\(\int \frac {(60-120 e x) \log (5)+(-40+60 e x) \log (5) \log (\frac {2-3 e x}{3 x})}{-2 x^3+3 e x^4} \, dx\) [3037]

Optimal result

Integrand size = 48, antiderivative size = 21 \[ \int \frac {(60-120 e x) \log (5)+(-40+60 e x) \log (5) \log \left (\frac {2-3 e x}{3 x}\right )}{-2 x^3+3 e x^4} \, dx=-\frac {10 \log (5) \left (-2+\log \left (-e+\frac {2}{3 x}\right )\right )}{x^2} \]

[Out]

-10*(ln(2/3/x-exp(1))-2)*ln(5)/x^2

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(73\) vs. \(2(21)=42\).

Time = 0.18 (sec) , antiderivative size = 73, normalized size of antiderivative = 3.48, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1607, 6874, 78, 2504, 2442, 45} \[ \int \frac {(60-120 e x) \log (5)+(-40+60 e x) \log (5) \log \left (\frac {2-3 e x}{3 x}\right )}{-2 x^3+3 e x^4} \, dx=-\frac {10 \log (5) \log \left (\frac {2}{3 x}-e\right )}{x^2}+\frac {20 \log (5)}{x^2}+\frac {45}{2} e^2 \log (5) \log \left (3 e-\frac {2}{x}\right )+\frac {45}{2} e^2 \log (5) \log (x)-\frac {45}{2} e^2 \log (5) \log (2-3 e x) \]

[In]

Int[((60 - 120*E*x)*Log[5] + (-40 + 60*E*x)*Log[5]*Log[(2 - 3*E*x)/(3*x)])/(-2*x^3 + 3*E*x^4),x]

[Out]

(20*Log[5])/x^2 + (45*E^2*Log[5]*Log[3*E - 2/x])/2 - (10*Log[5]*Log[-E + 2/(3*x)])/x^2 + (45*E^2*Log[5]*Log[x]
)/2 - (45*E^2*Log[5]*Log[2 - 3*E*x])/2

Rule 45

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 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2442

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[(f + g*
x)^(q + 1)*((a + b*Log[c*(d + e*x)^n])/(g*(q + 1))), x] - Dist[b*e*(n/(g*(q + 1))), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 2504

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {(60-120 e x) \log (5)+(-40+60 e x) \log (5) \log \left (\frac {2-3 e x}{3 x}\right )}{x^3 (-2+3 e x)} \, dx \\ & = \int \left (-\frac {60 (-1+2 e x) \log (5)}{x^3 (-2+3 e x)}+\frac {20 \log (5) \log \left (-e+\frac {2}{3 x}\right )}{x^3}\right ) \, dx \\ & = (20 \log (5)) \int \frac {\log \left (-e+\frac {2}{3 x}\right )}{x^3} \, dx-(60 \log (5)) \int \frac {-1+2 e x}{x^3 (-2+3 e x)} \, dx \\ & = -\left ((20 \log (5)) \text {Subst}\left (\int x \log \left (-e+\frac {2 x}{3}\right ) \, dx,x,\frac {1}{x}\right )\right )-(60 \log (5)) \int \left (\frac {1}{2 x^3}-\frac {e}{4 x^2}-\frac {3 e^2}{8 x}+\frac {9 e^3}{8 (-2+3 e x)}\right ) \, dx \\ & = \frac {15 \log (5)}{x^2}-\frac {15 e \log (5)}{x}-\frac {10 \log (5) \log \left (-e+\frac {2}{3 x}\right )}{x^2}+\frac {45}{2} e^2 \log (5) \log (x)-\frac {45}{2} e^2 \log (5) \log (2-3 e x)+\frac {1}{3} (20 \log (5)) \text {Subst}\left (\int \frac {x^2}{-e+\frac {2 x}{3}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {15 \log (5)}{x^2}-\frac {15 e \log (5)}{x}-\frac {10 \log (5) \log \left (-e+\frac {2}{3 x}\right )}{x^2}+\frac {45}{2} e^2 \log (5) \log (x)-\frac {45}{2} e^2 \log (5) \log (2-3 e x)+\frac {1}{3} (20 \log (5)) \text {Subst}\left (\int \left (\frac {9 e}{4}-\frac {27 e^2}{4 (3 e-2 x)}+\frac {3 x}{2}\right ) \, dx,x,\frac {1}{x}\right ) \\ & = \frac {20 \log (5)}{x^2}+\frac {45}{2} e^2 \log (5) \log \left (3 e-\frac {2}{x}\right )-\frac {10 \log (5) \log \left (-e+\frac {2}{3 x}\right )}{x^2}+\frac {45}{2} e^2 \log (5) \log (x)-\frac {45}{2} e^2 \log (5) \log (2-3 e x) \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(67\) vs. \(2(21)=42\).

