\(\int \frac {2^x}{a-2^{2 x} b} \, dx\) [484]

Optimal result

Integrand size = 16, antiderivative size = 30 \[ \int \frac {2^x}{a-2^{2 x} b} \, dx=\frac {\text {arctanh}\left (\frac {2^x \sqrt {b}}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (2)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2281, 214} \[ \int \frac {2^x}{a-2^{2 x} b} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} 2^x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (2)} \]

[In]

Int[2^x/(a - 2^(2*x)*b),x]

[Out]

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

Rule 214

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]

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}{a-b x^2} \, dx,x,2^x\right )}{\log (2)} \\ & = \frac {\tanh ^{-1}\left (\frac {2^x \sqrt {b}}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (2)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {2^x}{a-2^{2 x} b} \, dx=\frac {\text {arctanh}\left (\frac {2^x \sqrt {b}}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \log (2)} \]

[In]

Integrate[2^x/(a - 2^(2*x)*b),x]

[Out]

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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73

method	result	size
derivativedivides	\(\frac {\operatorname {arctanh}\left (\frac {b 2^{x}}{\sqrt {a b}}\right )}{\ln \left (2\right ) \sqrt {a b}}\)	\(22\)
default	\(\frac {\operatorname {arctanh}\left (\frac {b 2^{x}}{\sqrt {a b}}\right )}{\ln \left (2\right ) \sqrt {a b}}\)	\(22\)
risch	\(\frac {\ln \left (2^{x}+\frac {a}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, \ln \left (2\right )}-\frac {\ln \left (2^{x}-\frac {a}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, \ln \left (2\right )}\)	\(49\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 2.87 \[ \int \frac {2^x}{a-2^{2 x} b} \, dx=\left [\frac {\sqrt {a b} \log \left (\frac {2^{2 \, x} b + 2 \, \sqrt {a b} 2^{x} + a}{2^{2 \, x} b - a}\right )}{2 \, a b \log \left (2\right )}, -\frac {\sqrt {-a b} \arctan \left (\frac {\sqrt {-a b}}{2^{x} b}\right )}{a b \log \left (2\right )}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {2^x}{a-2^{2 x} b} \, dx=\frac {\operatorname {RootSum} {\left (4 z^{2} a b - 1, \left ( i \mapsto i \log {\left (2^{x} + 2 i a \right )} \right )\right )}}{\log {\left (2 \right )}} \]

[In]

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

[Out]

RootSum(4*_z**2*a*b - 1, Lambda(_i, _i*log(2**x + 2*_i*a)))/log(2)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (22) = 44\).

Time = 0.29 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.50 \[ \int \frac {2^x}{a-2^{2 x} b} \, dx=-\frac {\log \left (\frac {2^{x + 1} b - 2 \, \sqrt {a b}}{2^{x + 1} b + 2 \, \sqrt {a b}}\right )}{2 \, \sqrt {a b} \log \left (2\right )} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {2^x}{a-2^{2 x} b} \, dx=-\frac {\arctan \left (\frac {2^{x} b}{\sqrt {-a b}}\right )}{\sqrt {-a b} \log \left (2\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {2^x}{a-2^{2 x} b} \, dx=\frac {\mathrm {atanh}\left (\frac {2^x\,\sqrt {b}}{\sqrt {a}}\right )}{\sqrt {a}\,\sqrt {b}\,\ln \left (2\right )} \]

[In]

int(2^x/(a - 2^(2*x)*b),x)

[Out]

atanh((2^x*b^(1/2))/a^(1/2))/(a^(1/2)*b^(1/2)*log(2))