3.11.76 \(\int \frac {4+11 x+5 x^2+x^3+(20+55 x+25 x^2+5 x^3) \log (3)+(-4-7 x-2 x^2+(-20-35 x-10 x^2) \log (3)) \log (x)+(1+x+(5+5 x) \log (3)) \log ^2(x)}{12 x^2+10 x^3+2 x^4+(8 x-2 x^2-2 x^3) \log (x)+(-8 x-2 x^2) \log ^2(x)+2 x \log ^3(x)} \, dx\)

Optimal. Leaf size=30 \[ \frac {(x+5 x \log (3)) \log \left (x+\frac {x}{2+x-\log (x)}+\log (x)\right )}{2 x} \]

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Rubi [A]  time = 0.62, antiderivative size = 43, normalized size of antiderivative = 1.43, number of steps used = 6, number of rules used = 4, integrand size = 130, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {6688, 12, 6742, 6684} \begin {gather*} \frac {1}{2} (1+\log (243)) \log \left (x^2+3 x-\log ^2(x)+2 \log (x)\right )-\frac {1}{2} (1+\log (243)) \log (x-\log (x)+2) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4 + 11*x + 5*x^2 + x^3 + (20 + 55*x + 25*x^2 + 5*x^3)*Log[3] + (-4 - 7*x - 2*x^2 + (-20 - 35*x - 10*x^2)*
Log[3])*Log[x] + (1 + x + (5 + 5*x)*Log[3])*Log[x]^2)/(12*x^2 + 10*x^3 + 2*x^4 + (8*x - 2*x^2 - 2*x^3)*Log[x]
+ (-8*x - 2*x^2)*Log[x]^2 + 2*x*Log[x]^3),x]

[Out]

-1/2*((1 + Log[243])*Log[2 + x - Log[x]]) + ((1 + Log[243])*Log[3*x + x^2 + 2*Log[x] - Log[x]^2])/2

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 6688

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

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 {(1+\log (243)) \left (4+11 x+5 x^2+x^3-\left (4+7 x+2 x^2\right ) \log (x)+(1+x) \log ^2(x)\right )}{2 x \left (x \left (6+5 x+x^2\right )-\left (-4+x+x^2\right ) \log (x)-(4+x) \log ^2(x)+\log ^3(x)\right )} \, dx\\ &=\frac {1}{2} (1+\log (243)) \int \frac {4+11 x+5 x^2+x^3-\left (4+7 x+2 x^2\right ) \log (x)+(1+x) \log ^2(x)}{x \left (x \left (6+5 x+x^2\right )-\left (-4+x+x^2\right ) \log (x)-(4+x) \log ^2(x)+\log ^3(x)\right )} \, dx\\ &=\frac {1}{2} (1+\log (243)) \int \left (\frac {1-x}{x (2+x-\log (x))}+\frac {2+3 x+2 x^2-2 \log (x)}{x \left (3 x+x^2+2 \log (x)-\log ^2(x)\right )}\right ) \, dx\\ &=\frac {1}{2} (1+\log (243)) \int \frac {1-x}{x (2+x-\log (x))} \, dx+\frac {1}{2} (1+\log (243)) \int \frac {2+3 x+2 x^2-2 \log (x)}{x \left (3 x+x^2+2 \log (x)-\log ^2(x)\right )} \, dx\\ &=-\frac {1}{2} (1+\log (243)) \log (2+x-\log (x))+\frac {1}{2} (1+\log (243)) \log \left (3 x+x^2+2 \log (x)-\log ^2(x)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.35, size = 37, normalized size = 1.23 \begin {gather*} \frac {1}{2} (1+\log (243)) \left (-\log (2+x-\log (x))+\log \left (3 x+x^2+2 \log (x)-\log ^2(x)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 + 11*x + 5*x^2 + x^3 + (20 + 55*x + 25*x^2 + 5*x^3)*Log[3] + (-4 - 7*x - 2*x^2 + (-20 - 35*x - 10
*x^2)*Log[3])*Log[x] + (1 + x + (5 + 5*x)*Log[3])*Log[x]^2)/(12*x^2 + 10*x^3 + 2*x^4 + (8*x - 2*x^2 - 2*x^3)*L
og[x] + (-8*x - 2*x^2)*Log[x]^2 + 2*x*Log[x]^3),x]

[Out]

((1 + Log[243])*(-Log[2 + x - Log[x]] + Log[3*x + x^2 + 2*Log[x] - Log[x]^2]))/2

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fricas [A]  time = 0.64, size = 43, normalized size = 1.43 \begin {gather*} \frac {1}{2} \, {\left (5 \, \log \relax (3) + 1\right )} \log \left (-x^{2} + \log \relax (x)^{2} - 3 \, x - 2 \, \log \relax (x)\right ) - \frac {1}{2} \, {\left (5 \, \log \relax (3) + 1\right )} \log \left (-x + \log \relax (x) - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*x+5)*log(3)+x+1)*log(x)^2+((-10*x^2-35*x-20)*log(3)-2*x^2-7*x-4)*log(x)+(5*x^3+25*x^2+55*x+20)*
log(3)+x^3+5*x^2+11*x+4)/(2*x*log(x)^3+(-2*x^2-8*x)*log(x)^2+(-2*x^3-2*x^2+8*x)*log(x)+2*x^4+10*x^3+12*x^2),x,
algorithm="fricas")

[Out]

