3.97.60 \(\int \frac {-8+4 e^{6+2 x} x^2}{3 x^2} \, dx\)

Optimal. Leaf size=22 \[ -5+\frac {2 \left (4+\left (6+e^{6+2 x}\right ) x\right )}{3 x} \]

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Rubi [A]  time = 0.01, antiderivative size = 19, normalized size of antiderivative = 0.86, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 14, 2194} \begin {gather*} \frac {2}{3} e^{2 x+6}+\frac {8}{3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*E^(6 + 2*x))/3 + 8/(3*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} &=\frac {1}{3} \int \frac {-8+4 e^{6+2 x} x^2}{x^2} \, dx\\ &=\frac {1}{3} \int \left (4 e^{6+2 x}-\frac {8}{x^2}\right ) \, dx\\ &=\frac {8}{3 x}+\frac {4}{3} \int e^{6+2 x} \, dx\\ &=\frac {2}{3} e^{6+2 x}+\frac {8}{3 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 21, normalized size = 0.95 \begin {gather*} \frac {4}{3} \left (\frac {1}{2} e^{6+2 x}+\frac {2}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(E^(6 + 2*x)/2 + 2/x))/3

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fricas [A]  time = 0.54, size = 15, normalized size = 0.68 \begin {gather*} \frac {2 \, {\left (x e^{\left (2 \, x + 6\right )} + 4\right )}}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(4*x^2*exp(3)^2*exp(x)^2-8)/x^2,x, algorithm="fricas")

[Out]

2/3*(x*e^(2*x + 6) + 4)/x

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giac [A]  time = 0.14, size = 15, normalized size = 0.68 \begin {gather*} \frac {2 \, {\left (x e^{\left (2 \, x + 6\right )} + 4\right )}}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(4*x^2*exp(3)^2*exp(x)^2-8)/x^2,x, algorithm="giac")

[Out]

2/3*(x*e^(2*x + 6) + 4)/x

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maple [A]  time = 0.03, size = 15, normalized size = 0.68

method	result	size
risch	\(\frac {2 \,{\mathrm e}^{2 x +6}}{3}+\frac {8}{3 x}\)	\(15\)
default	\(\frac {2 \,{\mathrm e}^{6} {\mathrm e}^{2 x}}{3}+\frac {8}{3 x}\)	\(17\)
norman	\(\frac {\frac {8}{3}+\frac {2 x \,{\mathrm e}^{6} {\mathrm e}^{2 x}}{3}}{x}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*(4*x^2*exp(3)^2*exp(x)^2-8)/x^2,x,method=_RETURNVERBOSE)

[Out]

2/3*exp(2*x+6)+8/3/x

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maxima [A]  time = 0.36, size = 14, normalized size = 0.64 \begin {gather*} \frac {8}{3 \, x} + \frac {2}{3} \, e^{\left (2 \, x + 6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(4*x^2*exp(3)^2*exp(x)^2-8)/x^2,x, algorithm="maxima")

[Out]

8/3/x + 2/3*e^(2*x + 6)

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mupad [B]  time = 0.07, size = 14, normalized size = 0.64 \begin {gather*} \frac {2\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^6}{3}+\frac {8}{3\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^2*exp(2*x)*exp(6))/3 - 8/3)/x^2,x)

[Out]

(2*exp(2*x)*exp(6))/3 + 8/(3*x)

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sympy [A]  time = 0.09, size = 15, normalized size = 0.68 \begin {gather*} \frac {2 e^{6} e^{2 x}}{3} + \frac {8}{3 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(4*x**2*exp(3)**2*exp(x)**2-8)/x**2,x)

[Out]

2*exp(6)*exp(2*x)/3 + 8/(3*x)

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