\(\int e^x (a+b e^x)^n \, dx\) [5]

Optimal result

Integrand size = 13, antiderivative size = 20 \[ \int e^x \left (a+b e^x\right )^n \, dx=\frac {\left (a+b e^x\right )^{1+n}}{b (1+n)} \]

[Out]

(a+b*exp(x))^(1+n)/b/(1+n)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2278, 32} \[ \int e^x \left (a+b e^x\right )^n \, dx=\frac {\left (a+b e^x\right )^{n+1}}{b (n+1)} \]

[In]

Int[E^x*(a + b*E^x)^n,x]

[Out]

(a + b*E^x)^(1 + n)/(b*(1 + n))

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2278

Int[((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)*((a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.))^(p_.),
x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int[(a + b*x)^p, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b,
c, d, e, n, p}, x]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int (a+b x)^n \, dx,x,e^x\right ) \\ & = \frac {\left (a+b e^x\right )^{1+n}}{b (1+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int e^x \left (a+b e^x\right )^n \, dx=\frac {\left (a+b e^x\right )^{1+n}}{b+b n} \]

[In]

Integrate[E^x*(a + b*E^x)^n,x]

[Out]

(a + b*E^x)^(1 + n)/(b + b*n)

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(\frac {\left (a +b \,{\mathrm e}^{x}\right )^{1+n}}{b \left (1+n \right )}\)	\(20\)
default	\(\frac {\left (a +b \,{\mathrm e}^{x}\right )^{1+n}}{b \left (1+n \right )}\)	\(20\)
risch	\(\frac {\left (a +b \,{\mathrm e}^{x}\right ) \left (a +b \,{\mathrm e}^{x}\right )^{n}}{b \left (1+n \right )}\)	\(24\)
parallelrisch	\(\frac {{\mathrm e}^{x} \left (a +b \,{\mathrm e}^{x}\right )^{n} b +\left (a +b \,{\mathrm e}^{x}\right )^{n} a}{b \left (1+n \right )}\)	\(33\)
norman	\(\frac {{\mathrm e}^{x} {\mathrm e}^{n \ln \left (a +b \,{\mathrm e}^{x}\right )}}{1+n}+\frac {a \,{\mathrm e}^{n \ln \left (a +b \,{\mathrm e}^{x}\right )}}{b \left (1+n \right )}\)	\(40\)

[In]

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

[Out]

(a+b*exp(x))^(1+n)/b/(1+n)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int e^x \left (a+b e^x\right )^n \, dx=\frac {{\left (b e^{x} + a\right )} {\left (b e^{x} + a\right )}^{n}}{b n + b} \]

[In]

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

[Out]

(b*e^x + a)*(b*e^x + a)^n/(b*n + b)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (14) = 28\).

Time = 0.35 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.80 \[ \int e^x \left (a+b e^x\right )^n \, dx=\begin {cases} \frac {e^{x}}{a} & \text {for}\: b = 0 \wedge n = -1 \\a^{n} e^{x} & \text {for}\: b = 0 \\\frac {\log {\left (\frac {a}{b} + e^{x} \right )}}{b} & \text {for}\: n = -1 \\\frac {a \left (a + b e^{x}\right )^{n}}{b n + b} + \frac {b \left (a + b e^{x}\right )^{n} e^{x}}{b n + b} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((exp(x)/a, Eq(b, 0) & Eq(n, -1)), (a**n*exp(x), Eq(b, 0)), (log(a/b + exp(x))/b, Eq(n, -1)), (a*(a +
b*exp(x))**n/(b*n + b) + b*(a + b*exp(x))**n*exp(x)/(b*n + b), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int e^x \left (a+b e^x\right )^n \, dx=\frac {{\left (b e^{x} + a\right )}^{n + 1}}{b {\left (n + 1\right )}} \]

[In]

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

[Out]

(b*e^x + a)^(n + 1)/(b*(n + 1))

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int e^x \left (a+b e^x\right )^n \, dx=\frac {{\left (b e^{x} + a\right )}^{n + 1}}{b {\left (n + 1\right )}} \]

[In]

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

[Out]

(b*e^x + a)^(n + 1)/(b*(n + 1))

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int e^x \left (a+b e^x\right )^n \, dx=\frac {{\left (a+b\,{\mathrm {e}}^x\right )}^{n+1}}{b\,\left (n+1\right )} \]

[In]

int(exp(x)*(a + b*exp(x))^n,x)

[Out]

(a + b*exp(x))^(n + 1)/(b*(n + 1))