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

Optimal. Leaf size=22 \[ x-\frac{2 \log \left (2^x+1\right )}{\log (2)}+\frac{2^x}{\log (2)} \]

[Out]

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

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Rubi [A]  time = 0.0267354, antiderivative size = 22, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2282, 894} \[ x-\frac{2 \log \left (2^x+1\right )}{\log (2)}+\frac{2^x}{\log (2)} \]

Antiderivative was successfully verified.

[In]

Int[(1 + 4^x)/(1 + 2^x),x]

[Out]

x + 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 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps

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

Mathematica [A]  time = 0.0122895, size = 21, normalized size = 0.95 \[ \frac{2^x+x \log (2)-2 \log \left (2^x+1\right )}{\log (2)} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + 4^x)/(1 + 2^x),x]

[Out]

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

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Maple [A]  time = 0.027, size = 27, normalized size = 1.2 \begin{align*} x+{\frac{{{\rm e}^{x\ln \left ( 2 \right ) }}}{\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+2^x),x)

[Out]

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

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Maxima [A]  time = 1.45288, size = 30, normalized size = 1.36 \begin{align*} x + \frac{2^{x}}{\log \left (2\right )} - \frac{2 \, \log \left (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+2^x),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.651464, size = 57, normalized size = 2.59 \begin{align*} \frac{x \log \left (2\right ) + 2^{x} - 2 \, \log \left (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+2^x),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.289959, size = 29, normalized size = 1.32 \begin{align*} x + \frac{e^{\frac{x \log{\left (4 \right )}}{2}}}{\log{\left (2 \right )}} - \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+2**x),x)

[Out]

x + exp(x*log(4)/2)/log(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}{2^{x} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((4^x + 1)/(2^x + 1), x)