\(\int \frac {4^x}{\sqrt {a-2^{-x} b}} \, dx\) [503]

Optimal result

Integrand size = 18, antiderivative size = 96 \[ \int \frac {4^x}{\sqrt {a-2^{-x} b}} \, dx=\frac {2^{-1+2 x} \sqrt {a-2^{-x} b}}{a \log (2)}+\frac {3\ 2^{-2+x} b \sqrt {a-2^{-x} b}}{a^2 \log (2)}+\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {a-2^{-x} b}}{\sqrt {a}}\right )}{4 a^{5/2} \log (2)} \]

[Out]

3/4*b^2*arctanh((a-b/(2^x))^(1/2)/a^(1/2))/a^(5/2)/ln(2)+2^(-1+2*x)*(a-b/(2^x))^(1/2)/a/ln(2)+3*2^(-2+x)*b*(a-
b/(2^x))^(1/2)/a^2/ln(2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2280, 44, 65, 214} \[ \int \frac {4^x}{\sqrt {a-2^{-x} b}} \, dx=\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {a-b 2^{-x}}}{\sqrt {a}}\right )}{4 a^{5/2} \log (2)}+\frac {3 b 2^{x-2} \sqrt {a-b 2^{-x}}}{a^2 \log (2)}+\frac {2^{2 x-1} \sqrt {a-b 2^{-x}}}{a \log (2)} \]

[In]

Int[4^x/Sqrt[a - b/2^x],x]

[Out]

(2^(-1 + 2*x)*Sqrt[a - b/2^x])/(a*Log[2]) + (3*2^(-2 + x)*b*Sqrt[a - b/2^x])/(a^2*Log[2]) + (3*b^2*ArcTanh[Sqr
t[a - b/2^x]/Sqrt[a]])/(4*a^(5/2)*Log[2])

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

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

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 \sqrt {a-b x}} \, dx,x,2^{-x}\right )}{\log (2)} \\ & = \frac {2^{-1+2 x} \sqrt {a-2^{-x} b}}{a \log (2)}-\frac {(3 b) \text {Subst}\left (\int \frac {1}{x^2 \sqrt {a-b x}} \, dx,x,2^{-x}\right )}{4 a \log (2)} \\ & = \frac {2^{-1+2 x} \sqrt {a-2^{-x} b}}{a \log (2)}+\frac {3\ 2^{-2+x} b \sqrt {a-2^{-x} b}}{a^2 \log (2)}-\frac {\left (3 b^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a-b x}} \, dx,x,2^{-x}\right )}{8 a^2 \log (2)} \\ & = \frac {2^{-1+2 x} \sqrt {a-2^{-x} b}}{a \log (2)}+\frac {3\ 2^{-2+x} b \sqrt {a-2^{-x} b}}{a^2 \log (2)}+\frac {(3 b) \text {Subst}\left (\int \frac {1}{\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a-2^{-x} b}\right )}{4 a^2 \log (2)} \\ & = \frac {2^{-1+2 x} \sqrt {a-2^{-x} b}}{a \log (2)}+\frac {3\ 2^{-2+x} b \sqrt {a-2^{-x} b}}{a^2 \log (2)}+\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {a-2^{-x} b}}{\sqrt {a}}\right )}{4 a^{5/2} \log (2)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.20 \[ \int \frac {4^x}{\sqrt {a-2^{-x} b}} \, dx=\frac {2^{-2-\frac {x}{2}} \left (2^{x/2} \sqrt {a} \left (2^{1+2 x} a^2+2^x a b-3 b^2\right )+3 \sqrt {2^x a-b} b^2 \text {arctanh}\left (\frac {2^{x/2} \sqrt {a}}{\sqrt {2^x a-b}}\right )\right )}{a^{5/2} \sqrt {a-2^{-x} b} \log (2)} \]

[In]

Integrate[4^x/Sqrt[a - b/2^x],x]

[Out]

(2^(-2 - x/2)*(2^(x/2)*Sqrt[a]*(2^(1 + 2*x)*a^2 + 2^x*a*b - 3*b^2) + 3*Sqrt[2^x*a - b]*b^2*ArcTanh[(2^(x/2)*Sq
rt[a])/Sqrt[2^x*a - b]]))/(a^(5/2)*Sqrt[a - b/2^x]*Log[2])

Maple [F]

\[\int \frac {4^{x}}{\sqrt {a -b 2^{-x}}}d x\]

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.81 \[ \int \frac {4^x}{\sqrt {a-2^{-x} b}} \, dx=\left [\frac {3 \, \sqrt {a} b^{2} \log \left (-2 \cdot 2^{x} a - 2 \cdot 2^{x} \sqrt {a} \sqrt {\frac {2^{x} a - b}{2^{x}}} + b\right ) + 2 \, {\left (2 \cdot 2^{2 \, x} a^{2} + 3 \cdot 2^{x} a b\right )} \sqrt {\frac {2^{x} a - b}{2^{x}}}}{8 \, a^{3} \log \left (2\right )}, -\frac {3 \, \sqrt {-a} b^{2} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {2^{x} a - b}{2^{x}}}}{a}\right ) - {\left (2 \cdot 2^{2 \, x} a^{2} + 3 \cdot 2^{x} a b\right )} \sqrt {\frac {2^{x} a - b}{2^{x}}}}{4 \, a^{3} \log \left (2\right )}\right ] \]

[In]

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

[Out]

[1/8*(3*sqrt(a)*b^2*log(-2*2^x*a - 2*2^x*sqrt(a)*sqrt((2^x*a - b)/2^x) + b) + 2*(2*2^(2*x)*a^2 + 3*2^x*a*b)*sq
rt((2^x*a - b)/2^x))/(a^3*log(2)), -1/4*(3*sqrt(-a)*b^2*arctan(sqrt(-a)*sqrt((2^x*a - b)/2^x)/a) - (2*2^(2*x)*
a^2 + 3*2^x*a*b)*sqrt((2^x*a - b)/2^x))/(a^3*log(2))]

Sympy [F]

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

[In]

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

[Out]

Integral(4**x/sqrt(a - b/2**x), x)

Maxima [F]

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

[In]

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

[Out]

integrate(4^x/sqrt(a - b/2^x), x)

Giac [F]

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

[In]

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

[Out]

integrate(4^x/sqrt(a - b/2^x), x)

Mupad [F(-1)]

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

[In]

int(4^x/(a - b/2^x)^(1/2),x)

[Out]

int(4^x/(a - b/2^x)^(1/2), x)