3.48.38 \(\int \frac {5-25 x-108 x^2-180 x^3+36 x^2 \log (x)+(15-11 x-60 x^2+(-5+12 x) \log (x)) \log (3+5 x-\log (x))+(-3-5 x+\log (x)) \log ^2(3+5 x-\log (x))}{360 x^2+168 x^3-720 x^4+(-120 x^2+144 x^3) \log (x)+(60 x-44 x^2-240 x^3+(-20 x+48 x^2) \log (x)) \log (3+5 x-\log (x))+(-12 x-20 x^2+4 x \log (x)) \log ^2(3+5 x-\log (x))} \, dx\)

Optimal. Leaf size=30 \[ \frac {1}{4} \log \left (x-\frac {x}{x+\frac {1}{5} (x+\log (3+5 x-\log (x)))}\right ) \]

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Rubi [A]  time = 1.37, antiderivative size = 48, normalized size of antiderivative = 1.60, number of steps used = 6, number of rules used = 4, integrand size = 168, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {6741, 12, 6742, 6684} \begin {gather*} \frac {\log (x)}{4}+\frac {1}{4} \log (-6 x-\log (5 x-\log (x)+3)+5)-\frac {1}{4} \log (6 x+\log (5 x-\log (x)+3)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 - 25*x - 108*x^2 - 180*x^3 + 36*x^2*Log[x] + (15 - 11*x - 60*x^2 + (-5 + 12*x)*Log[x])*Log[3 + 5*x - Lo
g[x]] + (-3 - 5*x + Log[x])*Log[3 + 5*x - Log[x]]^2)/(360*x^2 + 168*x^3 - 720*x^4 + (-120*x^2 + 144*x^3)*Log[x
] + (60*x - 44*x^2 - 240*x^3 + (-20*x + 48*x^2)*Log[x])*Log[3 + 5*x - Log[x]] + (-12*x - 20*x^2 + 4*x*Log[x])*
Log[3 + 5*x - Log[x]]^2),x]

[Out]

Log[x]/4 + Log[5 - 6*x - Log[3 + 5*x - Log[x]]]/4 - Log[6*x + Log[3 + 5*x - Log[x]]]/4

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5-25 x-108 x^2-180 x^3+36 x^2 \log (x)+\left (15-11 x-60 x^2+(-5+12 x) \log (x)\right ) \log (3+5 x-\log (x))+(-3-5 x+\log (x)) \log ^2(3+5 x-\log (x))}{4 x (3+5 x-\log (x)) \left (30 x-36 x^2+5 \log (3+5 x-\log (x))-12 x \log (3+5 x-\log (x))-\log ^2(3+5 x-\log (x))\right )} \, dx\\ &=\frac {1}{4} \int \frac {5-25 x-108 x^2-180 x^3+36 x^2 \log (x)+\left (15-11 x-60 x^2+(-5+12 x) \log (x)\right ) \log (3+5 x-\log (x))+(-3-5 x+\log (x)) \log ^2(3+5 x-\log (x))}{x (3+5 x-\log (x)) \left (30 x-36 x^2+5 \log (3+5 x-\log (x))-12 x \log (3+5 x-\log (x))-\log ^2(3+5 x-\log (x))\right )} \, dx\\ &=\frac {1}{4} \int \left (\frac {1}{x}+\frac {-1+23 x+30 x^2-6 x \log (x)}{x (3+5 x-\log (x)) (-5+6 x+\log (3+5 x-\log (x)))}+\frac {1-23 x-30 x^2+6 x \log (x)}{x (3+5 x-\log (x)) (6 x+\log (3+5 x-\log (x)))}\right ) \, dx\\ &=\frac {\log (x)}{4}+\frac {1}{4} \int \frac {-1+23 x+30 x^2-6 x \log (x)}{x (3+5 x-\log (x)) (-5+6 x+\log (3+5 x-\log (x)))} \, dx+\frac {1}{4} \int \frac {1-23 x-30 x^2+6 x \log (x)}{x (3+5 x-\log (x)) (6 x+\log (3+5 x-\log (x)))} \, dx\\ &=\frac {\log (x)}{4}+\frac {1}{4} \log (5-6 x-\log (3+5 x-\log (x)))-\frac {1}{4} \log (6 x+\log (3+5 x-\log (x)))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 42, normalized size = 1.40 \begin {gather*} \frac {1}{4} (\log (x)+\log (5-6 x-\log (3+5 x-\log (x)))-\log (6 x+\log (3+5 x-\log (x)))) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 - 25*x - 108*x^2 - 180*x^3 + 36*x^2*Log[x] + (15 - 11*x - 60*x^2 + (-5 + 12*x)*Log[x])*Log[3 + 5*
x - Log[x]] + (-3 - 5*x + Log[x])*Log[3 + 5*x - Log[x]]^2)/(360*x^2 + 168*x^3 - 720*x^4 + (-120*x^2 + 144*x^3)
*Log[x] + (60*x - 44*x^2 - 240*x^3 + (-20*x + 48*x^2)*Log[x])*Log[3 + 5*x - Log[x]] + (-12*x - 20*x^2 + 4*x*Lo
g[x])*Log[3 + 5*x - Log[x]]^2),x]

