3.12.30 \(\int \frac {(-40+50 x+15 x^2+(4+x) \log (\frac {5}{2})) \log (4+x)+(-10 x+5 x^2+x \log (\frac {5}{2})) \log (\frac {1}{25} (100 x-100 x^2+25 x^3+(-20 x+10 x^2) \log (\frac {5}{2})+x \log ^2(\frac {5}{2})))}{-40 x+10 x^2+5 x^3+(4 x+x^2) \log (\frac {5}{2})} \, dx\) [1130]

Optimal. Leaf size=23 \[ \log (4+x) \log \left (x \left (2-x-\frac {1}{5} \log \left (\frac {5}{2}\right )\right )^2\right ) \]

[Out]

ln(x*(2-1/5*ln(5/2)-x)^2)*ln(4+x)

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Rubi [A]
time = 0.31, antiderivative size = 24, normalized size of antiderivative = 1.04, number of steps used = 15, number of rules used = 7, integrand size = 110, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.064, Rules used = {6820, 2465, 2439, 2438, 2441, 2440, 2580} \begin {gather*} \log (x+4) \log \left (\frac {1}{25} x \left (-5 x+10-\log \left (\frac {5}{2}\right )\right )^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-40 + 50*x + 15*x^2 + (4 + x)*Log[5/2])*Log[4 + x] + (-10*x + 5*x^2 + x*Log[5/2])*Log[(100*x - 100*x^2 +
25*x^3 + (-20*x + 10*x^2)*Log[5/2] + x*Log[5/2]^2)/25])/(-40*x + 10*x^2 + 5*x^3 + (4*x + x^2)*Log[5/2]),x]

[Out]

Log[4 + x]*Log[(x*(10 - 5*x - Log[5/2])^2)/25]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2439

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*d])*Log[x], x] + Dist[
b, Int[Log[1 + e*(x/d)]/x, x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[c*d, 0]

Rule 2440

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

Rule 2441

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

Rule 2465

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[
(a + b*Log[c*(d + e*x)^n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n}, x] && RationalFunct
ionQ[RFx, x] && IntegerQ[p]

Rule 2580

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]/((g_.) + (h_.)*(x_)), x_Sym
bol] :> Simp[Log[g + h*x]*(Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]/h), x] + (-Dist[b*p*(r/h), Int[Log[g + h*x]/(a
+ b*x), x], x] - Dist[d*q*(r/h), Int[Log[g + h*x]/(c + d*x), x], x]) /; FreeQ[{a, b, c, d, e, f, g, h, p, q,
r}, x] && NeQ[b*c - a*d, 0]

Rule 6820

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {\left (-10+15 x+\log \left (\frac {5}{2}\right )\right ) \log (4+x)}{x \left (-10+5 x+\log \left (\frac {5}{2}\right )\right )}+\frac {\log \left (\frac {1}{25} x \left (-10+5 x+\log \left (\frac {5}{2}\right )\right )^2\right )}{4+x}\right ) \, dx\\ &=\int \frac {\left (-10+15 x+\log \left (\frac {5}{2}\right )\right ) \log (4+x)}{x \left (-10+5 x+\log \left (\frac {5}{2}\right )\right )} \, dx+\int \frac {\log \left (\frac {1}{25} x \left (-10+5 x+\log \left (\frac {5}{2}\right )\right )^2\right )}{4+x} \, dx\\ &=\log (4+x) \log \left (\frac {1}{25} x \left (10-5 x-\log \left (\frac {5}{2}\right )\right )^2\right )-10 \int \frac {\log (4+x)}{-10+5 x+\log \left (\frac {5}{2}\right )} \, dx-\int \frac {\log (4+x)}{x} \, dx+\int \left (\frac {\log (4+x)}{x}+\frac {10 \log (4+x)}{-10+5 x+\log \left (\frac {5}{2}\right )}\right ) \, dx\\ &=-\log (4) \log (x)-2 \log (4+x) \log \left (\frac {10-5 x-\log \left (\frac {5}{2}\right )}{30-\log \left (\frac {5}{2}\right )}\right )+\log (4+x) \log \left (\frac {1}{25} x \left (10-5 x-\log \left (\frac {5}{2}\right )\right )^2\right )+2 \int \frac {\log \left (\frac {-10+5 x+\log \left (\frac {5}{2}\right )}{-30+\log \left (\frac {5}{2}\right )}\right )}{4+x} \, dx+10 \int \frac {\log (4+x)}{-10+5 x+\log \left (\frac {5}{2}\right )} \, dx-\int \frac {\log \left (1+\frac {x}{4}\right )}{x} \, dx+\int \frac {\log (4+x)}{x} \, dx\\ &=\log (4+x) \log \left (\frac {1}{25} x \left (10-5 x-\log \left (\frac {5}{2}\right )\right )^2\right )+\text {Li}_2\left (-\frac {x}{4}\right )-2 \int \frac {\log \left (\frac {-10+5 x+\log \left (\frac {5}{2}\right )}{-30+\log \left (\frac {5}{2}\right )}\right )}{4+x} \, dx+2 \text {Subst}\left (\int \frac {\log \left (1+\frac {5 x}{-30+\log \left (\frac {5}{2}\right )}\right )}{x} \, dx,x,4+x\right )+\int \frac {\log \left (1+\frac {x}{4}\right )}{x} \, dx\\ &=\log (4+x) \log \left (\frac {1}{25} x \left (10-5 x-\log \left (\frac {5}{2}\right )\right )^2\right )-2 \text {Li}_2\left (\frac {5 (4+x)}{30-\log \left (\frac {5}{2}\right )}\right )-2 \text {Subst}\left (\int \frac {\log \left (1+\frac {5 x}{-30+\log \left (\frac {5}{2}\right )}\right )}{x} \, dx,x,4+x\right )\\ &=\log (4+x) \log \left (\frac {1}{25} x \left (10-5 x-\log \left (\frac {5}{2}\right )\right )^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.04, size = 22, normalized size = 0.96 \begin {gather*} \log (4+x) \log \left (\frac {1}{25} x \left (-10+5 x+\log \left (\frac {5}{2}\right )\right )^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-40 + 50*x + 15*x^2 + (4 + x)*Log[5/2])*Log[4 + x] + (-10*x + 5*x^2 + x*Log[5/2])*Log[(100*x - 100
*x^2 + 25*x^3 + (-20*x + 10*x^2)*Log[5/2] + x*Log[5/2]^2)/25])/(-40*x + 10*x^2 + 5*x^3 + (4*x + x^2)*Log[5/2])
,x]

