\(\int \frac {1+2 x}{(15 x+e x+2 x^2+x \log (x)) \log (-15-e-2 x-\log (x))} \, dx\) [5400]

Optimal result

Integrand size = 39, antiderivative size = 14 \[ \int \frac {1+2 x}{\left (15 x+e x+2 x^2+x \log (x)\right ) \log (-15-e-2 x-\log (x))} \, dx=\log (\log (-15-e-2 x-\log (x))) \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {6, 6816} \[ \int \frac {1+2 x}{\left (15 x+e x+2 x^2+x \log (x)\right ) \log (-15-e-2 x-\log (x))} \, dx=\log (\log (-2 x-\log (x)-e-15)) \]

[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 6816

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

Rubi steps \begin{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {1+2 x}{\left (15 x+e x+2 x^2+x \log (x)\right ) \log (-15-e-2 x-\log (x))} \, dx=\log (\log (-15-e-2 x-\log (x))) \]

[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]]]

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.14

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

[In]

int((1+2*x)/(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))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07 \[ \int \frac {1+2 x}{\left (15 x+e x+2 x^2+x \log (x)\right ) \log (-15-e-2 x-\log (x))} \, dx=\log \left (\log \left (-2 \, x - e - \log \left (x\right ) - 15\right )\right ) \]

[In]

integrate((1+2*x)/(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))

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07 \[ \int \frac {1+2 x}{\left (15 x+e x+2 x^2+x \log (x)\right ) \log (-15-e-2 x-\log (x))} \, dx=\log {\left (\log {\left (- 2 x - \log {\left (x \right )} - 15 - e \right )} \right )} \]

[In]

integrate((1+2*x)/(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))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07 \[ \int \frac {1+2 x}{\left (15 x+e x+2 x^2+x \log (x)\right ) \log (-15-e-2 x-\log (x))} \, dx=\log \left (\log \left (-2 \, x - e - \log \left (x\right ) - 15\right )\right ) \]

[In]

integrate((1+2*x)/(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))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07 \[ \int \frac {1+2 x}{\left (15 x+e x+2 x^2+x \log (x)\right ) \log (-15-e-2 x-\log (x))} \, dx=\log \left (\log \left (-2 \, x - e - \log \left (x\right ) - 15\right )\right ) \]

[In]

integrate((1+2*x)/(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))

Mupad [B] (verification not implemented)

Time = 12.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07 \[ \int \frac {1+2 x}{\left (15 x+e x+2 x^2+x \log (x)\right ) \log (-15-e-2 x-\log (x))} \, dx=\ln \left (\ln \left (-2\,x-\mathrm {e}-\ln \left (x\right )-15\right )\right ) \]

[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))