3.500 \(\int \frac{2^{2 x}}{\sqrt{a-2^x b}} \, dx\)

Optimal. Leaf size=46 \[ \frac{2 \left (a-b 2^x\right )^{3/2}}{3 b^2 \log (2)}-\frac{2 a \sqrt{a-b 2^x}}{b^2 \log (2)} \]

[Out]

(-2*a*Sqrt[a - 2^x*b])/(b^2*Log[2]) + (2*(a - 2^x*b)^(3/2))/(3*b^2*Log[2])

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Rubi [A]  time = 0.0436841, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2248, 43} \[ \frac{2 \left (a-b 2^x\right )^{3/2}}{3 b^2 \log (2)}-\frac{2 a \sqrt{a-b 2^x}}{b^2 \log (2)} \]

Antiderivative was successfully verified.

[In]

Int[2^(2*x)/Sqrt[a - 2^x*b],x]

[Out]

(-2*a*Sqrt[a - 2^x*b])/(b^2*Log[2]) + (2*(a - 2^x*b)^(3/2))/(3*b^2*Log[2])

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]

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])

Rubi steps

\begin{align*} \int \frac{2^{2 x}}{\sqrt{a-2^x b}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x}{\sqrt{a-b x}} \, dx,x,2^x\right )}{\log (2)}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{a}{b \sqrt{a-b x}}-\frac{\sqrt{a-b x}}{b}\right ) \, dx,x,2^x\right )}{\log (2)}\\ &=-\frac{2 a \sqrt{a-2^x b}}{b^2 \log (2)}+\frac{2 \left (a-2^x b\right )^{3/2}}{3 b^2 \log (2)}\\ \end{align*}

Mathematica [A]  time = 0.0117797, size = 30, normalized size = 0.65 \[ -\frac{2 \sqrt{a-b 2^x} \left (2 a+b 2^x\right )}{b^2 \log (8)} \]

Antiderivative was successfully verified.

[In]

Integrate[2^(2*x)/Sqrt[a - 2^x*b],x]

[Out]

(-2*Sqrt[a - 2^x*b]*(2*a + 2^x*b))/(b^2*Log[8])

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Maple [A]  time = 0.01, size = 29, normalized size = 0.6 \begin{align*} -{\frac{2\,{2}^{x}b+4\,a}{3\,{b}^{2}\ln \left ( 2 \right ) }\sqrt{a-{2}^{x}b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2^(2*x)/(a-2^x*b)^(1/2),x)

[Out]

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

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Maxima [A]  time = 0.978245, size = 54, normalized size = 1.17 \begin{align*} \frac{2 \,{\left (-2^{x} b + a\right )}^{\frac{3}{2}}}{3 \, b^{2} \log \left (2\right )} - \frac{2 \, \sqrt{-2^{x} b + a} a}{b^{2} \log \left (2\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2^(2*x)/(a-2^x*b)^(1/2),x, algorithm="maxima")

[Out]

2/3*(-2^x*b + a)^(3/2)/(b^2*log(2)) - 2*sqrt(-2^x*b + a)*a/(b^2*log(2))

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Fricas [A]  time = 1.49874, size = 68, normalized size = 1.48 \begin{align*} -\frac{2 \,{\left (2^{x} b + 2 \, a\right )} \sqrt{-2^{x} b + a}}{3 \, b^{2} \log \left (2\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2^(2*x)/(a-2^x*b)^(1/2),x, algorithm="fricas")

[Out]

-2/3*(2^x*b + 2*a)*sqrt(-2^x*b + a)/(b^2*log(2))

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Sympy [A]  time = 0.952638, size = 60, normalized size = 1.3 \begin{align*} \begin{cases} - \frac{2 \cdot 2^{x} \sqrt{- 2^{x} b + a}}{3 b \log{\left (2 \right )}} - \frac{4 a \sqrt{- 2^{x} b + a}}{3 b^{2} \log{\left (2 \right )}} & \text{for}\: b \neq 0 \\\frac{2^{2 x}}{2 \sqrt{a} \log{\left (2 \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2**(2*x)/(a-2**x*b)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.15602, size = 45, normalized size = 0.98 \begin{align*} \frac{2 \,{\left ({\left (-2^{x} b + a\right )}^{\frac{3}{2}} - 3 \, \sqrt{-2^{x} b + a} a\right )}}{3 \, b^{2} \log \left (2\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2^(2*x)/(a-2^x*b)^(1/2),x, algorithm="giac")

[Out]

2/3*((-2^x*b + a)^(3/2) - 3*sqrt(-2^x*b + a)*a)/(b^2*log(2))