Integrand size = 20, antiderivative size = 96 \[ \int \frac {2^{2 x}}{\sqrt {a-2^{-x} b}} \, dx=\frac {2^{-1+2 x} \sqrt {a-2^{-x} b}}{a \log (2)}+\frac {3\ 2^{-2+x} b \sqrt {a-2^{-x} b}}{a^2 \log (2)}+\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {a-2^{-x} b}}{\sqrt {a}}\right )}{4 a^{5/2} \log (2)} \] Output:
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)+3/4*b^2*arctanh((a-b/(2^x))^(1/2)/a^(1/2))/a^(5/2)/ln(2)
Time = 0.00 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.20 \[ \int \frac {2^{2 x}}{\sqrt {a-2^{-x} b}} \, dx=\frac {2^{-2-\frac {x}{2}} \left (2^{x/2} \sqrt {a} \left (2^{1+2 x} a^2+2^x a b-3 b^2\right )+3 \sqrt {2^x a-b} b^2 \text {arctanh}\left (\frac {2^{x/2} \sqrt {a}}{\sqrt {2^x a-b}}\right )\right )}{a^{5/2} \sqrt {a-2^{-x} b} \log (2)} \] Input:
Integrate[2^(2*x)/Sqrt[a - b/2^x],x]
Output:
(2^(-2 - x/2)*(2^(x/2)*Sqrt[a]*(2^(1 + 2*x)*a^2 + 2^x*a*b - 3*b^2) + 3*Sqr t[2^x*a - b]*b^2*ArcTanh[(2^(x/2)*Sqrt[a])/Sqrt[2^x*a - b]]))/(a^(5/2)*Sqr t[a - b/2^x]*Log[2])
Time = 0.38 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 |
\(\displaystyle -\frac {\frac {3 b \left (-\frac {\int \frac {1}{\frac {a}{b}-\frac {2^{-2 x}}{b}}d\sqrt {a-2^{-x} b}}{a}-\frac {2^x \sqrt {a-b 2^{-x}}}{a}\right )}{4 a}-\frac {2^{2 x-1} \sqrt {a-b 2^{-x}}}{a}}{\log (2)}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {\frac {3 b \left (-\frac {b \text {arctanh}\left (\frac {\sqrt {a-b 2^{-x}}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {2^x \sqrt {a-b 2^{-x}}}{a}\right )}{4 a}-\frac {2^{2 x-1} \sqrt {a-b 2^{-x}}}{a}}{\log (2)}\) |
Input:
Int[2^(2*x)/Sqrt[a - b/2^x],x]
Output:
-((-((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])
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. \(197\) vs. \(2(84)=168\).
Time = 0.10 (sec) , antiderivative size = 198, normalized size of antiderivative = 2.06
method | result | size |
derivativedivides | \(\frac {\sqrt {2^{-x} \left (a 2^{x}-b \right )}\, 2^{x} \left (4 a^{\frac {5}{2}} \sqrt {a 2^{2 x}-2^{x} b}\, 2^{x}-2 a^{\frac {3}{2}} \sqrt {a 2^{2 x}-2^{x} b}\, b +8 a^{\frac {3}{2}} \sqrt {\left (a 2^{x}-b \right ) 2^{x}}\, b +4 a \ln \left (\frac {2 \sqrt {\left (a 2^{x}-b \right ) 2^{x}}\, \sqrt {a}+2 a 2^{x}-b}{2 \sqrt {a}}\right ) b^{2}-b^{2} \ln \left (\frac {2 \sqrt {a 2^{2 x}-2^{x} b}\, \sqrt {a}+2 a 2^{x}-b}{2 \sqrt {a}}\right ) a \right )}{8 \ln \left (2\right ) \sqrt {\left (a 2^{x}-b \right ) 2^{x}}\, a^{\frac {7}{2}}}\) | \(198\) |
default | \(\frac {\sqrt {2^{-x} \left (a 2^{x}-b \right )}\, 2^{x} \left (4 a^{\frac {5}{2}} \sqrt {a 2^{2 x}-2^{x} b}\, 2^{x}-2 a^{\frac {3}{2}} \sqrt {a 2^{2 x}-2^{x} b}\, b +8 a^{\frac {3}{2}} \sqrt {\left (a 2^{x}-b \right ) 2^{x}}\, b +4 a \ln \left (\frac {2 \sqrt {\left (a 2^{x}-b \right ) 2^{x}}\, \sqrt {a}+2 a 2^{x}-b}{2 \sqrt {a}}\right ) b^{2}-b^{2} \ln \left (\frac {2 \sqrt {a 2^{2 x}-2^{x} b}\, \sqrt {a}+2 a 2^{x}-b}{2 \sqrt {a}}\right ) a \right )}{8 \ln \left (2\right ) \sqrt {\left (a 2^{x}-b \right ) 2^{x}}\, a^{\frac {7}{2}}}\) | \(198\) |
Input:
int(2^(2*x)/(a-b/(2^x))^(1/2),x,method=_RETURNVERBOSE)
Output:
