3.505 \(\int (a^{k x}+a^{l x})^4 \, dx\)

Optimal. Leaf size=98 \[ \frac {3 a^{2 x (k+l)}}{\log (a) (k+l)}+\frac {4 a^{x (3 k+l)}}{\log (a) (3 k+l)}+\frac {4 a^{x (k+3 l)}}{\log (a) (k+3 l)}+\frac {a^{4 k x}}{4 k \log (a)}+\frac {a^{4 l x}}{4 l \log (a)} \]

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {6742, 2194} \[ \frac {3 a^{2 x (k+l)}}{\log (a) (k+l)}+\frac {4 a^{x (3 k+l)}}{\log (a) (3 k+l)}+\frac {4 a^{x (k+3 l)}}{\log (a) (k+3 l)}+\frac {a^{4 k x}}{4 k \log (a)}+\frac {a^{4 l x}}{4 l \log (a)} \]

Antiderivative was successfully verified.

[In]

Int[(a^(k*x) + a^(l*x))^4,x]

[Out]

a^(4*k*x)/(4*k*Log[a]) + a^(4*l*x)/(4*l*Log[a]) + (3*a^(2*(k + l)*x))/((k + l)*Log[a]) + (4*a^((3*k + l)*x))/(
(3*k + l)*Log[a]) + (4*a^((k + 3*l)*x))/((k + 3*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 )^4 \, dx &=\frac {\operatorname {Subst}\left (\int \left (e^{k x}+e^{l x}\right )^4 \, dx,x,x \log (a)\right )}{\log (a)}\\ &=\frac {\operatorname {Subst}\left (\int \left (e^{4 k x}+e^{4 l x}+6 e^{2 (k+l) x}+4 e^{(3 k+l) x}+4 e^{(k+3 l) x}\right ) \, dx,x,x \log (a)\right )}{\log (a)}\\ &=\frac {\operatorname {Subst}\left (\int e^{4 k x} \, dx,x,x \log (a)\right )}{\log (a)}+\frac {\operatorname {Subst}\left (\int e^{4 l x} \, dx,x,x \log (a)\right )}{\log (a)}+\frac {4 \operatorname {Subst}\left (\int e^{(3 k+l) x} \, dx,x,x \log (a)\right )}{\log (a)}+\frac {4 \operatorname {Subst}\left (\int e^{(k+3 l) x} \, dx,x,x \log (a)\right )}{\log (a)}+\frac {6 \operatorname {Subst}\left (\int e^{2 (k+l) x} \, dx,x,x \log (a)\right )}{\log (a)}\\ &=\frac {a^{4 k x}}{4 k \log (a)}+\frac {a^{4 l x}}{4 l \log (a)}+\frac {3 a^{2 (k+l) x}}{(k+l) \log (a)}+\frac {4 a^{(3 k+l) x}}{(3 k+l) \log (a)}+\frac {4 a^{(k+3 l) x}}{(k+3 l) \log (a)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.10, size = 80, normalized size = 0.82 \[ \frac {\frac {12 a^{2 x (k+l)}}{k+l}+\frac {16 a^{x (3 k+l)}}{3 k+l}+\frac {16 a^{x (k+3 l)}}{k+3 l}+\frac {a^{4 k x}}{k}+\frac {a^{4 l x}}{l}}{4 \log (a)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^(k*x) + a^(l*x))^4,x]

[Out]

(a^(4*k*x)/k + a^(4*l*x)/l + (12*a^(2*(k + l)*x))/(k + l) + (16*a^((3*k + l)*x))/(3*k + l) + (16*a^((k + 3*l)*
x))/(k + 3*l))/(4*Log[a])

________________________________________________________________________________________

IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a^{k x}+a^{l x}\right )^4 \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[(a^(k*x) + a^(l*x))^4,x]

[Out]

