3.6.9 \(\int (a^{k x}-a^{l x})^3 \, dx\) [509]

Optimal. Leaf size=79 \[ \frac {a^{3 k x}}{3 k \log (a)}-\frac {a^{3 l x}}{3 l \log (a)}-\frac {3 a^{(2 k+l) x}}{(2 k+l) \log (a)}+\frac {3 a^{(k+2 l) x}}{(k+2 l) \log (a)} \]

[Out]

1/3*a^(3*k*x)/k/ln(a)-1/3*a^(3*l*x)/l/ln(a)-3*a^((2*k+l)*x)/(2*k+l)/ln(a)+3*a^((k+2*l)*x)/(k+2*l)/ln(a)

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Rubi [A]
time = 0.06, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {6874, 2225} \begin {gather*} -\frac {3 a^{x (2 k+l)}}{\log (a) (2 k+l)}+\frac {3 a^{x (k+2 l)}}{\log (a) (k+2 l)}+\frac {a^{3 k x}}{3 k \log (a)}-\frac {a^{3 l x}}{3 l \log (a)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a^(k*x) - a^(l*x))^3,x]

[Out]

a^(3*k*x)/(3*k*Log[a]) - a^(3*l*x)/(3*l*Log[a]) - (3*a^((2*k + l)*x))/((2*k + l)*Log[a]) + (3*a^((k + 2*l)*x))
/((k + 2*l)*Log[a])

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 6874

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 )^3 \, dx &=\frac {\text {Subst}\left (\int \left (e^{k x}-e^{l x}\right )^3 \, dx,x,x \log (a)\right )}{\log (a)}\\ &=\frac {\text {Subst}\left (\int \left (e^{3 k x}-e^{3 l x}-3 e^{(2 k+l) x}+3 e^{(k+2 l) x}\right ) \, dx,x,x \log (a)\right )}{\log (a)}\\ &=\frac {\text {Subst}\left (\int e^{3 k x} \, dx,x,x \log (a)\right )}{\log (a)}-\frac {\text {Subst}\left (\int e^{3 l x} \, dx,x,x \log (a)\right )}{\log (a)}-\frac {3 \text {Subst}\left (\int e^{(2 k+l) x} \, dx,x,x \log (a)\right )}{\log (a)}+\frac {3 \text {Subst}\left (\int e^{(k+2 l) x} \, dx,x,x \log (a)\right )}{\log (a)}\\ &=\frac {a^{3 k x}}{3 k \log (a)}-\frac {a^{3 l x}}{3 l \log (a)}-\frac {3 a^{(2 k+l) x}}{(2 k+l) \log (a)}+\frac {3 a^{(k+2 l) x}}{(k+2 l) \log (a)}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 66, normalized size = 0.84 \begin {gather*} \frac {\frac {a^{3 k x}}{k}-\frac {a^{3 l x}}{l}-\frac {9 a^{(2 k+l) x}}{2 k+l}+\frac {9 a^{(k+2 l) x}}{k+2 l}}{3 \log (a)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a^(k*x) - a^(l*x))^3,x]

[Out]

(a^(3*k*x)/k - a^(3*l*x)/l - (9*a^((2*k + l)*x))/(2*k + l) + (9*a^((k + 2*l)*x))/(k + 2*l))/(3*Log[a])

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Maple [A]
time = 0.04, size = 84, normalized size = 1.06

