3.474 \(\int \frac {2^{2 x}}{a+2^x b} \, dx\)

Optimal. Leaf size=30 \[ \frac {2^x}{b \log (2)}-\frac {a \log \left (a+b 2^x\right )}{b^2 \log (2)} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 30, normalized size of antiderivative = 1.00, 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 = {2248, 43} \[ \frac {2^x}{b \log (2)}-\frac {a \log \left (a+b 2^x\right )}{b^2 \log (2)} \]

Antiderivative was successfully verified.

[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 43

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 2248

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*} \int \frac {2^{2 x}}{a+2^x b} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x}{a+b x} \, dx,x,2^x\right )}{\log (2)}\\ &=\frac {\operatorname {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*}

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Mathematica [A]  time = 0.01, size = 27, normalized size = 0.90 \[ \frac {\frac {2^x}{b}-\frac {a \log \left (a+b 2^x\right )}{b^2}}{\log (2)} \]

Antiderivative was successfully verified.

[In]

Integrate[2^(2*x)/(a + 2^x*b),x]

[Out]

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

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fricas [A]  time = 0.43, size = 25, normalized size = 0.83 \[ \frac {2^{x} b - a \log \left (2^{x} b + a\right )}{b^{2} \log \relax (2)} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 0.30, size = 31, normalized size = 1.03 \[ \frac {2^{x}}{b \log \relax (2)} - \frac {a \log \left ({\left | 2^{x} b + a \right |}\right )}{b^{2} \log \relax (2)} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [A]  time = 0.02, size = 35, normalized size = 1.17 \[ -\frac {a \ln \left (b \,{\mathrm e}^{\ln \relax (2) x}+a \right )}{\ln \relax (2) b^{2}}+\frac {{\mathrm e}^{\ln \relax (2) x}}{\ln \relax (2) b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.86, size = 30, normalized size = 1.00 \[ \frac {2^{x}}{b \log \relax (2)} - \frac {a \log \left (2^{x} b + a\right )}{b^{2} \log \relax (2)} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [B]  time = 3.61, size = 26, normalized size = 0.87 \[ -\frac {a\,\ln \left (a+2^x\,b\right )-2^x\,b}{b^2\,\ln \relax (2)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2^(2*x)/(a + 2^x*b),x)

[Out]

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

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sympy [A]  time = 0.16, size = 31, normalized size = 1.03 \[ - \frac {a \log {\left (2^{x} + \frac {a}{b} \right )}}{b^{2} \log {\relax (2 )}} + \begin {cases} \frac {2^{x}}{b \log {\relax (2 )}} & \text {for}\: b \log {\relax (2 )} \neq 0 \\\frac {x}{b} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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