Integrand size = 19, antiderivative size = 93 \[ \int \frac {2^{2 x}}{\sqrt {a+2^{-x} b}} \, dx=\frac {2^{-1+2 x} \sqrt {a+2^{-x} b}}{a \log (2)}-\frac {3\ 2^{-2+x} b \sqrt {a+2^{-x} b}}{a^2 \log (2)}+\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {a+2^{-x} b}}{\sqrt {a}}\right )}{4 a^{5/2} \log (2)} \]
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)
Time = 0.00 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.19 \[ \int \frac {2^{2 x}}{\sqrt {a+2^{-x} b}} \, dx=\frac {2^{-2-\frac {x}{2}} \left (2^{x/2} \sqrt {a} \left (2^{1+2 x} a^2-2^x a b-3 b^2\right )+3 b^2 \sqrt {2^x a+b} \text {arctanh}\left (\frac {2^{x/2} \sqrt {a}}{\sqrt {2^x a+b}}\right )\right )}{a^{5/2} \sqrt {a+2^{-x} b} \log (2)} \]
(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)*Sqrt[a])/Sqrt[2^x*a + b]]))/(a^(5/2)*Sqr t[a + b/2^x]*Log[2])
Time = 0.22 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2678, 52, 52, 73, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {2^{2 x}}{\sqrt {a+b 2^{-x}}} \, dx\) |
\(\Big \downarrow \) 2678 |
\(\displaystyle -\frac {\int \frac {2^{3 x}}{\sqrt {a+2^{-x} b}}d2^{-x}}{\log (2)}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle -\frac {-\frac {3 b \int \frac {2^{2 x}}{\sqrt {a+2^{-x} b}}d2^{-x}}{4 a}-\frac {2^{2 x-1} \sqrt {a+b 2^{-x}}}{a}}{\log (2)}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle -\frac {-\frac {3 b \left (-\frac {b \int \frac {2^x}{\sqrt {a+2^{-x} b}}d2^{-x}}{2 a}-\frac {2^x \sqrt {a+b 2^{-x}}}{a}\right )}{4 a}-\frac {2^{2 x-1} \sqrt {a+b 2^{-x}}}{a}}{\log (2)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {-\frac {3 b \left (-\frac {\int \frac {1}{\frac {2^{-2 x}}{b}-\frac {a}{b}}d\sqrt {a+2^{-x} b}}{a}-\frac {2^x \sqrt {a+b 2^{-x}}}{a}\right )}{4 a}-\frac {2^{2 x-1} \sqrt {a+b 2^{-x}}}{a}}{\log (2)}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {-\frac {3 b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a+b 2^{-x}}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {2^x \sqrt {a+b 2^{-x}}}{a}\right )}{4 a}-\frac {2^{2 x-1} \sqrt {a+b 2^{-x}}}{a}}{\log (2)}\) |
-((-((2^(-1 + 2*x)*Sqrt[a + b/2^x])/a) - (3*b*(-((2^x*Sqrt[a + b/2^x])/a) + (b*ArcTanh[Sqrt[a + b/2^x]/Sqrt[a]])/a^(3/2)))/(4*a))/Log[2])
3.6.2.3.1 Defintions of rubi rules used
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] - Simp[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] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_.)*(G_)^((h_.)*((f_ .) + (g_.)*(x_))), x_Symbol] :> With[{m = FullSimplify[g*h*(Log[G]/(d*e*Log [F]))]}, Simp[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)/De nominator[m]))], x] /; LeQ[m, -1] || GeQ[m, 1]] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x]
Leaf count of result is larger than twice the leaf count of optimal. \(181\) vs. \(2(81)=162\).
