3.508 \(\int (a^{k x}-a^{l x})^2 \, dx\)

Optimal. Leaf size=53 \[ -\frac {2 a^{x (k+l)}}{\log (a) (k+l)}+\frac {a^{2 k x}}{2 k \log (a)}+\frac {a^{2 l x}}{2 l \log (a)} \]

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Rubi [A]  time = 0.07, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {6742, 2194} \[ -\frac {2 a^{x (k+l)}}{\log (a) (k+l)}+\frac {a^{2 k x}}{2 k \log (a)}+\frac {a^{2 l x}}{2 l \log (a)} \]

Antiderivative was successfully verified.

[In]

Int[(a^(k*x) - a^(l*x))^2,x]

[Out]

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

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 6742

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

Rubi steps

\begin {align*} \int \left (a^{k x}-a^{l x}\right )^2 \, dx &=\frac {\operatorname {Subst}\left (\int \left (e^{k x}-e^{l x}\right )^2 \, dx,x,x \log (a)\right )}{\log (a)}\\ &=\frac {\operatorname {Subst}\left (\int \left (e^{2 k x}+e^{2 l x}-2 e^{(k+l) x}\right ) \, dx,x,x \log (a)\right )}{\log (a)}\\ &=\frac {\operatorname {Subst}\left (\int e^{2 k x} \, dx,x,x \log (a)\right )}{\log (a)}+\frac {\operatorname {Subst}\left (\int e^{2 l x} \, dx,x,x \log (a)\right )}{\log (a)}-\frac {2 \operatorname {Subst}\left (\int e^{(k+l) x} \, dx,x,x \log (a)\right )}{\log (a)}\\ &=\frac {a^{2 k x}}{2 k \log (a)}+\frac {a^{2 l x}}{2 l \log (a)}-\frac {2 a^{(k+l) x}}{(k+l) \log (a)}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 53, normalized size = 1.00 \[ -\frac {2 a^{x (k+l)}}{\log (a) (k+l)}+\frac {a^{2 k x}}{2 k \log (a)}+\frac {a^{2 l x}}{2 l \log (a)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^(k*x) - a^(l*x))^2,x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a^{k x}-a^{l x}\right )^2 \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[(a^(k*x) - a^(l*x))^2,x]

[Out]

Could not integrate

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fricas [A]  time = 1.38, size = 64, normalized size = 1.21 \[ -\frac {4 \, a^{k x} a^{l x} k l - {\left (k l + l^{2}\right )} a^{2 \, k x} - {\left (k^{2} + k l\right )} a^{2 \, l x}}{2 \, {\left (k^{2} l + k l^{2}\right )} \log \relax (a)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^(k*x)-a^(l*x))^2,x, algorithm="fricas")

[Out]

-1/2*(4*a^(k*x)*a^(l*x)*k*l - (k*l + l^2)*a^(2*k*x) - (k^2 + k*l)*a^(2*l*x))/((k^2*l + k*l^2)*log(a))

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giac [C]  time = 0.75, size = 691, normalized size = 13.04 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^(k*x)-a^(l*x))^2,x, algorithm="giac")

[Out]

