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

Optimal. Leaf size=29 \[ -\frac {9-\frac {\frac {3}{2}+x}{-e^{5/3}+\log (3)}}{x}+\log (x) \]

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Rubi [A]  time = 0.04, antiderivative size = 35, normalized size of antiderivative = 1.21, number of steps used = 5, number of rules used = 4, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {6, 12, 186, 43} \begin {gather*} \log (x)-\frac {3 \left (1+6 e^{5/3}-6 \log (3)\right )}{x \left (2 e^{5/3}-\log (9)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(1 + 6*E^(5/3) - 6*Log[3]))/(x*(2*E^(5/3) - Log[9])) + Log[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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 186

Int[(u_)^(m_.)*(v_)^(n_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*ExpandToSum[v, x]^n, x] /; FreeQ[{m, n}, x] &&
LinearQ[{u, v}, x] &&  !LinearMatchQ[{u, v}, x]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 50, normalized size = 1.72 \begin {gather*} \frac {-\frac {3 \left (1+6 e^{5/3}-6 \log (3)\right )}{x}+\left (2 e^{5/3}-\log (9)\right ) \log (x)}{2 \left (e^{5/3}-\log (3)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*(1 + 6*E^(5/3) - 6*Log[3]))/x + (2*E^(5/3) - Log[9])*Log[x])/(2*(E^(5/3) - Log[3]))

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fricas [A]  time = 0.67, size = 38, normalized size = 1.31 \begin {gather*} \frac {2 \, {\left (x e^{\frac {5}{3}} - x \log \relax (3)\right )} \log \relax (x) - 18 \, e^{\frac {5}{3}} + 18 \, \log \relax (3) - 3}{2 \, {\left (x e^{\frac {5}{3}} - x \log \relax (3)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.14, size = 28, normalized size = 0.97 \begin {gather*} -\frac {3 \, {\left (6 \, e^{\frac {5}{3}} - 6 \, \log \relax (3) + 1\right )}}{2 \, x {\left (e^{\frac {5}{3}} - \log \relax (3)\right )}} + \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.08, size = 28, normalized size = 0.97

method	result	size
norman	\(-\frac {3 \left (6 \ln \relax (3)-6 \,{\mathrm e}^{\frac {5}{3}}-1\right )}{2 \left (\ln \relax (3)-{\mathrm e}^{\frac {5}{3}}\right ) x}+\ln \relax (x )\)	\(28\)
default	\(\frac {-\frac {18 \ln \relax (3)-18 \,{\mathrm e}^{\frac {5}{3}}-3}{x}+\left (2 \ln \relax (3)-2 \,{\mathrm e}^{\frac {5}{3}}\right ) \ln \relax (x )}{2 \ln \relax (3)-2 \,{\mathrm e}^{\frac {5}{3}}}\)	\(40\)
risch	\(-\frac {9 \ln \relax (3)}{x \left (\ln \relax (3)-{\mathrm e}^{\frac {5}{3}}\right )}+\frac {9 \,{\mathrm e}^{\frac {5}{3}}}{x \left (\ln \relax (3)-{\mathrm e}^{\frac {5}{3}}\right )}+\frac {3}{2 x \left (\ln \relax (3)-{\mathrm e}^{\frac {5}{3}}\right )}+\ln \relax (x )\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.36, size = 27, normalized size = 0.93 \begin {gather*} -\frac {3 \, {\left (6 \, e^{\frac {5}{3}} - 6 \, \log \relax (3) + 1\right )}}{2 \, x {\left (e^{\frac {5}{3}} - \log \relax (3)\right )}} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.11, size = 29, normalized size = 1.00 \begin {gather*} \ln \relax (x)-\frac {18\,{\mathrm {e}}^{5/3}-18\,\ln \relax (3)+3}{x\,\left (2\,{\mathrm {e}}^{5/3}-\ln \relax (9)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x) - (18*exp(5/3) - 18*log(3) + 3)/(x*(2*exp(5/3) - log(9)))

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sympy [B]  time = 0.24, size = 39, normalized size = 1.34 \begin {gather*} \frac {2 \left (- \log {\relax (3 )} + e^{\frac {5}{3}}\right ) \log {\relax (x )} + \frac {- 18 e^{\frac {5}{3}} - 3 + 18 \log {\relax (3 )}}{x}}{- 2 \log {\relax (3 )} + 2 e^{\frac {5}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(-log(3) + exp(5/3))*log(x) + (-18*exp(5/3) - 3 + 18*log(3))/x)/(-2*log(3) + 2*exp(5/3))

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