3.60 \(\int (1+e^x)^2 x \, dx\)

Optimal. Leaf size=38 \[ \frac{x^2}{2}+2 e^x x+\frac{1}{2} e^{2 x} x-2 e^x-\frac{e^{2 x}}{4} \]

[Out]

-2*E^x - E^(2*x)/4 + 2*E^x*x + (E^(2*x)*x)/2 + x^2/2

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Rubi [A]  time = 0.0326818, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2183, 2176, 2194} \[ \frac{x^2}{2}+2 e^x x+\frac{1}{2} e^{2 x} x-2 e^x-\frac{e^{2 x}}{4} \]

Antiderivative was successfully verified.

[In]

Int[(1 + E^x)^2*x,x]

[Out]

-2*E^x - E^(2*x)/4 + 2*E^x*x + (E^(2*x)*x)/2 + x^2/2

Rule 2183

Int[((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> In
t[ExpandIntegrand[(c + d*x)^m, (a + b*(F^(g*(e + f*x)))^n)^p, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, m, n},
x] && IGtQ[p, 0]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps

\begin{align*} \int \left (1+e^x\right )^2 x \, dx &=\int \left (x+2 e^x x+e^{2 x} x\right ) \, dx\\ &=\frac{x^2}{2}+2 \int e^x x \, dx+\int e^{2 x} x \, dx\\ &=2 e^x x+\frac{1}{2} e^{2 x} x+\frac{x^2}{2}-\frac{1}{2} \int e^{2 x} \, dx-2 \int e^x \, dx\\ &=-2 e^x-\frac{e^{2 x}}{4}+2 e^x x+\frac{1}{2} e^{2 x} x+\frac{x^2}{2}\\ \end{align*}

Mathematica [A]  time = 0.0256854, size = 29, normalized size = 0.76 \[ \frac{1}{4} \left (2 x^2+8 e^x (x-1)+e^{2 x} (2 x-1)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + E^x)^2*x,x]

[Out]

(8*E^x*(-1 + x) + 2*x^2 + E^(2*x)*(-1 + 2*x))/4

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Maple [A]  time = 0.002, size = 29, normalized size = 0.8 \begin{align*}{\frac{{x}^{2}}{2}}+{\frac{ \left ({{\rm e}^{x}} \right ) ^{2}x}{2}}-{\frac{ \left ({{\rm e}^{x}} \right ) ^{2}}{4}}+2\,{{\rm e}^{x}}x-2\,{{\rm e}^{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+exp(x))^2*x,x)

[Out]

1/2*x^2+1/2*exp(x)^2*x-1/4*exp(x)^2+2*exp(x)*x-2*exp(x)

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Maxima [A]  time = 0.936797, size = 32, normalized size = 0.84 \begin{align*} \frac{1}{2} \, x^{2} + \frac{1}{4} \,{\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} + 2 \,{\left (x - 1\right )} e^{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+exp(x))^2*x,x, algorithm="maxima")

[Out]

1/2*x^2 + 1/4*(2*x - 1)*e^(2*x) + 2*(x - 1)*e^x

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Fricas [A]  time = 1.9211, size = 66, normalized size = 1.74 \begin{align*} \frac{1}{2} \, x^{2} + \frac{1}{4} \,{\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} + 2 \,{\left (x - 1\right )} e^{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+exp(x))^2*x,x, algorithm="fricas")

[Out]

1/2*x^2 + 1/4*(2*x - 1)*e^(2*x) + 2*(x - 1)*e^x

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Sympy [A]  time = 0.088345, size = 26, normalized size = 0.68 \begin{align*} \frac{x^{2}}{2} + \frac{\left (2 x - 1\right ) e^{2 x}}{4} + \frac{\left (8 x - 8\right ) e^{x}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+exp(x))**2*x,x)

[Out]

x**2/2 + (2*x - 1)*exp(2*x)/4 + (8*x - 8)*exp(x)/4

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Giac [A]  time = 1.06624, size = 32, normalized size = 0.84 \begin{align*} \frac{1}{2} \, x^{2} + \frac{1}{4} \,{\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} + 2 \,{\left (x - 1\right )} e^{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+exp(x))^2*x,x, algorithm="giac")

[Out]

1/2*x^2 + 1/4*(2*x - 1)*e^(2*x) + 2*(x - 1)*e^x