Time = 0.09 (sec) , antiderivative size = 67, normalized size of antiderivative = 3.19 \[ \int \frac {(60-120 e x) \log (5)+(-40+60 e x) \log (5) \log \left (\frac {2-3 e x}{3 x}\right )}{-2 x^3+3 e x^4} \, dx=20 \log (5) \left (\frac {1}{x^2}+\frac {9}{8} e^2 \log \left (3 e-\frac {2}{x}\right )-\frac {\log \left (-e+\frac {2}{3 x}\right )}{2 x^2}+\frac {9}{8} e^2 \log (x)-\frac {9}{8} e^2 \log (2-3 e x)\right ) \]

[In]

Integrate[((60 - 120*E*x)*Log[5] + (-40 + 60*E*x)*Log[5]*Log[(2 - 3*E*x)/(3*x)])/(-2*x^3 + 3*E*x^4),x]

[Out]

20*Log[5]*(x^(-2) + (9*E^2*Log[3*E - 2/x])/8 - Log[-E + 2/(3*x)]/(2*x^2) + (9*E^2*Log[x])/8 - (9*E^2*Log[2 - 3
*E*x])/8)

Maple [A] (verified)

Time = 0.71 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29

method	result	size
norman	\(\frac {-10 \ln \left (5\right ) \ln \left (\frac {-3 x \,{\mathrm e}+2}{3 x}\right )+20 \ln \left (5\right )}{x^{2}}\)	\(27\)
risch	\(-\frac {10 \ln \left (5\right ) \ln \left (\frac {-3 x \,{\mathrm e}+2}{3 x}\right )}{x^{2}}+\frac {20 \ln \left (5\right )}{x^{2}}\)	\(29\)
parallelrisch	\(-\frac {180 \,{\mathrm e}^{2} \ln \left (5\right ) x^{2}+40 \ln \left (5\right ) \ln \left (-\frac {3 x \,{\mathrm e}-2}{3 x}\right )-80 \ln \left (5\right )}{4 x^{2}}\)	\(39\)
parts	\(\frac {45 \,{\mathrm e}^{2} \ln \left (5\right ) \ln \left (\frac {2}{3 x}-{\mathrm e}\right )}{2}-\frac {10 \ln \left (5\right ) \ln \left (\frac {2}{3 x}-{\mathrm e}\right )}{x^{2}}-\frac {135 \,{\mathrm e}^{2} \ln \left (5\right )}{4}+\frac {15 \ln \left (5\right ) {\mathrm e}}{x}+\frac {5 \ln \left (5\right )}{x^{2}}-60 \ln \left (5\right ) \left (-\frac {3 \,{\mathrm e}^{2} \ln \left (x \right )}{8}+\frac {{\mathrm e}}{4 x}-\frac {1}{4 x^{2}}+\frac {3 \,{\mathrm e}^{2} \ln \left (3 x \,{\mathrm e}-2\right )}{8}\right )\)	\(100\)
derivativedivides	\(-45 \ln \left (5\right ) \left (\frac {\left (\frac {2}{3 x}-{\mathrm e}\right )^{2} \ln \left (\frac {2}{3 x}-{\mathrm e}\right )}{2}-\frac {\left (\frac {2}{3 x}-{\mathrm e}\right )^{2}}{4}\right )+45 \,{\mathrm e} \ln \left (5\right ) \left (\frac {2}{3 x}-{\mathrm e}\right )+\frac {135 \ln \left (5\right ) \left (\frac {2}{3 x}-{\mathrm e}\right )^{2}}{4}-45 \,{\mathrm e} \ln \left (5\right ) \left (\left (\frac {2}{3 x}-{\mathrm e}\right ) \ln \left (\frac {2}{3 x}-{\mathrm e}\right )+{\mathrm e}-\frac {2}{3 x}\right )-\frac {45 \,{\mathrm e}^{2} \ln \left (5\right ) \ln \left (\frac {2}{3 x}-{\mathrm e}\right )}{2}\)	\(133\)
default	\(-45 \ln \left (5\right ) \left (\frac {\left (\frac {2}{3 x}-{\mathrm e}\right )^{2} \ln \left (\frac {2}{3 x}-{\mathrm e}\right )}{2}-\frac {\left (\frac {2}{3 x}-{\mathrm e}\right )^{2}}{4}\right )+45 \,{\mathrm e} \ln \left (5\right ) \left (\frac {2}{3 x}-{\mathrm e}\right )+\frac {135 \ln \left (5\right ) \left (\frac {2}{3 x}-{\mathrm e}\right )^{2}}{4}-45 \,{\mathrm e} \ln \left (5\right ) \left (\left (\frac {2}{3 x}-{\mathrm e}\right ) \ln \left (\frac {2}{3 x}-{\mathrm e}\right )+{\mathrm e}-\frac {2}{3 x}\right )-\frac {45 \,{\mathrm e}^{2} \ln \left (5\right ) \ln \left (\frac {2}{3 x}-{\mathrm e}\right )}{2}\)	\(133\)

