\(\int \frac {2^{2 x}}{a+2^x b} \, dx\) [474]

Optimal result

Integrand size = 15, antiderivative size = 30 \[ \int \frac {2^{2 x}}{a+2^x b} \, dx=\frac {2^x}{b \log (2)}-\frac {a \log \left (a+2^x b\right )}{b^2 \log (2)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 30, 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 = {2280, 45} \[ \int \frac {2^{2 x}}{a+2^x b} \, dx=\frac {2^x}{b \log (2)}-\frac {a \log \left (a+b 2^x\right )}{b^2 \log (2)} \]

[In]

Int[2^(2*x)/(a + 2^x*b),x]

[Out]

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

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[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 {x}{a+b x} \, dx,x,2^x\right )}{\log (2)} \\ & = \frac {\text {Subst}\left (\int \left (\frac {1}{b}-\frac {a}{b (a+b x)}\right ) \, dx,x,2^x\right )}{\log (2)} \\ & = \frac {2^x}{b \log (2)}-\frac {a \log \left (a+2^x b\right )}{b^2 \log (2)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {2^{2 x}}{a+2^x b} \, dx=\frac {\frac {2^x}{b}-\frac {a \log \left (a+2^x b\right )}{b^2}}{\log (2)} \]

[In]

Integrate[2^(2*x)/(a + 2^x*b),x]

[Out]

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

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93

method	result	size
derivativedivides	\(\frac {\frac {2^{x}}{b}-\frac {a \ln \left (a +2^{x} b \right )}{b^{2}}}{\ln \left (2\right )}\)	\(28\)
default	\(\frac {\frac {2^{x}}{b}-\frac {a \ln \left (a +2^{x} b \right )}{b^{2}}}{\ln \left (2\right )}\)	\(28\)
risch	\(\frac {2^{x}}{b \ln \left (2\right )}-\frac {a \ln \left (2^{x}+\frac {a}{b}\right )}{\ln \left (2\right ) b^{2}}\)	\(33\)
norman	\(\frac {{\mathrm e}^{x \ln \left (2\right )}}{\ln \left (2\right ) b}-\frac {a \ln \left (a +{\mathrm e}^{x \ln \left (2\right )} b \right )}{\ln \left (2\right ) b^{2}}\)	\(35\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {2^{2 x}}{a+2^x b} \, dx=\frac {2^{x} b - a \log \left (2^{x} b + a\right )}{b^{2} \log \left (2\right )} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {2^{2 x}}{a+2^x b} \, dx=- \frac {a \log {\left (2^{x} + \frac {a}{b} \right )}}{b^{2} \log {\left (2 \right )}} + \begin {cases} \frac {2^{x}}{b \log {\left (2 \right )}} & \text {for}\: b \log {\left (2 \right )} \neq 0 \\\frac {x}{b} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {2^{2 x}}{a+2^x b} \, dx=\frac {2^{x}}{b \log \left (2\right )} - \frac {a \log \left (2^{x} b + a\right )}{b^{2} \log \left (2\right )} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {2^{2 x}}{a+2^x b} \, dx=\frac {2^{x}}{b \log \left (2\right )} - \frac {a \log \left ({\left | 2^{x} b + a \right |}\right )}{b^{2} \log \left (2\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {2^{2 x}}{a+2^x b} \, dx=-\frac {a\,\ln \left (a+2^x\,b\right )-2^x\,b}{b^2\,\ln \left (2\right )} \]

[In]

int(2^(2*x)/(a + 2^x*b),x)

[Out]

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