3.501 \(\int (a^{-4 x}-a^{2 x})^3 \, dx\)

Optimal. Leaf size=43 \[ -\frac {a^{-12 x}}{12 \log (a)}+\frac {a^{-6 x}}{2 \log (a)}-\frac {a^{6 x}}{6 \log (a)}+3 x \]

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Rubi [A]  time = 0.03, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2282, 266, 43} \[ -\frac {a^{-12 x}}{12 \log (a)}+\frac {a^{-6 x}}{2 \log (a)}-\frac {a^{6 x}}{6 \log (a)}+3 x \]

Antiderivative was successfully verified.

[In]

Int[(a^(-4*x) - a^(2*x))^3,x]

[Out]

3*x - 1/(12*a^(12*x)*Log[a]) + 1/(2*a^(6*x)*Log[a]) - a^(6*x)/(6*Log[a])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rubi steps

\begin {align*} \int \left (a^{-4 x}-a^{2 x}\right )^3 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1-x^3\right )^3}{x^7} \, dx,x,a^{2 x}\right )}{2 \log (a)}\\ &=\frac {\operatorname {Subst}\left (\int \frac {(1-x)^3}{x^3} \, dx,x,a^{6 x}\right )}{6 \log (a)}\\ &=\frac {\operatorname {Subst}\left (\int \left (-1+\frac {1}{x^3}-\frac {3}{x^2}+\frac {3}{x}\right ) \, dx,x,a^{6 x}\right )}{6 \log (a)}\\ &=3 x-\frac {a^{-12 x}}{12 \log (a)}+\frac {a^{-6 x}}{2 \log (a)}-\frac {a^{6 x}}{6 \log (a)}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 33, normalized size = 0.77 \[ -\frac {a^{-12 x}-6 a^{-6 x}+2 a^{6 x}-36 x \log (a)}{12 \log (a)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^(-4*x) - a^(2*x))^3,x]

[Out]

-1/12*(a^(-12*x) - 6/a^(6*x) + 2*a^(6*x) - 36*x*Log[a])/Log[a]

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a^{-4 x}-a^{2 x}\right )^3 \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[(a^(-4*x) - a^(2*x))^3,x]

[Out]

Could not integrate

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fricas [A]  time = 1.14, size = 39, normalized size = 0.91 \[ \frac {36 \, a^{12 \, x} x \log \relax (a) - 2 \, a^{18 \, x} + 6 \, a^{6 \, x} - 1}{12 \, a^{12 \, x} \log \relax (a)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/(a^(4*x))-a^(2*x))^3,x, algorithm="fricas")

[Out]

1/12*(36*a^(12*x)*x*log(a) - 2*a^(18*x) + 6*a^(6*x) - 1)/(a^(12*x)*log(a))

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giac [A]  time = 0.60, size = 46, normalized size = 1.07 \[ -\frac {2 \, a^{6 \, x} + \frac {9 \, a^{12 \, x} - 6 \, a^{6 \, x} + 1}{a^{12 \, x}} - 6 \, \log \left (a^{6 \, x}\right )}{12 \, \log \relax (a)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/(a^(4*x))-a^(2*x))^3,x, algorithm="giac")

[Out]

-1/12*(2*a^(6*x) + (9*a^(12*x) - 6*a^(6*x) + 1)/a^(12*x) - 6*log(a^(6*x)))/log(a)

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maple [A]  time = 0.07, size = 44, normalized size = 1.02

method	result	size
risch	\(3 x -\frac {a^{6 x}}{6 \ln \relax (a )}+\frac {a^{-6 x}}{2 \ln \relax (a )}-\frac {a^{-12 x}}{12 \ln \relax (a )}\)	\(44\)
norman	\(\left (-\frac {1}{12 \ln \relax (a )}+3 x \,{\mathrm e}^{12 x \ln \relax (a )}+\frac {{\mathrm e}^{6 x \ln \relax (a )}}{2 \ln \relax (a )}-\frac {{\mathrm e}^{18 x \ln \relax (a )}}{6 \ln \relax (a )}\right ) {\mathrm e}^{-12 x \ln \relax (a )}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/(a^(4*x))-a^(2*x))^3,x,method=_RETURNVERBOSE)

[Out]

3*x-1/6/ln(a)*(a^(2*x))^3+1/2/ln(a)/(a^(2*x))^3-1/12/ln(a)/(a^(2*x))^6

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maxima [A]  time = 0.56, size = 41, normalized size = 0.95 \[ 3 \, x - \frac {a^{6 \, x}}{6 \, \log \relax (a)} - \frac {1}{12 \, a^{12 \, x} \log \relax (a)} + \frac {1}{2 \, a^{6 \, x} \log \relax (a)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/(a^(4*x))-a^(2*x))^3,x, algorithm="maxima")

[Out]

3*x - 1/6*a^(6*x)/log(a) - 1/12/(a^(12*x)*log(a)) + 1/2/(a^(6*x)*log(a))

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mupad [B]  time = 0.41, size = 41, normalized size = 0.95 \[ 3\,x+\frac {1}{2\,a^{6\,x}\,\ln \relax (a)}-\frac {a^{6\,x}}{6\,\ln \relax (a)}-\frac {1}{12\,a^{12\,x}\,\ln \relax (a)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(a^(2*x) - 1/a^(4*x))^3,x)

[Out]

3*x + 1/(2*a^(6*x)*log(a)) - a^(6*x)/(6*log(a)) - 1/(12*a^(12*x)*log(a))

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sympy [A]  time = 0.22, size = 54, normalized size = 1.26 \[ 3 x + \begin {cases} \frac {- 24 a^{6 x} \log {\relax (a )}^{2} + 72 a^{- 6 x} \log {\relax (a )}^{2} - 12 a^{- 12 x} \log {\relax (a )}^{2}}{144 \log {\relax (a )}^{3}} & \text {for}\: 144 \log {\relax (a )}^{3} \neq 0 \\- 3 x & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/(a**(4*x))-a**(2*x))**3,x)

[Out]

3*x + Piecewise(((-24*a**(6*x)*log(a)**2 + 72*a**(-6*x)*log(a)**2 - 12*a**(-12*x)*log(a)**2)/(144*log(a)**3),
Ne(144*log(a)**3, 0)), (-3*x, True))

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