\(\int \frac {-34+40 x}{-35-17 x+10 x^2+7 \log (3)} \, dx\) [5045]

Optimal result

Integrand size = 22, antiderivative size = 21 \[ \int \frac {-34+40 x}{-35-17 x+10 x^2+7 \log (3)} \, dx=\log \left (\left (5+x+\frac {10}{7} \left (x-x^2\right )-\log (3)\right )^2\right ) \]

[Out]

ln((-10/7*x^2+17/7*x+5-ln(3))^2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {642} \[ \int \frac {-34+40 x}{-35-17 x+10 x^2+7 \log (3)} \, dx=2 \log \left (-10 x^2+17 x+7 (5-\log (3))\right ) \]

[In]

Int[(-34 + 40*x)/(-35 - 17*x + 10*x^2 + 7*Log[3]),x]

[Out]

2*Log[17*x - 10*x^2 + 7*(5 - Log[3])]

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]

Rubi steps \begin{align*} \text {integral}& = 2 \log \left (17 x-10 x^2+7 (5-\log (3))\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {-34+40 x}{-35-17 x+10 x^2+7 \log (3)} \, dx=2 \log \left (-17 x+10 x^2+7 (-5+\log (3))\right ) \]

[In]

Integrate[(-34 + 40*x)/(-35 - 17*x + 10*x^2 + 7*Log[3]),x]

[Out]

2*Log[-17*x + 10*x^2 + 7*(-5 + Log[3])]

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76

method	result	size
parallelrisch	\(2 \ln \left (\frac {7 \ln \left (3\right )}{10}+x^{2}-\frac {17 x}{10}-\frac {7}{2}\right )\)	\(16\)
default	\(2 \ln \left (7 \ln \left (3\right )+10 x^{2}-17 x -35\right )\)	\(18\)
norman	\(2 \ln \left (7 \ln \left (3\right )+10 x^{2}-17 x -35\right )\)	\(18\)
risch	\(2 \ln \left (7 \ln \left (3\right )+10 x^{2}-17 x -35\right )\)	\(18\)

[In]

int((40*x-34)/(7*ln(3)+10*x^2-17*x-35),x,method=_RETURNVERBOSE)

[Out]

2*ln(7/10*ln(3)+x^2-17/10*x-7/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {-34+40 x}{-35-17 x+10 x^2+7 \log (3)} \, dx=2 \, \log \left (10 \, x^{2} - 17 \, x + 7 \, \log \left (3\right ) - 35\right ) \]

[In]

integrate((40*x-34)/(7*log(3)+10*x^2-17*x-35),x, algorithm="fricas")

[Out]

2*log(10*x^2 - 17*x + 7*log(3) - 35)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {-34+40 x}{-35-17 x+10 x^2+7 \log (3)} \, dx=2 \log {\left (10 x^{2} - 17 x - 35 + 7 \log {\left (3 \right )} \right )} \]

[In]

integrate((40*x-34)/(7*ln(3)+10*x**2-17*x-35),x)

[Out]

2*log(10*x**2 - 17*x - 35 + 7*log(3))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {-34+40 x}{-35-17 x+10 x^2+7 \log (3)} \, dx=2 \, \log \left (10 \, x^{2} - 17 \, x + 7 \, \log \left (3\right ) - 35\right ) \]

[In]

integrate((40*x-34)/(7*log(3)+10*x^2-17*x-35),x, algorithm="maxima")

[Out]

2*log(10*x^2 - 17*x + 7*log(3) - 35)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {-34+40 x}{-35-17 x+10 x^2+7 \log (3)} \, dx=2 \, \log \left ({\left | 10 \, x^{2} - 17 \, x + 7 \, \log \left (3\right ) - 35 \right |}\right ) \]

[In]

integrate((40*x-34)/(7*log(3)+10*x^2-17*x-35),x, algorithm="giac")

[Out]

2*log(abs(10*x^2 - 17*x + 7*log(3) - 35))

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {-34+40 x}{-35-17 x+10 x^2+7 \log (3)} \, dx=2\,\ln \left (10\,x^2-17\,x+7\,\ln \left (3\right )-35\right ) \]

[In]

int(-(40*x - 34)/(17*x - 7*log(3) - 10*x^2 + 35),x)

[Out]

2*log(7*log(3) - 17*x + 10*x^2 - 35)