method	result	size
risch	\(\frac {a^{3 k x}}{3 k \ln \left (a \right )}-\frac {a^{3 l x}}{3 l \ln \left (a \right )}+\frac {3 a^{k x} a^{2 l x}}{\ln \left (a \right ) \left (k +2 l \right )}-\frac {3 a^{2 k x} a^{l x}}{\ln \left (a \right ) \left (2 k +l \right )}\)	\(84\)
norman	\(\frac {{\mathrm e}^{3 k x \ln \left (a \right )}}{3 k \ln \left (a \right )}-\frac {{\mathrm e}^{3 l x \ln \left (a \right )}}{3 l \ln \left (a \right )}+\frac {3 \,{\mathrm e}^{k x \ln \left (a \right )} {\mathrm e}^{2 l x \ln \left (a \right )}}{\ln \left (a \right ) \left (k +2 l \right )}-\frac {3 \,{\mathrm e}^{2 k x \ln \left (a \right )} {\mathrm e}^{l x \ln \left (a \right )}}{\ln \left (a \right ) \left (2 k +l \right )}\)	\(90\)
meijerg	\(-\frac {1-{\mathrm e}^{3 k x \ln \left (a \right )}}{3 k \ln \left (a \right )}+\frac {3-3 \,{\mathrm e}^{x k \ln \left (a \right ) \left (2+\frac {l}{k}\right )}}{k \ln \left (a \right ) \left (2+\frac {l}{k}\right )}-\frac {3 \left (1-{\mathrm e}^{x l \ln \left (a \right ) \left (1+\frac {k}{l}\right ) \left (1+\frac {1}{1+\frac {k}{l}}\right )}\right )}{l \ln \left (a \right ) \left (1+\frac {k}{l}\right ) \left (1+\frac {1}{1+\frac {k}{l}}\right )}+\frac {1-{\mathrm e}^{3 l x \ln \left (a \right )}}{3 l \ln \left (a \right )}\)	\(136\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^(k*x)-a^(l*x))^3,x,method=_RETURNVERBOSE)

[Out]

1/3/k/ln(a)*(a^(k*x))^3-1/3/l/ln(a)*(a^(l*x))^3+3/ln(a)/(k+2*l)*a^(k*x)*(a^(l*x))^2-3/ln(a)/(2*k+l)*(a^(k*x))^
2*a^(l*x)

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Maxima [A]
time = 1.26, size = 77, normalized size = 0.97 \begin {gather*} -\frac {3 \, a^{2 \, k x + l x}}{{\left (2 \, k + l\right )} \log \left (a\right )} + \frac {3 \, a^{k x + 2 \, l x}}{{\left (k + 2 \, l\right )} \log \left (a\right )} + \frac {a^{3 \, k x}}{3 \, k \log \left (a\right )} - \frac {a^{3 \, l x}}{3 \, l \log \left (a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^(k*x)-a^(l*x))^3,x, algorithm="maxima")

[Out]

-3*a^(2*k*x + l*x)/((2*k + l)*log(a)) + 3*a^(k*x + 2*l*x)/((k + 2*l)*log(a)) + 1/3*a^(3*k*x)/(k*log(a)) - 1/3*
a^(3*l*x)/(l*log(a))

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Fricas [A]
time = 0.57, size = 131, normalized size = 1.66 \begin {gather*} \frac {9 \, {\left (2 \, k^{2} l + k l^{2}\right )} a^{k x} a^{2 \, l x} - 9 \, {\left (k^{2} l + 2 \, k l^{2}\right )} a^{2 \, k x} a^{l x} + {\left (2 \, k^{2} l + 5 \, k l^{2} + 2 \, l^{3}\right )} a^{3 \, k x} - {\left (2 \, k^{3} + 5 \, k^{2} l + 2 \, k l^{2}\right )} a^{3 \, l x}}{3 \, {\left (2 \, k^{3} l + 5 \, k^{2} l^{2} + 2 \, k l^{3}\right )} \log \left (a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^(k*x)-a^(l*x))^3,x, algorithm="fricas")

[Out]

1/3*(9*(2*k^2*l + k*l^2)*a^(k*x)*a^(2*l*x) - 9*(k^2*l + 2*k*l^2)*a^(2*k*x)*a^(l*x) + (2*k^2*l + 5*k*l^2 + 2*l^
3)*a^(3*k*x) - (2*k^3 + 5*k^2*l + 2*k*l^2)*a^(3*l*x))/((2*k^3*l + 5*k^2*l^2 + 2*k*l^3)*log(a))

