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

Optimal. Leaf size=32 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} 2^x}{\sqrt {a-b 4^x}}\right )}{\sqrt {b} \log (2)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2249, 217, 203} \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} 2^x}{\sqrt {a-b 4^x}}\right )}{\sqrt {b} \log (2)} \]

Antiderivative was successfully verified.

[In]

Int[2^x/Sqrt[a - 2^(2*x)*b],x]

[Out]

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

Rule 203

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

Rule 217

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 2249

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*} \int \frac {2^x}{\sqrt {a-2^{2 x} b}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {a-b x^2}} \, dx,x,2^x\right )}{\log (2)}\\ &=\frac {\operatorname {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]  time = 0.00, size = 34, normalized size = 1.06 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} 2^x}{\sqrt {a-b 2^{2 x}}}\right )}{\sqrt {b} \log (2)} \]

Antiderivative was successfully verified.

[In]

Integrate[2^x/Sqrt[a - 2^(2*x)*b],x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.46, size = 92, normalized size = 2.88 \[ \left [-\frac {\sqrt {-b} \log \left (-2 \, \sqrt {-2^{2 \, x} b + a} 2^{x} \sqrt {-b} + 2 \cdot 2^{2 \, x} b - a\right )}{2 \, b \log \relax (2)}, -\frac {\arctan \left (\frac {\sqrt {-2^{2 \, x} b + a} 2^{x} \sqrt {b}}{2^{2 \, x} b - a}\right )}{\sqrt {b} \log \relax (2)}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2^x/(a-2^(2*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))]

________________________________________________________________________________________

giac [A]  time = 0.61, size = 36, normalized size = 1.12 \[ -\frac {\log \left ({\left | -2^{x} \sqrt {-b} + \sqrt {-2^{2 \, x} b + a} \right |}\right )}{\sqrt {-b} \log \relax (2)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {2^{x}}{\sqrt {-b 2^{2 x}+a}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 1.88, size = 22, normalized size = 0.69 \[ \frac {\arcsin \left (\frac {2^{x + 1} b}{2 \, \sqrt {a b}}\right )}{\sqrt {b} \log \relax (2)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 3.63, size = 33, normalized size = 1.03 \[ \frac {\ln \left (\sqrt {a-2^{2\,x}\,b}+2^x\,\sqrt {-b}\right )}{\sqrt {-b}\,\ln \relax (2)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2^x/(a - 2^(2*x)*b)^(1/2),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.97, size = 82, normalized size = 2.56 \[ \frac {\begin {cases} \frac {\sqrt {\frac {a}{b}} \operatorname {asin}{\left (2^{x} \sqrt {\frac {b}{a}} \right )}}{\sqrt {a}} & \text {for}\: a > 0 \wedge b > 0 \\\frac {\sqrt {- \frac {a}{b}} \operatorname {asinh}{\left (2^{x} \sqrt {- \frac {b}{a}} \right )}}{\sqrt {a}} & \text {for}\: a > 0 \wedge b < 0 \\\frac {\sqrt {\frac {a}{b}} \operatorname {acosh}{\left (2^{x} \sqrt {\frac {b}{a}} \right )}}{\sqrt {- a}} & \text {for}\: a < 0 \wedge b < 0 \end {cases}}{\log {\relax (2 )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((sqrt(a/b)*asin(2**x*sqrt(b/a))/sqrt(a), (a > 0) & (b > 0)), (sqrt(-a/b)*asinh(2**x*sqrt(-b/a))/sqrt
(a), (a > 0) & (b < 0)), (sqrt(a/b)*acosh(2**x*sqrt(b/a))/sqrt(-a), (a < 0) & (b < 0)))/log(2)

________________________________________________________________________________________