Could not integrate

________________________________________________________________________________________

fricas [B]  time = 1.34, size = 205, normalized size = 2.09 \[ \frac {16 \, {\left (3 \, k^{3} l + 4 \, k^{2} l^{2} + k l^{3}\right )} a^{k x} a^{3 \, l x} + 12 \, {\left (3 \, k^{3} l + 10 \, k^{2} l^{2} + 3 \, k l^{3}\right )} a^{2 \, k x} a^{2 \, l x} + 16 \, {\left (k^{3} l + 4 \, k^{2} l^{2} + 3 \, k l^{3}\right )} a^{3 \, k x} a^{l x} + {\left (3 \, k^{3} l + 13 \, k^{2} l^{2} + 13 \, k l^{3} + 3 \, l^{4}\right )} a^{4 \, k x} + {\left (3 \, k^{4} + 13 \, k^{3} l + 13 \, k^{2} l^{2} + 3 \, k l^{3}\right )} a^{4 \, l x}}{4 \, {\left (3 \, k^{4} l + 13 \, k^{3} l^{2} + 13 \, k^{2} l^{3} + 3 \, k l^{4}\right )} \log \relax (a)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^(k*x)+a^(l*x))^4,x, algorithm="fricas")

[Out]

1/4*(16*(3*k^3*l + 4*k^2*l^2 + k*l^3)*a^(k*x)*a^(3*l*x) + 12*(3*k^3*l + 10*k^2*l^2 + 3*k*l^3)*a^(2*k*x)*a^(2*l
*x) + 16*(k^3*l + 4*k^2*l^2 + 3*k*l^3)*a^(3*k*x)*a^(l*x) + (3*k^3*l + 13*k^2*l^2 + 13*k*l^3 + 3*l^4)*a^(4*k*x)
+ (3*k^4 + 13*k^3*l + 13*k^2*l^2 + 3*k*l^3)*a^(4*l*x))/((3*k^4*l + 13*k^3*l^2 + 13*k^2*l^3 + 3*k*l^4)*log(a))

________________________________________________________________________________________

giac [C]  time = 0.94, size = 1359, normalized size = 13.87 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^(k*x)+a^(l*x))^4,x, algorithm="giac")

[Out]