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 663 vs. \(2 (63) = 126\).
time = 2.50, size = 663, normalized size = 8.39 \begin {gather*} \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^{3 l x}}{3 l \log {\left (a \right )}} + \frac {3 a^{2 l x}}{2 l \log {\left (a \right )}} - \frac {3 a^{l x}}{l \log {\left (a \right )}} + x & \text {for}\: k = 0 \\- \frac {a^{3 l x}}{3 l \log {\left (a \right )}} + 3 x + \frac {a^{- 3 l x}}{l \log {\left (a \right )}} - \frac {a^{- 6 l x}}{6 l \log {\left (a \right )}} & \text {for}\: k = - 2 l \\\frac {2 a^{\frac {3 l x}{2}}}{l \log {\left (a \right )}} - \frac {a^{3 l x}}{3 l \log {\left (a \right )}} - 3 x - \frac {2 a^{- \frac {3 l x}{2}}}{3 l \log {\left (a \right )}} & \text {for}\: k = - \frac {l}{2} \\\frac {a^{3 k x}}{3 k \log {\left (a \right )}} - \frac {3 a^{2 k x}}{2 k \log {\left (a \right )}} + \frac {3 a^{k x}}{k \log {\left (a \right )}} - x & \text {for}\: l = 0 \\\frac {2 a^{3 k x} k^{2} l}{6 k^{3} l \log {\left (a \right )} + 15 k^{2} l^{2} \log {\left (a \right )} + 6 k l^{3} \log {\left (a \right )}} + \frac {5 a^{3 k x} k l^{2}}{6 k^{3} l \log {\left (a \right )} + 15 k^{2} l^{2} \log {\left (a \right )} + 6 k l^{3} \log {\left (a \right )}} + \frac {2 a^{3 k x} l^{3}}{6 k^{3} l \log {\left (a \right )} + 15 k^{2} l^{2} \log {\left (a \right )} + 6 k l^{3} \log {\left (a \right )}} - \frac {9 a^{2 k x} a^{l x} k^{2} l}{6 k^{3} l \log {\left (a \right )} + 15 k^{2} l^{2} \log {\left (a \right )} + 6 k l^{3} \log {\left (a \right )}} - \frac {18 a^{2 k x} a^{l x} k l^{2}}{6 k^{3} l \log {\left (a \right )} + 15 k^{2} l^{2} \log {\left (a \right )} + 6 k l^{3} \log {\left (a \right )}} + \frac {18 a^{k x} a^{2 l x} k^{2} l}{6 k^{3} l \log {\left (a \right )} + 15 k^{2} l^{2} \log {\left (a \right )} + 6 k l^{3} \log {\left (a \right )}} + \frac {9 a^{k x} a^{2 l x} k l^{2}}{6 k^{3} l \log {\left (a \right )} + 15 k^{2} l^{2} \log {\left (a \right )} + 6 k l^{3} \log {\left (a \right )}} - \frac {2 a^{3 l x} k^{3}}{6 k^{3} l \log {\left (a \right )} + 15 k^{2} l^{2} \log {\left (a \right )} + 6 k l^{3} \log {\left (a \right )}} - \frac {5 a^{3 l x} k^{2} l}{6 k^{3} l \log {\left (a \right )} + 15 k^{2} l^{2} \log {\left (a \right )} + 6 k l^{3} \log {\left (a \right )}} - \frac {2 a^{3 l x} k l^{2}}{6 k^{3} l \log {\left (a \right )} + 15 k^{2} l^{2} \log {\left (a \right )} + 6 k l^{3} \log {\left (a \right )}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a**(k*x)-a**(l*x))**3,x)

[Out]

