\(\int \frac {-40-80 x+20 x^2+(-3+x) \log (2)}{80+40 x+20 x^2+x \log (2)} \, dx\) [8773]

Optimal result

Integrand size = 33, antiderivative size = 21 \[ \int \frac {-40-80 x+20 x^2+(-3+x) \log (2)}{80+40 x+20 x^2+x \log (2)} \, dx=x-3 \log \left (1+\frac {1}{4} x \left (2+x+\frac {\log (2)}{20}\right )\right ) \]

[Out]

x-3*ln(x*(1/2+1/4*x+1/80*ln(2))+1)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {6, 1671, 642} \[ \int \frac {-40-80 x+20 x^2+(-3+x) \log (2)}{80+40 x+20 x^2+x \log (2)} \, dx=x-3 \log \left (20 x^2+x (40+\log (2))+80\right ) \]

[In]

Int[(-40 - 80*x + 20*x^2 + (-3 + x)*Log[2])/(80 + 40*x + 20*x^2 + x*Log[2]),x]

[Out]

x - 3*Log[80 + 20*x^2 + x*(40 + Log[2])]

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 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1671

Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x + c*x^2)^p, x
], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-40-80 x+20 x^2+(-3+x) \log (2)}{80+20 x^2+x (40+\log (2))} \, dx \\ & = \int \left (1-\frac {3 (40+40 x+\log (2))}{80+20 x^2+x (40+\log (2))}\right ) \, dx \\ & = x-3 \int \frac {40+40 x+\log (2)}{80+20 x^2+x (40+\log (2))} \, dx \\ & = x-3 \log \left (80+20 x^2+x (40+\log (2))\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {-40-80 x+20 x^2+(-3+x) \log (2)}{80+40 x+20 x^2+x \log (2)} \, dx=x-3 \log \left (80+40 x+20 x^2+x \log (2)\right ) \]

[In]

Integrate[(-40 - 80*x + 20*x^2 + (-3 + x)*Log[2])/(80 + 40*x + 20*x^2 + x*Log[2]),x]

[Out]

x - 3*Log[80 + 40*x + 20*x^2 + x*Log[2]]

Maple [A] (verified)

Time = 2.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90

method	result	size
risch	\(x -3 \ln \left (80+20 x^{2}+\left (\ln \left (2\right )+40\right ) x \right )\)	\(19\)
parallelrisch	\(x -3 \ln \left (\frac {x \ln \left (2\right )}{20}+x^{2}+2 x +4\right )\)	\(19\)
default	\(x -3 \ln \left (x \ln \left (2\right )+20 x^{2}+40 x +80\right )\)	\(20\)
norman	\(x -3 \ln \left (x \ln \left (2\right )+20 x^{2}+40 x +80\right )\)	\(20\)

[In]

int((ln(2)*(-3+x)+20*x^2-80*x-40)/(x*ln(2)+20*x^2+40*x+80),x,method=_RETURNVERBOSE)

[Out]

x-3*ln(80+20*x^2+(ln(2)+40)*x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {-40-80 x+20 x^2+(-3+x) \log (2)}{80+40 x+20 x^2+x \log (2)} \, dx=x - 3 \, \log \left (20 \, x^{2} + x \log \left (2\right ) + 40 \, x + 80\right ) \]

[In]

integrate((log(2)*(-3+x)+20*x^2-80*x-40)/(x*log(2)+20*x^2+40*x+80),x, algorithm="fricas")

[Out]

x - 3*log(20*x^2 + x*log(2) + 40*x + 80)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {-40-80 x+20 x^2+(-3+x) \log (2)}{80+40 x+20 x^2+x \log (2)} \, dx=x - 3 \log {\left (20 x^{2} + x \left (\log {\left (2 \right )} + 40\right ) + 80 \right )} \]

[In]

integrate((ln(2)*(-3+x)+20*x**2-80*x-40)/(x*ln(2)+20*x**2+40*x+80),x)

[Out]

x - 3*log(20*x**2 + x*(log(2) + 40) + 80)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {-40-80 x+20 x^2+(-3+x) \log (2)}{80+40 x+20 x^2+x \log (2)} \, dx=x - 3 \, \log \left (20 \, x^{2} + x {\left (\log \left (2\right ) + 40\right )} + 80\right ) \]

[In]

integrate((log(2)*(-3+x)+20*x^2-80*x-40)/(x*log(2)+20*x^2+40*x+80),x, algorithm="maxima")

[Out]

x - 3*log(20*x^2 + x*(log(2) + 40) + 80)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {-40-80 x+20 x^2+(-3+x) \log (2)}{80+40 x+20 x^2+x \log (2)} \, dx=x - 3 \, \log \left (20 \, x^{2} + x \log \left (2\right ) + 40 \, x + 80\right ) \]

[In]

integrate((log(2)*(-3+x)+20*x^2-80*x-40)/(x*log(2)+20*x^2+40*x+80),x, algorithm="giac")

[Out]

x - 3*log(20*x^2 + x*log(2) + 40*x + 80)

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {-40-80 x+20 x^2+(-3+x) \log (2)}{80+40 x+20 x^2+x \log (2)} \, dx=x-3\,\ln \left (\frac {x\,\left (\ln \left (2\right )+40\right )}{20}+x^2+4\right ) \]

[In]

int(-(80*x - log(2)*(x - 3) - 20*x^2 + 40)/(40*x + x*log(2) + 20*x^2 + 80),x)

[Out]

x - 3*log((x*(log(2) + 40))/20 + x^2 + 4)