\(\int a^{-x} b^{-x} (a^x-b^x)^2 \, dx\) [495]

Optimal result

Integrand size = 22, antiderivative size = 34 \[ \int a^{-x} b^{-x} \left (a^x-b^x\right )^2 \, dx=-2 x+\frac {a^x b^{-x}-a^{-x} b^x}{\log (a)-\log (b)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21, number of steps used = 9, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2325, 6874, 2225, 8} \[ \int a^{-x} b^{-x} \left (a^x-b^x\right )^2 \, dx=-\frac {a^{-x} b^x}{\log (a)-\log (b)}+\frac {a^x b^{-x}}{\log (a)-\log (b)}-2 x \]

[In]

Int[(a^x - b^x)^2/(a^x*b^x),x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \left (a^x-b^x\right )^2 e^{-x (\log (a)+\log (b))} \, dx \\ & = \int \left (a^{2 x} e^{-x (\log (a)+\log (b))}-2 a^x b^x e^{-x (\log (a)+\log (b))}+b^{2 x} e^{-x (\log (a)+\log (b))}\right ) \, dx \\ & = -\left (2 \int a^x b^x e^{-x (\log (a)+\log (b))} \, dx\right )+\int a^{2 x} e^{-x (\log (a)+\log (b))} \, dx+\int b^{2 x} e^{-x (\log (a)+\log (b))} \, dx \\ & = -(2 \int 1 \, dx)+\int e^{-x (\log (a)-\log (b))} \, dx+\int e^{x (\log (a)-\log (b))} \, dx \\ & = -2 x+\frac {a^x b^{-x}}{\log (a)-\log (b)}-\frac {a^{-x} b^x}{\log (a)-\log (b)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.35 \[ \int a^{-x} b^{-x} \left (a^x-b^x\right )^2 \, dx=-2 x+\frac {e^{x (\log (a)-\log (b))}}{\log (a)-\log (b)}+\frac {e^{x (-\log (a)+\log (b))}}{-\log (a)+\log (b)} \]

[In]

Integrate[(a^x - b^x)^2/(a^x*b^x),x]

[Out]

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

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.24

method	result	size
risch	\(-2 x -\frac {a^{-x} b^{x}}{\ln \left (a \right )-\ln \left (b \right )}+\frac {a^{x} b^{-x}}{\ln \left (a \right )-\ln \left (b \right )}\)	\(42\)
parallelrisch	\(-\frac {\left (2 x \,a^{x} b^{x} \ln \left (a \right )-2 x \,a^{x} b^{x} \ln \left (b \right )-a^{2 x}+b^{2 x}\right ) a^{-x} b^{-x}}{\ln \left (a \right )-\ln \left (b \right )}\)	\(57\)
norman	\(\left (\frac {{\mathrm e}^{2 x \ln \left (a \right )}}{\ln \left (a \right )-\ln \left (b \right )}-\frac {{\mathrm e}^{2 x \ln \left (b \right )}}{\ln \left (a \right )-\ln \left (b \right )}-2 x \,{\mathrm e}^{x \ln \left (a \right )} {\mathrm e}^{x \ln \left (b \right )}\right ) {\mathrm e}^{-x \ln \left (a \right )} {\mathrm e}^{-x \ln \left (b \right )}\)	\(65\)

[In]

int((a^x-b^x)^2/(a^x)/(b^x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.53 \[ \int a^{-x} b^{-x} \left (a^x-b^x\right )^2 \, dx=-\frac {2 \, {\left (x \log \left (a\right ) - x \log \left (b\right )\right )} a^{x} b^{x} - a^{2 \, x} + b^{2 \, x}}{a^{x} b^{x} {\left (\log \left (a\right ) - \log \left (b\right )\right )}} \]

[In]

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

[Out]

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

Sympy [F(-2)]

Exception generated. \[ \int a^{-x} b^{-x} \left (a^x-b^x\right )^2 \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> Invalid NaN comparison

Maxima [F(-2)]

Exception generated. \[ \int a^{-x} b^{-x} \left (a^x-b^x\right )^2 \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((a^x-b^x)^2/(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 = 0.33 (sec) , antiderivative size = 436, normalized size of antiderivative = 12.82 \[ \int a^{-x} b^{-x} \left (a^x-b^x\right )^2 \, dx=\text {Too large to display} \]

[In]

integrate((a^x-b^x)^2/(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))))
- 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(
abs(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))
)) - 2*x

Mupad [B] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int a^{-x} b^{-x} \left (a^x-b^x\right )^2 \, dx=\frac {\frac {a^x}{b^x}-\frac {b^x}{a^x}}{\ln \left (a\right )-\ln \left (b\right )}-2\,x \]

[In]

int((a^x - b^x)^2/(a^x*b^x),x)

[Out]

(a^x/b^x - b^x/a^x)/(log(a) - log(b)) - 2*x