b 2^{-x}+a}}\, dx \]

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)

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maxima [A]  time = 2.04, size = 124, normalized size = 1.33 \[ -\frac {3 \, b^{2} \log \left (\frac {\sqrt {a + \frac {b}{2^{x}}} - \sqrt {a}}{\sqrt {a + \frac {b}{2^{x}}} + \sqrt {a}}\right )}{8 \, a^{\frac {5}{2}} \log \relax (2)} - \frac {3 \, {\left (a + \frac {b}{2^{x}}\right )}^{\frac {3}{2}} b^{2} - 5 \, \sqrt {a + \frac {b}{2^{x}}} a b^{2}}{4 \, {\left ({\left (a + \frac {b}{2^{x}}\right )}^{2} a^{2} - 2 \, {\left (a + \frac {b}{2^{x}}\right )} a^{3} + a^{4}\right )} \log \relax (2)} \]

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]

-3/8*b^2*log((sqrt(a + b/2^x) - sqrt(a))/(sqrt(a + b/2^x) + sqrt(a)))/(a^(5/2)*log(2)) - 1/4*(3*(a + b/2^x)^(3
/2)*b^2 - 5*sqrt(a + b/2^x)*a*b^2)/(((a + b/2^x)^2*a^2 - 2*(a + b/2^x)*a^3 + a^4)*log(2))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {2^{2\,x}}{\sqrt {a+\frac {b}{2^x}}} \,d x \]

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)

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

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)

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