\(\int e^{d x} (a+b e^{c+d x})^n \, dx\) [6]

Optimal result

Integrand size = 19, antiderivative size = 32 \[ \int e^{d x} \left (a+b e^{c+d x}\right )^n \, dx=\frac {e^{-c} \left (a+b e^{c+d x}\right )^{1+n}}{b d (1+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2279, 2278, 32} \[ \int e^{d x} \left (a+b e^{c+d x}\right )^n \, dx=\frac {e^{-c} \left (a+b e^{c+d x}\right )^{n+1}}{b d (n+1)} \]

[In]

Int[E^(d*x)*(a + b*E^(c + d*x))^n,x]

[Out]

(a + b*E^(c + d*x))^(1 + n)/(b*d*E^c*(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]

Rule 2279

Int[((a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.))^(p_.)*((G_)^((h_.)*((f_.) + (g_.)*(x_))))^(m_.),
x_Symbol] :> Dist[(G^(h*(f + g*x)))^m/(F^(e*(c + d*x)))^n, Int[(F^(e*(c + d*x)))^n*(a + b*(F^(e*(c + d*x)))^n)
^p, x], x] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, m, n, p}, x] && EqQ[d*e*n*Log[F], g*h*m*Log[G]]

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int e^{d x} \left (a+b e^{c+d x}\right )^n \, dx=\frac {e^{-c} \left (a+b e^{c+d x}\right )^{1+n}}{b d+b d n} \]

[In]

Integrate[E^(d*x)*(a + b*E^(c + d*x))^n,x]

[Out]

(a + b*E^(c + d*x))^(1 + n)/(E^c*(b*d + b*d*n))

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97

method	result	size
default	\(\frac {\left (a +b \,{\mathrm e}^{d x} {\mathrm e}^{c}\right )^{1+n} {\mathrm e}^{-c}}{d b \left (1+n \right )}\)	\(31\)
risch	\(\frac {\left (a +b \,{\mathrm e}^{d x +c}\right ) {\mathrm e}^{-c} \left (a +b \,{\mathrm e}^{d x +c}\right )^{n}}{b d \left (1+n \right )}\)	\(39\)
norman	\(\frac {{\mathrm e}^{d x} {\mathrm e}^{n \ln \left (a +b \,{\mathrm e}^{d x} {\mathrm e}^{c}\right )}}{d \left (1+n \right )}+\frac {{\mathrm e}^{-c} a \,{\mathrm e}^{n \ln \left (a +b \,{\mathrm e}^{d x} {\mathrm e}^{c}\right )}}{b d \left (1+n \right )}\)	\(60\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int e^{d x} \left (a+b e^{c+d x}\right )^n \, dx=\frac {{\left (b e^{\left (d x\right )} + a e^{\left (-c\right )}\right )} {\left (b e^{\left (d x + c\right )} + a\right )}^{n}}{b d n + b d} \]

[In]

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

[Out]

(b*e^(d*x) + a*e^(-c))*(b*e^(d*x + c) + a)^n/(b*d*n + b*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 114 vs. \(2 (22) = 44\).

Time = 2.71 (sec) , antiderivative size = 114, normalized size of antiderivative = 3.56 \[ \int e^{d x} \left (a+b e^{c+d x}\right )^n \, dx=\begin {cases} \frac {x}{a} & \text {for}\: b = 0 \wedge d = 0 \wedge n = -1 \\\frac {a^{n} e^{d x}}{d} & \text {for}\: b = 0 \\x \left (a + b e^{c}\right )^{n} & \text {for}\: d = 0 \\\frac {e^{- c} \log {\left (\frac {a}{b} + e^{c} e^{d x} \right )}}{b d} & \text {for}\: n = -1 \\\frac {a \left (a + b e^{c} e^{d x}\right )^{n}}{b d n e^{c} + b d e^{c}} + \frac {b \left (a + b e^{c} e^{d x}\right )^{n} e^{c} e^{d x}}{b d n e^{c} + b d e^{c}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \[ \int e^{d x} \left (a+b e^{c+d x}\right )^n \, dx=\frac {{\left (b e^{\left (d x + c\right )} + a\right )}^{n + 1} e^{\left (-c\right )}}{b d {\left (n + 1\right )}} \]

[In]

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

[Out]

(b*e^(d*x + c) + a)^(n + 1)*e^(-c)/(b*d*(n + 1))

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \[ \int e^{d x} \left (a+b e^{c+d x}\right )^n \, dx=\frac {{\left (b e^{\left (d x + c\right )} + a\right )}^{n + 1} e^{\left (-c\right )}}{b d {\left (n + 1\right )}} \]

[In]

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

[Out]

(b*e^(d*x + c) + a)^(n + 1)*e^(-c)/(b*d*(n + 1))

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.38 \[ \int e^{d x} \left (a+b e^{c+d x}\right )^n \, dx={\left (a+b\,{\mathrm {e}}^{c+d\,x}\right )}^n\,\left (\frac {{\mathrm {e}}^{d\,x}}{d\,\left (n+1\right )}+\frac {a\,{\mathrm {e}}^{-c}}{b\,d\,\left (n+1\right )}\right ) \]

[In]

int(exp(d*x)*(a + b*exp(c + d*x))^n,x)

[Out]

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