\(\int \frac {1+4^x}{1+2^{-x}} \, dx\) [725]

Optimal result

Integrand size = 15, antiderivative size = 34 \[ \int \frac {1+4^x}{1+2^{-x}} \, dx=-\frac {2^x}{\log (2)}+\frac {2^{-1+2 x}}{\log (2)}+\frac {2 \log \left (1+2^x\right )}{\log (2)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2320, 711} \[ \int \frac {1+4^x}{1+2^{-x}} \, dx=\frac {2 \log \left (2^x+1\right )}{\log (2)}-\frac {2^x}{\log (2)}+\frac {2^{2 x-1}}{\log (2)} \]

[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 711

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]

Rule 2320

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]]]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1+x^2}{1+x} \, dx,x,2^x\right )}{\log (2)} \\ & = \frac {\text {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] (verified)

Time = 0.08 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.68 \[ \int \frac {1+4^x}{1+2^{-x}} \, dx=\frac {2^x \left (-2+2^x\right )+4 \log \left (1+2^x\right )}{\log (4)} \]

[In]

Integrate[(1 + 4^x)/(1 + 2^(-x)),x]

[Out]

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

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00

method	result	size
risch	\(-\frac {2^{x}}{\ln \left (2\right )}+\frac {2^{2 x}}{2 \ln \left (2\right )}+\frac {2 \ln \left (1+2^{x}\right )}{\ln \left (2\right )}\)	\(34\)
norman	\(-\frac {{\mathrm e}^{x \ln \left (2\right )}}{\ln \left (2\right )}+\frac {{\mathrm e}^{2 x \ln \left (2\right )}}{2 \ln \left (2\right )}+\frac {2 \ln \left (1+{\mathrm e}^{x \ln \left (2\right )}\right )}{\ln \left (2\right )}\)	\(40\)

[In]

int((1+4^x)/(1+1/(2^x)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74 \[ \int \frac {1+4^x}{1+2^{-x}} \, dx=\frac {2^{2 \, x} - 2 \cdot 2^{x} + 4 \, \log \left (2^{x} + 1\right )}{2 \, \log \left (2\right )} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.24 \[ \int \frac {1+4^x}{1+2^{-x}} \, dx=\frac {4^{x} \log {\left (2 \right )} - 2 e^{\frac {x \log {\left (4 \right )}}{2}} \log {\left (2 \right )}}{2 \log {\left (2 \right )}^{2}} + \frac {2 \log {\left (e^{\frac {x \log {\left (4 \right )}}{2}} + 1 \right )}}{\log {\left (2 \right )}} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.18 \[ \int \frac {1+4^x}{1+2^{-x}} \, dx=2 \, x - \frac {2^{2 \, x - 1} {\left (2^{-x + 1} - 1\right )}}{\log \left (2\right )} + \frac {2 \, \log \left (\frac {1}{2^{x}} + 1\right )}{\log \left (2\right )} \]

[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)

Giac [F]

\[ \int \frac {1+4^x}{1+2^{-x}} \, dx=\int { \frac {4^{x} + 1}{\frac {1}{2^{x}} + 1} \,d x } \]

[In]

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

[Out]

integrate((4^x + 1)/(1/2^x + 1), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1+4^x}{1+2^{-x}} \, dx=\int \frac {4^x+1}{\frac {1}{2^x}+1} \,d x \]

[In]

int((4^x + 1)/(1/2^x + 1),x)

[Out]

int((4^x + 1)/(1/2^x + 1), x)