\(\int \frac {1}{2+5 x-3 x^2} \, dx\) [82]

Optimal result

Integrand size = 12, antiderivative size = 21 \[ \int \frac {1}{2+5 x-3 x^2} \, dx=-\frac {1}{7} \log (2-x)+\frac {1}{7} \log (1+3 x) \]

[Out]

-1/7*ln(2-x)+1/7*ln(1+3*x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {630, 31} \[ \int \frac {1}{2+5 x-3 x^2} \, dx=\frac {1}{7} \log (3 x+1)-\frac {1}{7} \log (2-x) \]

[In]

Int[(2 + 5*x - 3*x^2)^(-1),x]

[Out]

-1/7*Log[2 - x] + Log[1 + 3*x]/7

Rule 31

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

Rule 630

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

Rubi steps \begin{align*} \text {integral}& = -\left (\frac {3}{7} \int \frac {1}{-1-3 x} \, dx\right )+\frac {3}{7} \int \frac {1}{6-3 x} \, dx \\ & = -\frac {1}{7} \log (2-x)+\frac {1}{7} \log (1+3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {1}{2+5 x-3 x^2} \, dx=-\frac {1}{7} \log (2-x)+\frac {1}{7} \log (1+3 x) \]

[In]

Integrate[(2 + 5*x - 3*x^2)^(-1),x]

[Out]

-1/7*Log[2 - x] + Log[1 + 3*x]/7

Maple [A] (verified)

Time = 2.18 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67

method	result	size
parallelrisch	\(-\frac {\ln \left (-2+x \right )}{7}+\frac {\ln \left (x +\frac {1}{3}\right )}{7}\)	\(14\)
default	\(-\frac {\ln \left (-2+x \right )}{7}+\frac {\ln \left (3 x +1\right )}{7}\)	\(16\)
norman	\(-\frac {\ln \left (-2+x \right )}{7}+\frac {\ln \left (3 x +1\right )}{7}\)	\(16\)
risch	\(-\frac {\ln \left (-2+x \right )}{7}+\frac {\ln \left (3 x +1\right )}{7}\)	\(16\)

[In]

int(1/(-3*x^2+5*x+2),x,method=_RETURNVERBOSE)

[Out]

-1/7*ln(-2+x)+1/7*ln(x+1/3)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {1}{2+5 x-3 x^2} \, dx=\frac {1}{7} \, \log \left (3 \, x + 1\right ) - \frac {1}{7} \, \log \left (x - 2\right ) \]

[In]

integrate(1/(-3*x^2+5*x+2),x, algorithm="fricas")

[Out]

1/7*log(3*x + 1) - 1/7*log(x - 2)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67 \[ \int \frac {1}{2+5 x-3 x^2} \, dx=- \frac {\log {\left (x - 2 \right )}}{7} + \frac {\log {\left (x + \frac {1}{3} \right )}}{7} \]

[In]

integrate(1/(-3*x**2+5*x+2),x)

[Out]

-log(x - 2)/7 + log(x + 1/3)/7

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {1}{2+5 x-3 x^2} \, dx=\frac {1}{7} \, \log \left (3 \, x + 1\right ) - \frac {1}{7} \, \log \left (x - 2\right ) \]

[In]

integrate(1/(-3*x^2+5*x+2),x, algorithm="maxima")

[Out]

1/7*log(3*x + 1) - 1/7*log(x - 2)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {1}{2+5 x-3 x^2} \, dx=\frac {1}{7} \, \log \left ({\left | 3 \, x + 1 \right |}\right ) - \frac {1}{7} \, \log \left ({\left | x - 2 \right |}\right ) \]

[In]

integrate(1/(-3*x^2+5*x+2),x, algorithm="giac")

[Out]

1/7*log(abs(3*x + 1)) - 1/7*log(abs(x - 2))

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.38 \[ \int \frac {1}{2+5 x-3 x^2} \, dx=\frac {2\,\mathrm {atanh}\left (\frac {6\,x}{7}-\frac {5}{7}\right )}{7} \]

[In]

int(1/(5*x - 3*x^2 + 2),x)

[Out]

(2*atanh((6*x)/7 - 5/7))/7