[Out]

(Log[x] + Log[5 - 6*x - Log[3 + 5*x - Log[x]]] - Log[6*x + Log[3 + 5*x - Log[x]]])/4

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fricas [A]  time = 0.77, size = 40, normalized size = 1.33 \begin {gather*} -\frac {1}{4} \, \log \left (6 \, x + \log \left (5 \, x - \log \relax (x) + 3\right )\right ) + \frac {1}{4} \, \log \left (6 \, x + \log \left (5 \, x - \log \relax (x) + 3\right ) - 5\right ) + \frac {1}{4} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((log(x)-5*x-3)*log(-log(x)+5*x+3)^2+((12*x-5)*log(x)-60*x^2-11*x+15)*log(-log(x)+5*x+3)+36*x^2*log(
x)-180*x^3-108*x^2-25*x+5)/((4*x*log(x)-20*x^2-12*x)*log(-log(x)+5*x+3)^2+((48*x^2-20*x)*log(x)-240*x^3-44*x^2
+60*x)*log(-log(x)+5*x+3)+(144*x^3-120*x^2)*log(x)-720*x^4+168*x^3+360*x^2),x, algorithm="fricas")

[Out]

-1/4*log(6*x + log(5*x - log(x) + 3)) + 1/4*log(6*x + log(5*x - log(x) + 3) - 5) + 1/4*log(x)

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giac [A]  time = 0.52, size = 40, normalized size = 1.33 \begin {gather*} -\frac {1}{4} \, \log \left (6 \, x + \log \left (5 \, x - \log \relax (x) + 3\right )\right ) + \frac {1}{4} \, \log \left (6 \, x + \log \left (5 \, x - \log \relax (x) + 3\right ) - 5\right ) + \frac {1}{4} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((log(x)-5*x-3)*log(-log(x)+5*x+3)^2+((12*x-5)*log(x)-60*x^2-11*x+15)*log(-log(x)+5*x+3)+36*x^2*log(
x)-180*x^3-108*x^2-25*x+5)/((4*x*log(x)-20*x^2-12*x)*log(-log(x)+5*x+3)^2+((48*x^2-20*x)*log(x)-240*x^3-44*x^2
+60*x)*log(-log(x)+5*x+3)+(144*x^3-120*x^2)*log(x)-720*x^4+168*x^3+360*x^2),x, algorithm="giac")

[Out]

-1/4*log(6*x + log(5*x - log(x) + 3)) + 1/4*log(6*x + log(5*x - log(x) + 3) - 5) + 1/4*log(x)

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maple [A]  time = 0.05, size = 41, normalized size = 1.37

method	result	size
risch	\(\frac {\ln \relax (x )}{4}+\frac {\ln \left (-5+6 x +\ln \left (-\ln \relax (x )+5 x +3\right )\right )}{4}-\frac {\ln \left (6 x +\ln \left (-\ln \relax (x )+5 x +3\right )\right )}{4}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((ln(x)-5*x-3)*ln(-ln(x)+5*x+3)^2+((12*x-5)*ln(x)-60*x^2-11*x+15)*ln(-ln(x)+5*x+3)+36*x^2*ln(x)-180*x^3-10
8*x^2-25*x+5)/((4*x*ln(x)-20*x^2-12*x)*ln(-ln(x)+5*x+3)^2+((48*x^2-20*x)*ln(x)-240*x^3-44*x^2+60*x)*ln(-ln(x)+
5*x+3)+(144*x^3-120*x^2)*ln(x)-720*x^4+168*x^3+360*x^2),x,method=_RETURNVERBOSE)

