3.489 \(\int \frac {2^x}{\sqrt {a+4^x b}} \, dx\)

Optimal. Leaf size=31 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {b} 2^x}{\sqrt {a+b 4^x}}\right )}{\sqrt {b} \log (2)} \]

[Out]

arctanh(2^x*b^(1/2)/(a+4^x*b)^(1/2))/ln(2)/b^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2249, 217, 206} \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {b} 2^x}{\sqrt {a+b 4^x}}\right )}{\sqrt {b} \log (2)} \]

Antiderivative was successfully verified.

[In]

Int[2^x/Sqrt[a + 4^x*b],x]

[Out]

ArcTanh[(2^x*Sqrt[b])/Sqrt[a + 4^x*b]]/(Sqrt[b]*Log[2])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 2249

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_.)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Wit
h[{m = FullSimplify[(d*e*Log[F])/(g*h*Log[G])]}, Dist[Denominator[m]/(g*h*Log[G]), Subst[Int[x^(Denominator[m]
- 1)*(a + b*F^(c*e - (d*e*f)/g)*x^Numerator[m])^p, x], x, G^((h*(f + g*x))/Denominator[m])], x] /; LtQ[m, -1]
|| GtQ[m, 1]] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x]

Rubi steps

\begin {align*} \int \frac {2^x}{\sqrt {a+4^x b}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,2^x\right )}{\log (2)}\\ &=\frac {\operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {2^x}{\sqrt {a+4^x b}}\right )}{\log (2)}\\ &=\frac {\tanh ^{-1}\left (\frac {2^x \sqrt {b}}{\sqrt {a+4^x b}}\right )}{\sqrt {b} \log (2)}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 33, normalized size = 1.06 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {b} 2^x}{\sqrt {a+b 2^{2 x}}}\right )}{\sqrt {b} \log (2)} \]

Antiderivative was successfully verified.

[In]

Integrate[2^x/Sqrt[a + 4^x*b],x]

[Out]

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

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fricas [A]  time = 0.43, size = 77, normalized size = 2.48 \[ \left [\frac {\log \left (-2 \, \sqrt {2^{2 \, x} b + a} 2^{x} \sqrt {b} - 2 \cdot 2^{2 \, x} b - a\right )}{2 \, \sqrt {b} \log \relax (2)}, -\frac {\sqrt {-b} \arctan \left (\frac {2^{x} \sqrt {-b}}{\sqrt {2^{2 \, x} b + a}}\right )}{b \log \relax (2)}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2^x/(a+4^x*b)^(1/2),x, algorithm="fricas")

[Out]

[1/2*log(-2*sqrt(2^(2*x)*b + a)*2^x*sqrt(b) - 2*2^(2*x)*b - a)/(sqrt(b)*log(2)), -sqrt(-b)*arctan(2^x*sqrt(-b)
/sqrt(2^(2*x)*b + a))/(b*log(2))]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {2^{x}}{\sqrt {4^{x} b + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(2^x/sqrt(4^x*b + a), x)

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {2^{x}}{\sqrt {b 4^{x}+a}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2^x/(a+4^x*b)^(1/2),x)

[Out]

int(2^x/(a+4^x*b)^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {2^{x}}{\sqrt {4^{x} b + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2^x/(a+4^x*b)^(1/2),x, algorithm="maxima")

[Out]

integrate(2^x/sqrt(4^x*b + a), x)

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mupad [B]  time = 3.68, size = 28, normalized size = 0.90 \[ \frac {\ln \left (\sqrt {a+2^{2\,x}\,b}+2^x\,\sqrt {b}\right )}{\sqrt {b}\,\ln \relax (2)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2^x/(a + 4^x*b)^(1/2),x)

[Out]

log((a + 2^(2*x)*b)^(1/2) + 2^x*b^(1/2))/(b^(1/2)*log(2))

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sympy [A]  time = 0.82, size = 85, normalized size = 2.74 \[ \frac {\begin {cases} \frac {\sqrt {- \frac {a}{b}} \operatorname {asin}{\left (2^{x} \sqrt {- \frac {b}{a}} \right )}}{\sqrt {a}} & \text {for}\: a > 0 \wedge b < 0 \\\frac {\sqrt {\frac {a}{b}} \operatorname {asinh}{\left (2^{x} \sqrt {\frac {b}{a}} \right )}}{\sqrt {a}} & \text {for}\: a > 0 \wedge b > 0 \\\frac {\sqrt {- \frac {a}{b}} \operatorname {acosh}{\left (2^{x} \sqrt {- \frac {b}{a}} \right )}}{\sqrt {- a}} & \text {for}\: b > 0 \wedge a < 0 \end {cases}}{\log {\relax (2 )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2**x/(a+4**x*b)**(1/2),x)

[Out]

Piecewise((sqrt(-a/b)*asin(2**x*sqrt(-b/a))/sqrt(a), (a > 0) & (b < 0)), (sqrt(a/b)*asinh(2**x*sqrt(b/a))/sqrt
(a), (a > 0) & (b > 0)), (sqrt(-a/b)*acosh(2**x*sqrt(-b/a))/sqrt(-a), (b > 0) & (a < 0)))/log(2)

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