\(\int (a+b e^x)^3 \, dx\) [696]

Optimal result

Integrand size = 9, antiderivative size = 40 \[ \int \left (a+b e^x\right )^3 \, dx=3 a^2 b e^x+\frac {3}{2} a b^2 e^{2 x}+\frac {1}{3} b^3 e^{3 x}+a^3 x \]

[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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2320, 45} \[ \int \left (a+b e^x\right )^3 \, dx=a^3 x+3 a^2 b e^x+\frac {3}{2} a b^2 e^{2 x}+\frac {1}{3} b^3 e^{3 x} \]

[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 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*} \text {integral}& = \text {Subst}\left (\int \frac {(a+b x)^3}{x} \, dx,x,e^x\right ) \\ & = \text {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] (verified)

Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00 \[ \int \left (a+b e^x\right )^3 \, dx=\frac {1}{6} b e^x \left (18 a^2+9 a b e^x+2 b^2 e^{2 x}\right )+a^3 \log \left (e^x\right ) \]

[In]

Integrate[(a + b*E^x)^3,x]

[Out]

(b*E^x*(18*a^2 + 9*a*b*E^x + 2*b^2*E^(2*x)))/6 + a^3*Log[E^x]

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.85

method	result	size
norman	\(3 a^{2} b \,{\mathrm e}^{x}+\frac {3 a \,b^{2} {\mathrm e}^{2 x}}{2}+\frac {b^{3} {\mathrm e}^{3 x}}{3}+a^{3} x\)	\(34\)
risch	\(3 a^{2} b \,{\mathrm e}^{x}+\frac {3 a \,b^{2} {\mathrm e}^{2 x}}{2}+\frac {b^{3} {\mathrm e}^{3 x}}{3}+a^{3} x\)	\(34\)
parallelrisch	\(3 a^{2} b \,{\mathrm e}^{x}+\frac {3 a \,b^{2} {\mathrm e}^{2 x}}{2}+\frac {b^{3} {\mathrm e}^{3 x}}{3}+a^{3} x\)	\(34\)
parts	\(3 a^{2} b \,{\mathrm e}^{x}+\frac {3 a \,b^{2} {\mathrm e}^{2 x}}{2}+\frac {b^{3} {\mathrm e}^{3 x}}{3}+a^{3} x\)	\(34\)
derivativedivides	\(\frac {b^{3} {\mathrm e}^{3 x}}{3}+\frac {3 a \,b^{2} {\mathrm e}^{2 x}}{2}+3 a^{2} b \,{\mathrm e}^{x}+a^{3} \ln \left ({\mathrm e}^{x}\right )\)	\(36\)
default	\(\frac {b^{3} {\mathrm e}^{3 x}}{3}+\frac {3 a \,b^{2} {\mathrm e}^{2 x}}{2}+3 a^{2} b \,{\mathrm e}^{x}+a^{3} \ln \left ({\mathrm e}^{x}\right )\)	\(36\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.82 \[ \int \left (a+b e^x\right )^3 \, dx=a^{3} x + \frac {1}{3} \, b^{3} e^{\left (3 \, x\right )} + \frac {3}{2} \, a b^{2} e^{\left (2 \, x\right )} + 3 \, a^{2} b e^{x} \]

[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

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.92 \[ \int \left (a+b e^x\right )^3 \, dx=a^{3} x + 3 a^{2} b e^{x} + \frac {3 a b^{2} e^{2 x}}{2} + \frac {b^{3} e^{3 x}}{3} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.82 \[ \int \left (a+b e^x\right )^3 \, dx=a^{3} x + \frac {1}{3} \, b^{3} e^{\left (3 \, x\right )} + \frac {3}{2} \, a b^{2} e^{\left (2 \, x\right )} + 3 \, a^{2} b e^{x} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.82 \[ \int \left (a+b e^x\right )^3 \, dx=a^{3} x + \frac {1}{3} \, b^{3} e^{\left (3 \, x\right )} + \frac {3}{2} \, a b^{2} e^{\left (2 \, x\right )} + 3 \, a^{2} b e^{x} \]

[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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.82 \[ \int \left (a+b e^x\right )^3 \, dx=x\,a^3+3\,{\mathrm {e}}^x\,a^2\,b+\frac {3\,{\mathrm {e}}^{2\,x}\,a\,b^2}{2}+\frac {{\mathrm {e}}^{3\,x}\,b^3}{3} \]

[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