3.725 \(\int \frac{1+4^x}{1+2^{-x}} \, dx\)

Optimal. Leaf size=34 \[ \frac{2 \log \left (2^x+1\right )}{\log (2)}-\frac{2^x}{\log (2)}+\frac{2^{2 x-1}}{\log (2)} \]

[Out]

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

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Rubi [A]  time = 0.0312803, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2282, 697} \[ \frac{2 \log \left (2^x+1\right )}{\log (2)}-\frac{2^x}{\log (2)}+\frac{2^{2 x-1}}{\log (2)} \]

Antiderivative was successfully verified.

[In]

Int[(1 + 4^x)/(1 + 2^(-x)),x]

[Out]

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

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 697

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \frac{1+4^x}{1+2^{-x}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1+x^2}{1+x} \, dx,x,2^x\right )}{\log (2)}\\ &=\frac{\operatorname{Subst}\left (\int \left (-1+x+\frac{2}{1+x}\right ) \, dx,x,2^x\right )}{\log (2)}\\ &=-\frac{2^x}{\log (2)}+\frac{2^{-1+2 x}}{\log (2)}+\frac{2 \log \left (1+2^x\right )}{\log (2)}\\ \end{align*}

Mathematica [A]  time = 0.0214097, size = 23, normalized size = 0.68 \[ \frac{2^x \left (2^x-2\right )+4 \log \left (2^x+1\right )}{\log (4)} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + 4^x)/(1 + 2^(-x)),x]

[Out]

(2^x*(-2 + 2^x) + 4*Log[1 + 2^x])/Log[4]

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Maple [A]  time = 0.03, size = 40, normalized size = 1.2 \begin{align*} -{\frac{{{\rm e}^{x\ln \left ( 2 \right ) }}}{\ln \left ( 2 \right ) }}+{\frac{ \left ({{\rm e}^{x\ln \left ( 2 \right ) }} \right ) ^{2}}{2\,\ln \left ( 2 \right ) }}+2\,{\frac{\ln \left ( 1+{{\rm e}^{x\ln \left ( 2 \right ) }} \right ) }{\ln \left ( 2 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+4^x)/(1+1/(2^x)),x)

[Out]

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

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Maxima [A]  time = 1.45459, size = 54, normalized size = 1.59 \begin{align*} 2 \, x - \frac{2^{2 \, x - 1}{\left (2^{-x + 1} - 1\right )}}{\log \left (2\right )} + \frac{2 \, \log \left (\frac{1}{2^{x}} + 1\right )}{\log \left (2\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+4^x)/(1+1/(2^x)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.843412, size = 63, normalized size = 1.85 \begin{align*} \frac{2^{2 \, x} - 2 \cdot 2^{x} + 4 \, \log \left (2^{x} + 1\right )}{2 \, \log \left (2\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+4^x)/(1+1/(2^x)),x, algorithm="fricas")

[Out]

1/2*(2^(2*x) - 2*2^x + 4*log(2^x + 1))/log(2)

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Sympy [A]  time = 0.316153, size = 42, normalized size = 1.24 \begin{align*} \frac{4^{x} \log{\left (2 \right )} - 2 e^{\frac{x \log{\left (4 \right )}}{2}} \log{\left (2 \right )}}{2 \log{\left (2 \right )}^{2}} + \frac{2 \log{\left (e^{\frac{x \log{\left (4 \right )}}{2}} + 1 \right )}}{\log{\left (2 \right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+4**x)/(1+1/(2**x)),x)

[Out]

(4**x*log(2) - 2*exp(x*log(4)/2)*log(2))/(2*log(2)**2) + 2*log(exp(x*log(4)/2) + 1)/log(2)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{4^{x} + 1}{\frac{1}{2^{x}} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+4^x)/(1+1/(2^x)),x, algorithm="giac")

[Out]

integrate((4^x + 1)/(1/2^x + 1), x)