3.515 \(\int (1+a^{m x})^4 \, dx\)

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

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Rubi [A]  time = 0.02, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2282, 43} \[ \frac {4 a^{m x}}{m \log (a)}+\frac {3 a^{2 m x}}{m \log (a)}+\frac {4 a^{3 m x}}{3 m \log (a)}+\frac {a^{4 m x}}{4 m \log (a)}+x \]

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 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 )^4 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {(1+x)^4}{x} \, dx,x,a^{m x}\right )}{m \log (a)}\\ &=\frac {\operatorname {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 = 49, normalized size = 0.75 \[ \frac {4 a^{m x}+3 a^{2 m x}+\frac {4}{3} a^{3 m x}+\frac {1}{4} a^{4 m x}+m x \log (a)}{m \log (a)} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + a^(m*x))^4,x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (1+a^{m x}\right )^4 \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[(1 + a^(m*x))^4,x]

[Out]

Could not integrate

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fricas [A]  time = 1.38, size = 47, normalized size = 0.72 \[ \frac {12 \, m x \log \relax (a) + 3 \, a^{4 \, m x} + 16 \, a^{3 \, m x} + 36 \, a^{2 \, m x} + 48 \, a^{m x}}{12 \, m \log \relax (a)} \]

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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giac [A]  time = 0.58, size = 48, normalized size = 0.74 \[ \frac {12 \, m x \log \left ({\left | a \right |}\right ) + 3 \, a^{4 \, m x} + 16 \, a^{3 \, m x} + 36 \, a^{2 \, m x} + 48 \, a^{m x}}{12 \, m \log \relax (a)} \]

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

method	result	size
derivativedivides	\(\frac {\frac {a^{4 m x}}{4}+\frac {4 a^{3 m x}}{3}+3 a^{2 m x}+4 a^{m x}+\ln \left (a^{m x}\right )}{m \ln \relax (a )}\)	\(50\)
default	\(\frac {\frac {a^{4 m x}}{4}+\frac {4 a^{3 m x}}{3}+3 a^{2 m x}+4 a^{m x}+\ln \left (a^{m x}\right )}{m \ln \relax (a )}\)	\(50\)
risch	\(x +\frac {4 a^{m x}}{m \ln \relax (a )}+\frac {3 a^{2 m x}}{m \ln \relax (a )}+\frac {4 a^{3 m x}}{3 m \ln \relax (a )}+\frac {a^{4 m x}}{4 m \ln \relax (a )}\)	\(65\)
norman	\(x +\frac {4 \,{\mathrm e}^{m x \ln \relax (a )}}{m \ln \relax (a )}+\frac {3 \,{\mathrm e}^{2 m x \ln \relax (a )}}{m \ln \relax (a )}+\frac {4 \,{\mathrm e}^{3 m x \ln \relax (a )}}{3 m \ln \relax (a )}+\frac {{\mathrm e}^{4 m x \ln \relax (a )}}{4 m \ln \relax (a )}\)	\(69\)
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))^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 = 0.51, size = 61, normalized size = 0.94 \[ x + \frac {a^{4 \, m x}}{4 \, m \log \relax (a)} + \frac {4 \, a^{3 \, m x}}{3 \, m \log \relax (a)} + \frac {3 \, a^{2 \, m x}}{m \log \relax (a)} + \frac {4 \, a^{m x}}{m \log \relax (a)} \]

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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mupad [B]  time = 0.33, size = 42, normalized size = 0.65 \[ x+\frac {4\,a^{m\,x}+3\,a^{2\,m\,x}+\frac {4\,a^{3\,m\,x}}{3}+\frac {a^{4\,m\,x}}{4}}{m\,\ln \relax (a)} \]

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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sympy [A]  time = 0.16, size = 88, normalized size = 1.35 \[ x + \begin {cases} \frac {3 a^{4 m x} m^{3} \log {\relax (a )}^{3} + 16 a^{3 m x} m^{3} \log {\relax (a )}^{3} + 36 a^{2 m x} m^{3} \log {\relax (a )}^{3} + 48 a^{m x} m^{3} \log {\relax (a )}^{3}}{12 m^{4} \log {\relax (a )}^{4}} & \text {for}\: 12 m^{4} \log {\relax (a )}^{4} \neq 0 \\15 x & \text {otherwise} \end {cases} \]

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(12*m**4*log(a)**4, 0)), (15*x, True))

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