3.516 \(\int (1+a^{m x})^n \, dx\)

Optimal. Leaf size=40 \[ -\frac{\left (a^{m x}+1\right )^{n+1} \, _2F_1\left (1,n+1;n+2;a^{m x}+1\right )}{m (n+1) \log (a)} \]

[Out]

-(((1 + a^(m*x))^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, 1 + a^(m*x)])/(m*(1 + n)*Log[a]))

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Rubi [A]  time = 0.0210817, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2282, 65} \[ -\frac{\left (a^{m x}+1\right )^{n+1} \text{Hypergeometric2F1}\left (1,n+1,n+2,a^{m x}+1\right )}{m (n+1) \log (a)} \]

Antiderivative was successfully verified.

[In]

Int[(1 + a^(m*x))^n,x]

[Out]

-(((1 + a^(m*x))^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, 1 + a^(m*x)])/(m*(1 + n)*Log[a]))

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rubi steps

\begin{align*} \int \left (1+a^{m x}\right )^n \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(1+x)^n}{x} \, dx,x,a^{m x}\right )}{m \log (a)}\\ &=-\frac{\left (1+a^{m x}\right )^{1+n} \, _2F_1\left (1,1+n;2+n;1+a^{m x}\right )}{m (1+n) \log (a)}\\ \end{align*}

Mathematica [A]  time = 0.0142504, size = 40, normalized size = 1. \[ -\frac{\left (a^{m x}+1\right )^{n+1} \, _2F_1\left (1,n+1;n+2;a^{m x}+1\right )}{m (n+1) \log (a)} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + a^(m*x))^n,x]

[Out]

-(((1 + a^(m*x))^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, 1 + a^(m*x)])/(m*(1 + n)*Log[a]))

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Maple [F]  time = 0.029, size = 0, normalized size = 0. \begin{align*} \int \left ( 1+{a}^{mx} \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+a^(m*x))^n,x)

[Out]

int((1+a^(m*x))^n,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a^{m x} + 1\right )}^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+a^(m*x))^n,x, algorithm="maxima")

[Out]

integrate((a^(m*x) + 1)^n, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a^{m x} + 1\right )}^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+a^(m*x))^n,x, algorithm="fricas")

[Out]

integral((a^(m*x) + 1)^n, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a^{m x} + 1\right )^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+a**(m*x))**n,x)

[Out]

Integral((a**(m*x) + 1)**n, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a^{m x} + 1\right )}^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+a^(m*x))^n,x, algorithm="giac")

[Out]

integrate((a^(m*x) + 1)^n, x)