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

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

[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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Rubi [A]  time = 0.0195108, antiderivative size = 40, normalized size of antiderivative = 1., 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 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]]]

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])

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*}

Mathematica [A]  time = 0.0124967, size = 40, normalized size = 1. \[ 3 a^2 b e^x+a^3 x+\frac{3}{2} a b^2 e^{2 x}+\frac{1}{3} b^3 e^{3 x} \]

Antiderivative was successfully verified.

[In]

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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Maple [A]  time = 0.039, size = 36, normalized size = 0.9 \begin{align*}{\frac{{b}^{3} \left ({{\rm e}^{x}} \right ) ^{3}}{3}}+{\frac{3\,a{b}^{2} \left ({{\rm e}^{x}} \right ) ^{2}}{2}}+3\,{a}^{2}b{{\rm e}^{x}}+{a}^{3}\ln \left ({{\rm e}^{x}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.955045, size = 45, normalized size = 1.12 \begin{align*} 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} \end{align*}

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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Fricas [A]  time = 0.768507, size = 80, normalized size = 2. \begin{align*} 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} \end{align*}

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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Sympy [A]  time = 0.141257, size = 37, normalized size = 0.92 \begin{align*} 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} \end{align*}

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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Giac [A]  time = 1.2367, size = 45, normalized size = 1.12 \begin{align*} 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} \end{align*}

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