3.60.12 \(\int \frac {48-8 x^2-14 x^3-3 x^4+e^{5-x} (4 x^2+6 x^3+x^4)}{3 x^2} \, dx\)

Optimal. Leaf size=32 \[ \frac {\left (-4+x-e^{5-x} x-x^2\right ) \left (-4+(4+x)^2\right )}{3 x} \]

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Rubi [A]  time = 0.09, antiderivative size = 60, normalized size of antiderivative = 1.88, number of steps used = 13, number of rules used = 5, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.109, Rules used = {12, 14, 2196, 2194, 2176} \begin {gather*} -\frac {x^3}{3}-\frac {1}{3} e^{5-x} x^2-\frac {7 x^2}{3}-\frac {8}{3} e^{5-x} x-\frac {8 x}{3}-4 e^{5-x}-\frac {16}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(48 - 8*x^2 - 14*x^3 - 3*x^4 + E^(5 - x)*(4*x^2 + 6*x^3 + x^4))/(3*x^2),x]

[Out]

-4*E^(5 - x) - 16/x - (8*x)/3 - (8*E^(5 - x)*x)/3 - (7*x^2)/3 - (E^(5 - x)*x^2)/3 - x^3/3

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 \frac {48-8 x^2-14 x^3-3 x^4+e^{5-x} \left (4 x^2+6 x^3+x^4\right )}{x^2} \, dx\\ &=\frac {1}{3} \int \left (e^{5-x} \left (4+6 x+x^2\right )+\frac {48-8 x^2-14 x^3-3 x^4}{x^2}\right ) \, dx\\ &=\frac {1}{3} \int e^{5-x} \left (4+6 x+x^2\right ) \, dx+\frac {1}{3} \int \frac {48-8 x^2-14 x^3-3 x^4}{x^2} \, dx\\ &=\frac {1}{3} \int \left (-8+\frac {48}{x^2}-14 x-3 x^2\right ) \, dx+\frac {1}{3} \int \left (4 e^{5-x}+6 e^{5-x} x+e^{5-x} x^2\right ) \, dx\\ &=-\frac {16}{x}-\frac {8 x}{3}-\frac {7 x^2}{3}-\frac {x^3}{3}+\frac {1}{3} \int e^{5-x} x^2 \, dx+\frac {4}{3} \int e^{5-x} \, dx+2 \int e^{5-x} x \, dx\\ &=-\frac {4 e^{5-x}}{3}-\frac {16}{x}-\frac {8 x}{3}-2 e^{5-x} x-\frac {7 x^2}{3}-\frac {1}{3} e^{5-x} x^2-\frac {x^3}{3}+\frac {2}{3} \int e^{5-x} x \, dx+2 \int e^{5-x} \, dx\\ &=-\frac {10 e^{5-x}}{3}-\frac {16}{x}-\frac {8 x}{3}-\frac {8}{3} e^{5-x} x-\frac {7 x^2}{3}-\frac {1}{3} e^{5-x} x^2-\frac {x^3}{3}+\frac {2}{3} \int e^{5-x} \, dx\\ &=-4 e^{5-x}-\frac {16}{x}-\frac {8 x}{3}-\frac {8}{3} e^{5-x} x-\frac {7 x^2}{3}-\frac {1}{3} e^{5-x} x^2-\frac {x^3}{3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 49, normalized size = 1.53 \begin {gather*} \frac {1}{3} \left (-\frac {48}{x}-8 x-7 x^2-x^3+e^{-x} \left (-12 e^5-8 e^5 x-e^5 x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(48 - 8*x^2 - 14*x^3 - 3*x^4 + E^(5 - x)*(4*x^2 + 6*x^3 + x^4))/(3*x^2),x]

[Out]

(-48/x - 8*x - 7*x^2 - x^3 + (-12*E^5 - 8*E^5*x - E^5*x^2)/E^x)/3

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fricas [A]  time = 0.64, size = 39, normalized size = 1.22 \begin {gather*} -\frac {x^{4} + 7 \, x^{3} + 8 \, x^{2} + {\left (x^{3} + 8 \, x^{2} + 12 \, x\right )} e^{\left (-x + 5\right )} + 48}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((x^4+6*x^3+4*x^2)*exp(5-x)-3*x^4-14*x^3-8*x^2+48)/x^2,x, algorithm="fricas")

[Out]

-1/3*(x^4 + 7*x^3 + 8*x^2 + (x^3 + 8*x^2 + 12*x)*e^(-x + 5) + 48)/x

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giac [A]  time = 0.23, size = 50, normalized size = 1.56 \begin {gather*} -\frac {x^{4} + x^{3} e^{\left (-x + 5\right )} + 7 \, x^{3} + 8 \, x^{2} e^{\left (-x + 5\right )} + 8 \, x^{2} + 12 \, x e^{\left (-x + 5\right )} + 48}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((x^4+6*x^3+4*x^2)*exp(5-x)-3*x^4-14*x^3-8*x^2+48)/x^2,x, algorithm="giac")