Piecewise((0, Eq(a, 1) & (Eq(a, 1) | Eq(k, 0)) & (Eq(a, 1) | Eq(l, 0))), (-a**(3*l*x)/(3*l*log(a)) + 3*a**(2*l
*x)/(2*l*log(a)) - 3*a**(l*x)/(l*log(a)) + x, Eq(k, 0)), (-a**(3*l*x)/(3*l*log(a)) + 3*x + 1/(a**(3*l*x)*l*log
(a)) - 1/(6*a**(6*l*x)*l*log(a)), Eq(k, -2*l)), (2*a**(3*l*x/2)/(l*log(a)) - a**(3*l*x)/(3*l*log(a)) - 3*x - 2
/(3*a**(3*l*x/2)*l*log(a)), Eq(k, -l/2)), (a**(3*k*x)/(3*k*log(a)) - 3*a**(2*k*x)/(2*k*log(a)) + 3*a**(k*x)/(k
*log(a)) - x, Eq(l, 0)), (2*a**(3*k*x)*k**2*l/(6*k**3*l*log(a) + 15*k**2*l**2*log(a) + 6*k*l**3*log(a)) + 5*a*
*(3*k*x)*k*l**2/(6*k**3*l*log(a) + 15*k**2*l**2*log(a) + 6*k*l**3*log(a)) + 2*a**(3*k*x)*l**3/(6*k**3*l*log(a)
+ 15*k**2*l**2*log(a) + 6*k*l**3*log(a)) - 9*a**(2*k*x)*a**(l*x)*k**2*l/(6*k**3*l*log(a) + 15*k**2*l**2*log(a
) + 6*k*l**3*log(a)) - 18*a**(2*k*x)*a**(l*x)*k*l**2/(6*k**3*l*log(a) + 15*k**2*l**2*log(a) + 6*k*l**3*log(a))
+ 18*a**(k*x)*a**(2*l*x)*k**2*l/(6*k**3*l*log(a) + 15*k**2*l**2*log(a) + 6*k*l**3*log(a)) + 9*a**(k*x)*a**(2*
l*x)*k*l**2/(6*k**3*l*log(a) + 15*k**2*l**2*log(a) + 6*k*l**3*log(a)) - 2*a**(3*l*x)*k**3/(6*k**3*l*log(a) + 1
5*k**2*l**2*log(a) + 6*k*l**3*log(a)) - 5*a**(3*l*x)*k**2*l/(6*k**3*l*log(a) + 15*k**2*l**2*log(a) + 6*k*l**3*
log(a)) - 2*a**(3*l*x)*k*l**2/(6*k**3*l*log(a) + 15*k**2*l**2*log(a) + 6*k*l**3*log(a)), True))

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Giac [C] Result contains complex when optimal does not.
time = 1.00, size = 1033, normalized size = 13.08 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^(k*x)-a^(l*x))^3,x, algorithm="giac")

[Out]