[In]

int(((60*x*exp(1)-40)*ln(5)*ln(1/3*(-3*x*exp(1)+2)/x)+(-120*x*exp(1)+60)*ln(5))/(3*x^4*exp(1)-2*x^3),x,method=
_RETURNVERBOSE)

[Out]

(-10*ln(5)*ln(1/3*(-3*x*exp(1)+2)/x)+20*ln(5))/x^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {(60-120 e x) \log (5)+(-40+60 e x) \log (5) \log \left (\frac {2-3 e x}{3 x}\right )}{-2 x^3+3 e x^4} \, dx=-\frac {10 \, {\left (\log \left (5\right ) \log \left (-\frac {3 \, x e - 2}{3 \, x}\right ) - 2 \, \log \left (5\right )\right )}}{x^{2}} \]

[In]

integrate(((60*x*exp(1)-40)*log(5)*log(1/3*(-3*x*exp(1)+2)/x)+(-120*x*exp(1)+60)*log(5))/(3*x^4*exp(1)-2*x^3),
x, algorithm="fricas")

[Out]

-10*(log(5)*log(-1/3*(3*x*e - 2)/x) - 2*log(5))/x^2

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int \frac {(60-120 e x) \log (5)+(-40+60 e x) \log (5) \log \left (\frac {2-3 e x}{3 x}\right )}{-2 x^3+3 e x^4} \, dx=- \frac {10 \log {\left (5 \right )} \log {\left (\frac {- e x + \frac {2}{3}}{x} \right )}}{x^{2}} + \frac {20 \log {\left (5 \right )}}{x^{2}} \]

[In]

integrate(((60*x*exp(1)-40)*ln(5)*ln(1/3*(-3*x*exp(1)+2)/x)+(-120*x*exp(1)+60)*ln(5))/(3*x**4*exp(1)-2*x**3),x
)

[Out]

-10*log(5)*log((-E*x + 2/3)/x)/x**2 + 20*log(5)/x**2

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 295 vs. \(2 (20) = 40\).