1/2*(2*k*cos(-2*pi*k*x*sgn(a) + 2*pi*k*x)*log(abs(a))/(4*k^2*log(abs(a))^2 + (pi*k*sgn(a) - pi*k)^2) - (pi*k*s
gn(a) - pi*k)*sin(-2*pi*k*x*sgn(a) + 2*pi*k*x)/(4*k^2*log(abs(a))^2 + (pi*k*sgn(a) - pi*k)^2))*abs(a)^(4*k*x)
+ 1/2*(2*l*cos(-2*pi*l*x*sgn(a) + 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(-2*pi*l*x*sgn(a) + 2*pi*l*x)/(4*l^2*log(abs(a))^2 + (pi*l*sgn(a) - pi*l)^2))*abs(a)^(4*l*x
) - 1/2*I*abs(a)^(4*k*x)*(-I*e^(2*I*pi*k*x*sgn(a) - 2*I*pi*k*x)/(2*I*pi*k*sgn(a) - 2*I*pi*k + 4*k*log(abs(a)))
+ I*e^(-2*I*pi*k*x*sgn(a) + 2*I*pi*k*x)/(-2*I*pi*k*sgn(a) + 2*I*pi*k + 4*k*log(abs(a)))) - 1/2*I*abs(a)^(4*l*
x)*(-I*e^(2*I*pi*l*x*sgn(a) - 2*I*pi*l*x)/(2*I*pi*l*sgn(a) - 2*I*pi*l + 4*l*log(abs(a))) + I*e^(-2*I*pi*l*x*sg
n(a) + 2*I*pi*l*x)/(-2*I*pi*l*sgn(a) + 2*I*pi*l + 4*l*log(abs(a)))) + 8*(2*(3*k*log(abs(a)) + l*log(abs(a)))*c
os(-3/2*pi*k*x*sgn(a) - 1/2*pi*l*x*sgn(a) + 3/2*pi*k*x + 1/2*pi*l*x)/((3*pi*k*sgn(a) + pi*l*sgn(a) - 3*pi*k -
pi*l)^2 + 4*(3*k*log(abs(a)) + l*log(abs(a)))^2) - (3*pi*k*sgn(a) + pi*l*sgn(a) - 3*pi*k - pi*l)*sin(-3/2*pi*k
*x*sgn(a) - 1/2*pi*l*x*sgn(a) + 3/2*pi*k*x + 1/2*pi*l*x)/((3*pi*k*sgn(a) + pi*l*sgn(a) - 3*pi*k - pi*l)^2 + 4*
(3*k*log(abs(a)) + l*log(abs(a)))^2))*e^((3*k*log(abs(a)) + l*log(abs(a)))*x) - 1/2*I*(-8*I*e^(3/2*I*pi*k*x*sg
n(a) + 1/2*I*pi*l*x*sgn(a) - 3/2*I*pi*k*x - 1/2*I*pi*l*x)/(3*I*pi*k*sgn(a) + I*pi*l*sgn(a) - 3*I*pi*k - I*pi*l
+ 6*k*log(abs(a)) + 2*l*log(abs(a))) + 8*I*e^(-3/2*I*pi*k*x*sgn(a) - 1/2*I*pi*l*x*sgn(a) + 3/2*I*pi*k*x + 1/2
*I*pi*l*x)/(-3*I*pi*k*sgn(a) - I*pi*l*sgn(a) + 3*I*pi*k + I*pi*l + 6*k*log(abs(a)) + 2*l*log(abs(a))))*e^((3*k
*log(abs(a)) + l*log(abs(a)))*x) + 8*(2*(k*log(abs(a)) + 3*l*log(abs(a)))*cos(-1/2*pi*k*x*sgn(a) - 3/2*pi*l*x*
sgn(a) + 1/2*pi*k*x + 3/2*pi*l*x)/((pi*k*sgn(a) + 3*pi*l*sgn(a) - pi*k - 3*pi*l)^2 + 4*(k*log(abs(a)) + 3*l*lo
g(abs(a)))^2) - (pi*k*sgn(a) + 3*pi*l*sgn(a) - pi*k - 3*pi*l)*sin(-1/2*pi*k*x*sgn(a) - 3/2*pi*l*x*sgn(a) + 1/2
*pi*k*x + 3/2*pi*l*x)/((pi*k*sgn(a) + 3*pi*l*sgn(a) - pi*k - 3*pi*l)^2 + 4*(k*log(abs(a)) + 3*l*log(abs(a)))^2
))*e^((k*log(abs(a)) + 3*l*log(abs(a)))*x) - 1/2*I*(-8*I*e^(1/2*I*pi*k*x*sgn(a) + 3/2*I*pi*l*x*sgn(a) - 1/2*I*
pi*k*x - 3/2*I*pi*l*x)/(I*pi*k*sgn(a) + 3*I*pi*l*sgn(a) - I*pi*k - 3*I*pi*l + 2*k*log(abs(a)) + 6*l*log(abs(a)
)) + 8*I*e^(-1/2*I*pi*k*x*sgn(a) - 3/2*I*pi*l*x*sgn(a) + 1/2*I*pi*k*x + 3/2*I*pi*l*x)/(-I*pi*k*sgn(a) - 3*I*pi
*l*sgn(a) + I*pi*k + 3*I*pi*l + 2*k*log(abs(a)) + 6*l*log(abs(a))))*e^((k*log(abs(a)) + 3*l*log(abs(a)))*x) +
6*(2*(k*log(abs(a)) + l*log(abs(a)))*cos(-pi*k*x*sgn(a) - pi*l*x*sgn(a) + pi*k*x + 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)*s
in(-pi*k*x*sgn(a) - pi*l*x*sgn(a) + pi*k*x + pi*l*x)/((pi*k*sgn(a) + pi*l*sgn(a) - pi*k - pi*l)^2 + 4*(k*log(a
bs(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) + I*pi*l*
x*sgn(a) - I*pi*k*x - 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(a
bs(a))) + 6*I*e^(-I*pi*k*x*sgn(a) - I*pi*l*x*sgn(a) + I*pi*k*x + 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^(2*(k*log(abs(a)) + l*log(abs(a)))*x)