[Out]

1/4*ln(x)+1/4*ln(-5+6*x+ln(-ln(x)+5*x+3))-1/4*ln(6*x+ln(-ln(x)+5*x+3))

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maxima [A]  time = 0.41, size = 40, normalized size = 1.33 \begin {gather*} -\frac {1}{4} \, \log \left (6 \, x + \log \left (5 \, x - \log \relax (x) + 3\right )\right ) + \frac {1}{4} \, \log \left (6 \, x + \log \left (5 \, x - \log \relax (x) + 3\right ) - 5\right ) + \frac {1}{4} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((log(x)-5*x-3)*log(-log(x)+5*x+3)^2+((12*x-5)*log(x)-60*x^2-11*x+15)*log(-log(x)+5*x+3)+36*x^2*log(
x)-180*x^3-108*x^2-25*x+5)/((4*x*log(x)-20*x^2-12*x)*log(-log(x)+5*x+3)^2+((48*x^2-20*x)*log(x)-240*x^3-44*x^2
+60*x)*log(-log(x)+5*x+3)+(144*x^3-120*x^2)*log(x)-720*x^4+168*x^3+360*x^2),x, algorithm="maxima")

[Out]

-1/4*log(6*x + log(5*x - log(x) + 3)) + 1/4*log(6*x + log(5*x - log(x) + 3) - 5) + 1/4*log(x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {25\,x-36\,x^2\,\ln \relax (x)+{\ln \left (5\,x-\ln \relax (x)+3\right )}^2\,\left (5\,x-\ln \relax (x)+3\right )+\ln \left (5\,x-\ln \relax (x)+3\right )\,\left (11\,x-\ln \relax (x)\,\left (12\,x-5\right )+60\,x^2-15\right )+108\,x^2+180\,x^3-5}{\ln \relax (x)\,\left (120\,x^2-144\,x^3\right )+\ln \left (5\,x-\ln \relax (x)+3\right )\,\left (\ln \relax (x)\,\left (20\,x-48\,x^2\right )-60\,x+44\,x^2+240\,x^3\right )+{\ln \left (5\,x-\ln \relax (x)+3\right )}^2\,\left (12\,x-4\,x\,\ln \relax (x)+20\,x^2\right )-360\,x^2-168\,x^3+720\,x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((25*x - 36*x^2*log(x) + log(5*x - log(x) + 3)^2*(5*x - log(x) + 3) + log(5*x - log(x) + 3)*(11*x - log(x)*
(12*x - 5) + 60*x^2 - 15) + 108*x^2 + 180*x^3 - 5)/(log(x)*(120*x^2 - 144*x^3) + log(5*x - log(x) + 3)*(log(x)
*(20*x - 48*x^2) - 60*x + 44*x^2 + 240*x^3) + log(5*x - log(x) + 3)^2*(12*x - 4*x*log(x) + 20*x^2) - 360*x^2 -
168*x^3 + 720*x^4),x)

[Out]

int((25*x - 36*x^2*log(x) + log(5*x - log(x) + 3)^2*(5*x - log(x) + 3) + log(5*x - log(x) + 3)*(11*x - log(x)*
(12*x - 5) + 60*x^2 - 15) + 108*x^2 + 180*x^3 - 5)/(log(x)*(120*x^2 - 144*x^3) + log(5*x - log(x) + 3)*(log(x)
*(20*x - 48*x^2) - 60*x + 44*x^2 + 240*x^3) + log(5*x - log(x) + 3)^2*(12*x - 4*x*log(x) + 20*x^2) - 360*x^2 -
168*x^3 + 720*x^4), x)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: PolynomialError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((ln(x)-5*x-3)*ln(-ln(x)+5*x+3)**2+((12*x-5)*ln(x)-60*x**2-11*x+15)*ln(-ln(x)+5*x+3)+36*x**2*ln(x)-1
80*x**3-108*x**2-25*x+5)/((4*x*ln(x)-20*x**2-12*x)*ln(-ln(x)+5*x+3)**2+((48*x**2-20*x)*ln(x)-240*x**3-44*x**2+
60*x)*ln(-ln(x)+5*x+3)+(144*x**3-120*x**2)*ln(x)-720*x**4+168*x**3+360*x**2),x)

[Out]

Exception raised: PolynomialError

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