3.86.84 \(\int \frac {18+2 x+e^{e^5} (18-x^2)-9 \log (x^2)}{3 x^2} \, dx\)

Optimal. Leaf size=31 \[ \log (5)-\frac {\left (3+\frac {x}{3}\right ) \left (e^{e^5} (2+x)-\log \left (x^2\right )\right )}{x} \]

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Rubi [A]  time = 0.05, antiderivative size = 43, normalized size of antiderivative = 1.39, number of steps used = 6, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {12, 14, 2304} \begin {gather*} \frac {3 \log \left (x^2\right )}{x}-\frac {1}{3} e^{e^5} x-\frac {6 \left (1+e^{e^5}\right )}{x}+\frac {6}{x}+\frac {2 \log (x)}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(18 + 2*x + E^E^5*(18 - x^2) - 9*Log[x^2])/(3*x^2),x]

[Out]

6/x - (6*(1 + E^E^5))/x - (E^E^5*x)/3 + (2*Log[x])/3 + (3*Log[x^2])/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 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int \frac {18+2 x+e^{e^5} \left (18-x^2\right )-9 \log \left (x^2\right )}{x^2} \, dx\\ &=\frac {1}{3} \int \left (\frac {18 \left (1+e^{e^5}\right )+2 x-e^{e^5} x^2}{x^2}-\frac {9 \log \left (x^2\right )}{x^2}\right ) \, dx\\ &=\frac {1}{3} \int \frac {18 \left (1+e^{e^5}\right )+2 x-e^{e^5} x^2}{x^2} \, dx-3 \int \frac {\log \left (x^2\right )}{x^2} \, dx\\ &=\frac {6}{x}+\frac {3 \log \left (x^2\right )}{x}+\frac {1}{3} \int \left (-e^{e^5}+\frac {18 \left (1+e^{e^5}\right )}{x^2}+\frac {2}{x}\right ) \, dx\\ &=\frac {6}{x}-\frac {6 \left (1+e^{e^5}\right )}{x}-\frac {e^{e^5} x}{3}+\frac {2 \log (x)}{3}+\frac {3 \log \left (x^2\right )}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 30, normalized size = 0.97 \begin {gather*} -\frac {e^{e^5} \left (18+x^2\right )-2 x \log (x)-9 \log \left (x^2\right )}{3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(18 + 2*x + E^E^5*(18 - x^2) - 9*Log[x^2])/(3*x^2),x]

[Out]

-1/3*(E^E^5*(18 + x^2) - 2*x*Log[x] - 9*Log[x^2])/x

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fricas [A]  time = 1.36, size = 24, normalized size = 0.77 \begin {gather*} -\frac {{\left (x^{2} + 18\right )} e^{\left (e^{5}\right )} - {\left (x + 9\right )} \log \left (x^{2}\right )}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-9*log(x^2)+(-x^2+18)*exp(exp(5))+2*x+18)/x^2,x, algorithm="fricas")

[Out]

-1/3*((x^2 + 18)*e^(e^5) - (x + 9)*log(x^2))/x

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giac [A]  time = 0.21, size = 29, normalized size = 0.94 \begin {gather*} -\frac {x^{2} e^{\left (e^{5}\right )} - 2 \, x \log \relax (x) + 18 \, e^{\left (e^{5}\right )} - 9 \, \log \left (x^{2}\right )}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-9*log(x^2)+(-x^2+18)*exp(exp(5))+2*x+18)/x^2,x, algorithm="giac")

[Out]

-1/3*(x^2*e^(e^5) - 2*x*log(x) + 18*e^(e^5) - 9*log(x^2))/x

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maple [A]  time = 0.06, size = 29, normalized size = 0.94

method	result	size
default	\(-\frac {x \,{\mathrm e}^{{\mathrm e}^{5}}}{3}+\frac {2 \ln \relax (x )}{3}-\frac {6 \,{\mathrm e}^{{\mathrm e}^{5}}}{x}+\frac {3 \ln \left (x^{2}\right )}{x}\)	\(29\)
norman	\(\frac {\frac {x \ln \left (x^{2}\right )}{3}-\frac {x^{2} {\mathrm e}^{{\mathrm e}^{5}}}{3}-6 \,{\mathrm e}^{{\mathrm e}^{5}}+3 \ln \left (x^{2}\right )}{x}\)	\(32\)
risch	\(\frac {3 \ln \left (x^{2}\right )}{x}+\frac {-x^{2} {\mathrm e}^{{\mathrm e}^{5}}+2 x \ln \relax (x )-18 \,{\mathrm e}^{{\mathrm e}^{5}}}{3 x}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*(-9*ln(x^2)+(-x^2+18)*exp(exp(5))+2*x+18)/x^2,x,method=_RETURNVERBOSE)

[Out]

-1/3*x*exp(exp(5))+2/3*ln(x)-6/x*exp(exp(5))+3*ln(x^2)/x

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maxima [A]  time = 0.36, size = 28, normalized size = 0.90 \begin {gather*} -\frac {1}{3} \, x e^{\left (e^{5}\right )} - \frac {6 \, e^{\left (e^{5}\right )}}{x} + \frac {3 \, \log \left (x^{2}\right )}{x} + \frac {2}{3} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-9*log(x^2)+(-x^2+18)*exp(exp(5))+2*x+18)/x^2,x, algorithm="maxima")

[Out]

-1/3*x*e^(e^5) - 6*e^(e^5)/x + 3*log(x^2)/x + 2/3*log(x)

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mupad [B]  time = 5.54, size = 29, normalized size = 0.94 \begin {gather*} \frac {\ln \left (x^2\right )}{3}+\frac {3\,\ln \left (x^2\right )-6\,{\mathrm {e}}^{{\mathrm {e}}^5}}{x}-\frac {x\,{\mathrm {e}}^{{\mathrm {e}}^5}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x)/3 - 3*log(x^2) - (exp(exp(5))*(x^2 - 18))/3 + 6)/x^2,x)

[Out]

log(x^2)/3 + (3*log(x^2) - 6*exp(exp(5)))/x - (x*exp(exp(5)))/3

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sympy [A]  time = 0.16, size = 31, normalized size = 1.00 \begin {gather*} - \frac {x e^{e^{5}}}{3} + \frac {2 \log {\relax (x )}}{3} + \frac {3 \log {\left (x^{2} \right )}}{x} - \frac {6 e^{e^{5}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-9*ln(x**2)+(-x**2+18)*exp(exp(5))+2*x+18)/x**2,x)

[Out]

-x*exp(exp(5))/3 + 2*log(x)/3 + 3*log(x**2)/x - 6*exp(exp(5))/x

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