Time = 0.33 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.96
method | result | size |
derivativedivides | \(-\frac {\sqrt {\left (a 2^{x}+b \right ) 2^{-x}}\, 2^{x} \left (-4 a^{\frac {5}{2}} \sqrt {a 2^{2 x}+2^{x} b}\, 2^{x}-4 \ln \left (\frac {2 \sqrt {\left (a 2^{x}+b \right ) 2^{x}}\, \sqrt {a}+2 a 2^{x}+b}{2 \sqrt {a}}\right ) a \,b^{2}+8 \sqrt {\left (a 2^{x}+b \right ) 2^{x}}\, a^{\frac {3}{2}} b -2 a^{\frac {3}{2}} \sqrt {a 2^{2 x}+2^{x} b}\, b +b^{2} \ln \left (\frac {2 \sqrt {a 2^{2 x}+2^{x} b}\, \sqrt {a}+2 a 2^{x}+b}{2 \sqrt {a}}\right ) a \right )}{8 \ln \left (2\right ) \sqrt {\left (a 2^{x}+b \right ) 2^{x}}\, a^{\frac {7}{2}}}\) | \(182\) |
default | \(-\frac {\sqrt {\left (a 2^{x}+b \right ) 2^{-x}}\, 2^{x} \left (-4 a^{\frac {5}{2}} \sqrt {a 2^{2 x}+2^{x} b}\, 2^{x}-4 \ln \left (\frac {2 \sqrt {\left (a 2^{x}+b \right ) 2^{x}}\, \sqrt {a}+2 a 2^{x}+b}{2 \sqrt {a}}\right ) a \,b^{2}+8 \sqrt {\left (a 2^{x}+b \right ) 2^{x}}\, a^{\frac {3}{2}} b -2 a^{\frac {3}{2}} \sqrt {a 2^{2 x}+2^{x} b}\, b +b^{2} \ln \left (\frac {2 \sqrt {a 2^{2 x}+2^{x} b}\, \sqrt {a}+2 a 2^{x}+b}{2 \sqrt {a}}\right ) a \right )}{8 \ln \left (2\right ) \sqrt {\left (a 2^{x}+b \right ) 2^{x}}\, a^{\frac {7}{2}}}\) | \(182\) |
-1/8/ln(2)*((a*2^x+b)/(2^x))^(1/2)*2^x*(-4*a^(5/2)*(a*(2^x)^2+2^x*b)^(1/2) *2^x-4*ln(1/2*(2*((a*2^x+b)*2^x)^(1/2)*a^(1/2)+2*a*2^x+b)/a^(1/2))*a*b^2+8 *((a*2^x+b)*2^x)^(1/2)*a^(3/2)*b-2*a^(3/2)*(a*(2^x)^2+2^x*b)^(1/2)*b+b^2*l n(1/2*(2*(a*(2^x)^2+2^x*b)^(1/2)*a^(1/2)+2*a*2^x+b)/a^(1/2))*a)/((a*2^x+b) *2^x)^(1/2)/a^(7/2)
Time = 0.32 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.78 \[ \int \frac {2^{2 x}}{\sqrt {a+2^{-x} b}} \, dx=\left [\frac {3 \, \sqrt {a} b^{2} \log \left (2 \cdot 2^{x} a + 2 \cdot 2^{x} \sqrt {a} \sqrt {\frac {2^{x} a + b}{2^{x}}} + b\right ) + 2 \, {\left (2 \cdot 2^{2 \, x} a^{2} - 3 \cdot 2^{x} a b\right )} \sqrt {\frac {2^{x} a + b}{2^{x}}}}{8 \, a^{3} \log \left (2\right )}, -\frac {3 \, \sqrt {-a} b^{2} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {2^{x} a + b}{2^{x}}}}{a}\right ) - {\left (2 \cdot 2^{2 \, x} a^{2} - 3 \cdot 2^{x} a b\right )} \sqrt {\frac {2^{x} a + b}{2^{x}}}}{4 \, a^{3} \log \left (2\right )}\right ] \]
[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)*sqrt((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))]
\[ \int \frac {2^{2 x}}{\sqrt {a+2^{-x} b}} \, dx=\int \frac {2^{2 x}}{\sqrt {a + 2^{- x} b}}\, dx \]
Time = 0.27 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.33 \[ \int \frac {2^{2 x}}{\sqrt {a+2^{-x} b}} \, dx=-\frac {3 \, b^{2} \log \left (\frac {\sqrt {a + \frac {b}{2^{x}}} - \sqrt {a}}{\sqrt {a + \frac {b}{2^{x}}} + \sqrt {a}}\right )}{8 \, a^{\frac {5}{2}} \log \left (2\right )} - \frac {3 \, {\left (a + \frac {b}{2^{x}}\right )}^{\frac {3}{2}} b^{2} - 5 \, \sqrt {a + \frac {b}{2^{x}}} a b^{2}}{4 \, {\left ({\left (a + \frac {b}{2^{x}}\right )}^{2} a^{2} - 2 \, {\left (a + \frac {b}{2^{x}}\right )} a^{3} + a^{4}\right )} \log \left (2\right )} \]
-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))
Time = 0.46 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.99 \[ \int \frac {2^{2 x}}{\sqrt {a+2^{-x} b}} \, dx=\frac {2 \, \sqrt {2^{2 \, x} a + 2^{x} b} {\left (\frac {2 \cdot 2^{x}}{a} - \frac {3 \, b}{a^{2}}\right )} - \frac {3 \, b^{2} \log \left ({\left | 2 \, {\left (2^{x} \sqrt {a} - \sqrt {2^{2 \, x} a + 2^{x} b}\right )} \sqrt {a} + b \right |}\right )}{a^{\frac {5}{2}}} + \frac {3 \, b^{2} \log \left ({\left | b \right |}\right )}{a^{\frac {5}{2}}}}{8 \, \log \left (2\right )} \]
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)
Timed out. \[ \int \frac {2^{2 x}}{\sqrt {a+2^{-x} b}} \, dx=\int \frac {2^{2\,x}}{\sqrt {a+\frac {b}{2^x}}} \,d x \]