\(\int e^{c+d x} (a+b e^{c+d x})^n \, dx\) [7]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [B] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 21, antiderivative size = 27 \[ \int e^{c+d x} \left (a+b e^{c+d x}\right )^n \, dx=\frac {\left (a+b e^{c+d x}\right )^{1+n}}{b d (1+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2278, 32} \[ \int e^{c+d x} \left (a+b e^{c+d x}\right )^n \, dx=\frac {\left (a+b e^{c+d x}\right )^{n+1}}{b d (n+1)} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int e^{c+d x} \left (a+b e^{c+d x}\right )^n \, dx=\frac {\left (a+b e^{c+d x}\right )^{1+n}}{b d+b d n} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00

method result size
derivativedivides \(\frac {\left (a +b \,{\mathrm e}^{d x +c}\right )^{1+n}}{b d \left (1+n \right )}\) \(27\)
default \(\frac {\left (a +b \,{\mathrm e}^{d x +c}\right )^{1+n}}{b d \left (1+n \right )}\) \(27\)
risch \(\frac {\left (a +b \,{\mathrm e}^{d x +c}\right ) \left (a +b \,{\mathrm e}^{d x +c}\right )^{n}}{b d \left (1+n \right )}\) \(35\)
parallelrisch \(\frac {{\mathrm e}^{d x +c} \left (a +b \,{\mathrm e}^{d x +c}\right )^{n} b +\left (a +b \,{\mathrm e}^{d x +c}\right )^{n} a}{b d \left (1+n \right )}\) \(48\)
norman \(\frac {{\mathrm e}^{d x +c} {\mathrm e}^{n \ln \left (a +b \,{\mathrm e}^{d x +c}\right )}}{d \left (1+n \right )}+\frac {a \,{\mathrm e}^{n \ln \left (a +b \,{\mathrm e}^{d x +c}\right )}}{b d \left (1+n \right )}\) \(58\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int e^{c+d x} \left (a+b e^{c+d x}\right )^n \, dx=\frac {{\left (b e^{\left (d x + c\right )} + a\right )} {\left (b e^{\left (d x + c\right )} + a\right )}^{n}}{b d n + b d} \]

[In]

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

[Out]

(b*e^(d*x + c) + a)*(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. 107 vs. \(2 (19) = 38\).

Time = 6.05 (sec) , antiderivative size = 107, normalized size of antiderivative = 3.96 \[ \int e^{c+d x} \left (a+b e^{c+d x}\right )^n \, dx=\begin {cases} \frac {x e^{c}}{a} & \text {for}\: b = 0 \wedge d = 0 \wedge n = -1 \\\frac {a^{n} e^{c} e^{d x}}{d} & \text {for}\: b = 0 \\x \left (a + b e^{c}\right )^{n} e^{c} & \text {for}\: d = 0 \\\frac {\log {\left (\frac {a e^{- c}}{b} + e^{d x} \right )}}{b d} & \text {for}\: n = -1 \\\frac {a \left (a + b e^{c} e^{d x}\right )^{n}}{b d n + b d} + \frac {b \left (a + b e^{c} e^{d x}\right )^{n} e^{c} e^{d x}}{b d n + b d} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int e^{c+d x} \left (a+b e^{c+d x}\right )^n \, dx=\frac {{\left (b e^{\left (d x + c\right )} + a\right )}^{n + 1}}{b d {\left (n + 1\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int e^{c+d x} \left (a+b e^{c+d x}\right )^n \, dx=\frac {{\left (b e^{\left (d x + c\right )} + a\right )}^{n + 1}}{b d {\left (n + 1\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int e^{c+d x} \left (a+b e^{c+d x}\right )^n \, dx=\frac {{\left (a+b\,{\mathrm {e}}^{c+d\,x}\right )}^{n+1}}{b\,d\,\left (n+1\right )} \]

[In]

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

[Out]

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