(2*k*cos(-pi*k*x*sgn(a) + pi*k*x)*log(abs(a))/(4*k^2*log(abs(a))^2 + (pi*k*sgn(a) - pi*k)^2) - (pi*k*sgn(a) -
pi*k)*sin(-pi*k*x*sgn(a) + pi*k*x)/(4*k^2*log(abs(a))^2 + (pi*k*sgn(a) - pi*k)^2))*abs(a)^(2*k*x) + (2*l*cos(-
pi*l*x*sgn(a) + pi*l*x)*log(abs(a))/(4*l^2*log(abs(a))^2 + (pi*l*sgn(a) - pi*l)^2) - (pi*l*sgn(a) - pi*l)*sin(
-pi*l*x*sgn(a) + pi*l*x)/(4*l^2*log(abs(a))^2 + (pi*l*sgn(a) - pi*l)^2))*abs(a)^(2*l*x) - 1/2*I*abs(a)^(2*k*x)
*(-I*e^(I*pi*k*x*sgn(a) - I*pi*k*x)/(I*pi*k*sgn(a) - I*pi*k + 2*k*log(abs(a))) + I*e^(-I*pi*k*x*sgn(a) + I*pi*
k*x)/(-I*pi*k*sgn(a) + I*pi*k + 2*k*log(abs(a)))) - 1/2*I*abs(a)^(2*l*x)*(-I*e^(I*pi*l*x*sgn(a) - I*pi*l*x)/(I
*pi*l*sgn(a) - I*pi*l + 2*l*log(abs(a))) + I*e^(-I*pi*l*x*sgn(a) + I*pi*l*x)/(-I*pi*l*sgn(a) + I*pi*l + 2*l*lo
g(abs(a)))) - 4*(2*(k*log(abs(a)) + l*log(abs(a)))*cos(-1/2*pi*k*x*sgn(a) - 1/2*pi*l*x*sgn(a) + 1/2*pi*k*x + 1
/2*pi*l*x)/((pi*k*sgn(a) + pi*l*sgn(a) - pi*k - pi*l)^2 + 4*(k*log(abs(a)) + l*log(abs(a)))^2) - (pi*k*sgn(a)
+ pi*l*sgn(a) - pi*k - pi*l)*sin(-1/2*pi*k*x*sgn(a) - 1/2*pi*l*x*sgn(a) + 1/2*pi*k*x + 1/2*pi*l*x)/((pi*k*sgn(
a) + pi*l*sgn(a) - pi*k - pi*l)^2 + 4*(k*log(abs(a)) + l*log(abs(a)))^2))*e^((k*log(abs(a)) + l*log(abs(a)))*x
) - 1/2*I*(4*I*e^(1/2*I*pi*k*x*sgn(a) + 1/2*I*pi*l*x*sgn(a) - 1/2*I*pi*k*x - 1/2*I*pi*l*x)/(I*pi*k*sgn(a) + I*
pi*l*sgn(a) - I*pi*k - I*pi*l + 2*k*log(abs(a)) + 2*l*log(abs(a))) - 4*I*e^(-1/2*I*pi*k*x*sgn(a) - 1/2*I*pi*l*
x*sgn(a) + 1/2*I*pi*k*x + 1/2*I*pi*l*x)/(-I*pi*k*sgn(a) - I*pi*l*sgn(a) + I*pi*k + I*pi*l + 2*k*log(abs(a)) +
2*l*log(abs(a))))*e^((k*log(abs(a)) + l*log(abs(a)))*x)

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maple [A]  time = 0.04, size = 55, normalized size = 1.04

method	result	size
risch	\(\frac {a^{2 k x}}{2 k \ln \relax (a )}+\frac {a^{2 l x}}{2 l \ln \relax (a )}-\frac {2 a^{k x} a^{l x}}{\ln \relax (a ) \left (k +l \right )}\)	\(55\)
norman	\(\frac {{\mathrm e}^{2 k x \ln \relax (a )}}{2 k \ln \relax (a )}+\frac {{\mathrm e}^{2 l x \ln \relax (a )}}{2 l \ln \relax (a )}-\frac {2 \,{\mathrm e}^{k x \ln \relax (a )} {\mathrm e}^{l x \ln \relax (a )}}{\ln \relax (a ) \left (k +l \right )}\)	\(59\)
meijerg	error in int/gbinthm/express: improper op or subscript selector\	N/A

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^(k*x)-a^(l*x))^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.55, size = 51, normalized size = 0.96 \[ -\frac {2 \, a^{k x + l x}}{{\left (k + l\right )} \log \relax (a)} + \frac {a^{2 \, k x}}{2 \, k \log \relax (a)} + \frac {a^{2 \, l x}}{2 \, l \log \relax (a)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^(k*x)-a^(l*x))^2,x, algorithm="maxima")

