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

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

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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, 2194} $$\begin{array}{c}2x^3+2x^2+\frac{8x}{5}-32e^x+4e^{2x}\end{array}$$

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 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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{1}{5}\int (8-160e^x+40e^{2x}+20x+30x^2)dx\\ & =\frac{8x}{5}+2x^2+2x^3+8\int e^{2x}dx-32\int e^{x}dx\\ & =-32e^x+4e^{2x}+\frac{8x}{5}+2x^2+2x^3\end{array}\end{array}$$

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

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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fricas [A]  time = 0.78, size = 24, normalized size = 0.92 $$\begin{array}{c}2x^3+2x^2+\frac{8}{5}x+4e^{\left(2x\right)}-32e^x\end{array}$$

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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giac [A]  time = 0.18, size = 24, normalized size = 0.92 $$\begin{array}{c}2x^3+2x^2+\frac{8}{5}x+4e^{\left(2x\right)}-32e^x\end{array}$$

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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maple [A]  time = 0.02, size = 25, normalized size = 0.96

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

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 = 0.69, size = 24, normalized size = 0.92 $$\begin{array}{c}2x^3+2x^2+\frac{8}{5}x+4e^{\left(2x\right)}-32e^x\end{array}$$

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

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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sympy [A]  time = 0.10, size = 26, normalized size = 1.00 $$\begin{array}{c}2x^3+2x^2+\frac{8x}{5}+4e^{2x}-32e^x\end{array}$$

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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