\(\int \frac {31+5 x}{11-4 x+3 x^2} \, dx\) [472]

Optimal result

Integrand size = 18, antiderivative size = 37 \[ \int \frac {31+5 x}{11-4 x+3 x^2} \, dx=-\frac {103 \arctan \left (\frac {2-3 x}{\sqrt {29}}\right )}{3 \sqrt {29}}+\frac {5}{6} \log \left (11-4 x+3 x^2\right ) \]

[Out]

5/6*ln(3*x^2-4*x+11)-103/87*arctan(1/29*(2-3*x)*29^(1/2))*29^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {648, 632, 210, 642} \[ \int \frac {31+5 x}{11-4 x+3 x^2} \, dx=\frac {5}{6} \log \left (3 x^2-4 x+11\right )-\frac {103 \arctan \left (\frac {2-3 x}{\sqrt {29}}\right )}{3 \sqrt {29}} \]

[In]

Int[(31 + 5*x)/(11 - 4*x + 3*x^2),x]

[Out]

(-103*ArcTan[(2 - 3*x)/Sqrt[29]])/(3*Sqrt[29]) + (5*Log[11 - 4*x + 3*x^2])/6

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 632

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

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 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), 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 {5}{6} \int \frac {-4+6 x}{11-4 x+3 x^2} \, dx+\frac {103}{3} \int \frac {1}{11-4 x+3 x^2} \, dx \\ & = \frac {5}{6} \log \left (11-4 x+3 x^2\right )-\frac {206}{3} \text {Subst}\left (\int \frac {1}{-116-x^2} \, dx,x,-4+6 x\right ) \\ & = -\frac {103 \tan ^{-1}\left (\frac {2-3 x}{\sqrt {29}}\right )}{3 \sqrt {29}}+\frac {5}{6} \log \left (11-4 x+3 x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int \frac {31+5 x}{11-4 x+3 x^2} \, dx=\frac {103 \arctan \left (\frac {-2+3 x}{\sqrt {29}}\right )}{3 \sqrt {29}}+\frac {5}{6} \log \left (11-4 x+3 x^2\right ) \]

[In]

Integrate[(31 + 5*x)/(11 - 4*x + 3*x^2),x]

[Out]

(103*ArcTan[(-2 + 3*x)/Sqrt[29]])/(3*Sqrt[29]) + (5*Log[11 - 4*x + 3*x^2])/6

Maple [A] (verified)

Time = 1.37 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.84

method	result	size
default	\(\frac {5 \ln \left (3 x^{2}-4 x +11\right )}{6}+\frac {103 \sqrt {29}\, \arctan \left (\frac {\left (6 x -4\right ) \sqrt {29}}{58}\right )}{87}\)	\(31\)
risch	\(\frac {5 \ln \left (9 x^{2}-12 x +33\right )}{6}+\frac {103 \sqrt {29}\, \arctan \left (\frac {\left (3 x -2\right ) \sqrt {29}}{29}\right )}{87}\)	\(31\)

[In]

int((31+5*x)/(3*x^2-4*x+11),x,method=_RETURNVERBOSE)

[Out]

5/6*ln(3*x^2-4*x+11)+103/87*29^(1/2)*arctan(1/58*(6*x-4)*29^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.81 \[ \int \frac {31+5 x}{11-4 x+3 x^2} \, dx=\frac {103}{87} \, \sqrt {29} \arctan \left (\frac {1}{29} \, \sqrt {29} {\left (3 \, x - 2\right )}\right ) + \frac {5}{6} \, \log \left (3 \, x^{2} - 4 \, x + 11\right ) \]

[In]

integrate((31+5*x)/(3*x^2-4*x+11),x, algorithm="fricas")

[Out]

103/87*sqrt(29)*arctan(1/29*sqrt(29)*(3*x - 2)) + 5/6*log(3*x^2 - 4*x + 11)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.19 \[ \int \frac {31+5 x}{11-4 x+3 x^2} \, dx=\frac {5 \log {\left (x^{2} - \frac {4 x}{3} + \frac {11}{3} \right )}}{6} + \frac {103 \sqrt {29} \operatorname {atan}{\left (\frac {3 \sqrt {29} x}{29} - \frac {2 \sqrt {29}}{29} \right )}}{87} \]

[In]

integrate((31+5*x)/(3*x**2-4*x+11),x)

[Out]

5*log(x**2 - 4*x/3 + 11/3)/6 + 103*sqrt(29)*atan(3*sqrt(29)*x/29 - 2*sqrt(29)/29)/87

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.81 \[ \int \frac {31+5 x}{11-4 x+3 x^2} \, dx=\frac {103}{87} \, \sqrt {29} \arctan \left (\frac {1}{29} \, \sqrt {29} {\left (3 \, x - 2\right )}\right ) + \frac {5}{6} \, \log \left (3 \, x^{2} - 4 \, x + 11\right ) \]

[In]

integrate((31+5*x)/(3*x^2-4*x+11),x, algorithm="maxima")

[Out]

103/87*sqrt(29)*arctan(1/29*sqrt(29)*(3*x - 2)) + 5/6*log(3*x^2 - 4*x + 11)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.81 \[ \int \frac {31+5 x}{11-4 x+3 x^2} \, dx=\frac {103}{87} \, \sqrt {29} \arctan \left (\frac {1}{29} \, \sqrt {29} {\left (3 \, x - 2\right )}\right ) + \frac {5}{6} \, \log \left (3 \, x^{2} - 4 \, x + 11\right ) \]

[In]

integrate((31+5*x)/(3*x^2-4*x+11),x, algorithm="giac")

[Out]

103/87*sqrt(29)*arctan(1/29*sqrt(29)*(3*x - 2)) + 5/6*log(3*x^2 - 4*x + 11)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.81 \[ \int \frac {31+5 x}{11-4 x+3 x^2} \, dx=\frac {5\,\ln \left (x^2-\frac {4\,x}{3}+\frac {11}{3}\right )}{6}+\frac {103\,\sqrt {29}\,\mathrm {atan}\left (\frac {3\,\sqrt {29}\,x}{29}-\frac {2\,\sqrt {29}}{29}\right )}{87} \]

[In]

int((5*x + 31)/(3*x^2 - 4*x + 11),x)

[Out]

(5*log(x^2 - (4*x)/3 + 11/3))/6 + (103*29^(1/2)*atan((3*29^(1/2)*x)/29 - (2*29^(1/2))/29))/87