3.28 \(\int e^{-n x} (a+b e^{n x})^2 \, dx\)

Optimal. Leaf size=32 \[ -\frac{a^2 e^{-n x}}{n}+2 a b x+\frac{b^2 e^{n x}}{n} \]

[Out]

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

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Rubi [A]  time = 0.0342845, antiderivative size = 32, 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} \[ -\frac{a^2 e^{-n x}}{n}+2 a b x+\frac{b^2 e^{n x}}{n} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*E^(n*x))^2/E^(n*x),x]

[Out]

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

Mathematica [A]  time = 0.0178364, size = 31, normalized size = 0.97 \[ \frac{a^2 \left (-e^{-n x}\right )+2 a b n x+b^2 e^{n x}}{n} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*E^(n*x))^2/E^(n*x),x]

[Out]

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

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Maple [A]  time = 0.006, size = 39, normalized size = 1.2 \begin{align*}{\frac{{b}^{2}{{\rm e}^{nx}}}{n}}+2\,{\frac{ab\ln \left ({{\rm e}^{nx}} \right ) }{n}}-{\frac{{a}^{2}}{n{{\rm e}^{nx}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*exp(n*x))^2/exp(n*x),x)

[Out]

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

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Maxima [A]  time = 1.15055, size = 41, normalized size = 1.28 \begin{align*} 2 \, a b x + \frac{b^{2} e^{\left (n x\right )}}{n} - \frac{a^{2} 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))^2/exp(n*x),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.48498, size = 73, normalized size = 2.28 \begin{align*} \frac{{\left (2 \, a b n x e^{\left (n x\right )} + b^{2} e^{\left (2 \, n x\right )} - a^{2}\right )} 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))^2/exp(n*x),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.183723, size = 39, normalized size = 1.22 \begin{align*} 2 a b x + \begin{cases} \frac{- a^{2} n e^{- n x} + b^{2} n e^{n x}}{n^{2}} & \text{for}\: n^{2} \neq 0 \\x \left (a^{2} + b^{2}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*exp(n*x))**2/exp(n*x),x)

[Out]

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

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Giac [A]  time = 1.27072, size = 41, normalized size = 1.28 \begin{align*} 2 \, a b x + \frac{b^{2} e^{\left (n x\right )}}{n} - \frac{a^{2} 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))^2/exp(n*x),x, algorithm="giac")

[Out]

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