3.504 \(\int (a^{k x}+a^{l x})^3 \, dx\)
Optimal. Leaf size=79 \[ \frac{3 a^{x (2 k+l)}}{\log (a) (2 k+l)}+\frac{3 a^{x (k+2 l)}}{\log (a) (k+2 l)}+\frac{a^{3 k x}}{3 k \log (a)}+\frac{a^{3 l x}}{3 l \log (a)} \]
[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])
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Rubi [A] time = 0.100182, antiderivative size = 79, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used =
{6742, 2194} \[ \frac{3 a^{x (2 k+l)}}{\log (a) (2 k+l)}+\frac{3 a^{x (k+2 l)}}{\log (a) (k+2 l)}+\frac{a^{3 k x}}{3 k \log (a)}+\frac{a^{3 l x}}{3 l \log (a)} \]
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 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]
Rubi steps
\begin{align*} \int \left (a^{k x}+a^{l x}\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int \left (e^{k x}+e^{l x}\right )^3 \, dx,x,x \log (a)\right )}{\log (a)}\\ &=\frac{\operatorname{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{\operatorname{Subst}\left (\int e^{3 k x} \, dx,x,x \log (a)\right )}{\log (a)}+\frac{\operatorname{Subst}\left (\int e^{3 l x} \, dx,x,x \log (a)\right )}{\log (a)}+\frac{3 \operatorname{Subst}\left (\int e^{(2 k+l) x} \, dx,x,x \log (a)\right )}{\log (a)}+\frac{3 \operatorname{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*}
Mathematica [A] time = 0.087738, size = 65, normalized size = 0.82 \[ \frac{\frac{9 a^{x (2 k+l)}}{2 k+l}+\frac{9 a^{x (k+2 l)}}{k+2 l}+\frac{a^{3 k x}}{k}+\frac{a^{3 l x}}{l}}{3 \log (a)} \]
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 = 0.03, size = 90, normalized size = 1.1 \begin{align*}{\frac{ \left ({{\rm e}^{kx\ln \left ( a \right ) }} \right ) ^{3}}{3\,k\ln \left ( a \right ) }}+{\frac{ \left ({{\rm e}^{lx\ln \left ( a \right ) }} \right ) ^{3}}{3\,l\ln \left ( a \right ) }}+3\,{\frac{{{\rm e}^{kx\ln \left ( a \right ) }} \left ({{\rm e}^{lx\ln \left ( a \right ) }} \right ) ^{2}}{\ln \left ( a \right ) \left ( k+2\,l \right ) }}+3\,{\frac{ \left ({{\rm e}^{kx\ln \left ( a \right ) }} \right ) ^{2}{{\rm e}^{lx\ln \left ( a \right ) }}}{\ln \left ( a \right ) \left ( 2\,k+l \right ) }} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a^(k*x)+a^(l*x))^3,x)
[Out]
1/3/k/ln(a)*exp(k*x*ln(a))^3+1/3/l/ln(a)*exp(l*x*ln(a))^3+3/ln(a)/(k+2*l)*exp(k*x*ln(a))*exp(l*x*ln(a))^2+3/ln
(a)/(2*k+l)*exp(k*x*ln(a))^2*exp(l*x*ln(a))
________________________________________________________________________________________
Maxima [A] time = 0.942141, size = 104, normalized size = 1.32 \begin{align*} \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{align*}
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 = 1.7881, size = 278, normalized size = 3.52 \begin{align*} \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{align*}
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 [A] time = 30.8855, size = 665, normalized size = 8.42 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a**(k*x)+a**(l*x))**3,x)
[Out]
Piecewise((8*x, 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 - a**(-3*l*x)/(l*log(
a)) - a**(-6*l*x)/(6*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*a
**(-3*l*x/2)/(3*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) + 15
*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*l
og(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] time = 1.20619, size = 1395, normalized size = 17.66 \begin{align*} \text{result too large to display} \end{align*}
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) - 1/2*I*abs(a)^(3*k*x)*(-2*I*e^(3/2*I*pi*k*x*sgn(a) - 3/2*I*pi*k*x)/(3*I*pi*k*sgn(a) - 3*I*p
i*k + 6*k*log(abs(a))) + 2*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(ab
s(a)))) - 1/2*I*abs(a)^(3*l*x)*(-2*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))) + 2*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) - 1/2*I*
(-6*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))) + 6*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*lo
g(abs(a)))^2) - (pi*k*sgn(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*sgn(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) - 1/2*I*(-6*I*e^(1/2*I*pi*k*x*sgn(a) + I*pi*l*x*sgn(a) - 1/2*I*pi*k*x - I*p
i*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))) + 6*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))))*e^((k*log(abs(a)) + 2*l*log(abs(a)))*x)