3.6.20 \(\int (1-a^{m x})^4 \, dx\) [520]

Optimal. Leaf size=65 \[ x-\frac {4 a^{m x}}{m \log (a)}+\frac {3 a^{2 m x}}{m \log (a)}-\frac {4 a^{3 m x}}{3 m \log (a)}+\frac {a^{4 m x}}{4 m \log (a)} \]

[Out]

x-4*a^(m*x)/m/ln(a)+3*a^(2*m*x)/m/ln(a)-4/3*a^(3*m*x)/m/ln(a)+1/4*a^(4*m*x)/m/ln(a)

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Rubi [A]
time = 0.01, antiderivative size = 65, 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 = {2320, 45} \begin {gather*} -\frac {4 a^{m x}}{m \log (a)}+\frac {3 a^{2 m x}}{m \log (a)}-\frac {4 a^{3 m x}}{3 m \log (a)}+\frac {a^{4 m x}}{4 m \log (a)}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - a^(m*x))^4,x]

[Out]

x - (4*a^(m*x))/(m*Log[a]) + (3*a^(2*m*x))/(m*Log[a]) - (4*a^(3*m*x))/(3*m*Log[a]) + a^(4*m*x)/(4*m*Log[a])

Rule 45

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 2320

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

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Mathematica [A]
time = 0.02, size = 53, normalized size = 0.82 \begin {gather*} \frac {\frac {a^{m x} \left (-48+36 a^{m x}-16 a^{2 m x}+3 a^{3 m x}\right )}{12 m}+\frac {\log \left (a^{m x}\right )}{m}}{\log (a)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - a^(m*x))^4,x]

[Out]

((a^(m*x)*(-48 + 36*a^(m*x) - 16*a^(2*m*x) + 3*a^(3*m*x)))/(12*m) + Log[a^(m*x)]/m)/Log[a]

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Maple [A]
time = 0.04, size = 50, normalized size = 0.77

method	result	size
derivativedivides	\(\frac {\frac {a^{4 m x}}{4}-\frac {4 a^{3 m x}}{3}+3 a^{2 m x}-4 a^{m x}+\ln \left (a^{m x}\right )}{m \ln \left (a \right )}\)	\(50\)
default	\(\frac {\frac {a^{4 m x}}{4}-\frac {4 a^{3 m x}}{3}+3 a^{2 m x}-4 a^{m x}+\ln \left (a^{m x}\right )}{m \ln \left (a \right )}\)	\(50\)
risch	\(x -\frac {4 a^{m x}}{m \ln \left (a \right )}+\frac {3 a^{2 m x}}{m \ln \left (a \right )}-\frac {4 a^{3 m x}}{3 m \ln \left (a \right )}+\frac {a^{4 m x}}{4 m \ln \left (a \right )}\)	\(65\)
norman	\(x -\frac {4 \,{\mathrm e}^{m x \ln \left (a \right )}}{m \ln \left (a \right )}+\frac {3 \,{\mathrm e}^{2 m x \ln \left (a \right )}}{m \ln \left (a \right )}-\frac {4 \,{\mathrm e}^{3 m x \ln \left (a \right )}}{3 m \ln \left (a \right )}+\frac {{\mathrm e}^{4 m x \ln \left (a \right )}}{4 m \ln \left (a \right )}\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-a^(m*x))^4,x,method=_RETURNVERBOSE)

[Out]

1/m/ln(a)*(1/4*(a^(m*x))^4-4/3*(a^(m*x))^3+3*(a^(m*x))^2-4*a^(m*x)+ln(a^(m*x)))

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Maxima [A]
time = 5.66, size = 61, normalized size = 0.94 \begin {gather*} x + \frac {a^{4 \, m x}}{4 \, m \log \left (a\right )} - \frac {4 \, a^{3 \, m x}}{3 \, m \log \left (a\right )} + \frac {3 \, a^{2 \, m x}}{m \log \left (a\right )} - \frac {4 \, a^{m x}}{m \log \left (a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-a^(m*x))^4,x, algorithm="maxima")

[Out]

x + 1/4*a^(4*m*x)/(m*log(a)) - 4/3*a^(3*m*x)/(m*log(a)) + 3*a^(2*m*x)/(m*log(a)) - 4*a^(m*x)/(m*log(a))

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Fricas [A]
time = 0.44, size = 47, normalized size = 0.72 \begin {gather*} \frac {12 \, m x \log \left (a\right ) + 3 \, a^{4 \, m x} - 16 \, a^{3 \, m x} + 36 \, a^{2 \, m x} - 48 \, a^{m x}}{12 \, m \log \left (a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-a^(m*x))^4,x, algorithm="fricas")

[Out]

1/12*(12*m*x*log(a) + 3*a^(4*m*x) - 16*a^(3*m*x) + 36*a^(2*m*x) - 48*a^(m*x))/(m*log(a))

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Sympy [A]
time = 0.06, size = 87, normalized size = 1.34 \begin {gather*} x + \begin {cases} \frac {3 a^{4 m x} m^{3} \log {\left (a \right )}^{3} - 16 a^{3 m x} m^{3} \log {\left (a \right )}^{3} + 36 a^{2 m x} m^{3} \log {\left (a \right )}^{3} - 48 a^{m x} m^{3} \log {\left (a \right )}^{3}}{12 m^{4} \log {\left (a \right )}^{4}} & \text {for}\: m^{4} \log {\left (a \right )}^{4} \neq 0 \\- x & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-a**(m*x))**4,x)

[Out]

x + Piecewise(((3*a**(4*m*x)*m**3*log(a)**3 - 16*a**(3*m*x)*m**3*log(a)**3 + 36*a**(2*m*x)*m**3*log(a)**3 - 48
*a**(m*x)*m**3*log(a)**3)/(12*m**4*log(a)**4), Ne(m**4*log(a)**4, 0)), (-x, True))

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Giac [A]
time = 1.59, size = 48, normalized size = 0.74 \begin {gather*} \frac {12 \, m x \log \left ({\left | a \right |}\right ) + 3 \, a^{4 \, m x} - 16 \, a^{3 \, m x} + 36 \, a^{2 \, m x} - 48 \, a^{m x}}{12 \, m \log \left (a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-a^(m*x))^4,x, algorithm="giac")

[Out]

1/12*(12*m*x*log(abs(a)) + 3*a^(4*m*x) - 16*a^(3*m*x) + 36*a^(2*m*x) - 48*a^(m*x))/(m*log(a))

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Mupad [B]
time = 0.31, size = 43, normalized size = 0.66 \begin {gather*} x-\frac {4\,a^{m\,x}-3\,a^{2\,m\,x}+\frac {4\,a^{3\,m\,x}}{3}-\frac {a^{4\,m\,x}}{4}}{m\,\ln \left (a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^(m*x) - 1)^4,x)

[Out]

x - (4*a^(m*x) - 3*a^(2*m*x) + (4*a^(3*m*x))/3 - a^(4*m*x)/4)/(m*log(a))

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