3.14 \(\int (e^x)^n (a+b (e^x)^n)^p \, dx\)

Optimal. Leaf size=25 \[ \frac {\left (a+b \left (e^x\right )^n\right )^{p+1}}{b n (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2246, 32} \[ \frac {\left (a+b \left (e^x\right )^n\right )^{p+1}}{b n (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[(E^x)^n*(a + b*(E^x)^n)^p,x]

[Out]

(a + b*(E^x)^n)^(1 + p)/(b*n*(1 + p))

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]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 24, normalized size = 0.96 \[ \frac {\left (a+b \left (e^x\right )^n\right )^{p+1}}{b n p+b n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a + b*(E^x)^n)^(1 + p)/(b*n + b*n*p)

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fricas [A]  time = 0.43, size = 29, normalized size = 1.16 \[ \frac {{\left (b e^{\left (n x\right )} + a\right )} {\left (b e^{\left (n x\right )} + a\right )}^{p}}{b n p + b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*e^(n*x) + a)*(b*e^(n*x) + a)^p/(b*n*p + b*n)

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giac [A]  time = 0.46, size = 24, normalized size = 0.96 \[ \frac {{\left (b e^{\left (n x\right )} + a\right )}^{p + 1}}{b n {\left (p + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*e^(n*x) + a)^(p + 1)/(b*n*(p + 1))

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maple [A]  time = 0.00, size = 25, normalized size = 1.00 \[ \frac {\left (b \left ({\mathrm e}^{x}\right )^{n}+a \right )^{p +1}}{\left (p +1\right ) b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.44, size = 24, normalized size = 0.96 \[ \frac {{\left (b e^{\left (n x\right )} + a\right )}^{p + 1}}{b n {\left (p + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*e^(n*x) + a)^(p + 1)/(b*n*(p + 1))

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mupad [B]  time = 3.51, size = 24, normalized size = 0.96 \[ \frac {{\left (a+b\,{\mathrm {e}}^{n\,x}\right )}^{p+1}}{b\,n\,\left (p+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)^n*(a + b*exp(x)^n)^p,x)

[Out]

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

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sympy [A]  time = 2.38, size = 80, normalized size = 3.20 \[ \begin {cases} \frac {x}{a} & \text {for}\: b = 0 \wedge n = 0 \wedge p = -1 \\\frac {a^{p} \left (e^{x}\right )^{n}}{n} & \text {for}\: b = 0 \\x \left (a + b\right )^{p} & \text {for}\: n = 0 \\\frac {\log {\left (\frac {a}{b} + \left (e^{x}\right )^{n} \right )}}{b n} & \text {for}\: p = -1 \\\frac {a \left (a + b \left (e^{x}\right )^{n}\right )^{p}}{b n p + b n} + \frac {b \left (a + b \left (e^{x}\right )^{n}\right )^{p} \left (e^{x}\right )^{n}}{b n p + b n} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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