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

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giac [A]  time = 0.71, size = 43, normalized size = 1.43 \begin {gather*} \frac {1}{2} \, {\left (5 \, \log \relax (3) + 1\right )} \log \left (x^{2} - \log \relax (x)^{2} + 3 \, x + 2 \, \log \relax (x)\right ) - \frac {1}{2} \, {\left (5 \, \log \relax (3) + 1\right )} \log \left (x - \log \relax (x) + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*x+5)*log(3)+x+1)*log(x)^2+((-10*x^2-35*x-20)*log(3)-2*x^2-7*x-4)*log(x)+(5*x^3+25*x^2+55*x+20)*
log(3)+x^3+5*x^2+11*x+4)/(2*x*log(x)^3+(-2*x^2-8*x)*log(x)^2+(-2*x^3-2*x^2+8*x)*log(x)+2*x^4+10*x^3+12*x^2),x,
algorithm="giac")

[Out]

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

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maple [A]  time = 0.27, size = 42, normalized size = 1.40

method	result	size
norman	\(\left (-\frac {5 \ln \relax (3)}{2}-\frac {1}{2}\right ) \ln \left (2-\ln \relax (x )+x \right )+\left (\frac {5 \ln \relax (3)}{2}+\frac {1}{2}\right ) \ln \left (x^{2}-\ln \relax (x )^{2}+3 x +2 \ln \relax (x )\right )\)	\(42\)
default	\(-\frac {\ln \left (2-\ln \relax (x )+x \right )}{2}+\frac {\ln \left (x^{2}-\ln \relax (x )^{2}+3 x +2 \ln \relax (x )\right )}{2}-\frac {5 \ln \relax (3) \ln \left (2-\ln \relax (x )+x \right )}{2}+\frac {5 \ln \relax (3) \ln \left (x^{2}-\ln \relax (x )^{2}+3 x +2 \ln \relax (x )\right )}{2}\)	\(66\)
risch	\(-\frac {\ln \left (\ln \relax (x )-x -2\right )}{2}-\frac {5 \ln \left (\ln \relax (x )-x -2\right ) \ln \relax (3)}{2}+\frac {5 \ln \left (-x^{2}+\ln \relax (x )^{2}-3 x -2 \ln \relax (x )\right ) \ln \relax (3)}{2}+\frac {\ln \left (-x^{2}+\ln \relax (x )^{2}-3 x -2 \ln \relax (x )\right )}{2}\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((5*x+5)*ln(3)+x+1)*ln(x)^2+((-10*x^2-35*x-20)*ln(3)-2*x^2-7*x-4)*ln(x)+(5*x^3+25*x^2+55*x+20)*ln(3)+x^3+
5*x^2+11*x+4)/(2*x*ln(x)^3+(-2*x^2-8*x)*ln(x)^2+(-2*x^3-2*x^2+8*x)*ln(x)+2*x^4+10*x^3+12*x^2),x,method=_RETURN
VERBOSE)

[Out]

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

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maxima [A]  time = 0.56, size = 43, normalized size = 1.43 \begin {gather*} \frac {1}{2} \, {\left (5 \, \log \relax (3) + 1\right )} \log \left (-x^{2} + \log \relax (x)^{2} - 3 \, x - 2 \, \log \relax (x)\right ) - \frac {1}{2} \, {\left (5 \, \log \relax (3) + 1\right )} \log \left (-x + \log \relax (x) - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*x+5)*log(3)+x+1)*log(x)^2+((-10*x^2-35*x-20)*log(3)-2*x^2-7*x-4)*log(x)+(5*x^3+25*x^2+55*x+20)*
log(3)+x^3+5*x^2+11*x+4)/(2*x*log(x)^3+(-2*x^2-8*x)*log(x)^2+(-2*x^3-2*x^2+8*x)*log(x)+2*x^4+10*x^3+12*x^2),x,
algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.04, size = 36, normalized size = 1.20 \begin {gather*} \left (\frac {\ln \left (243\right )}{2}+\frac {1}{2}\right )\,\left (\ln \left (x^2+3\,x-{\ln \relax (x)}^2+2\,\ln \relax (x)\right )-\ln \left (x-\ln \relax (x)+2\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((11*x + log(3)*(55*x + 25*x^2 + 5*x^3 + 20) + log(x)^2*(x + log(3)*(5*x + 5) + 1) + 5*x^2 + x^3 - log(x)*(
7*x + log(3)*(35*x + 10*x^2 + 20) + 2*x^2 + 4) + 4)/(2*x*log(x)^3 - log(x)^2*(8*x + 2*x^2) + 12*x^2 + 10*x^3 +
2*x^4 - log(x)*(2*x^2 - 8*x + 2*x^3)),x)

[Out]

(log(243)/2 + 1/2)*(log(3*x + 2*log(x) - log(x)^2 + x^2) - log(x - log(x) + 2))

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sympy [A]  time = 0.64, size = 42, normalized size = 1.40 \begin {gather*} - \frac {\left (1 + 5 \log {\relax (3 )}\right ) \log {\left (- x + \log {\relax (x )} - 2 \right )}}{2} + \frac {\left (1 + 5 \log {\relax (3 )}\right ) \log {\left (- x^{2} - 3 x + \log {\relax (x )}^{2} - 2 \log {\relax (x )} \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*x+5)*ln(3)+x+1)*ln(x)**2+((-10*x**2-35*x-20)*ln(3)-2*x**2-7*x-4)*ln(x)+(5*x**3+25*x**2+55*x+20)
*ln(3)+x**3+5*x**2+11*x+4)/(2*x*ln(x)**3+(-2*x**2-8*x)*ln(x)**2+(-2*x**3-2*x**2+8*x)*ln(x)+2*x**4+10*x**3+12*x
**2),x)

[Out]

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

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