[Out]

-2*a^(k*x + l*x)/((k + l)*log(a)) + 1/2*a^(2*k*x)/(k*log(a)) + 1/2*a^(2*l*x)/(l*log(a))

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mupad [B]  time = 0.36, size = 69, normalized size = 1.30 \[ \frac {a^{2\,k\,x}}{2\,k\,\ln \relax (a)}+\frac {\frac {a^{2\,l\,x}\,k^2}{2}-l\,\left (2\,a^{k\,x+l\,x}\,k-\frac {a^{2\,l\,x}\,k}{2}\right )}{k\,l\,\ln \relax (a)\,\left (k+l\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^(k*x) - a^(l*x))^2,x)

[Out]

a^(2*k*x)/(2*k*log(a)) + ((a^(2*l*x)*k^2)/2 - l*(2*a^(k*x + l*x)*k - (a^(2*l*x)*k)/2))/(k*l*log(a)*(k + l))

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sympy [A]  time = 1.98, size = 248, normalized size = 4.68 \[ \begin {cases} 0 & \text {for}\: a = 1 \wedge \left (a = 1 \vee k = 0\right ) \wedge \left (a = 1 \vee l = 0\right ) \\\frac {a^{2 l x}}{2 l \log {\relax (a )}} - \frac {2 a^{l x}}{l \log {\relax (a )}} + x & \text {for}\: k = 0 \\\frac {a^{2 l x}}{2 l \log {\relax (a )}} - 2 x - \frac {a^{- 2 l x}}{2 l \log {\relax (a )}} & \text {for}\: k = - l \\\frac {a^{2 k x}}{2 k \log {\relax (a )}} - \frac {2 a^{k x}}{k \log {\relax (a )}} + x & \text {for}\: l = 0 \\\frac {a^{2 k x} k l}{2 k^{2} l \log {\relax (a )} + 2 k l^{2} \log {\relax (a )}} + \frac {a^{2 k x} l^{2}}{2 k^{2} l \log {\relax (a )} + 2 k l^{2} \log {\relax (a )}} - \frac {4 a^{k x} a^{l x} k l}{2 k^{2} l \log {\relax (a )} + 2 k l^{2} \log {\relax (a )}} + \frac {a^{2 l x} k^{2}}{2 k^{2} l \log {\relax (a )} + 2 k l^{2} \log {\relax (a )}} + \frac {a^{2 l x} k l}{2 k^{2} l \log {\relax (a )} + 2 k l^{2} \log {\relax (a )}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a**(k*x)-a**(l*x))**2,x)

[Out]

Piecewise((0, Eq(a, 1) & (Eq(a, 1) | Eq(k, 0)) & (Eq(a, 1) | Eq(l, 0))), (a**(2*l*x)/(2*l*log(a)) - 2*a**(l*x)
/(l*log(a)) + x, Eq(k, 0)), (a**(2*l*x)/(2*l*log(a)) - 2*x - a**(-2*l*x)/(2*l*log(a)), Eq(k, -l)), (a**(2*k*x)
/(2*k*log(a)) - 2*a**(k*x)/(k*log(a)) + x, Eq(l, 0)), (a**(2*k*x)*k*l/(2*k**2*l*log(a) + 2*k*l**2*log(a)) + a*
*(2*k*x)*l**2/(2*k**2*l*log(a) + 2*k*l**2*log(a)) - 4*a**(k*x)*a**(l*x)*k*l/(2*k**2*l*log(a) + 2*k*l**2*log(a)
) + a**(2*l*x)*k**2/(2*k**2*l*log(a) + 2*k*l**2*log(a)) + a**(2*l*x)*k*l/(2*k**2*l*log(a) + 2*k*l**2*log(a)),
True))

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