3.696 \(\int (a+b e^x)^3 \, dx\)

Optimal. Leaf size=40 \[ a^3 x+3 a^2 b e^x+\frac {3}{2} a b^2 e^{2 x}+\frac {1}{3} b^3 e^{3 x} \]

[Out]

3*a^2*b*exp(x)+3/2*a*b^2*exp(2*x)+1/3*b^3*exp(3*x)+a^3*x

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Rubi [A]  time = 0.02, antiderivative size = 40, 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} \[ 3 a^2 b e^x+a^3 x+\frac {3}{2} a b^2 e^{2 x}+\frac {1}{3} b^3 e^{3 x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*E^x)^3,x]

[Out]

3*a^2*b*E^x + (3*a*b^2*E^(2*x))/2 + (b^3*E^(3*x))/3 + a^3*x

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 (a+b e^x\right )^3 \, dx &=\operatorname {Subst}\left (\int \frac {(a+b x)^3}{x} \, dx,x,e^x\right )\\ &=\operatorname {Subst}\left (\int \left (3 a^2 b+\frac {a^3}{x}+3 a b^2 x+b^3 x^2\right ) \, dx,x,e^x\right )\\ &=3 a^2 b e^x+\frac {3}{2} a b^2 e^{2 x}+\frac {1}{3} b^3 e^{3 x}+a^3 x\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 40, normalized size = 1.00 \[ a^3 x+3 a^2 b e^x+\frac {3}{2} a b^2 e^{2 x}+\frac {1}{3} b^3 e^{3 x} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*E^x)^3,x]

[Out]

3*a^2*b*E^x + (3*a*b^2*E^(2*x))/2 + (b^3*E^(3*x))/3 + a^3*x

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fricas [A]  time = 0.40, size = 33, normalized size = 0.82 \[ a^{3} x + \frac {1}{3} \, b^{3} e^{\left (3 \, x\right )} + \frac {3}{2} \, a b^{2} e^{\left (2 \, x\right )} + 3 \, a^{2} b e^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*exp(x))^3,x, algorithm="fricas")

[Out]

a^3*x + 1/3*b^3*e^(3*x) + 3/2*a*b^2*e^(2*x) + 3*a^2*b*e^x

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giac [A]  time = 0.21, size = 33, normalized size = 0.82 \[ a^{3} x + \frac {1}{3} \, b^{3} e^{\left (3 \, x\right )} + \frac {3}{2} \, a b^{2} e^{\left (2 \, x\right )} + 3 \, a^{2} b e^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*exp(x))^3,x, algorithm="giac")

[Out]

a^3*x + 1/3*b^3*e^(3*x) + 3/2*a*b^2*e^(2*x) + 3*a^2*b*e^x

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maple [A]  time = 0.02, size = 36, normalized size = 0.90 \[ a^{3} \ln \left ({\mathrm e}^{x}\right )+3 a^{2} b \,{\mathrm e}^{x}+\frac {3 a \,b^{2} {\mathrm e}^{2 x}}{2}+\frac {b^{3} {\mathrm e}^{3 x}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*exp(x)+a)^3,x)

[Out]

1/3*exp(x)^3*b^3+3/2*exp(x)^2*a*b^2+3*a^2*b*exp(x)+a^3*ln(exp(x))

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maxima [A]  time = 0.99, size = 33, normalized size = 0.82 \[ a^{3} x + \frac {1}{3} \, b^{3} e^{\left (3 \, x\right )} + \frac {3}{2} \, a b^{2} e^{\left (2 \, x\right )} + 3 \, a^{2} b e^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*exp(x))^3,x, algorithm="maxima")

[Out]

a^3*x + 1/3*b^3*e^(3*x) + 3/2*a*b^2*e^(2*x) + 3*a^2*b*e^x

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mupad [B]  time = 0.07, size = 33, normalized size = 0.82 \[ x\,a^3+3\,{\mathrm {e}}^x\,a^2\,b+\frac {3\,{\mathrm {e}}^{2\,x}\,a\,b^2}{2}+\frac {{\mathrm {e}}^{3\,x}\,b^3}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*exp(x))^3,x)

[Out]

(b^3*exp(3*x))/3 + a^3*x + 3*a^2*b*exp(x) + (3*a*b^2*exp(2*x))/2

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sympy [A]  time = 0.14, size = 37, normalized size = 0.92 \[ a^{3} x + 3 a^{2} b e^{x} + \frac {3 a b^{2} e^{2 x}}{2} + \frac {b^{3} e^{3 x}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*exp(x))**3,x)

[Out]

a**3*x + 3*a**2*b*exp(x) + 3*a*b**2*exp(2*x)/2 + b**3*exp(3*x)/3

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