[Out]

Log[4 + x]*Log[(x*(-10 + 5*x + Log[5/2])^2)/25]

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(47\) vs. \(2(19)=38\).
time = 0.20, size = 48, normalized size = 2.09

method	result	size
default	\(\ln \left (4+x \right ) \ln \left (x \ln \left (\frac {5}{2}\right )^{2}+10 x^{2} \ln \left (\frac {5}{2}\right )-20 x \ln \left (\frac {5}{2}\right )+25 x^{3}-100 x^{2}+100 x \right )-2 \ln \left (5\right ) \ln \left (4+x \right )\)	\(48\)
risch	\(2 \ln \left (4+x \right ) \ln \left (\ln \left (5\right )-\ln \left (2\right )+5 x -10\right )+\ln \left (x \right ) \ln \left (4+x \right )-\frac {i \pi \ln \left (4+x \right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (\ln \left (\frac {5}{2}\right )+5 x -10\right )^{2}\right ) \mathrm {csgn}\left (i x \left (\ln \left (\frac {5}{2}\right )+5 x -10\right )^{2}\right )}{2}+\frac {i \pi \ln \left (4+x \right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left (\ln \left (\frac {5}{2}\right )+5 x -10\right )^{2}\right )^{2}}{2}-\frac {i \pi \ln \left (4+x \right ) \mathrm {csgn}\left (i \left (\ln \left (\frac {5}{2}\right )+5 x -10\right )\right )^{2} \mathrm {csgn}\left (i \left (\ln \left (\frac {5}{2}\right )+5 x -10\right )^{2}\right )}{2}+i \pi \ln \left (4+x \right ) \mathrm {csgn}\left (i \left (\ln \left (\frac {5}{2}\right )+5 x -10\right )\right ) \mathrm {csgn}\left (i \left (\ln \left (\frac {5}{2}\right )+5 x -10\right )^{2}\right )^{2}-\frac {i \pi \ln \left (4+x \right ) \mathrm {csgn}\left (i \left (\ln \left (\frac {5}{2}\right )+5 x -10\right )^{2}\right )^{3}}{2}+\frac {i \pi \ln \left (4+x \right ) \mathrm {csgn}\left (i \left (\ln \left (\frac {5}{2}\right )+5 x -10\right )^{2}\right ) \mathrm {csgn}\left (i x \left (\ln \left (\frac {5}{2}\right )+5 x -10\right )^{2}\right )^{2}}{2}-\frac {i \pi \ln \left (4+x \right ) \mathrm {csgn}\left (i x \left (\ln \left (\frac {5}{2}\right )+5 x -10\right )^{2}\right )^{3}}{2}-2 \ln \left (5\right ) \ln \left (4+x \right )\)	\(256\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*ln(5/2)+5*x^2-10*x)*ln(1/25*x*ln(5/2)^2+1/25*(10*x^2-20*x)*ln(5/2)+x^3-4*x^2+4*x)+((4+x)*ln(5/2)+15*x^
2+50*x-40)*ln(4+x))/((x^2+4*x)*ln(5/2)+5*x^3+10*x^2-40*x),x,method=_RETURNVERBOSE)

[Out]

ln(4+x)*ln(x*ln(5/2)^2+10*x^2*ln(5/2)-20*x*ln(5/2)+25*x^3-100*x^2+100*x)-2*ln(5)*ln(4+x)

