3.48.96 \(\int \frac {1}{5} (1-48 x+12 e^5 x+e^x (-12 x-6 x^2)) \, dx\)

Optimal. Leaf size=27 \[ \frac {1}{5} \left (-5-e^3+x+6 \left (-4+e^5-e^x\right ) x^2\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 30, normalized size of antiderivative = 1.11, number of steps used = 11, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {6, 12, 1593, 2196, 2176, 2194} \begin {gather*} -\frac {6}{5} e^x x^2-\frac {6}{5} \left (4-e^5\right ) x^2+\frac {x}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - 48*x + 12*E^5*x + E^x*(-12*x - 6*x^2))/5,x]

[Out]

x/5 - (6*E^x*x^2)/5 - (6*(4 - E^5)*x^2)/5

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

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

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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} &=\int \frac {1}{5} \left (1+\left (-48+12 e^5\right ) x+e^x \left (-12 x-6 x^2\right )\right ) \, dx\\ &=\frac {1}{5} \int \left (1+\left (-48+12 e^5\right ) x+e^x \left (-12 x-6 x^2\right )\right ) \, dx\\ &=\frac {x}{5}-\frac {6}{5} \left (4-e^5\right ) x^2+\frac {1}{5} \int e^x \left (-12 x-6 x^2\right ) \, dx\\ &=\frac {x}{5}-\frac {6}{5} \left (4-e^5\right ) x^2+\frac {1}{5} \int e^x (-12-6 x) x \, dx\\ &=\frac {x}{5}-\frac {6}{5} \left (4-e^5\right ) x^2+\frac {1}{5} \int \left (-12 e^x x-6 e^x x^2\right ) \, dx\\ &=\frac {x}{5}-\frac {6}{5} \left (4-e^5\right ) x^2-\frac {6}{5} \int e^x x^2 \, dx-\frac {12}{5} \int e^x x \, dx\\ &=\frac {x}{5}-\frac {12 e^x x}{5}-\frac {6 e^x x^2}{5}-\frac {6}{5} \left (4-e^5\right ) x^2+\frac {12 \int e^x \, dx}{5}+\frac {12}{5} \int e^x x \, dx\\ &=\frac {12 e^x}{5}+\frac {x}{5}-\frac {6 e^x x^2}{5}-\frac {6}{5} \left (4-e^5\right ) x^2-\frac {12 \int e^x \, dx}{5}\\ &=\frac {x}{5}-\frac {6 e^x x^2}{5}-\frac {6}{5} \left (4-e^5\right ) x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 21, normalized size = 0.78 \begin {gather*} \frac {1}{5} \left (x-6 \left (4-e^5+e^x\right ) x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - 48*x + 12*E^5*x + E^x*(-12*x - 6*x^2))/5,x]

[Out]

(x - 6*(4 - E^5 + E^x)*x^2)/5

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fricas [A]  time = 1.02, size = 23, normalized size = 0.85 \begin {gather*} \frac {6}{5} \, x^{2} e^{5} - \frac {6}{5} \, x^{2} e^{x} - \frac {24}{5} \, x^{2} + \frac {1}{5} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-6*x^2-12*x)*exp(x)+12/5*x*exp(5)-48/5*x+1/5,x, algorithm="fricas")

[Out]

6/5*x^2*e^5 - 6/5*x^2*e^x - 24/5*x^2 + 1/5*x

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giac [A]  time = 0.14, size = 23, normalized size = 0.85 \begin {gather*} \frac {6}{5} \, x^{2} e^{5} - \frac {6}{5} \, x^{2} e^{x} - \frac {24}{5} \, x^{2} + \frac {1}{5} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-6*x^2-12*x)*exp(x)+12/5*x*exp(5)-48/5*x+1/5,x, algorithm="giac")

[Out]

6/5*x^2*e^5 - 6/5*x^2*e^x - 24/5*x^2 + 1/5*x

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maple [A]  time = 0.03, size = 22, normalized size = 0.81

method	result	size
norman	\(\left (\frac {6 \,{\mathrm e}^{5}}{5}-\frac {24}{5}\right ) x^{2}+\frac {x}{5}-\frac {6 \,{\mathrm e}^{x} x^{2}}{5}\)	\(22\)
default	\(\frac {x}{5}-\frac {24 x^{2}}{5}+\frac {6 x^{2} {\mathrm e}^{5}}{5}-\frac {6 \,{\mathrm e}^{x} x^{2}}{5}\)	\(24\)
risch	\(\frac {x}{5}-\frac {24 x^{2}}{5}+\frac {6 x^{2} {\mathrm e}^{5}}{5}-\frac {6 \,{\mathrm e}^{x} x^{2}}{5}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(-6*x^2-12*x)*exp(x)+12/5*x*exp(5)-48/5*x+1/5,x,method=_RETURNVERBOSE)

[Out]

(6/5*exp(5)-24/5)*x^2+1/5*x-6/5*exp(x)*x^2

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maxima [A]  time = 0.35, size = 23, normalized size = 0.85 \begin {gather*} \frac {6}{5} \, x^{2} e^{5} - \frac {6}{5} \, x^{2} e^{x} - \frac {24}{5} \, x^{2} + \frac {1}{5} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-6*x^2-12*x)*exp(x)+12/5*x*exp(5)-48/5*x+1/5,x, algorithm="maxima")

[Out]

6/5*x^2*e^5 - 6/5*x^2*e^x - 24/5*x^2 + 1/5*x

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mupad [B]  time = 3.40, size = 21, normalized size = 0.78 \begin {gather*} \frac {x}{5}-\frac {6\,x^2\,{\mathrm {e}}^x}{5}+x^2\,\left (\frac {6\,{\mathrm {e}}^5}{5}-\frac {24}{5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((12*x*exp(5))/5 - (48*x)/5 - (exp(x)*(12*x + 6*x^2))/5 + 1/5,x)

[Out]

x/5 - (6*x^2*exp(x))/5 + x^2*((6*exp(5))/5 - 24/5)

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sympy [A]  time = 0.11, size = 26, normalized size = 0.96 \begin {gather*} - \frac {6 x^{2} e^{x}}{5} + x^{2} \left (- \frac {24}{5} + \frac {6 e^{5}}{5}\right ) + \frac {x}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-6*x**2-12*x)*exp(x)+12/5*x*exp(5)-48/5*x+1/5,x)

[Out]

-6*x**2*exp(x)/5 + x**2*(-24/5 + 6*exp(5)/5) + x/5

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