3.23.71 \(\int \frac {x^2+(6 x+2 x^2) \log (3+x)}{(30+10 x+(3 x^2+x^3) \log (3+x)) \log (\frac {1}{5} (10+x^2 \log (3+x)))} \, dx\)

Optimal. Leaf size=17 \[ 5+\log \left (\log \left (2+\frac {1}{5} x^2 \log (3+x)\right )\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 16, normalized size of antiderivative = 0.94, number of steps used = 1, number of rules used = 1, integrand size = 57, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.018, Rules used = {6684} \begin {gather*} \log \left (\log \left (\frac {1}{5} \left (x^2 \log (x+3)+10\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2 + (6*x + 2*x^2)*Log[3 + x])/((30 + 10*x + (3*x^2 + x^3)*Log[3 + x])*Log[(10 + x^2*Log[3 + x])/5]),x]

[Out]

Log[Log[(10 + x^2*Log[3 + x])/5]]

Rule 6684

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log \left (\log \left (\frac {1}{5} \left (10+x^2 \log (3+x)\right )\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.34, size = 15, normalized size = 0.88 \begin {gather*} \log \left (\log \left (2+\frac {1}{5} x^2 \log (3+x)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2 + (6*x + 2*x^2)*Log[3 + x])/((30 + 10*x + (3*x^2 + x^3)*Log[3 + x])*Log[(10 + x^2*Log[3 + x])/5
]),x]

[Out]

Log[Log[2 + (x^2*Log[3 + x])/5]]

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fricas [A]  time = 1.16, size = 13, normalized size = 0.76 \begin {gather*} \log \left (\log \left (\frac {1}{5} \, x^{2} \log \left (x + 3\right ) + 2\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+6*x)*log(3+x)+x^2)/((x^3+3*x^2)*log(3+x)+10*x+30)/log(1/5*log(3+x)*x^2+2),x, algorithm="fric
as")

[Out]

log(log(1/5*x^2*log(x + 3) + 2))

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giac [A]  time = 0.26, size = 17, normalized size = 1.00 \begin {gather*} \log \left (\log \relax (5) - \log \left (x^{2} \log \left (x + 3\right ) + 10\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+6*x)*log(3+x)+x^2)/((x^3+3*x^2)*log(3+x)+10*x+30)/log(1/5*log(3+x)*x^2+2),x, algorithm="giac
")

[Out]

log(log(5) - log(x^2*log(x + 3) + 10))

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maple [A]  time = 0.03, size = 14, normalized size = 0.82

method	result	size
risch	\(\ln \left (\ln \left (\frac {\ln \left (3+x \right ) x^{2}}{5}+2\right )\right )\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^2+6*x)*ln(3+x)+x^2)/((x^3+3*x^2)*ln(3+x)+10*x+30)/ln(1/5*ln(3+x)*x^2+2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.88, size = 17, normalized size = 1.00 \begin {gather*} \log \left (-\log \relax (5) + \log \left (x^{2} \log \left (x + 3\right ) + 10\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+6*x)*log(3+x)+x^2)/((x^3+3*x^2)*log(3+x)+10*x+30)/log(1/5*log(3+x)*x^2+2),x, algorithm="maxi
ma")

[Out]

log(-log(5) + log(x^2*log(x + 3) + 10))

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mupad [B]  time = 2.08, size = 13, normalized size = 0.76 \begin {gather*} \ln \left (\ln \left (\frac {x^2\,\ln \left (x+3\right )}{5}+2\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x + 3)*(6*x + 2*x^2) + x^2)/(log((x^2*log(x + 3))/5 + 2)*(10*x + log(x + 3)*(3*x^2 + x^3) + 30)),x)

[Out]

log(log((x^2*log(x + 3))/5 + 2))

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sympy [A]  time = 0.64, size = 14, normalized size = 0.82 \begin {gather*} \log {\left (\log {\left (\frac {x^{2} \log {\left (x + 3 \right )}}{5} + 2 \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**2+6*x)*ln(3+x)+x**2)/((x**3+3*x**2)*ln(3+x)+10*x+30)/ln(1/5*ln(3+x)*x**2+2),x)

[Out]

log(log(x**2*log(x + 3)/5 + 2))

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