\(\int \frac {2^x}{\sqrt {a+2^{2 x} b}} \, dx\) [490]

Optimal result

Integrand size = 17, antiderivative size = 31 \[ \int \frac {2^x}{\sqrt {a+2^{2 x} b}} \, dx=\frac {\text {arctanh}\left (\frac {2^x \sqrt {b}}{\sqrt {a+4^x b}}\right )}{\sqrt {b} \log (2)} \]

[Out]

arctanh(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 = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2281, 223, 212} \[ \int \frac {2^x}{\sqrt {a+2^{2 x} b}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} 2^x}{\sqrt {a+b 4^x}}\right )}{\sqrt {b} \log (2)} \]

[In]

Int[2^x/Sqrt[a + 2^(2*x)*b],x]

[Out]

ArcTanh[(2^x*Sqrt[b])/Sqrt[a + 4^x*b]]/(Sqrt[b]*Log[2])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[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 {\tanh ^{-1}\left (\frac {2^x \sqrt {b}}{\sqrt {a+4^x b}}\right )}{\sqrt {b} \log (2)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06 \[ \int \frac {2^x}{\sqrt {a+2^{2 x} b}} \, dx=\frac {\text {arctanh}\left (\frac {2^x \sqrt {b}}{\sqrt {a+2^{2 x} b}}\right )}{\sqrt {b} \log (2)} \]

[In]

Integrate[2^x/Sqrt[a + 2^(2*x)*b],x]

[Out]

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

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.48 \[ \int \frac {2^x}{\sqrt {a+2^{2 x} b}} \, dx=\left [\frac {\log \left (-2 \, \sqrt {2^{2 \, x} b + a} 2^{x} \sqrt {b} - 2 \cdot 2^{2 \, x} b - a\right )}{2 \, \sqrt {b} \log \left (2\right )}, -\frac {\sqrt {-b} \arctan \left (\frac {2^{x} \sqrt {-b}}{\sqrt {2^{2 \, x} b + a}}\right )}{b \log \left (2\right )}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.97 \[ \int \frac {2^x}{\sqrt {a+2^{2 x} b}} \, dx=\frac {\begin {cases} \frac {\log {\left (2 \cdot 2^{x} b + 2 \sqrt {b} \sqrt {2^{2 x} b + a} \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \wedge b \neq 0 \\\frac {2^{x} \log {\left (2^{x} \right )}}{\sqrt {2^{2 x} b}} & \text {for}\: b \neq 0 \\\frac {2^{x}}{\sqrt {a}} & \text {otherwise} \end {cases}}{\log {\left (2 \right )}} \]

[In]

integrate(2**x/(a+2**(2*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)/sqrt(
2**(2*x)*b), Ne(b, 0)), (2**x/sqrt(a), True))/log(2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71 \[ \int \frac {2^x}{\sqrt {a+2^{2 x} b}} \, dx=\frac {\operatorname {arsinh}\left (\frac {2^{x + 1} b}{2 \, \sqrt {a b}}\right )}{\sqrt {b} \log \left (2\right )} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {2^x}{\sqrt {a+2^{2 x} b}} \, dx=-\frac {\log \left ({\left | -2^{x} \sqrt {b} + \sqrt {2^{2 \, x} b + a} \right |}\right )}{\sqrt {b} \log \left (2\right )} \]

[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))

Mupad [B] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {2^x}{\sqrt {a+2^{2 x} b}} \, dx=\frac {\ln \left (\sqrt {a+2^{2\,x}\,b}+2^x\,\sqrt {b}\right )}{\sqrt {b}\,\ln \left (2\right )} \]

[In]

int(2^x/(a + 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))