\(\int \frac {2^{2 x}}{\sqrt {a-2^x b}} \, dx\) [500]

Optimal result

Integrand size = 18, antiderivative size = 46 \[ \int \frac {2^{2 x}}{\sqrt {a-2^x b}} \, dx=-\frac {2 a \sqrt {a-2^x b}}{b^2 \log (2)}+\frac {2 \left (a-2^x b\right )^{3/2}}{3 b^2 \log (2)} \]

[Out]

2/3*(a-2^x*b)^(3/2)/b^2/ln(2)-2*a*(a-2^x*b)^(1/2)/b^2/ln(2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2280, 45} \[ \int \frac {2^{2 x}}{\sqrt {a-2^x b}} \, dx=\frac {2 \left (a-b 2^x\right )^{3/2}}{3 b^2 \log (2)}-\frac {2 a \sqrt {a-b 2^x}}{b^2 \log (2)} \]

[In]

Int[2^(2*x)/Sqrt[a - 2^x*b],x]

[Out]

(-2*a*Sqrt[a - 2^x*b])/(b^2*Log[2]) + (2*(a - 2^x*b)^(3/2))/(3*b^2*Log[2])

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2280

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*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x}{\sqrt {a-b x}} \, dx,x,2^x\right )}{\log (2)} \\ & = \frac {\text {Subst}\left (\int \left (\frac {a}{b \sqrt {a-b x}}-\frac {\sqrt {a-b x}}{b}\right ) \, dx,x,2^x\right )}{\log (2)} \\ & = -\frac {2 a \sqrt {a-2^x b}}{b^2 \log (2)}+\frac {2 \left (a-2^x b\right )^{3/2}}{3 b^2 \log (2)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.65 \[ \int \frac {2^{2 x}}{\sqrt {a-2^x b}} \, dx=-\frac {2 \sqrt {a-2^x b} \left (2 a+2^x b\right )}{b^2 \log (8)} \]

[In]

Integrate[2^(2*x)/Sqrt[a - 2^x*b],x]

[Out]

(-2*Sqrt[a - 2^x*b]*(2*a + 2^x*b))/(b^2*Log[8])

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.63

method	result	size
risch	\(-\frac {2 \left (2^{x} b +2 a \right ) \sqrt {a -2^{x} b}}{3 b^{2} \ln \left (2\right )}\)	\(29\)
derivativedivides	\(\frac {\frac {2 \left (a -2^{x} b \right )^{\frac {3}{2}}}{3}-2 \sqrt {a -2^{x} b}\, a}{b^{2} \ln \left (2\right )}\)	\(36\)
default	\(\frac {\frac {2 \left (a -2^{x} b \right )^{\frac {3}{2}}}{3}-2 \sqrt {a -2^{x} b}\, a}{b^{2} \ln \left (2\right )}\)	\(36\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.61 \[ \int \frac {2^{2 x}}{\sqrt {a-2^x b}} \, dx=-\frac {2 \, {\left (2^{x} b + 2 \, a\right )} \sqrt {-2^{x} b + a}}{3 \, b^{2} \log \left (2\right )} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.30 \[ \int \frac {2^{2 x}}{\sqrt {a-2^x b}} \, dx=\begin {cases} - \frac {2 \cdot 2^{x} \sqrt {- 2^{x} b + a}}{3 b \log {\left (2 \right )}} - \frac {4 a \sqrt {- 2^{x} b + a}}{3 b^{2} \log {\left (2 \right )}} & \text {for}\: b \neq 0 \\\frac {2^{2 x}}{2 \sqrt {a} \log {\left (2 \right )}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-2*2**x*sqrt(-2**x*b + a)/(3*b*log(2)) - 4*a*sqrt(-2**x*b + a)/(3*b**2*log(2)), Ne(b, 0)), (2**(2*x
)/(2*sqrt(a)*log(2)), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.87 \[ \int \frac {2^{2 x}}{\sqrt {a-2^x b}} \, dx=\frac {2 \, {\left (-2^{x} b + a\right )}^{\frac {3}{2}}}{3 \, b^{2} \log \left (2\right )} - \frac {2 \, \sqrt {-2^{x} b + a} a}{b^{2} \log \left (2\right )} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.72 \[ \int \frac {2^{2 x}}{\sqrt {a-2^x b}} \, dx=\frac {2 \, {\left ({\left (-2^{x} b + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {-2^{x} b + a} a\right )}}{3 \, b^{2} \log \left (2\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.61 \[ \int \frac {2^{2 x}}{\sqrt {a-2^x b}} \, dx=-\frac {2\,\sqrt {a-2^x\,b}\,\left (2\,a+2^x\,b\right )}{3\,b^2\,\ln \left (2\right )} \]

[In]

int(2^(2*x)/(a - 2^x*b)^(1/2),x)

[Out]

-(2*(a - 2^x*b)^(1/2)*(2*a + 2^x*b))/(3*b^2*log(2))