3.478 \(\int \frac{2^{2 x}}{a+2^{-x} b} \, dx\)

Optimal. Leaf size=58 \[ \frac{b^2 x}{a^3}+\frac{b^2 \log \left (a+b 2^{-x}\right )}{a^3 \log (2)}-\frac{b 2^x}{a^2 \log (2)}+\frac{2^{2 x-1}}{a \log (2)} \]

[Out]

(b^2*x)/a^3 + 2^(-1 + 2*x)/(a*Log[2]) - (2^x*b)/(a^2*Log[2]) + (b^2*Log[a + b/2^x])/(a^3*Log[2])

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Rubi [A]  time = 0.0508679, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2248, 44} \[ \frac{b^2 x}{a^3}+\frac{b^2 \log \left (a+b 2^{-x}\right )}{a^3 \log (2)}-\frac{b 2^x}{a^2 \log (2)}+\frac{2^{2 x-1}}{a \log (2)} \]

Antiderivative was successfully verified.

[In]

Int[2^(2*x)/(a + b/2^x),x]

[Out]

(b^2*x)/a^3 + 2^(-1 + 2*x)/(a*Log[2]) - (2^x*b)/(a^2*Log[2]) + (b^2*Log[a + b/2^x])/(a^3*Log[2])

Rule 2248

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]

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{2^{2 x}}{a+2^{-x} b} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{1}{x^3 (a+b x)} \, dx,x,2^{-x}\right )}{\log (2)}\\ &=-\frac{\operatorname{Subst}\left (\int \left (\frac{1}{a x^3}-\frac{b}{a^2 x^2}+\frac{b^2}{a^3 x}-\frac{b^3}{a^3 (a+b x)}\right ) \, dx,x,2^{-x}\right )}{\log (2)}\\ &=\frac{b^2 x}{a^3}+\frac{2^{-1+2 x}}{a \log (2)}-\frac{2^x b}{a^2 \log (2)}+\frac{b^2 \log \left (a+2^{-x} b\right )}{a^3 \log (2)}\\ \end{align*}

Mathematica [A]  time = 0.0176555, size = 36, normalized size = 0.62 \[ \frac{2 b^2 \log \left (a 2^x+b\right )+a 2^x \left (a 2^x-2 b\right )}{a^3 \log (4)} \]

Antiderivative was successfully verified.

[In]

Integrate[2^(2*x)/(a + b/2^x),x]

[Out]

(2^x*a*(2^x*a - 2*b) + 2*b^2*Log[2^x*a + b])/(a^3*Log[4])

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Maple [A]  time = 0.01, size = 54, normalized size = 0.9 \begin{align*}{\frac{ \left ({{\rm e}^{x\ln \left ( 2 \right ) }} \right ) ^{2}}{2\,a\ln \left ( 2 \right ) }}-{\frac{{{\rm e}^{x\ln \left ( 2 \right ) }}b}{\ln \left ( 2 \right ){a}^{2}}}+{\frac{{b}^{2}\ln \left ( a{{\rm e}^{x\ln \left ( 2 \right ) }}+b \right ) }{{a}^{3}\ln \left ( 2 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2^(2*x)/(a+b/(2^x)),x)

[Out]

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

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Maxima [A]  time = 0.992382, size = 80, normalized size = 1.38 \begin{align*} \frac{b^{2} x}{a^{3}} - \frac{{\left (2^{-x + 1} b - a\right )} 2^{2 \, x - 1}}{a^{2} \log \left (2\right )} + \frac{b^{2} \log \left (a + \frac{b}{2^{x}}\right )}{a^{3} \log \left (2\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.57711, size = 90, normalized size = 1.55 \begin{align*} \frac{2^{2 \, x} a^{2} - 2 \cdot 2^{x} a b + 2 \, b^{2} \log \left (2^{x} a + b\right )}{2 \, a^{3} \log \left (2\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.192232, size = 68, normalized size = 1.17 \begin{align*} \begin{cases} \frac{2^{2 x} a^{2} \log{\left (2 \right )} - 2 \cdot 2^{x} a b \log{\left (2 \right )}}{2 a^{3} \log{\left (2 \right )}^{2}} & \text{for}\: 2 a^{3} \log{\left (2 \right )}^{2} \neq 0 \\\frac{x \left (a - b\right )}{a^{2}} & \text{otherwise} \end{cases} + \frac{b^{2} \log{\left (2^{x} + \frac{b}{a} \right )}}{a^{3} \log{\left (2 \right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.21713, size = 65, normalized size = 1.12 \begin{align*} \frac{b^{2} \log \left ({\left | 2^{x} a + b \right |}\right )}{a^{3} \log \left (2\right )} + \frac{2^{2 \, x} a \log \left (2\right ) - 2 \cdot 2^{x} b \log \left (2\right )}{2 \, a^{2} \log \left (2\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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