1/8/ln(2)*(1/(2^x)*(a*2^x-b))^(1/2)*2^x*(4*a^(5/2)*(a*(2^x)^2-2^x*b)^(1/2) *2^x-2*a^(3/2)*(a*(2^x)^2-2^x*b)^(1/2)*b+8*a^(3/2)*((a*2^x-b)*2^x)^(1/2)*b +4*a*ln(1/2*(2*((a*2^x-b)*2^x)^(1/2)*a^(1/2)+2*a*2^x-b)/a^(1/2))*b^2-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 = 0.09 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.93 \[ \int \frac {2^{2 x}}{\sqrt {a-2^{-x} b}} \, dx=\left [\frac {3 \, \sqrt {a} b^{2} \log \left (-2 \cdot 2^{x} a - 2 \cdot 2^{x} \sqrt {a} \sqrt {\frac {2^{x} a - b}{2^{x}}} + b\right ) + 2 \, {\left (2 \cdot 2^{2 \, x} a^{2} + 3 \cdot 2^{x} a b\right )} \sqrt {\frac {2^{x} a - b}{2^{x}}}}{8 \, a^{3} \log \left (2\right )}, -\frac {3 \, \sqrt {-a} b^{2} \arctan \left (\frac {2^{x} \sqrt {-a} \sqrt {\frac {2^{x} a - b}{2^{x}}}}{2^{x} a - b}\right ) - {\left (2 \cdot 2^{2 \, x} a^{2} + 3 \cdot 2^{x} a b\right )} \sqrt {\frac {2^{x} a - b}{2^{x}}}}{4 \, a^{3} \log \left (2\right )}\right ] \] Input:
integrate(2^(2*x)/(a-b/(2^x))^(1/2),x, algorithm="fricas")
Output:
[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(2^x*sqrt(-a)*sqrt((2^x*a - b)/2^x)/(2^x*a - b)) - (2*2^(2*x)*a^2 + 3*2^x*a*b)*sqrt((2^x*a - b)/2^x))/(a^3*log(2))]
\[ \int \frac {2^{2 x}}{\sqrt {a-2^{-x} b}} \, dx=\int \frac {2^{2 x}}{\sqrt {a - 2^{- x} b}}\, dx \] Input:
integrate(2**(2*x)/(a-b/(2**x))**(1/2),x)
Output:
Integral(2**(2*x)/sqrt(a - b/2**x), x)
Time = 0.11 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.35 \[ \int \frac {2^{2 x}}{\sqrt {a-2^{-x} b}} \, dx=-\frac {3 \, b^{2} \log \left (\frac {\sqrt {a - \frac {b}{2^{x}}} - \sqrt {a}}{\sqrt {a - \frac {b}{2^{x}}} + \sqrt {a}}\right )}{8 \, a^{\frac {5}{2}} \log \left (2\right )} - \frac {3 \, {\left (a - \frac {b}{2^{x}}\right )}^{\frac {3}{2}} b^{2} - 5 \, \sqrt {a - \frac {b}{2^{x}}} a b^{2}}{4 \, {\left ({\left (a - \frac {b}{2^{x}}\right )}^{2} a^{2} - 2 \, {\left (a - \frac {b}{2^{x}}\right )} a^{3} + a^{4}\right )} \log \left (2\right )} \] Input:
integrate(2^(2*x)/(a-b/(2^x))^(1/2),x, algorithm="maxima")
Output:
-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 = 0.16 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00 \[ \int \frac {2^{2 x}}{\sqrt {a-2^{-x} b}} \, dx=\frac {2 \, \sqrt {2^{2 \, x} a - 2^{x} b} {\left (\frac {2 \cdot 2^{x}}{a} + \frac {3 \, b}{a^{2}}\right )} - \frac {3 \, b^{2} \log \left ({\left | 2 \, {\left (2^{x} \sqrt {a} - \sqrt {2^{2 \, x} a - 2^{x} b}\right )} \sqrt {a} - b \right |}\right )}{a^{\frac {5}{2}}} + \frac {3 \, b^{2} \log \left ({\left | b \right |}\right )}{a^{\frac {5}{2}}}}{8 \, \log \left (2\right )} \] Input:
integrate(2^(2*x)/(a-b/(2^x))^(1/2),x, algorithm="giac")
Output:
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. \[ \int \frac {2^{2 x}}{\sqrt {a-2^{-x} b}} \, dx=\int \frac {2^{2\,x}}{\sqrt {a-\frac {b}{2^x}}} \,d x \] Input:
int(2^(2*x)/(a - b/2^x)^(1/2),x)
Output:
int(2^(2*x)/(a - b/2^x)^(1/2), x)
\[ \int \frac {2^{2 x}}{\sqrt {a-2^{-x} b}} \, dx=\frac {4 \sqrt {2^{x} a -b}\, 2^{x} 2^{\frac {x}{2}} a +6 \sqrt {2^{x} a -b}\, 2^{\frac {x}{2}} b +3 \left (\int \frac {\sqrt {2^{x} a -b}\, 2^{\frac {x}{2}}}{2^{x} a -b}d x \right ) \mathrm {log}\left (2\right ) b^{2}}{8 \,\mathrm {log}\left (2\right ) a^{2}} \] Input:
int(2^(2*x)/(a-b/(2^x))^(1/2),x)
Output:
(4*sqrt(2**x*a - b)*2**x*2**(x/2)*a + 6*sqrt(2**x*a - b)*2**(x/2)*b + 3*in t((sqrt(2**x*a - b)*2**(x/2))/(2**x*a - b),x)*log(2)*b**2)/(8*log(2)*a**2)