[Out]

-1/3*(x^4 + x^3*e^(-x + 5) + 7*x^3 + 8*x^2*e^(-x + 5) + 8*x^2 + 12*x*e^(-x + 5) + 48)/x

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maple [A]  time = 0.10, size = 38, normalized size = 1.19

method	result	size
risch	\(-\frac {x^{3}}{3}-\frac {7 x^{2}}{3}-\frac {8 x}{3}-\frac {16}{x}+\frac {\left (-x^{2}-8 x -12\right ) {\mathrm e}^{5-x}}{3}\)	\(38\)
norman	\(\frac {-16-\frac {8 x^{2}}{3}-\frac {7 x^{3}}{3}-\frac {x^{4}}{3}-4 x \,{\mathrm e}^{5-x}-\frac {8 \,{\mathrm e}^{5-x} x^{2}}{3}-\frac {{\mathrm e}^{5-x} x^{3}}{3}}{x}\)	\(53\)
derivativedivides	\(-\frac {{\mathrm e}^{5-x} \left (5-x \right )^{2}}{3}+6 \,{\mathrm e}^{5-x} \left (5-x \right )-\frac {77 \,{\mathrm e}^{5-x}}{3}-\frac {16}{x}+255-51 x -\frac {22 \left (5-x \right )^{2}}{3}+\frac {\left (5-x \right )^{3}}{3}\)	\(65\)
default	\(-\frac {{\mathrm e}^{5-x} \left (5-x \right )^{2}}{3}+6 \,{\mathrm e}^{5-x} \left (5-x \right )-\frac {77 \,{\mathrm e}^{5-x}}{3}-\frac {16}{x}+255-51 x -\frac {22 \left (5-x \right )^{2}}{3}+\frac {\left (5-x \right )^{3}}{3}\)	\(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((x^4+6*x^3+4*x^2)*exp(5-x)-3*x^4-14*x^3-8*x^2+48)/x^2,x,method=_RETURNVERBOSE)

[Out]

-1/3*x^3-7/3*x^2-8/3*x-16/x+1/3*(-x^2-8*x-12)*exp(5-x)

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maxima [B]  time = 0.36, size = 62, normalized size = 1.94 \begin {gather*} -\frac {1}{3} \, x^{3} - \frac {7}{3} \, x^{2} - \frac {1}{3} \, {\left (x^{2} e^{5} + 2 \, x e^{5} + 2 \, e^{5}\right )} e^{\left (-x\right )} - 2 \, {\left (x e^{5} + e^{5}\right )} e^{\left (-x\right )} - \frac {8}{3} \, x - \frac {16}{x} - \frac {4}{3} \, e^{\left (-x + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((x^4+6*x^3+4*x^2)*exp(5-x)-3*x^4-14*x^3-8*x^2+48)/x^2,x, algorithm="maxima")

[Out]

-1/3*x^3 - 7/3*x^2 - 1/3*(x^2*e^5 + 2*x*e^5 + 2*e^5)*e^(-x) - 2*(x*e^5 + e^5)*e^(-x) - 8/3*x - 16/x - 4/3*e^(-
x + 5)

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mupad [B]  time = 0.10, size = 47, normalized size = 1.47 \begin {gather*} -\frac {8\,x}{3}-4\,{\mathrm {e}}^{5-x}-\frac {8\,x\,{\mathrm {e}}^{5-x}}{3}-\frac {x^2\,{\mathrm {e}}^{5-x}}{3}-\frac {16}{x}-\frac {7\,x^2}{3}-\frac {x^3}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((8*x^2)/3 - (exp(5 - x)*(4*x^2 + 6*x^3 + x^4))/3 + (14*x^3)/3 + x^4 - 16)/x^2,x)

[Out]

- (8*x)/3 - 4*exp(5 - x) - (8*x*exp(5 - x))/3 - (x^2*exp(5 - x))/3 - 16/x - (7*x^2)/3 - x^3/3

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sympy [A]  time = 0.13, size = 36, normalized size = 1.12 \begin {gather*} - \frac {x^{3}}{3} - \frac {7 x^{2}}{3} - \frac {8 x}{3} + \frac {\left (- x^{2} - 8 x - 12\right ) e^{5 - x}}{3} - \frac {16}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((x**4+6*x**3+4*x**2)*exp(5-x)-3*x**4-14*x**3-8*x**2+48)/x**2,x)

[Out]

-x**3/3 - 7*x**2/3 - 8*x/3 + (-x**2 - 8*x - 12)*exp(5 - x)/3 - 16/x

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