3.160 \(\int a^x b^{-x} \, dx\)

Optimal. Leaf size=18 \[ \frac {a^x b^{-x}}{\log (a)-\log (b)} \]

[Out]

a^x/(b^x)/(ln(a)-ln(b))

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Rubi [A]  time = 0.02, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2287, 2194} \[ \frac {a^x b^{-x}}{\log (a)-\log (b)} \]

Antiderivative was successfully verified.

[In]

Int[a^x/b^x,x]

[Out]

a^x/(b^x*(Log[a] - Log[b]))

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2287

Int[(u_.)*(F_)^(v_)*(G_)^(w_), x_Symbol] :> With[{z = v*Log[F] + w*Log[G]}, Int[u*NormalizeIntegrand[E^z, x],
x] /; BinomialQ[z, x] || (PolynomialQ[z, x] && LeQ[Exponent[z, x], 2])] /; FreeQ[{F, G}, x]

Rubi steps

\begin {align*} \int a^x b^{-x} \, dx &=\int e^{x (\log (a)-\log (b))} \, dx\\ &=\frac {a^x b^{-x}}{\log (a)-\log (b)}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 1.00 \[ \frac {a^x b^{-x}}{\log (a)-\log (b)} \]

Antiderivative was successfully verified.

[In]

Integrate[a^x/b^x,x]

[Out]

a^x/(b^x*(Log[a] - Log[b]))

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fricas [A]  time = 0.42, size = 18, normalized size = 1.00 \[ \frac {a^{x}}{b^{x} {\left (\log \relax (a) - \log \relax (b)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^x/(b^x*(log(a) - log(b)))

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giac [C]  time = 1.44, size = 216, normalized size = 12.00 \[ 2 \, {\left (\frac {2 \, {\left (\log \left ({\left | a \right |}\right ) - \log \left ({\left | b \right |}\right )\right )} \cos \left (-\frac {1}{2} \, \pi x \mathrm {sgn}\relax (a) + \frac {1}{2} \, \pi x \mathrm {sgn}\relax (b)\right )}{{\left (\pi \mathrm {sgn}\relax (a) - \pi \mathrm {sgn}\relax (b)\right )}^{2} + 4 \, {\left (\log \left ({\left | a \right |}\right ) - \log \left ({\left | b \right |}\right )\right )}^{2}} - \frac {{\left (\pi \mathrm {sgn}\relax (a) - \pi \mathrm {sgn}\relax (b)\right )} \sin \left (-\frac {1}{2} \, \pi x \mathrm {sgn}\relax (a) + \frac {1}{2} \, \pi x \mathrm {sgn}\relax (b)\right )}{{\left (\pi \mathrm {sgn}\relax (a) - \pi \mathrm {sgn}\relax (b)\right )}^{2} + 4 \, {\left (\log \left ({\left | a \right |}\right ) - \log \left ({\left | b \right |}\right )\right )}^{2}}\right )} e^{\left (x {\left (\log \left ({\left | a \right |}\right ) - \log \left ({\left | b \right |}\right )\right )}\right )} - \frac {1}{2} i \, {\left (-\frac {2 i \, e^{\left (\frac {1}{2} i \, \pi x \mathrm {sgn}\relax (a) - \frac {1}{2} i \, \pi x \mathrm {sgn}\relax (b)\right )}}{i \, \pi \mathrm {sgn}\relax (a) - i \, \pi \mathrm {sgn}\relax (b) + 2 \, \log \left ({\left | a \right |}\right ) - 2 \, \log \left ({\left | b \right |}\right )} + \frac {2 i \, e^{\left (-\frac {1}{2} i \, \pi x \mathrm {sgn}\relax (a) + \frac {1}{2} i \, \pi x \mathrm {sgn}\relax (b)\right )}}{-i \, \pi \mathrm {sgn}\relax (a) + i \, \pi \mathrm {sgn}\relax (b) + 2 \, \log \left ({\left | a \right |}\right ) - 2 \, \log \left ({\left | b \right |}\right )}\right )} e^{\left (x {\left (\log \left ({\left | a \right |}\right ) - \log \left ({\left | b \right |}\right )\right )}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(2*(log(abs(a)) - log(abs(b)))*cos(-1/2*pi*x*sgn(a) + 1/2*pi*x*sgn(b))/((pi*sgn(a) - pi*sgn(b))^2 + 4*(log(a
bs(a)) - log(abs(b)))^2) - (pi*sgn(a) - pi*sgn(b))*sin(-1/2*pi*x*sgn(a) + 1/2*pi*x*sgn(b))/((pi*sgn(a) - pi*sg
n(b))^2 + 4*(log(abs(a)) - log(abs(b)))^2))*e^(x*(log(abs(a)) - log(abs(b)))) - 1/2*I*(-2*I*e^(1/2*I*pi*x*sgn(
a) - 1/2*I*pi*x*sgn(b))/(I*pi*sgn(a) - I*pi*sgn(b) + 2*log(abs(a)) - 2*log(abs(b))) + 2*I*e^(-1/2*I*pi*x*sgn(a
) + 1/2*I*pi*x*sgn(b))/(-I*pi*sgn(a) + I*pi*sgn(b) + 2*log(abs(a)) - 2*log(abs(b))))*e^(x*(log(abs(a)) - log(a
bs(b))))

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maple [A]  time = 0.01, size = 19, normalized size = 1.06 \[ \frac {a^{x} b^{-x}}{\ln \relax (a )-\ln \relax (b )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(a^x/(b^x),x)

[Out]

a^x/(b^x)/(ln(a)-ln(b))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(-log(b)/log(a)>0)', see `assum
e?` for more details)Is -log(b)/log(a) equal to -1?

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mupad [B]  time = 0.20, size = 18, normalized size = 1.00 \[ \frac {a^x}{b^x\,\left (\ln \relax (a)-\ln \relax (b)\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(a^x/b^x,x)

[Out]

a^x/(b^x*(log(a) - log(b)))

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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