3.17.84 \(\int \frac {1}{5} (8-160 e^x+40 e^{2 x}+20 x+30 x^2) \, dx\) [1684]

Optimal. Leaf size=26 \[ 4 \left (-5+\left (-4+e^x\right )^2+\frac {2 x}{5}+\frac {1}{2} x \left (x+x^2\right )\right ) \]

[Out]

8/5*x-20+2*x*(x^2+x)+4*(exp(x)-4)^2

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Rubi [A]
time = 0.01, antiderivative size = 28, normalized size of antiderivative = 1.08, number of steps used = 4, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {12, 2225} \begin {gather*} 2 x^3+2 x^2+\frac {8 x}{5}-32 e^x+4 e^{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(8 - 160*E^x + 40*E^(2*x) + 20*x + 30*x^2)/5,x]

[Out]

-32*E^x + 4*E^(2*x) + (8*x)/5 + 2*x^2 + 2*x^3

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2225

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 {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \left (8-160 e^x+40 e^{2 x}+20 x+30 x^2\right ) \, dx\\ &=\frac {8 x}{5}+2 x^2+2 x^3+8 \int e^{2 x} \, dx-32 \int e^x \, dx\\ &=-32 e^x+4 e^{2 x}+\frac {8 x}{5}+2 x^2+2 x^3\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.01, size = 28, normalized size = 1.08 \begin {gather*} -32 e^x+4 e^{2 x}+\frac {8 x}{5}+2 x^2+2 x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8 - 160*E^x + 40*E^(2*x) + 20*x + 30*x^2)/5,x]

[Out]

-32*E^x + 4*E^(2*x) + (8*x)/5 + 2*x^2 + 2*x^3

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Maple [A]
time = 0.03, size = 25, normalized size = 0.96

method	result	size
default	\(\frac {8 x}{5}+2 x^{2}+2 x^{3}+4 \,{\mathrm e}^{2 x}-32 \,{\mathrm e}^{x}\)	\(25\)
norman	\(\frac {8 x}{5}+2 x^{2}+2 x^{3}+4 \,{\mathrm e}^{2 x}-32 \,{\mathrm e}^{x}\)	\(25\)
risch	\(\frac {8 x}{5}+2 x^{2}+2 x^{3}+4 \,{\mathrm e}^{2 x}-32 \,{\mathrm e}^{x}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(8*exp(x)^2-32*exp(x)+6*x^2+4*x+8/5,x,method=_RETURNVERBOSE)

[Out]

8/5*x+2*x^2+2*x^3+4*exp(x)^2-32*exp(x)

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Maxima [A]
time = 0.30, size = 24, normalized size = 0.92 \begin {gather*} 2 \, x^{3} + 2 \, x^{2} + \frac {8}{5} \, x + 4 \, e^{\left (2 \, x\right )} - 32 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(8*exp(x)^2-32*exp(x)+6*x^2+4*x+8/5,x, algorithm="maxima")

[Out]

2*x^3 + 2*x^2 + 8/5*x + 4*e^(2*x) - 32*e^x

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Fricas [A]
time = 0.37, size = 24, normalized size = 0.92 \begin {gather*} 2 \, x^{3} + 2 \, x^{2} + \frac {8}{5} \, x + 4 \, e^{\left (2 \, x\right )} - 32 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(8*exp(x)^2-32*exp(x)+6*x^2+4*x+8/5,x, algorithm="fricas")

[Out]

2*x^3 + 2*x^2 + 8/5*x + 4*e^(2*x) - 32*e^x

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Sympy [A]
time = 0.04, size = 26, normalized size = 1.00 \begin {gather*} 2 x^{3} + 2 x^{2} + \frac {8 x}{5} + 4 e^{2 x} - 32 e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(8*exp(x)**2-32*exp(x)+6*x**2+4*x+8/5,x)

[Out]

2*x**3 + 2*x**2 + 8*x/5 + 4*exp(2*x) - 32*exp(x)

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Giac [A]
time = 0.39, size = 24, normalized size = 0.92 \begin {gather*} 2 \, x^{3} + 2 \, x^{2} + \frac {8}{5} \, x + 4 \, e^{\left (2 \, x\right )} - 32 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(8*exp(x)^2-32*exp(x)+6*x^2+4*x+8/5,x, algorithm="giac")

[Out]

2*x^3 + 2*x^2 + 8/5*x + 4*e^(2*x) - 32*e^x

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Mupad [B]
time = 1.13, size = 24, normalized size = 0.92 \begin {gather*} \frac {8\,x}{5}+4\,{\mathrm {e}}^{2\,x}-32\,{\mathrm {e}}^x+2\,x^2+2\,x^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(4*x + 8*exp(2*x) - 32*exp(x) + 6*x^2 + 8/5,x)

[Out]

(8*x)/5 + 4*exp(2*x) - 32*exp(x) + 2*x^2 + 2*x^3

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