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

Optimal. Leaf size=32 \[ \frac {e^{-c} \left (a+b e^{c+d x}\right )^{n+1}}{b d (n+1)} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2247, 2246, 32} \[ \frac {e^{-c} \left (a+b e^{c+d x}\right )^{n+1}}{b d (n+1)} \]

Antiderivative was successfully verified.

[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 2246

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 2247

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*} \int e^{d x} \left (a+b e^{c+d x}\right )^n \, dx &=e^{-c} \int e^{c+d x} \left (a+b e^{c+d x}\right )^n \, dx\\ &=\frac {e^{-c} \operatorname {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*}

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Mathematica [A]  time = 0.04, size = 31, normalized size = 0.97 \[ \frac {e^{-c} \left (a+b e^{c+d x}\right )^{n+1}}{b d n+b d} \]

Antiderivative was successfully verified.

[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))

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fricas [A]  time = 0.43, size = 36, normalized size = 1.12 \[ \frac {{\left (b e^{\left (d x\right )} + a e^{\left (-c\right )}\right )} {\left (b e^{\left (d x + c\right )} + a\right )}^{n}}{b d n + b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.39, size = 38, normalized size = 1.19 \[ \frac {{\left (b e^{\left (d x + c\right )} + a\right )} {\left (b e^{\left (d x + c\right )} + a\right )}^{n} e^{\left (-c\right )}}{b d {\left (n + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 31, normalized size = 0.97 \[ \frac {\left (b \,{\mathrm e}^{c} {\mathrm e}^{d x}+a \right )^{n +1} {\mathrm e}^{-c}}{\left (n +1\right ) b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.44, size = 30, normalized size = 0.94 \[ \frac {{\left (b e^{\left (d x + c\right )} + a\right )}^{n + 1} e^{\left (-c\right )}}{b d {\left (n + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [B]  time = 3.55, size = 44, normalized size = 1.38 \[ {\left (a+b\,{\mathrm {e}}^{c+d\,x}\right )}^n\,\left (\frac {{\mathrm {e}}^{d\,x}}{d\,\left (n+1\right )}+\frac {a\,{\mathrm {e}}^{-c}}{b\,d\,\left (n+1\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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sympy [A]  time = 14.33, size = 114, normalized size = 3.56 \[ \begin {cases} \frac {x}{a} & \text {for}\: b = 0 \wedge d = 0 \wedge n = -1 \\\frac {a^{n} e^{d x}}{d} & \text {for}\: b = 0 \\x \left (a + b e^{c}\right )^{n} & \text {for}\: d = 0 \\\frac {e^{- c} \log {\left (\frac {a}{b} + e^{c} e^{d x} \right )}}{b d} & \text {for}\: n = -1 \\\frac {a \left (a + b e^{c} e^{d x}\right )^{n}}{b d n e^{c} + b d e^{c}} + \frac {b \left (a + b e^{c} e^{d x}\right )^{n} e^{c} e^{d x}}{b d n e^{c} + b d e^{c}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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