\(\int \frac {7-3 x}{-5+2 x+x^2} \, dx\) [2233]

Optimal result

Integrand size = 16, antiderivative size = 47 \[ \int \frac {7-3 x}{-5+2 x+x^2} \, dx=-\frac {1}{6} \left (9-5 \sqrt {6}\right ) \log \left (1-\sqrt {6}+x\right )-\frac {1}{6} \left (9+5 \sqrt {6}\right ) \log \left (1+\sqrt {6}+x\right ) \]

[Out]

-1/6*ln(1+x-6^(1/2))*(9-5*6^(1/2))-1/6*ln(1+x+6^(1/2))*(9+5*6^(1/2))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {646, 31} \[ \int \frac {7-3 x}{-5+2 x+x^2} \, dx=-\frac {1}{6} \left (9-5 \sqrt {6}\right ) \log \left (x-\sqrt {6}+1\right )-\frac {1}{6} \left (9+5 \sqrt {6}\right ) \log \left (x+\sqrt {6}+1\right ) \]

[In]

Int[(7 - 3*x)/(-5 + 2*x + x^2),x]

[Out]

-1/6*((9 - 5*Sqrt[6])*Log[1 - Sqrt[6] + x]) - ((9 + 5*Sqrt[6])*Log[1 + Sqrt[6] + x])/6

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 646

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && NiceSqrtQ[b^2 - 4*a*
c]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{6} \left (-9+5 \sqrt {6}\right ) \int \frac {1}{1-\sqrt {6}+x} \, dx-\frac {1}{6} \left (9+5 \sqrt {6}\right ) \int \frac {1}{1+\sqrt {6}+x} \, dx \\ & = -\frac {1}{6} \left (9-5 \sqrt {6}\right ) \log \left (1-\sqrt {6}+x\right )-\frac {1}{6} \left (9+5 \sqrt {6}\right ) \log \left (1+\sqrt {6}+x\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int \frac {7-3 x}{-5+2 x+x^2} \, dx=\frac {1}{6} \left (-9+5 \sqrt {6}\right ) \log \left (-1+\sqrt {6}-x\right )+\frac {1}{6} \left (-9-5 \sqrt {6}\right ) \log \left (1+\sqrt {6}+x\right ) \]

[In]

Integrate[(7 - 3*x)/(-5 + 2*x + x^2),x]

[Out]

((-9 + 5*Sqrt[6])*Log[-1 + Sqrt[6] - x])/6 + ((-9 - 5*Sqrt[6])*Log[1 + Sqrt[6] + x])/6

Maple [A] (verified)

Time = 21.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.62

method	result	size
default	\(-\frac {3 \ln \left (x^{2}+2 x -5\right )}{2}-\frac {5 \sqrt {6}\, \operatorname {arctanh}\left (\frac {\left (2+2 x \right ) \sqrt {6}}{12}\right )}{3}\)	\(29\)
risch	\(-\frac {3 \ln \left (1+x -\sqrt {6}\right )}{2}+\frac {5 \ln \left (1+x -\sqrt {6}\right ) \sqrt {6}}{6}-\frac {3 \ln \left (1+x +\sqrt {6}\right )}{2}-\frac {5 \ln \left (1+x +\sqrt {6}\right ) \sqrt {6}}{6}\)	\(48\)

[In]

int((7-3*x)/(x^2+2*x-5),x,method=_RETURNVERBOSE)

[Out]

-3/2*ln(x^2+2*x-5)-5/3*6^(1/2)*arctanh(1/12*(2+2*x)*6^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.15 \[ \int \frac {7-3 x}{-5+2 x+x^2} \, dx=\frac {5}{6} \, \sqrt {3} \sqrt {2} \log \left (-\frac {2 \, \sqrt {3} \sqrt {2} {\left (x + 1\right )} - x^{2} - 2 \, x - 7}{x^{2} + 2 \, x - 5}\right ) - \frac {3}{2} \, \log \left (x^{2} + 2 \, x - 5\right ) \]

[In]

integrate((7-3*x)/(x^2+2*x-5),x, algorithm="fricas")

[Out]

5/6*sqrt(3)*sqrt(2)*log(-(2*sqrt(3)*sqrt(2)*(x + 1) - x^2 - 2*x - 7)/(x^2 + 2*x - 5)) - 3/2*log(x^2 + 2*x - 5)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.94 \[ \int \frac {7-3 x}{-5+2 x+x^2} \, dx=- \left (\frac {3}{2} + \frac {5 \sqrt {6}}{6}\right ) \log {\left (x + 1 + \sqrt {6} \right )} - \left (\frac {3}{2} - \frac {5 \sqrt {6}}{6}\right ) \log {\left (x - \sqrt {6} + 1 \right )} \]

[In]

integrate((7-3*x)/(x**2+2*x-5),x)

[Out]

-(3/2 + 5*sqrt(6)/6)*log(x + 1 + sqrt(6)) - (3/2 - 5*sqrt(6)/6)*log(x - sqrt(6) + 1)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.74 \[ \int \frac {7-3 x}{-5+2 x+x^2} \, dx=\frac {5}{6} \, \sqrt {6} \log \left (\frac {x - \sqrt {6} + 1}{x + \sqrt {6} + 1}\right ) - \frac {3}{2} \, \log \left (x^{2} + 2 \, x - 5\right ) \]

[In]

integrate((7-3*x)/(x^2+2*x-5),x, algorithm="maxima")

[Out]

5/6*sqrt(6)*log((x - sqrt(6) + 1)/(x + sqrt(6) + 1)) - 3/2*log(x^2 + 2*x - 5)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.94 \[ \int \frac {7-3 x}{-5+2 x+x^2} \, dx=\frac {5}{6} \, \sqrt {6} \log \left (\frac {{\left | 2 \, x - 2 \, \sqrt {6} + 2 \right |}}{{\left | 2 \, x + 2 \, \sqrt {6} + 2 \right |}}\right ) - \frac {3}{2} \, \log \left ({\left | x^{2} + 2 \, x - 5 \right |}\right ) \]

[In]

integrate((7-3*x)/(x^2+2*x-5),x, algorithm="giac")

[Out]

5/6*sqrt(6)*log(abs(2*x - 2*sqrt(6) + 2)/abs(2*x + 2*sqrt(6) + 2)) - 3/2*log(abs(x^2 + 2*x - 5))

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.72 \[ \int \frac {7-3 x}{-5+2 x+x^2} \, dx=\ln \left (x-\sqrt {6}+1\right )\,\left (\frac {5\,\sqrt {6}}{6}-\frac {3}{2}\right )-\ln \left (x+\sqrt {6}+1\right )\,\left (\frac {5\,\sqrt {6}}{6}+\frac {3}{2}\right ) \]

[In]

int(-(3*x - 7)/(2*x + x^2 - 5),x)

[Out]

log(x - 6^(1/2) + 1)*((5*6^(1/2))/6 - 3/2) - log(x + 6^(1/2) + 1)*((5*6^(1/2))/6 + 3/2)