\(\int \frac {2^x}{\sqrt {a-4^x b}} \, dx\) [491]

Optimal result

Integrand size = 16, antiderivative size = 32 \[ \int \frac {2^x}{\sqrt {a-4^x b}} \, dx=\frac {\arctan \left (\frac {2^x \sqrt {b}}{\sqrt {a-4^x b}}\right )}{\sqrt {b} \log (2)} \]

[Out]

arctan(2^x*b^(1/2)/(a-4^x*b)^(1/2))/ln(2)/b^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2281, 223, 209} \[ \int \frac {2^x}{\sqrt {a-4^x b}} \, dx=\frac {\arctan \left (\frac {\sqrt {b} 2^x}{\sqrt {a-b 4^x}}\right )}{\sqrt {b} \log (2)} \]

[In]

Int[2^x/Sqrt[a - 4^x*b],x]

[Out]

ArcTan[(2^x*Sqrt[b])/Sqrt[a - 4^x*b]]/(Sqrt[b]*Log[2])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 2281

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_.)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Wit
h[{m = FullSimplify[d*e*(Log[F]/(g*h*Log[G]))]}, Dist[Denominator[m]/(g*h*Log[G]), Subst[Int[x^(Denominator[m]
- 1)*(a + b*F^(c*e - d*e*(f/g))*x^Numerator[m])^p, x], x, G^(h*((f + g*x)/Denominator[m]))], x] /; LtQ[m, -1]
|| GtQ[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 {1}{\sqrt {a-b x^2}} \, dx,x,2^x\right )}{\log (2)} \\ & = \frac {\text {Subst}\left (\int \frac {1}{1+b x^2} \, dx,x,\frac {2^x}{\sqrt {a-4^x b}}\right )}{\log (2)} \\ & = \frac {\tan ^{-1}\left (\frac {2^x \sqrt {b}}{\sqrt {a-4^x b}}\right )}{\sqrt {b} \log (2)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {2^x}{\sqrt {a-4^x b}} \, dx=\frac {\arctan \left (\frac {2^x \sqrt {b}}{\sqrt {a-2^{2 x} b}}\right )}{\sqrt {b} \log (2)} \]

[In]

Integrate[2^x/Sqrt[a - 4^x*b],x]

[Out]

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

Maple [F]

\[\int \frac {2^{x}}{\sqrt {a -4^{x} b}}d x\]

[In]

int(2^x/(a-4^x*b)^(1/2),x)

[Out]

int(2^x/(a-4^x*b)^(1/2),x)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 92, normalized size of antiderivative = 2.88 \[ \int \frac {2^x}{\sqrt {a-4^x b}} \, dx=\left [-\frac {\sqrt {-b} \log \left (-2 \, \sqrt {-2^{2 \, x} b + a} 2^{x} \sqrt {-b} + 2 \cdot 2^{2 \, x} b - a\right )}{2 \, b \log \left (2\right )}, -\frac {\arctan \left (\frac {\sqrt {-2^{2 \, x} b + a} 2^{x} \sqrt {b}}{2^{2 \, x} b - a}\right )}{\sqrt {b} \log \left (2\right )}\right ] \]

[In]

integrate(2^x/(a-4^x*b)^(1/2),x, algorithm="fricas")

[Out]

[-1/2*sqrt(-b)*log(-2*sqrt(-2^(2*x)*b + a)*2^x*sqrt(-b) + 2*2^(2*x)*b - a)/(b*log(2)), -arctan(sqrt(-2^(2*x)*b
+ a)*2^x*sqrt(b)/(2^(2*x)*b - a))/(sqrt(b)*log(2))]

Sympy [A] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.06 \[ \int \frac {2^x}{\sqrt {a-4^x b}} \, dx=\frac {\begin {cases} \frac {\log {\left (- 2 \cdot 2^{x} b + 2 \sqrt {- b} \sqrt {- 2^{2 x} b + a} \right )}}{\sqrt {- b}} & \text {for}\: a \neq 0 \wedge b \neq 0 \\\frac {2^{x} \log {\left (2^{x} \right )}}{\sqrt {- 2^{2 x} b}} & \text {for}\: b \neq 0 \\\frac {2^{x}}{\sqrt {a}} & \text {otherwise} \end {cases}}{\log {\left (2 \right )}} \]

[In]

integrate(2**x/(a-4**x*b)**(1/2),x)

[Out]

Piecewise((log(-2*2**x*b + 2*sqrt(-b)*sqrt(-2**(2*x)*b + a))/sqrt(-b), Ne(a, 0) & Ne(b, 0)), (2**x*log(2**x)/s
qrt(-2**(2*x)*b), Ne(b, 0)), (2**x/sqrt(a), True))/log(2)

Maxima [F]

\[ \int \frac {2^x}{\sqrt {a-4^x b}} \, dx=\int { \frac {2^{x}}{\sqrt {-4^{x} b + a}} \,d x } \]

[In]

integrate(2^x/(a-4^x*b)^(1/2),x, algorithm="maxima")

[Out]

integrate(2^x/sqrt(-4^x*b + a), x)

Giac [F]

\[ \int \frac {2^x}{\sqrt {a-4^x b}} \, dx=\int { \frac {2^{x}}{\sqrt {-4^{x} b + a}} \,d x } \]

[In]

integrate(2^x/(a-4^x*b)^(1/2),x, algorithm="giac")

[Out]

integrate(2^x/sqrt(-4^x*b + a), x)

Mupad [B] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03 \[ \int \frac {2^x}{\sqrt {a-4^x b}} \, dx=\frac {\ln \left (\sqrt {a-2^{2\,x}\,b}+2^x\,\sqrt {-b}\right )}{\sqrt {-b}\,\ln \left (2\right )} \]

[In]

int(2^x/(a - 4^x*b)^(1/2),x)

[Out]

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