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Maxima [A]
time = 0.53, size = 34, normalized size = 1.48 \begin {gather*} -{\left (2 \, \log \left (5\right ) - \log \left (x\right )\right )} \log \left (x + 4\right ) + 2 \, \log \left (5 \, x + \log \left (5\right ) - \log \left (2\right ) - 10\right ) \log \left (x + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(5/2)+5*x^2-10*x)*log(1/25*x*log(5/2)^2+1/25*(10*x^2-20*x)*log(5/2)+x^3-4*x^2+4*x)+((4+x)*log
(5/2)+15*x^2+50*x-40)*log(4+x))/((x^2+4*x)*log(5/2)+5*x^3+10*x^2-40*x),x, algorithm="maxima")

[Out]

-(2*log(5) - log(x))*log(x + 4) + 2*log(5*x + log(5) - log(2) - 10)*log(x + 4)

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Fricas [A]
time = 0.33, size = 36, normalized size = 1.57 \begin {gather*} \log \left (x^{3} + \frac {1}{25} \, x \log \left (\frac {5}{2}\right )^{2} - 4 \, x^{2} + \frac {2}{5} \, {\left (x^{2} - 2 \, x\right )} \log \left (\frac {5}{2}\right ) + 4 \, x\right ) \log \left (x + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(5/2)+5*x^2-10*x)*log(1/25*x*log(5/2)^2+1/25*(10*x^2-20*x)*log(5/2)+x^3-4*x^2+4*x)+((4+x)*log
(5/2)+15*x^2+50*x-40)*log(4+x))/((x^2+4*x)*log(5/2)+5*x^3+10*x^2-40*x),x, algorithm="fricas")

[Out]

log(x^3 + 1/25*x*log(5/2)^2 - 4*x^2 + 2/5*(x^2 - 2*x)*log(5/2) + 4*x)*log(x + 4)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (19) = 38\).
time = 0.24, size = 44, normalized size = 1.91 \begin {gather*} \log {\left (x + 4 \right )} \log {\left (x^{3} - 4 x^{2} + \frac {x \log {\left (\frac {5}{2} \right )}^{2}}{25} + 4 x + \left (\frac {2 x^{2}}{5} - \frac {4 x}{5}\right ) \log {\left (\frac {5}{2} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*ln(5/2)+5*x**2-10*x)*ln(1/25*x*ln(5/2)**2+1/25*(10*x**2-20*x)*ln(5/2)+x**3-4*x**2+4*x)+((4+x)*ln
(5/2)+15*x**2+50*x-40)*ln(4+x))/((x**2+4*x)*ln(5/2)+5*x**3+10*x**2-40*x),x)

[Out]

log(x + 4)*log(x**3 - 4*x**2 + x*log(5/2)**2/25 + 4*x + (2*x**2/5 - 4*x/5)*log(5/2))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (18) = 36\).
time = 0.45, size = 47, normalized size = 2.04 \begin {gather*} -2 \, \log \left (5\right ) \log \left (x + 4\right ) + \log \left (25 \, x^{3} + 10 \, x^{2} \log \left (\frac {5}{2}\right ) + x \log \left (\frac {5}{2}\right )^{2} - 100 \, x^{2} - 20 \, x \log \left (\frac {5}{2}\right ) + 100 \, x\right ) \log \left (x + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(5/2)+5*x^2-10*x)*log(1/25*x*log(5/2)^2+1/25*(10*x^2-20*x)*log(5/2)+x^3-4*x^2+4*x)+((4+x)*log
(5/2)+15*x^2+50*x-40)*log(4+x))/((x^2+4*x)*log(5/2)+5*x^3+10*x^2-40*x),x, algorithm="giac")

[Out]

-2*log(5)*log(x + 4) + log(25*x^3 + 10*x^2*log(5/2) + x*log(5/2)^2 - 100*x^2 - 20*x*log(5/2) + 100*x)*log(x +
4)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {\ln \left (4\,x-\frac {\ln \left (\frac {5}{2}\right )\,\left (20\,x-10\,x^2\right )}{25}+\frac {x\,{\ln \left (\frac {5}{2}\right )}^2}{25}-4\,x^2+x^3\right )\,\left (x\,\ln \left (\frac {5}{2}\right )-10\,x+5\,x^2\right )+\ln \left (x+4\right )\,\left (50\,x+\ln \left (\frac {5}{2}\right )\,\left (x+4\right )+15\,x^2-40\right )}{10\,x^2-40\,x+5\,x^3+\ln \left (\frac {5}{2}\right )\,\left (x^2+4\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(4*x - (log(5/2)*(20*x - 10*x^2))/25 + (x*log(5/2)^2)/25 - 4*x^2 + x^3)*(x*log(5/2) - 10*x + 5*x^2) +
log(x + 4)*(50*x + log(5/2)*(x + 4) + 15*x^2 - 40))/(10*x^2 - 40*x + 5*x^3 + log(5/2)*(4*x + x^2)),x)

[Out]

int((log(4*x - (log(5/2)*(20*x - 10*x^2))/25 + (x*log(5/2)^2)/25 - 4*x^2 + x^3)*(x*log(5/2) - 10*x + 5*x^2) +
log(x + 4)*(50*x + log(5/2)*(x + 4) + 15*x^2 - 40))/(10*x^2 - 40*x + 5*x^3 + log(5/2)*(4*x + x^2)), x)

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