3.29.89 \(\int \frac {-e-48 x^3+6 e^2 x^3+18 e^x x^3}{6 x^3} \, dx\)

Optimal. Leaf size=28 \[ -3 \left (2-e^x\right )+\frac {e}{12 x^2}-\left (8-e^2\right ) x \]

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Rubi [A]  time = 0.03, antiderivative size = 24, normalized size of antiderivative = 0.86, number of steps used = 7, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6, 12, 14, 2194} \begin {gather*} \frac {e}{12 x^2}-\left (\left (8-e^2\right ) x\right )+3 e^x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-E - 48*x^3 + 6*E^2*x^3 + 18*E^x*x^3)/(6*x^3),x]

[Out]

3*E^x + E/(12*x^2) - (8 - E^2)*x

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 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 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 {gather*} \begin {aligned} \text {integral} &=\int \frac {-e+18 e^x x^3+\left (-48+6 e^2\right ) x^3}{6 x^3} \, dx\\ &=\frac {1}{6} \int \frac {-e+18 e^x x^3+\left (-48+6 e^2\right ) x^3}{x^3} \, dx\\ &=\frac {1}{6} \int \left (18 e^x+\frac {-e-6 \left (8-e^2\right ) x^3}{x^3}\right ) \, dx\\ &=\frac {1}{6} \int \frac {-e-6 \left (8-e^2\right ) x^3}{x^3} \, dx+3 \int e^x \, dx\\ &=3 e^x+\frac {1}{6} \int \left (6 \left (-8+e^2\right )-\frac {e}{x^3}\right ) \, dx\\ &=3 e^x+\frac {e}{12 x^2}-\left (8-e^2\right ) x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 22, normalized size = 0.79 \begin {gather*} 3 e^x+\frac {e}{12 x^2}-8 x+e^2 x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-E - 48*x^3 + 6*E^2*x^3 + 18*E^x*x^3)/(6*x^3),x]

[Out]

3*E^x + E/(12*x^2) - 8*x + E^2*x

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fricas [A]  time = 0.86, size = 27, normalized size = 0.96 \begin {gather*} \frac {12 \, x^{3} e^{2} - 96 \, x^{3} + 36 \, x^{2} e^{x} + e}{12 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(18*exp(x)*x^3+6*x^3*exp(2)-exp(1)-48*x^3)/x^3,x, algorithm="fricas")

[Out]

1/12*(12*x^3*e^2 - 96*x^3 + 36*x^2*e^x + e)/x^2

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giac [A]  time = 0.19, size = 27, normalized size = 0.96 \begin {gather*} \frac {12 \, x^{3} e^{2} - 96 \, x^{3} + 36 \, x^{2} e^{x} + e}{12 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(18*exp(x)*x^3+6*x^3*exp(2)-exp(1)-48*x^3)/x^3,x, algorithm="giac")

[Out]

1/12*(12*x^3*e^2 - 96*x^3 + 36*x^2*e^x + e)/x^2

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maple [A]  time = 0.04, size = 20, normalized size = 0.71

method	result	size
default	\(-8 x +\frac {{\mathrm e}}{12 x^{2}}+{\mathrm e}^{2} x +3 \,{\mathrm e}^{x}\)	\(20\)
risch	\(-8 x +\frac {{\mathrm e}}{12 x^{2}}+{\mathrm e}^{2} x +3 \,{\mathrm e}^{x}\)	\(20\)
norman	\(\frac {\left (-8+{\mathrm e}^{2}\right ) x^{3}+3 \,{\mathrm e}^{x} x^{2}+\frac {{\mathrm e}}{12}}{x^{2}}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/6*(18*exp(x)*x^3+6*x^3*exp(2)-exp(1)-48*x^3)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.39, size = 19, normalized size = 0.68 \begin {gather*} x e^{2} - 8 \, x + \frac {e}{12 \, x^{2}} + 3 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(18*exp(x)*x^3+6*x^3*exp(2)-exp(1)-48*x^3)/x^3,x, algorithm="maxima")

[Out]

x*e^2 - 8*x + 1/12*e/x^2 + 3*e^x

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mupad [B]  time = 0.08, size = 18, normalized size = 0.64 \begin {gather*} 3\,{\mathrm {e}}^x+x\,\left ({\mathrm {e}}^2-8\right )+\frac {\mathrm {e}}{12\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(1)/6 - 3*x^3*exp(x) - x^3*exp(2) + 8*x^3)/x^3,x)

[Out]

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

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sympy [A]  time = 0.12, size = 22, normalized size = 0.79 \begin {gather*} - \frac {x \left (48 - 6 e^{2}\right )}{6} + 3 e^{x} + \frac {e}{12 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(18*exp(x)*x**3+6*x**3*exp(2)-exp(1)-48*x**3)/x**3,x)

[Out]

-x*(48 - 6*exp(2))/6 + 3*exp(x) + E/(12*x**2)

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