3.519 \(\int (1-a^{m x})^3 \, dx\)

Optimal. Leaf size=50 \[ -\frac {3 a^{m x}}{m \log (a)}+\frac {3 a^{2 m x}}{2 m \log (a)}-\frac {a^{3 m x}}{3 m \log (a)}+x \]

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Rubi [A]  time = 0.02, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2282, 43} \[ -\frac {3 a^{m x}}{m \log (a)}+\frac {3 a^{2 m x}}{2 m \log (a)}-\frac {a^{3 m x}}{3 m \log (a)}+x \]

Antiderivative was successfully verified.

[In]

Int[(1 - a^(m*x))^3,x]

[Out]

x - (3*a^(m*x))/(m*Log[a]) + (3*a^(2*m*x))/(2*m*Log[a]) - a^(3*m*x)/(3*m*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 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 (1-a^{m x}\right )^3 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {(1-x)^3}{x} \, dx,x,a^{m x}\right )}{m \log (a)}\\ &=\frac {\operatorname {Subst}\left (\int \left (-3+\frac {1}{x}+3 x-x^2\right ) \, dx,x,a^{m x}\right )}{m \log (a)}\\ &=x-\frac {3 a^{m x}}{m \log (a)}+\frac {3 a^{2 m x}}{2 m \log (a)}-\frac {a^{3 m x}}{3 m \log (a)}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 35, normalized size = 0.70 \[ x-\frac {a^{m x} \left (-9 a^{m x}+2 a^{2 m x}+18\right )}{6 m \log (a)} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - a^(m*x))^3,x]

[Out]

x - (a^(m*x)*(18 - 9*a^(m*x) + 2*a^(2*m*x)))/(6*m*Log[a])

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (1-a^{m x}\right )^3 \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[(1 - a^(m*x))^3,x]

[Out]

Could not integrate

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fricas [A]  time = 1.32, size = 39, normalized size = 0.78 \[ \frac {6 \, m x \log \relax (a) - 2 \, a^{3 \, m x} + 9 \, a^{2 \, m x} - 18 \, a^{m x}}{6 \, m \log \relax (a)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-a^(m*x))^3,x, algorithm="fricas")

[Out]

1/6*(6*m*x*log(a) - 2*a^(3*m*x) + 9*a^(2*m*x) - 18*a^(m*x))/(m*log(a))

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giac [A]  time = 0.57, size = 40, normalized size = 0.80 \[ \frac {6 \, m x \log \left ({\left | a \right |}\right ) - 2 \, a^{3 \, m x} + 9 \, a^{2 \, m x} - 18 \, a^{m x}}{6 \, m \log \relax (a)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-a^(m*x))^3,x, algorithm="giac")

[Out]

1/6*(6*m*x*log(abs(a)) - 2*a^(3*m*x) + 9*a^(2*m*x) - 18*a^(m*x))/(m*log(a))

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maple [A]  time = 0.04, size = 41, normalized size = 0.82

method	result	size
derivativedivides	\(\frac {-\frac {a^{3 m x}}{3}+\frac {3 a^{2 m x}}{2}-3 a^{m x}+\ln \left (a^{m x}\right )}{m \ln \relax (a )}\)	\(41\)
default	\(\frac {-\frac {a^{3 m x}}{3}+\frac {3 a^{2 m x}}{2}-3 a^{m x}+\ln \left (a^{m x}\right )}{m \ln \relax (a )}\)	\(41\)
risch	\(x -\frac {3 a^{m x}}{m \ln \relax (a )}+\frac {3 a^{2 m x}}{2 m \ln \relax (a )}-\frac {a^{3 m x}}{3 m \ln \relax (a )}\)	\(49\)
norman	\(x -\frac {3 \,{\mathrm e}^{m x \ln \relax (a )}}{m \ln \relax (a )}+\frac {3 \,{\mathrm e}^{2 m x \ln \relax (a )}}{2 m \ln \relax (a )}-\frac {{\mathrm e}^{3 m x \ln \relax (a )}}{3 m \ln \relax (a )}\)	\(52\)
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((1-a^(m*x))^3,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.50, size = 46, normalized size = 0.92 \[ x - \frac {a^{3 \, m x}}{3 \, m \log \relax (a)} + \frac {3 \, a^{2 \, m x}}{2 \, m \log \relax (a)} - \frac {3 \, a^{m x}}{m \log \relax (a)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-a^(m*x))^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.32, size = 35, normalized size = 0.70 \[ x-\frac {3\,a^{m\,x}-\frac {3\,a^{2\,m\,x}}{2}+\frac {a^{3\,m\,x}}{3}}{m\,\ln \relax (a)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(a^(m*x) - 1)^3,x)

[Out]

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

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sympy [A]  time = 0.14, size = 71, normalized size = 1.42 \[ x + \begin {cases} \frac {- 2 a^{3 m x} m^{2} \log {\relax (a )}^{2} + 9 a^{2 m x} m^{2} \log {\relax (a )}^{2} - 18 a^{m x} m^{2} \log {\relax (a )}^{2}}{6 m^{3} \log {\relax (a )}^{3}} & \text {for}\: 6 m^{3} \log {\relax (a )}^{3} \neq 0 \\- x & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-a**(m*x))**3,x)

[Out]

x + Piecewise(((-2*a**(3*m*x)*m**2*log(a)**2 + 9*a**(2*m*x)*m**2*log(a)**2 - 18*a**(m*x)*m**2*log(a)**2)/(6*m*
*3*log(a)**3), Ne(6*m**3*log(a)**3, 0)), (-x, True))

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