3.6.1 \(\int \frac {1}{4} (1+80 x+24 x^2+e^x (-20-28 x-4 x^2)) \, dx\)

Optimal. Leaf size=21 \[ x (5+x) \left (-e^x+\frac {1}{4 x}+2 x\right ) \]

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Rubi [A]  time = 0.05, antiderivative size = 30, normalized size of antiderivative = 1.43, number of steps used = 10, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 2196, 2194, 2176} \begin {gather*} 2 x^3-e^x x^2+10 x^2-5 e^x x+\frac {x}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + 80*x + 24*x^2 + E^x*(-20 - 28*x - 4*x^2))/4,x]

[Out]

x/4 - 5*E^x*x + 10*x^2 - E^x*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 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}{4} \int \left (1+80 x+24 x^2+e^x \left (-20-28 x-4 x^2\right )\right ) \, dx\\ &=\frac {x}{4}+10 x^2+2 x^3+\frac {1}{4} \int e^x \left (-20-28 x-4 x^2\right ) \, dx\\ &=\frac {x}{4}+10 x^2+2 x^3+\frac {1}{4} \int \left (-20 e^x-28 e^x x-4 e^x x^2\right ) \, dx\\ &=\frac {x}{4}+10 x^2+2 x^3-5 \int e^x \, dx-7 \int e^x x \, dx-\int e^x x^2 \, dx\\ &=-5 e^x+\frac {x}{4}-7 e^x x+10 x^2-e^x x^2+2 x^3+2 \int e^x x \, dx+7 \int e^x \, dx\\ &=2 e^x+\frac {x}{4}-5 e^x x+10 x^2-e^x x^2+2 x^3-2 \int e^x \, dx\\ &=\frac {x}{4}-5 e^x x+10 x^2-e^x x^2+2 x^3\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 28, normalized size = 1.33 \begin {gather*} \frac {x}{4}+10 x^2+2 x^3-e^x \left (5 x+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + 80*x + 24*x^2 + E^x*(-20 - 28*x - 4*x^2))/4,x]

[Out]

x/4 + 10*x^2 + 2*x^3 - E^x*(5*x + x^2)

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fricas [A]  time = 0.55, size = 25, normalized size = 1.19 \begin {gather*} 2 \, x^{3} + 10 \, x^{2} - {\left (x^{2} + 5 \, x\right )} e^{x} + \frac {1}{4} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-4*x^2-28*x-20)*exp(x)+6*x^2+20*x+1/4,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.45, size = 25, normalized size = 1.19 \begin {gather*} 2 \, x^{3} + 10 \, x^{2} - {\left (x^{2} + 5 \, x\right )} e^{x} + \frac {1}{4} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-4*x^2-28*x-20)*exp(x)+6*x^2+20*x+1/4,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.02, size = 27, normalized size = 1.29

method	result	size
default	\(\frac {x}{4}+10 x^{2}+2 x^{3}-{\mathrm e}^{x} x^{2}-5 \,{\mathrm e}^{x} x\)	\(27\)
norman	\(\frac {x}{4}+10 x^{2}+2 x^{3}-{\mathrm e}^{x} x^{2}-5 \,{\mathrm e}^{x} x\)	\(27\)
risch	\(\frac {\left (-4 x^{2}-20 x \right ) {\mathrm e}^{x}}{4}+2 x^{3}+10 x^{2}+\frac {x}{4}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(-4*x^2-28*x-20)*exp(x)+6*x^2+20*x+1/4,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.45, size = 25, normalized size = 1.19 \begin {gather*} 2 \, x^{3} + 10 \, x^{2} - {\left (x^{2} + 5 \, x\right )} e^{x} + \frac {1}{4} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-4*x^2-28*x-20)*exp(x)+6*x^2+20*x+1/4,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.44, size = 22, normalized size = 1.05 \begin {gather*} \frac {x\,\left (40\,x-20\,{\mathrm {e}}^x-4\,x\,{\mathrm {e}}^x+8\,x^2+1\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(20*x - (exp(x)*(28*x + 4*x^2 + 20))/4 + 6*x^2 + 1/4,x)

[Out]

(x*(40*x - 20*exp(x) - 4*x*exp(x) + 8*x^2 + 1))/4

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sympy [A]  time = 0.10, size = 24, normalized size = 1.14 \begin {gather*} 2 x^{3} + 10 x^{2} + \frac {x}{4} + \left (- x^{2} - 5 x\right ) e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-4*x**2-28*x-20)*exp(x)+6*x**2+20*x+1/4,x)

[Out]

2*x**3 + 10*x**2 + x/4 + (-x**2 - 5*x)*exp(x)

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