\(\int \frac {4^x}{a-2^{-x} b} \, dx\) [479]

Optimal result

Integrand size = 16, antiderivative size = 58 \[ \int \frac {4^x}{a-2^{-x} b} \, dx=\frac {b^2 x}{a^3}+\frac {2^{-1+2 x}}{a \log (2)}+\frac {2^x b}{a^2 \log (2)}+\frac {b^2 \log \left (a-2^{-x} b\right )}{a^3 \log (2)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2280, 46} \[ \int \frac {4^x}{a-2^{-x} b} \, dx=\frac {b^2 x}{a^3}+\frac {b^2 \log \left (a-b 2^{-x}\right )}{a^3 \log (2)}+\frac {b 2^x}{a^2 \log (2)}+\frac {2^{2 x-1}}{a \log (2)} \]

[In]

Int[4^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 46

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] && Lt
Q[m + n + 2, 0])

Rule 2280

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {1}{x^3 (a-b x)} \, dx,x,2^{-x}\right )}{\log (2)} \\ & = -\frac {\text {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] (verified)

Time = 0.10 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.66 \[ \int \frac {4^x}{a-2^{-x} b} \, dx=\frac {2^x a \left (2^x a+2 b\right )+2 b^2 \log \left (2^x a-b\right )}{a^3 \log (4)} \]

[In]

Integrate[4^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])

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.86

method	result	size
risch	\(\frac {2^{2 x}}{2 \ln \left (2\right ) a}+\frac {2^{x} b}{a^{2} \ln \left (2\right )}+\frac {b^{2} \ln \left (2^{x}-\frac {b}{a}\right )}{\ln \left (2\right ) a^{3}}\)	\(50\)
norman	\(\frac {b \,{\mathrm e}^{x \ln \left (2\right )}}{\ln \left (2\right ) a^{2}}+\frac {{\mathrm e}^{2 x \ln \left (2\right )}}{2 \ln \left (2\right ) a}+\frac {b^{2} \ln \left (a \,{\mathrm e}^{x \ln \left (2\right )}-b \right )}{\ln \left (2\right ) a^{3}}\)	\(55\)

[In]

int(4^x/(a-b/(2^x)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.71 \[ \int \frac {4^x}{a-2^{-x} b} \, dx=\frac {2^{2 \, x} a^{2} + 2 \cdot 2^{x} a b + 2 \, b^{2} \log \left (2^{x} a - b\right )}{2 \, a^{3} \log \left (2\right )} \]

[In]

integrate(4^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))

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.29 \[ \int \frac {4^x}{a-2^{-x} b} \, dx=\begin {cases} \frac {4^{x} a^{2} \log {\left (2 \right )} + 2 a b e^{\frac {x \log {\left (4 \right )}}{2}} \log {\left (2 \right )}}{2 a^{3} \log {\left (2 \right )}^{2}} & \text {for}\: a^{3} \log {\left (2 \right )}^{2} \neq 0 \\\frac {x \left (a + b\right )}{a^{2}} & \text {otherwise} \end {cases} + \frac {b^{2} \log {\left (e^{\frac {x \log {\left (4 \right )}}{2}} - \frac {b}{a} \right )}}{a^{3} \log {\left (2 \right )}} \]

[In]

integrate(4**x/(a-b/(2**x)),x)

[Out]

Piecewise(((4**x*a**2*log(2) + 2*a*b*exp(x*log(4)/2)*log(2))/(2*a**3*log(2)**2), Ne(a**3*log(2)**2, 0)), (x*(a
+ b)/a**2, True)) + b**2*log(exp(x*log(4)/2) - b/a)/(a**3*log(2))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00 \[ \int \frac {4^x}{a-2^{-x} b} \, dx=\frac {b^{2} x}{a^{3}} + \frac {{\left (2^{-x + 1} b + a\right )} 2^{2 \, x - 1}}{a^{2} \log \left (2\right )} + \frac {b^{2} \log \left (-a + \frac {b}{2^{x}}\right )}{a^{3} \log \left (2\right )} \]

[In]

integrate(4^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))

Giac [F]

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

[In]

integrate(4^x/(a-b/(2^x)),x, algorithm="giac")

[Out]

integrate(4^x/(a - b/2^x), x)

Mupad [F(-1)]

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

[In]

int(4^x/(a - b/2^x),x)

[Out]

int(4^x/(a - b/2^x), x)