3.54.100 \(\int \frac {1+2 x}{(15 x+e x+2 x^2+x \log (x)) \log (-15-e-2 x-\log (x))} \, dx\)

Optimal. Leaf size=14 \[ \log (\log (-15-e-2 x-\log (x))) \]

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Rubi [A]  time = 0.07, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {6, 6684} \begin {gather*} \log (\log (-2 x-\log (x)-e-15)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + 2*x)/((15*x + E*x + 2*x^2 + x*Log[x])*Log[-15 - E - 2*x - Log[x]]),x]

[Out]

Log[Log[-15 - E - 2*x - Log[x]]]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

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} &=\int \frac {1+2 x}{\left ((15+e) x+2 x^2+x \log (x)\right ) \log (-15-e-2 x-\log (x))} \, dx\\ &=\log (\log (-15-e-2 x-\log (x)))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 14, normalized size = 1.00 \begin {gather*} \log (\log (-15-e-2 x-\log (x))) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + 2*x)/((15*x + E*x + 2*x^2 + x*Log[x])*Log[-15 - E - 2*x - Log[x]]),x]

[Out]

Log[Log[-15 - E - 2*x - Log[x]]]

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fricas [A]  time = 0.64, size = 15, normalized size = 1.07 \begin {gather*} \log \left (\log \left (-2 \, x - e - \log \relax (x) - 15\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x+1)/(x*log(x)+x*exp(1)+2*x^2+15*x)/log(-log(x)-exp(1)-2*x-15),x, algorithm="fricas")

[Out]

log(log(-2*x - e - log(x) - 15))

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giac [A]  time = 0.20, size = 15, normalized size = 1.07 \begin {gather*} \log \left (\log \left (-2 \, x - e - \log \relax (x) - 15\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x+1)/(x*log(x)+x*exp(1)+2*x^2+15*x)/log(-log(x)-exp(1)-2*x-15),x, algorithm="giac")

[Out]

log(log(-2*x - e - log(x) - 15))

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maple [A]  time = 0.14, size = 16, normalized size = 1.14

method	result	size
norman	\(\ln \left (\ln \left (-\ln \relax (x )-{\mathrm e}-2 x -15\right )\right )\)	\(16\)
risch	\(\ln \left (\ln \left (-\ln \relax (x )-{\mathrm e}-2 x -15\right )\right )\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x+1)/(x*ln(x)+x*exp(1)+2*x^2+15*x)/ln(-ln(x)-exp(1)-2*x-15),x,method=_RETURNVERBOSE)

[Out]

ln(ln(-ln(x)-exp(1)-2*x-15))

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maxima [A]  time = 0.43, size = 15, normalized size = 1.07 \begin {gather*} \log \left (\log \left (-2 \, x - e - \log \relax (x) - 15\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x+1)/(x*log(x)+x*exp(1)+2*x^2+15*x)/log(-log(x)-exp(1)-2*x-15),x, algorithm="maxima")

[Out]

log(log(-2*x - e - log(x) - 15))

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mupad [B]  time = 4.15, size = 15, normalized size = 1.07 \begin {gather*} \ln \left (\ln \left (-2\,x-\mathrm {e}-\ln \relax (x)-15\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + 1)/(log(- 2*x - exp(1) - log(x) - 15)*(15*x + x*exp(1) + x*log(x) + 2*x^2)),x)

[Out]

log(log(- 2*x - exp(1) - log(x) - 15))

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sympy [A]  time = 0.40, size = 15, normalized size = 1.07 \begin {gather*} \log {\left (\log {\left (- 2 x - \log {\relax (x )} - 15 - e \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x+1)/(x*ln(x)+x*exp(1)+2*x**2+15*x)/ln(-ln(x)-exp(1)-2*x-15),x)

[Out]

log(log(-2*x - log(x) - 15 - E))

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