3.41.13 \(\int \frac {-x^2+(-40 x-2 x^2) \log (20+x) \log (\log (20+x))}{(240+12 x) \log (20+x)} \, dx\)

Optimal. Leaf size=12 \[ -\frac {1}{12} x^2 \log (\log (20+x)) \]

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Rubi [F]  time = 0.24, 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 {-x^2+\left (-40 x-2 x^2\right ) \log (20+x) \log (\log (20+x))}{(240+12 x) \log (20+x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-x^2 + (-40*x - 2*x^2)*Log[20 + x]*Log[Log[20 + x]])/((240 + 12*x)*Log[20 + x]),x]

[Out]

-1/12*ExpIntegralEi[2*Log[20 + x]] - (100*Log[Log[20 + x]])/3 + (10*LogIntegral[20 + x])/3 - Defer[Int][x*Log[
Log[20 + x]], x]/6

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1}{12} x \left (-\frac {x}{(20+x) \log (20+x)}-2 \log (\log (20+x))\right ) \, dx\\ &=\frac {1}{12} \int x \left (-\frac {x}{(20+x) \log (20+x)}-2 \log (\log (20+x))\right ) \, dx\\ &=\frac {1}{12} \int \left (-\frac {x^2}{(20+x) \log (20+x)}-2 x \log (\log (20+x))\right ) \, dx\\ &=-\left (\frac {1}{12} \int \frac {x^2}{(20+x) \log (20+x)} \, dx\right )-\frac {1}{6} \int x \log (\log (20+x)) \, dx\\ &=-\left (\frac {1}{12} \operatorname {Subst}\left (\int \frac {(-20+x)^2}{x \log (x)} \, dx,x,20+x\right )\right )-\frac {1}{6} \int x \log (\log (20+x)) \, dx\\ &=-\left (\frac {1}{12} \operatorname {Subst}\left (\int \left (-\frac {40}{\log (x)}+\frac {400}{x \log (x)}+\frac {x}{\log (x)}\right ) \, dx,x,20+x\right )\right )-\frac {1}{6} \int x \log (\log (20+x)) \, dx\\ &=-\left (\frac {1}{12} \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,20+x\right )\right )-\frac {1}{6} \int x \log (\log (20+x)) \, dx+\frac {10}{3} \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,20+x\right )-\frac {100}{3} \operatorname {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,20+x\right )\\ &=\frac {10 \text {li}(20+x)}{3}-\frac {1}{12} \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (20+x)\right )-\frac {1}{6} \int x \log (\log (20+x)) \, dx-\frac {100}{3} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (20+x)\right )\\ &=-\frac {1}{12} \text {Ei}(2 \log (20+x))-\frac {100}{3} \log (\log (20+x))+\frac {10 \text {li}(20+x)}{3}-\frac {1}{6} \int x \log (\log (20+x)) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 12, normalized size = 1.00 \begin {gather*} -\frac {1}{12} x^2 \log (\log (20+x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-x^2 + (-40*x - 2*x^2)*Log[20 + x]*Log[Log[20 + x]])/((240 + 12*x)*Log[20 + x]),x]

[Out]

-1/12*(x^2*Log[Log[20 + x]])

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fricas [A]  time = 0.64, size = 10, normalized size = 0.83 \begin {gather*} -\frac {1}{12} \, x^{2} \log \left (\log \left (x + 20\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2-40*x)*log(20+x)*log(log(20+x))-x^2)/(12*x+240)/log(20+x),x, algorithm="fricas")

[Out]

-1/12*x^2*log(log(x + 20))

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giac [A]  time = 0.27, size = 10, normalized size = 0.83 \begin {gather*} -\frac {1}{12} \, x^{2} \log \left (\log \left (x + 20\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2-40*x)*log(20+x)*log(log(20+x))-x^2)/(12*x+240)/log(20+x),x, algorithm="giac")

[Out]

-1/12*x^2*log(log(x + 20))

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maple [A]  time = 0.09, size = 11, normalized size = 0.92

method	result	size
norman	\(-\frac {x^{2} \ln \left (\ln \left (20+x \right )\right )}{12}\)	\(11\)
risch	\(-\frac {x^{2} \ln \left (\ln \left (20+x \right )\right )}{12}\)	\(11\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^2-40*x)*ln(20+x)*ln(ln(20+x))-x^2)/(12*x+240)/ln(20+x),x,method=_RETURNVERBOSE)

[Out]

-1/12*x^2*ln(ln(20+x))

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maxima [A]  time = 0.40, size = 10, normalized size = 0.83 \begin {gather*} -\frac {1}{12} \, x^{2} \log \left (\log \left (x + 20\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2-40*x)*log(20+x)*log(log(20+x))-x^2)/(12*x+240)/log(20+x),x, algorithm="maxima")

[Out]

-1/12*x^2*log(log(x + 20))

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mupad [B]  time = 3.78, size = 10, normalized size = 0.83 \begin {gather*} -\frac {x^2\,\ln \left (\ln \left (x+20\right )\right )}{12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^2 + log(x + 20)*log(log(x + 20))*(40*x + 2*x^2))/(log(x + 20)*(12*x + 240)),x)

[Out]

-(x^2*log(log(x + 20)))/12

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sympy [A]  time = 0.51, size = 24, normalized size = 2.00 \begin {gather*} \left (\frac {100}{9} - \frac {x^{2}}{12}\right ) \log {\left (\log {\left (x + 20 \right )} \right )} - \frac {100 \log {\left (\log {\left (x + 20 \right )} \right )}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**2-40*x)*ln(20+x)*ln(ln(20+x))-x**2)/(12*x+240)/ln(20+x),x)

[Out]

(100/9 - x**2/12)*log(log(x + 20)) - 100*log(log(x + 20))/9

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