3.29 \(\int e^{-n x} (a+b e^{n x})^3 \, dx\)

Optimal. Leaf size=52 \[ 3 a^2 b x-\frac{a^3 e^{-n x}}{n}+\frac{3 a b^2 e^{n x}}{n}+\frac{b^3 e^{2 n x}}{2 n} \]

[Out]

-(a^3/(E^(n*x)*n)) + (3*a*b^2*E^(n*x))/n + (b^3*E^(2*n*x))/(2*n) + 3*a^2*b*x

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Rubi [A]  time = 0.0416476, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2248, 43} \[ 3 a^2 b x-\frac{a^3 e^{-n x}}{n}+\frac{3 a b^2 e^{n x}}{n}+\frac{b^3 e^{2 n x}}{2 n} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*E^(n*x))^3/E^(n*x),x]

[Out]

-(a^3/(E^(n*x)*n)) + (3*a*b^2*E^(n*x))/n + (b^3*E^(2*n*x))/(2*n) + 3*a^2*b*x

Rule 2248

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_.)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Wit
h[{m = FullSimplify[(g*h*Log[G])/(d*e*Log[F])]}, Dist[(Denominator[m]*G^(f*h - (c*g*h)/d))/(d*e*Log[F]), Subst
[Int[x^(Numerator[m] - 1)*(a + b*x^Denominator[m])^p, x], x, F^((e*(c + d*x))/Denominator[m])], x] /; LeQ[m, -
1] || GeQ[m, 1]] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int e^{-n x} \left (a+b e^{n x}\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^3}{x^2} \, dx,x,e^{n x}\right )}{n}\\ &=\frac{\operatorname{Subst}\left (\int \left (3 a b^2+\frac{a^3}{x^2}+\frac{3 a^2 b}{x}+b^3 x\right ) \, dx,x,e^{n x}\right )}{n}\\ &=-\frac{a^3 e^{-n x}}{n}+\frac{3 a b^2 e^{n x}}{n}+\frac{b^3 e^{2 n x}}{2 n}+3 a^2 b x\\ \end{align*}

Mathematica [A]  time = 0.0249187, size = 48, normalized size = 0.92 \[ \frac{3 a^2 b n x+a^3 \left (-e^{-n x}\right )+3 a b^2 e^{n x}+\frac{1}{2} b^3 e^{2 n x}}{n} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*E^(n*x))^3/E^(n*x),x]

[Out]

(-(a^3/E^(n*x)) + 3*a*b^2*E^(n*x) + (b^3*E^(2*n*x))/2 + 3*a^2*b*n*x)/n

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Maple [A]  time = 0.006, size = 57, normalized size = 1.1 \begin{align*}{\frac{{b}^{3} \left ({{\rm e}^{nx}} \right ) ^{2}}{2\,n}}+3\,{\frac{a{b}^{2}{{\rm e}^{nx}}}{n}}+3\,{\frac{{a}^{2}b\ln \left ({{\rm e}^{nx}} \right ) }{n}}-{\frac{{a}^{3}}{n{{\rm e}^{nx}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*exp(n*x))^3/exp(n*x),x)

[Out]

1/2/n*b^3*exp(n*x)^2+3*a*b^2*exp(n*x)/n+3/n*a^2*b*ln(exp(n*x))-a^3/exp(n*x)/n

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Maxima [A]  time = 1.15863, size = 63, normalized size = 1.21 \begin{align*} 3 \, a^{2} b x + \frac{b^{3} e^{\left (2 \, n x\right )}}{2 \, n} + \frac{3 \, a b^{2} e^{\left (n x\right )}}{n} - \frac{a^{3} e^{\left (-n x\right )}}{n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*exp(n*x))^3/exp(n*x),x, algorithm="maxima")

[Out]

3*a^2*b*x + 1/2*b^3*e^(2*n*x)/n + 3*a*b^2*e^(n*x)/n - a^3*e^(-n*x)/n

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Fricas [A]  time = 1.49152, size = 111, normalized size = 2.13 \begin{align*} \frac{{\left (6 \, a^{2} b n x e^{\left (n x\right )} + b^{3} e^{\left (3 \, n x\right )} + 6 \, a b^{2} e^{\left (2 \, n x\right )} - 2 \, a^{3}\right )} e^{\left (-n x\right )}}{2 \, n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*exp(n*x))^3/exp(n*x),x, algorithm="fricas")

[Out]

1/2*(6*a^2*b*n*x*e^(n*x) + b^3*e^(3*n*x) + 6*a*b^2*e^(2*n*x) - 2*a^3)*e^(-n*x)/n

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Sympy [A]  time = 0.190626, size = 73, normalized size = 1.4 \begin{align*} 3 a^{2} b x + \begin{cases} \frac{- 2 a^{3} n^{2} e^{- n x} + 6 a b^{2} n^{2} e^{n x} + b^{3} n^{2} e^{2 n x}}{2 n^{3}} & \text{for}\: 2 n^{3} \neq 0 \\x \left (a^{3} + 3 a b^{2} + b^{3}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*exp(n*x))**3/exp(n*x),x)

[Out]

3*a**2*b*x + Piecewise(((-2*a**3*n**2*exp(-n*x) + 6*a*b**2*n**2*exp(n*x) + b**3*n**2*exp(2*n*x))/(2*n**3), Ne(
2*n**3, 0)), (x*(a**3 + 3*a*b**2 + b**3), True))

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Giac [A]  time = 1.30001, size = 63, normalized size = 1.21 \begin{align*} 3 \, a^{2} b x + \frac{b^{3} e^{\left (2 \, n x\right )}}{2 \, n} + \frac{3 \, a b^{2} e^{\left (n x\right )}}{n} - \frac{a^{3} e^{\left (-n x\right )}}{n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*exp(n*x))^3/exp(n*x),x, algorithm="giac")

[Out]

3*a^2*b*x + 1/2*b^3*e^(2*n*x)/n + 3*a*b^2*e^(n*x)/n - a^3*e^(-n*x)/n