3.495 \(\int a^{-x} b^{-x} (a^x-b^x)^2 \, dx\)

Optimal. Leaf size=34 \[ \frac{a^x b^{-x}-a^{-x} b^x}{\log (a)-\log (b)}-2 x \]

[Out]

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

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Rubi [A]  time = 0.207676, antiderivative size = 41, normalized size of antiderivative = 1.21, number of steps used = 9, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {2287, 6742, 2194, 8} \[ -\frac{a^{-x} b^x}{\log (a)-\log (b)}+\frac{a^x b^{-x}}{\log (a)-\log (b)}-2 x \]

Antiderivative was successfully verified.

[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 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]

Rule 6742

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

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 8

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

Rubi steps

\begin{align*} \int a^{-x} b^{-x} \left (a^x-b^x\right )^2 \, dx &=\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]  time = 0.0499614, size = 46, normalized size = 1.35 \[ \frac{e^{x (\log (a)-\log (b))}}{\log (a)-\log (b)}+\frac{e^{x (\log (b)-\log (a))}}{\log (b)-\log (a)}-2 x \]

Antiderivative was successfully verified.

[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])

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Maple [A]  time = 0.017, size = 65, normalized size = 1.9 \begin{align*}{\frac{1}{{{\rm e}^{x\ln \left ( a \right ) }}{{\rm e}^{x\ln \left ( b \right ) }}} \left ({\frac{ \left ({{\rm e}^{x\ln \left ( a \right ) }} \right ) ^{2}}{\ln \left ( a \right ) -\ln \left ( b \right ) }}-{\frac{ \left ({{\rm e}^{x\ln \left ( b \right ) }} \right ) ^{2}}{\ln \left ( a \right ) -\ln \left ( b \right ) }}-2\,x{{\rm e}^{x\ln \left ( a \right ) }}{{\rm e}^{x\ln \left ( b \right ) }} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.82862, size = 113, normalized size = 3.32 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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Giac [C]  time = 1.30936, size = 589, normalized size = 17.32 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))) - 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)))) - 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*sgn(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*p
i*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