________________________________________________________________________________________

maple [A]  time = 0.03, size = 109, normalized size = 1.11

method	result	size
risch	\(\frac {a^{4 k x}}{4 k \ln \relax (a )}+\frac {4 a^{3 k x} a^{l x}}{\ln \relax (a ) \left (3 k +l \right )}+\frac {3 a^{2 k x} a^{2 l x}}{\ln \relax (a ) \left (k +l \right )}+\frac {4 a^{k x} a^{3 l x}}{\ln \relax (a ) \left (k +3 l \right )}+\frac {a^{4 l x}}{4 l \ln \relax (a )}\)	\(109\)
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))^4,x,method=_RETURNVERBOSE)

[Out]

1/4/ln(a)/k*(a^(k*x))^4+4*(a^(k*x))^3/ln(a)/(3*k+l)*a^(l*x)+3*(a^(k*x))^2/ln(a)/(k+l)*(a^(l*x))^2+4*a^(k*x)/ln
(a)/(k+3*l)*(a^(l*x))^3+1/4/ln(a)/l*(a^(l*x))^4

________________________________________________________________________________________

maxima [A]  time = 0.55, size = 99, normalized size = 1.01 \[ \frac {4 \, a^{3 \, k x + l x}}{{\left (3 \, k + l\right )} \log \relax (a)} + \frac {4 \, a^{k x + 3 \, l x}}{{\left (k + 3 \, l\right )} \log \relax (a)} + \frac {3 \, a^{2 \, k x + 2 \, l x}}{{\left (k + l\right )} \log \relax (a)} + \frac {a^{4 \, k x}}{4 \, k \log \relax (a)} + \frac {a^{4 \, l x}}{4 \, l \log \relax (a)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^(k*x)+a^(l*x))^4,x, algorithm="maxima")

[Out]

4*a^(3*k*x + l*x)/((3*k + l)*log(a)) + 4*a^(k*x + 3*l*x)/((k + 3*l)*log(a)) + 3*a^(2*k*x + 2*l*x)/((k + l)*log
(a)) + 1/4*a^(4*k*x)/(k*log(a)) + 1/4*a^(4*l*x)/(l*log(a))

________________________________________________________________________________________

mupad [B]  time = 0.42, size = 106, normalized size = 1.08 \[ \frac {3\,a^{2\,k\,x}\,a^{2\,l\,x}}{k\,\ln \relax (a)+l\,\ln \relax (a)}+\frac {4\,a^{k\,x}\,a^{3\,l\,x}}{k\,\ln \relax (a)+3\,l\,\ln \relax (a)}+\frac {4\,a^{3\,k\,x}\,a^{l\,x}}{3\,k\,\ln \relax (a)+l\,\ln \relax (a)}+\frac {a^{4\,k\,x}}{4\,k\,\ln \relax (a)}+\frac {a^{4\,l\,x}}{4\,l\,\ln \relax (a)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^(k*x) + a^(l*x))^4,x)

[Out]

(3*a^(2*k*x)*a^(2*l*x))/(k*log(a) + l*log(a)) + (4*a^(k*x)*a^(3*l*x))/(k*log(a) + 3*l*log(a)) + (4*a^(3*k*x)*a
^(l*x))/(3*k*log(a) + l*log(a)) + a^(4*k*x)/(4*k*log(a)) + a^(4*l*x)/(4*l*log(a))

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a**(k*x)+a**(l*x))**4,x)

[Out]

Timed out

________________________________________________________________________________________