Time = 0.22 (sec) , antiderivative size = 295, normalized size of antiderivative = 14.05 \[ \int \frac {(60-120 e x) \log (5)+(-40+60 e x) \log (5) \log \left (\frac {2-3 e x}{3 x}\right )}{-2 x^3+3 e x^4} \, dx=15 \, {\left (3 \, e \log \left (3 \, x e - 2\right ) - 3 \, e \log \left (x\right ) + \frac {2}{x}\right )} e \log \left (5\right ) \log \left (\frac {2}{3 \, x} - e\right ) - 30 \, {\left (3 \, e \log \left (3 \, x e - 2\right ) - 3 \, e \log \left (x\right ) + \frac {2}{x}\right )} e \log \left (5\right ) - 5 \, {\left (9 \, e^{2} \log \left (3 \, x e - 2\right ) - 9 \, e^{2} \log \left (x\right ) + \frac {2 \, {\left (3 \, x e + 1\right )}}{x^{2}}\right )} \log \left (5\right ) \log \left (\frac {2}{3 \, x} - e\right ) + \frac {15}{2} \, {\left (9 \, e^{2} \log \left (3 \, x e - 2\right ) - 9 \, e^{2} \log \left (x\right ) + \frac {2 \, {\left (3 \, x e + 1\right )}}{x^{2}}\right )} \log \left (5\right ) - \frac {15 \, {\left (3 \, x e \log \left (3 \, x e - 2\right )^{2} + 3 \, x e \log \left (x\right )^{2} - 6 \, x e \log \left (x\right ) - 6 \, {\left (x e \log \left (x\right ) - x e\right )} \log \left (3 \, x e - 2\right ) + 4\right )} e \log \left (5\right )}{2 \, x} + \frac {5 \, {\left (9 \, x^{2} e^{2} \log \left (3 \, x e - 2\right )^{2} + 9 \, x^{2} e^{2} \log \left (x\right )^{2} - 27 \, x^{2} e^{2} \log \left (x\right ) + 18 \, x e - 9 \, {\left (2 \, x^{2} e^{2} \log \left (x\right ) - 3 \, x^{2} e^{2}\right )} \log \left (3 \, x e - 2\right ) + 2\right )} \log \left (5\right )}{2 \, x^{2}} \]

[In]

integrate(((60*x*exp(1)-40)*log(5)*log(1/3*(-3*x*exp(1)+2)/x)+(-120*x*exp(1)+60)*log(5))/(3*x^4*exp(1)-2*x^3),
x, algorithm="maxima")

[Out]

15*(3*e*log(3*x*e - 2) - 3*e*log(x) + 2/x)*e*log(5)*log(2/3/x - e) - 30*(3*e*log(3*x*e - 2) - 3*e*log(x) + 2/x
)*e*log(5) - 5*(9*e^2*log(3*x*e - 2) - 9*e^2*log(x) + 2*(3*x*e + 1)/x^2)*log(5)*log(2/3/x - e) + 15/2*(9*e^2*l
og(3*x*e - 2) - 9*e^2*log(x) + 2*(3*x*e + 1)/x^2)*log(5) - 15/2*(3*x*e*log(3*x*e - 2)^2 + 3*x*e*log(x)^2 - 6*x
*e*log(x) - 6*(x*e*log(x) - x*e)*log(3*x*e - 2) + 4)*e*log(5)/x + 5/2*(9*x^2*e^2*log(3*x*e - 2)^2 + 9*x^2*e^2*
log(x)^2 - 27*x^2*e^2*log(x) + 18*x*e - 9*(2*x^2*e^2*log(x) - 3*x^2*e^2)*log(3*x*e - 2) + 2)*log(5)/x^2

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.52 \[ \int \frac {(60-120 e x) \log (5)+(-40+60 e x) \log (5) \log \left (\frac {2-3 e x}{3 x}\right )}{-2 x^3+3 e x^4} \, dx=\frac {10 \, {\left (\log \left (5\right ) \log \left (3\right ) - \log \left (5\right ) \log \left (-3 \, x e + 2\right ) + \log \left (5\right ) \log \left (x\right ) + 2 \, \log \left (5\right )\right )}}{x^{2}} \]

[In]

integrate(((60*x*exp(1)-40)*log(5)*log(1/3*(-3*x*exp(1)+2)/x)+(-120*x*exp(1)+60)*log(5))/(3*x^4*exp(1)-2*x^3),
x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 11.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {(60-120 e x) \log (5)+(-40+60 e x) \log (5) \log \left (\frac {2-3 e x}{3 x}\right )}{-2 x^3+3 e x^4} \, dx=-\frac {5\,\ln \left (5\right )\,\left (2\,\ln \left (-\frac {x\,\mathrm {e}-\frac {2}{3}}{x}\right )-4\right )}{x^2} \]

[In]

int(-(log(5)*(120*x*exp(1) - 60) - log(-(x*exp(1) - 2/3)/x)*log(5)*(60*x*exp(1) - 40))/(3*x^4*exp(1) - 2*x^3),
x)

[Out]

-(5*log(5)*(2*log(-(x*exp(1) - 2/3)/x) - 4))/x^2