3.6.40 \(\int \frac {-1-3 x-x^2 \log (3)+(-18-3 x+(-12 x-2 x^2) \log (3)) \log (6+x) \log (\log (6+x))}{(30+5 x) \log (6+x)} \, dx\)

Optimal. Leaf size=23 \[ \frac {1}{5} (-1-x-x (2+x \log (3))) \log (\log (6+x)) \]

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Rubi [F]  time = 0.39, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-1-3 x-x^2 \log (3)+\left (-18-3 x+\left (-12 x-2 x^2\right ) \log (3)\right ) \log (6+x) \log (\log (6+x))}{(30+5 x) \log (6+x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-1 - 3*x - x^2*Log[3] + (-18 - 3*x + (-12*x - 2*x^2)*Log[3])*Log[6 + x]*Log[Log[6 + x]])/((30 + 5*x)*Log[
6 + x]),x]

[Out]

-1/5*(ExpIntegralEi[2*Log[6 + x]]*Log[3]) - (3*(6 + x)*Log[Log[6 + x]])/5 + ((17 - 36*Log[3])*Log[Log[6 + x]])
/5 + (3*LogIntegral[6 + x])/5 + (6*Log[3]*LogIntegral[6 + x])/5 - (3*(1 - Log[9])*LogIntegral[6 + x])/5 - (Log
[9]*Defer[Int][x*Log[Log[6 + x]], x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {-1-3 x-x^2 \log (3)}{5 (6+x) \log (6+x)}-\frac {1}{5} (3+x \log (9)) \log (\log (6+x))\right ) \, dx\\ &=\frac {1}{5} \int \frac {-1-3 x-x^2 \log (3)}{(6+x) \log (6+x)} \, dx-\frac {1}{5} \int (3+x \log (9)) \log (\log (6+x)) \, dx\\ &=\frac {1}{5} \int \left (\frac {17-36 \log (3)}{(6+x) \log (6+x)}-\frac {x \log (3)}{\log (6+x)}-\frac {3 (1-\log (9))}{\log (6+x)}\right ) \, dx-\frac {1}{5} \int (3 \log (\log (6+x))+x \log (9) \log (\log (6+x))) \, dx\\ &=-\left (\frac {3}{5} \int \log (\log (6+x)) \, dx\right )+\frac {1}{5} (17-36 \log (3)) \int \frac {1}{(6+x) \log (6+x)} \, dx-\frac {1}{5} \log (3) \int \frac {x}{\log (6+x)} \, dx-\frac {1}{5} (3 (1-\log (9))) \int \frac {1}{\log (6+x)} \, dx-\frac {1}{5} \log (9) \int x \log (\log (6+x)) \, dx\\ &=-\left (\frac {3}{5} \operatorname {Subst}(\int \log (\log (x)) \, dx,x,6+x)\right )+\frac {1}{5} (17-36 \log (3)) \operatorname {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,6+x\right )-\frac {1}{5} \log (3) \int \left (-\frac {6}{\log (6+x)}+\frac {6+x}{\log (6+x)}\right ) \, dx-\frac {1}{5} (3 (1-\log (9))) \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,6+x\right )-\frac {1}{5} \log (9) \int x \log (\log (6+x)) \, dx\\ &=-\frac {3}{5} (6+x) \log (\log (6+x))-\frac {3}{5} (1-\log (9)) \text {li}(6+x)+\frac {3}{5} \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,6+x\right )+\frac {1}{5} (17-36 \log (3)) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (6+x)\right )-\frac {1}{5} \log (3) \int \frac {6+x}{\log (6+x)} \, dx+\frac {1}{5} (6 \log (3)) \int \frac {1}{\log (6+x)} \, dx-\frac {1}{5} \log (9) \int x \log (\log (6+x)) \, dx\\ &=-\frac {3}{5} (6+x) \log (\log (6+x))+\frac {1}{5} (17-36 \log (3)) \log (\log (6+x))+\frac {3 \text {li}(6+x)}{5}-\frac {3}{5} (1-\log (9)) \text {li}(6+x)-\frac {1}{5} \log (3) \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,6+x\right )+\frac {1}{5} (6 \log (3)) \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,6+x\right )-\frac {1}{5} \log (9) \int x \log (\log (6+x)) \, dx\\ &=-\frac {3}{5} (6+x) \log (\log (6+x))+\frac {1}{5} (17-36 \log (3)) \log (\log (6+x))+\frac {3 \text {li}(6+x)}{5}+\frac {6}{5} \log (3) \text {li}(6+x)-\frac {3}{5} (1-\log (9)) \text {li}(6+x)-\frac {1}{5} \log (3) \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (6+x)\right )-\frac {1}{5} \log (9) \int x \log (\log (6+x)) \, dx\\ &=-\frac {1}{5} \text {Ei}(2 \log (6+x)) \log (3)-\frac {3}{5} (6+x) \log (\log (6+x))+\frac {1}{5} (17-36 \log (3)) \log (\log (6+x))+\frac {3 \text {li}(6+x)}{5}+\frac {6}{5} \log (3) \text {li}(6+x)-\frac {3}{5} (1-\log (9)) \text {li}(6+x)-\frac {1}{5} \log (9) \int x \log (\log (6+x)) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 20, normalized size = 0.87 \begin {gather*} -\frac {1}{5} \left (1+3 x+x^2 \log (3)\right ) \log (\log (6+x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 - 3*x - x^2*Log[3] + (-18 - 3*x + (-12*x - 2*x^2)*Log[3])*Log[6 + x]*Log[Log[6 + x]])/((30 + 5*x
)*Log[6 + x]),x]