2/3*(2*k*cos(-3/2*pi*k*x*sgn(a) + 3/2*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(-3/2*pi*k*x*sgn(a) + 3/2*pi*k*x)/(4*k^2*log(abs(a))^2 + (pi*k*sgn(a) - pi*k)^2))*abs(a)^
(3*k*x) - 2/3*(2*l*cos(-3/2*pi*l*x*sgn(a) + 3/2*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(-3/2*pi*l*x*sgn(a) + 3/2*pi*l*x)/(4*l^2*log(abs(a))^2 + (pi*l*sgn(a) - pi*l)^2
))*abs(a)^(3*l*x) + I*abs(a)^(3*k*x)*(I*e^(3/2*I*pi*k*x*sgn(a) - 3/2*I*pi*k*x)/(3*I*pi*k*sgn(a) - 3*I*pi*k + 6
*k*log(abs(a))) - I*e^(-3/2*I*pi*k*x*sgn(a) + 3/2*I*pi*k*x)/(-3*I*pi*k*sgn(a) + 3*I*pi*k + 6*k*log(abs(a)))) +
I*abs(a)^(3*l*x)*(-I*e^(3/2*I*pi*l*x*sgn(a) - 3/2*I*pi*l*x)/(3*I*pi*l*sgn(a) - 3*I*pi*l + 6*l*log(abs(a))) +
I*e^(-3/2*I*pi*l*x*sgn(a) + 3/2*I*pi*l*x)/(-3*I*pi*l*sgn(a) + 3*I*pi*l + 6*l*log(abs(a)))) - 6*(2*(2*k*log(abs
(a)) + l*log(abs(a)))*cos(-pi*k*x*sgn(a) - 1/2*pi*l*x*sgn(a) + pi*k*x + 1/2*pi*l*x)/((2*pi*k*sgn(a) + pi*l*sgn
(a) - 2*pi*k - pi*l)^2 + 4*(2*k*log(abs(a)) + l*log(abs(a)))^2) - (2*pi*k*sgn(a) + pi*l*sgn(a) - 2*pi*k - pi*l
)*sin(-pi*k*x*sgn(a) - 1/2*pi*l*x*sgn(a) + pi*k*x + 1/2*pi*l*x)/((2*pi*k*sgn(a) + pi*l*sgn(a) - 2*pi*k - pi*l)
^2 + 4*(2*k*log(abs(a)) + l*log(abs(a)))^2))*e^((2*k*log(abs(a)) + l*log(abs(a)))*x) + 3*I*(-I*e^(I*pi*k*x*sgn
(a) + 1/2*I*pi*l*x*sgn(a) - I*pi*k*x - 1/2*I*pi*l*x)/(2*I*pi*k*sgn(a) + I*pi*l*sgn(a) - 2*I*pi*k - I*pi*l + 4*
k*log(abs(a)) + 2*l*log(abs(a))) + I*e^(-I*pi*k*x*sgn(a) - 1/2*I*pi*l*x*sgn(a) + I*pi*k*x + 1/2*I*pi*l*x)/(-2*
I*pi*k*sgn(a) - I*pi*l*sgn(a) + 2*I*pi*k + I*pi*l + 4*k*log(abs(a)) + 2*l*log(abs(a))))*e^((2*k*log(abs(a)) +
l*log(abs(a)))*x) + 6*(2*(k*log(abs(a)) + 2*l*log(abs(a)))*cos(-1/2*pi*k*x*sgn(a) - pi*l*x*sgn(a) + 1/2*pi*k*x
+ pi*l*x)/((pi*k*sgn(a) + 2*pi*l*sgn(a) - pi*k - 2*pi*l)^2 + 4*(k*log(abs(a)) + 2*l*log(abs(a)))^2) - (pi*k*s
gn(a) + 2*pi*l*sgn(a) - pi*k - 2*pi*l)*sin(-1/2*pi*k*x*sgn(a) - pi*l*x*sgn(a) + 1/2*pi*k*x + pi*l*x)/((pi*k*sg
n(a) + 2*pi*l*sgn(a) - pi*k - 2*pi*l)^2 + 4*(k*log(abs(a)) + 2*l*log(abs(a)))^2))*e^((k*log(abs(a)) + 2*l*log(
abs(a)))*x) + 3*I*(I*e^(1/2*I*pi*k*x*sgn(a) + I*pi*l*x*sgn(a) - 1/2*I*pi*k*x - I*pi*l*x)/(I*pi*k*sgn(a) + 2*I*
pi*l*sgn(a) - I*pi*k - 2*I*pi*l + 2*k*log(abs(a)) + 4*l*log(abs(a))) - I*e^(-1/2*I*pi*k*x*sgn(a) - I*pi*l*x*sg
n(a) + 1/2*I*pi*k*x + I*pi*l*x)/(-I*pi*k*sgn(a) - 2*I*pi*l*sgn(a) + I*pi*k + 2*I*pi*l + 2*k*log(abs(a)) + 4*l*
log(abs(a))))*e^((k*log(abs(a)) + 2*l*log(abs(a)))*x)

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Mupad [B]
time = 0.36, size = 81, normalized size = 1.03 \begin {gather*} \frac {3\,a^{k\,x}\,a^{2\,l\,x}}{k\,\ln \left (a\right )+2\,l\,\ln \left (a\right )}-\frac {3\,a^{2\,k\,x}\,a^{l\,x}}{2\,k\,\ln \left (a\right )+l\,\ln \left (a\right )}+\frac {a^{3\,k\,x}}{3\,k\,\ln \left (a\right )}-\frac {a^{3\,l\,x}}{3\,l\,\ln \left (a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^(k*x) - a^(l*x))^3,x)

[Out]

(3*a^(k*x)*a^(2*l*x))/(k*log(a) + 2*l*log(a)) - (3*a^(2*k*x)*a^(l*x))/(2*k*log(a) + l*log(a)) + a^(3*k*x)/(3*k
*log(a)) - a^(3*l*x)/(3*l*log(a))

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