3.2.21 \(\int \frac {1}{3} (120+e^{5+\frac {x}{3}} (10-10 x)-60 x+e^x (-120-60 x+30 x^2)) \, dx\)

Optimal. Leaf size=24 \[ 10 (4-x) \left (e^{5+\frac {x}{3}}+x-e^x x\right ) \]

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Rubi [B]  time = 0.06, antiderivative size = 50, normalized size of antiderivative = 2.08, number of steps used = 12, number of rules used = 4, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {12, 2176, 2194, 2196} \begin {gather*} 10 e^x x^2-10 x^2-40 e^x x+40 x+30 e^{\frac {x}{3}+5}+10 e^{\frac {x}{3}+5} (1-x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(120 + E^(5 + x/3)*(10 - 10*x) - 60*x + E^x*(-120 - 60*x + 30*x^2))/3,x]

[Out]

30*E^(5 + x/3) + 10*E^(5 + x/3)*(1 - x) + 40*x - 40*E^x*x - 10*x^2 + 10*E^x*x^2

Rule 12

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

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]

Rule 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int \left (120+e^{5+\frac {x}{3}} (10-10 x)-60 x+e^x \left (-120-60 x+30 x^2\right )\right ) \, dx\\ &=40 x-10 x^2+\frac {1}{3} \int e^{5+\frac {x}{3}} (10-10 x) \, dx+\frac {1}{3} \int e^x \left (-120-60 x+30 x^2\right ) \, dx\\ &=10 e^{5+\frac {x}{3}} (1-x)+40 x-10 x^2+\frac {1}{3} \int \left (-120 e^x-60 e^x x+30 e^x x^2\right ) \, dx+10 \int e^{5+\frac {x}{3}} \, dx\\ &=30 e^{5+\frac {x}{3}}+10 e^{5+\frac {x}{3}} (1-x)+40 x-10 x^2+10 \int e^x x^2 \, dx-20 \int e^x x \, dx-40 \int e^x \, dx\\ &=30 e^{5+\frac {x}{3}}-40 e^x+10 e^{5+\frac {x}{3}} (1-x)+40 x-20 e^x x-10 x^2+10 e^x x^2+20 \int e^x \, dx-20 \int e^x x \, dx\\ &=30 e^{5+\frac {x}{3}}-20 e^x+10 e^{5+\frac {x}{3}} (1-x)+40 x-40 e^x x-10 x^2+10 e^x x^2+20 \int e^x \, dx\\ &=30 e^{5+\frac {x}{3}}+10 e^{5+\frac {x}{3}} (1-x)+40 x-40 e^x x-10 x^2+10 e^x x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 25, normalized size = 1.04 \begin {gather*} 10 (-4+x) \left (-e^{5+\frac {x}{3}}-x+e^x x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(120 + E^(5 + x/3)*(10 - 10*x) - 60*x + E^x*(-120 - 60*x + 30*x^2))/3,x]

[Out]

10*(-4 + x)*(-E^(5 + x/3) - x + E^x*x)

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fricas [A]  time = 0.61, size = 38, normalized size = 1.58 \begin {gather*} -10 \, {\left ({\left (x^{2} - 4 \, x\right )} e^{15} - {\left (x^{2} - 4 \, x\right )} e^{\left (x + 15\right )} + {\left (x - 4\right )} e^{\left (\frac {1}{3} \, x + 20\right )}\right )} e^{\left (-15\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(30*x^2-60*x-120)*exp(x)+1/3*(-10*x+10)*exp(5)*exp(1/3*x)-20*x+40,x, algorithm="fricas")

[Out]

-10*((x^2 - 4*x)*e^15 - (x^2 - 4*x)*e^(x + 15) + (x - 4)*e^(1/3*x + 20))*e^(-15)

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giac [A]  time = 0.56, size = 31, normalized size = 1.29 \begin {gather*} -10 \, x^{2} + 10 \, {\left (x^{2} - 4 \, x\right )} e^{x} - 10 \, {\left (x - 4\right )} e^{\left (\frac {1}{3} \, x + 5\right )} + 40 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(30*x^2-60*x-120)*exp(x)+1/3*(-10*x+10)*exp(5)*exp(1/3*x)-20*x+40,x, algorithm="giac")

[Out]

-10*x^2 + 10*(x^2 - 4*x)*e^x - 10*(x - 4)*e^(1/3*x + 5) + 40*x

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maple [A]  time = 0.03, size = 36, normalized size = 1.50

method	result	size
risch	\(\frac {\left (30 x^{2}-120 x \right ) {\mathrm e}^{x}}{3}+\frac {\left (-30 x +120\right ) {\mathrm e}^{\frac {x}{3}+5}}{3}-10 x^{2}+40 x\)	\(36\)
default	\(-10 x^{2}+40 x +10 \,{\mathrm e}^{x} x^{2}-40 \,{\mathrm e}^{x} x +\frac {10 \,{\mathrm e}^{5} \left (-3 x \,{\mathrm e}^{\frac {x}{3}}+12 \,{\mathrm e}^{\frac {x}{3}}\right )}{3}\)	\(40\)
norman	\(-10 \,{\mathrm e}^{5} {\mathrm e}^{\frac {x}{3}} x +10 \,{\mathrm e}^{x} x^{2}+40 \,{\mathrm e}^{5} {\mathrm e}^{\frac {x}{3}}-40 \,{\mathrm e}^{x} x -10 x^{2}+40 x\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*(30*x^2-60*x-120)*exp(x)+1/3*(-10*x+10)*exp(5)*exp(1/3*x)-20*x+40,x,method=_RETURNVERBOSE)

[Out]

1/3*(30*x^2-120*x)*exp(x)+1/3*(-30*x+120)*exp(1/3*x+5)-10*x^2+40*x

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maxima [A]  time = 0.51, size = 35, normalized size = 1.46 \begin {gather*} -10 \, x^{2} - 10 \, {\left (x e^{5} - 4 \, e^{5}\right )} e^{\left (\frac {1}{3} \, x\right )} + 10 \, {\left (x^{2} - 4 \, x\right )} e^{x} + 40 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(30*x^2-60*x-120)*exp(x)+1/3*(-10*x+10)*exp(5)*exp(1/3*x)-20*x+40,x, algorithm="maxima")

[Out]

-10*x^2 - 10*(x*e^5 - 4*e^5)*e^(1/3*x) + 10*(x^2 - 4*x)*e^x + 40*x

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mupad [B]  time = 0.26, size = 18, normalized size = 0.75 \begin {gather*} -10\,\left (x-4\right )\,\left (x+{\mathrm {e}}^{\frac {x}{3}+5}-x\,{\mathrm {e}}^x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(40 - (exp(x)*(60*x - 30*x^2 + 120))/3 - (exp(x/3)*exp(5)*(10*x - 10))/3 - 20*x,x)

[Out]

-10*(x - 4)*(x + exp(x/3 + 5) - x*exp(x))

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sympy [A]  time = 0.14, size = 36, normalized size = 1.50 \begin {gather*} - 10 x^{2} + 40 x + \left (10 x^{2} - 40 x\right ) e^{x} + \left (- 10 x e^{5} + 40 e^{5}\right ) e^{\frac {x}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(30*x**2-60*x-120)*exp(x)+1/3*(-10*x+10)*exp(5)*exp(1/3*x)-20*x+40,x)

[Out]

-10*x**2 + 40*x + (10*x**2 - 40*x)*exp(x) + (-10*x*exp(5) + 40*exp(5))*exp(x/3)

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