[Out]

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

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fricas [A]  time = 0.64, size = 18, normalized size = 0.78 \begin {gather*} -\frac {1}{5} \, {\left (x^{2} \log \relax (3) + 3 \, x + 1\right )} \log \left (\log \left (x + 6\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2-12*x)*log(3)-3*x-18)*log(x+6)*log(log(x+6))-x^2*log(3)-3*x-1)/(5*x+30)/log(x+6),x, algorit
hm="fricas")

[Out]

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

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giac [A]  time = 0.33, size = 25, normalized size = 1.09 \begin {gather*} -\frac {1}{5} \, {\left (x^{2} \log \relax (3) + 3 \, x\right )} \log \left (\log \left (x + 6\right )\right ) - \frac {1}{5} \, \log \left (\log \left (x + 6\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2-12*x)*log(3)-3*x-18)*log(x+6)*log(log(x+6))-x^2*log(3)-3*x-1)/(5*x+30)/log(x+6),x, algorit
hm="giac")

[Out]

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

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maple [A]  time = 0.29, size = 26, normalized size = 1.13

method	result	size
risch	\(\left (-\frac {x^{2} \ln \relax (3)}{5}-\frac {3 x}{5}\right ) \ln \left (\ln \left (x +6\right )\right )-\frac {\ln \left (\ln \left (x +6\right )\right )}{5}\)	\(26\)
default	\(-\frac {x^{2} \ln \relax (3) \ln \left (\ln \left (x +6\right )\right )}{5}-\frac {\ln \left (\ln \left (x +6\right )\right )}{5}-\frac {3 x \ln \left (\ln \left (x +6\right )\right )}{5}\)	\(29\)
norman	\(-\frac {x^{2} \ln \relax (3) \ln \left (\ln \left (x +6\right )\right )}{5}-\frac {\ln \left (\ln \left (x +6\right )\right )}{5}-\frac {3 x \ln \left (\ln \left (x +6\right )\right )}{5}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-2*x^2-12*x)*ln(3)-3*x-18)*ln(x+6)*ln(ln(x+6))-x^2*ln(3)-3*x-1)/(5*x+30)/ln(x+6),x,method=_RETURNVERBOS
E)

[Out]

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

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maxima [A]  time = 0.51, size = 25, normalized size = 1.09 \begin {gather*} -\frac {1}{5} \, {\left (x^{2} \log \relax (3) + 3 \, x\right )} \log \left (\log \left (x + 6\right )\right ) - \frac {1}{5} \, \log \left (\log \left (x + 6\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2-12*x)*log(3)-3*x-18)*log(x+6)*log(log(x+6))-x^2*log(3)-3*x-1)/(5*x+30)/log(x+6),x, algorit
hm="maxima")

[Out]

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

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mupad [B]  time = 0.83, size = 18, normalized size = 0.78 \begin {gather*} -\frac {\ln \left (\ln \left (x+6\right )\right )\,\left (\ln \relax (3)\,x^2+3\,x+1\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*x + x^2*log(3) + log(x + 6)*log(log(x + 6))*(3*x + log(3)*(12*x + 2*x^2) + 18) + 1)/(log(x + 6)*(5*x +
30)),x)

[Out]

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

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sympy [A]  time = 0.67, size = 46, normalized size = 2.00 \begin {gather*} \left (- \frac {x^{2} \log {\relax (3 )}}{5} - \frac {3 x}{5} - \frac {9}{5} + \frac {12 \log {\relax (3 )}}{5}\right ) \log {\left (\log {\left (x + 6 \right )} \right )} - \frac {4 \left (-2 + 3 \log {\relax (3 )}\right ) \log {\left (\log {\left (x + 6 \right )} \right )}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x**2-12*x)*ln(3)-3*x-18)*ln(x+6)*ln(ln(x+6))-x**2*ln(3)-3*x-1)/(5*x+30)/ln(x+6),x)

[Out]

(-x**2*log(3)/5 - 3*x/5 - 9/5 + 12*log(3)/5)*log(log(x + 6)) - 4*(-2 + 3*log(3))*log(log(x + 6))/5

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