\(\int a^x b^{-x} \, dx\) [160]

Optimal result

Integrand size = 9, antiderivative size = 18 \[ \int a^x b^{-x} \, dx=\frac {a^x b^{-x}}{\log (a)-\log (b)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2325, 2225} \[ \int a^x b^{-x} \, dx=\frac {a^x b^{-x}}{\log (a)-\log (b)} \]

[In]

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

[Out]

a^x/(b^x*(Log[a] - Log[b]))

Rule 2225

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 2325

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int a^x b^{-x} \, dx=\frac {a^x b^{-x}}{\log (a)-\log (b)} \]

[In]

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

[Out]

a^x/(b^x*(Log[a] - Log[b]))

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06

method	result	size
gosper	\(\frac {a^{x} b^{-x}}{\ln \left (a \right )-\ln \left (b \right )}\)	\(19\)
risch	\(\frac {a^{x} b^{-x}}{\ln \left (a \right )-\ln \left (b \right )}\)	\(19\)
parallelrisch	\(\frac {a^{x} b^{-x}}{\ln \left (a \right )-\ln \left (b \right )}\)	\(19\)
norman	\(\frac {{\mathrm e}^{x \ln \left (a \right )} {\mathrm e}^{-x \ln \left (b \right )}}{\ln \left (a \right )-\ln \left (b \right )}\)	\(23\)
meijerg	\(-\frac {1-{\mathrm e}^{x \ln \left (a \right ) \left (1-\frac {\ln \left (b \right )}{\ln \left (a \right )}\right )}}{\ln \left (a \right ) \left (1-\frac {\ln \left (b \right )}{\ln \left (a \right )}\right )}\)	\(38\)

[In]

int(a^x/(b^x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int a^x b^{-x} \, dx=\frac {a^{x}}{b^{x} {\left (\log \left (a\right ) - \log \left (b\right )\right )}} \]

[In]

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

[Out]

a^x/(b^x*(log(a) - log(b)))

Sympy [F(-2)]

Exception generated. \[ \int a^x b^{-x} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> Invalid NaN comparison

Maxima [F(-2)]

Exception generated. \[ \int a^x b^{-x} \, dx=\text {Exception raised: ValueError} \]

[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

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 216, normalized size of antiderivative = 12.00 \[ \int a^x b^{-x} \, dx=2 \, {\left (\frac {2 \, {\left (\log \left ({\left | a \right |}\right ) - \log \left ({\left | b \right |}\right )\right )} \cos \left (-\frac {1}{2} \, \pi x \mathrm {sgn}\left (a\right ) + \frac {1}{2} \, \pi x \mathrm {sgn}\left (b\right )\right )}{{\left (\pi \mathrm {sgn}\left (a\right ) - \pi \mathrm {sgn}\left (b\right )\right )}^{2} + 4 \, {\left (\log \left ({\left | a \right |}\right ) - \log \left ({\left | b \right |}\right )\right )}^{2}} - \frac {{\left (\pi \mathrm {sgn}\left (a\right ) - \pi \mathrm {sgn}\left (b\right )\right )} \sin \left (-\frac {1}{2} \, \pi x \mathrm {sgn}\left (a\right ) + \frac {1}{2} \, \pi x \mathrm {sgn}\left (b\right )\right )}{{\left (\pi \mathrm {sgn}\left (a\right ) - \pi \mathrm {sgn}\left (b\right )\right )}^{2} + 4 \, {\left (\log \left ({\left | a \right |}\right ) - \log \left ({\left | b \right |}\right )\right )}^{2}}\right )} e^{\left (x {\left (\log \left ({\left | a \right |}\right ) - \log \left ({\left | b \right |}\right )\right )}\right )} + i \, {\left (\frac {i \, e^{\left (\frac {1}{2} i \, \pi x \mathrm {sgn}\left (a\right ) - \frac {1}{2} i \, \pi x \mathrm {sgn}\left (b\right )\right )}}{i \, \pi \mathrm {sgn}\left (a\right ) - i \, \pi \mathrm {sgn}\left (b\right ) + 2 \, \log \left ({\left | a \right |}\right ) - 2 \, \log \left ({\left | b \right |}\right )} - \frac {i \, e^{\left (-\frac {1}{2} i \, \pi x \mathrm {sgn}\left (a\right ) + \frac {1}{2} i \, \pi x \mathrm {sgn}\left (b\right )\right )}}{-i \, \pi \mathrm {sgn}\left (a\right ) + i \, \pi \mathrm {sgn}\left (b\right ) + 2 \, \log \left ({\left | a \right |}\right ) - 2 \, \log \left ({\left | b \right |}\right )}\right )} e^{\left (x {\left (\log \left ({\left | a \right |}\right ) - \log \left ({\left | b \right |}\right )\right )}\right )} \]

[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)))) + I*(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))) - 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(abs(b))))

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int a^x b^{-x} \, dx=\frac {a^x}{b^x\,\left (\ln \left (a\right )-\ln \left (b\right )\right )} \]

[In]

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

[Out]

a^x/(b^x*(log(a) - log(b)))