3.200 \(\int \frac {7+5 x+4 x^2}{5+4 x+4 x^2} \, dx\)

Optimal. Leaf size=27 \[ \frac {1}{8} \log \left (4 x^2+4 x+5\right )+x+\frac {3}{8} \tan ^{-1}\left (x+\frac {1}{2}\right ) \]

[Out]

x+3/8*arctan(1/2+x)+1/8*ln(4*x^2+4*x+5)

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {1657, 634, 618, 204, 628} \[ \frac {1}{8} \log \left (4 x^2+4 x+5\right )+x+\frac {3}{8} \tan ^{-1}\left (x+\frac {1}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(7 + 5*x + 4*x^2)/(5 + 4*x + 4*x^2),x]

[Out]

x + (3*ArcTan[1/2 + x])/8 + Log[5 + 4*x + 4*x^2]/8

Rule 204

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

Rule 618

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 628

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 634

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]

Rule 1657

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*} \int \frac {7+5 x+4 x^2}{5+4 x+4 x^2} \, dx &=\int \left (1+\frac {2+x}{5+4 x+4 x^2}\right ) \, dx\\ &=x+\int \frac {2+x}{5+4 x+4 x^2} \, dx\\ &=x+\frac {1}{8} \int \frac {4+8 x}{5+4 x+4 x^2} \, dx+\frac {3}{2} \int \frac {1}{5+4 x+4 x^2} \, dx\\ &=x+\frac {1}{8} \log \left (5+4 x+4 x^2\right )-3 \operatorname {Subst}\left (\int \frac {1}{-64-x^2} \, dx,x,4+8 x\right )\\ &=x+\frac {3}{8} \tan ^{-1}\left (\frac {1}{2}+x\right )+\frac {1}{8} \log \left (5+4 x+4 x^2\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 31, normalized size = 1.15 \[ \frac {1}{8} \log \left (4 x^2+4 x+5\right )+x+\frac {3}{8} \tan ^{-1}\left (\frac {1}{2} (2 x+1)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(7 + 5*x + 4*x^2)/(5 + 4*x + 4*x^2),x]

[Out]

x + (3*ArcTan[(1 + 2*x)/2])/8 + Log[5 + 4*x + 4*x^2]/8

________________________________________________________________________________________

fricas [A]  time = 0.40, size = 21, normalized size = 0.78 \[ x + \frac {3}{8} \, \arctan \left (x + \frac {1}{2}\right ) + \frac {1}{8} \, \log \left (4 \, x^{2} + 4 \, x + 5\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2+5*x+7)/(4*x^2+4*x+5),x, algorithm="fricas")

[Out]

x + 3/8*arctan(x + 1/2) + 1/8*log(4*x^2 + 4*x + 5)

________________________________________________________________________________________

giac [A]  time = 0.94, size = 21, normalized size = 0.78 \[ x + \frac {3}{8} \, \arctan \left (x + \frac {1}{2}\right ) + \frac {1}{8} \, \log \left (4 \, x^{2} + 4 \, x + 5\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2+5*x+7)/(4*x^2+4*x+5),x, algorithm="giac")

[Out]

x + 3/8*arctan(x + 1/2) + 1/8*log(4*x^2 + 4*x + 5)

________________________________________________________________________________________

maple [A]  time = 0.00, size = 22, normalized size = 0.81 \[ x +\frac {3 \arctan \left (x +\frac {1}{2}\right )}{8}+\frac {\ln \left (4 x^{2}+4 x +5\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^2+5*x+7)/(4*x^2+4*x+5),x)

[Out]

x+3/8*arctan(x+1/2)+1/8*ln(4*x^2+4*x+5)

________________________________________________________________________________________

maxima [A]  time = 1.19, size = 21, normalized size = 0.78 \[ x + \frac {3}{8} \, \arctan \left (x + \frac {1}{2}\right ) + \frac {1}{8} \, \log \left (4 \, x^{2} + 4 \, x + 5\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2+5*x+7)/(4*x^2+4*x+5),x, algorithm="maxima")

[Out]

x + 3/8*arctan(x + 1/2) + 1/8*log(4*x^2 + 4*x + 5)

________________________________________________________________________________________

mupad [B]  time = 0.04, size = 17, normalized size = 0.63 \[ x+\frac {\ln \left (x^2+x+\frac {5}{4}\right )}{8}+\frac {3\,\mathrm {atan}\left (x+\frac {1}{2}\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x + 4*x^2 + 7)/(4*x + 4*x^2 + 5),x)

[Out]

x + log(x + x^2 + 5/4)/8 + (3*atan(x + 1/2))/8

________________________________________________________________________________________

sympy [A]  time = 0.12, size = 22, normalized size = 0.81 \[ x + \frac {\log {\left (x^{2} + x + \frac {5}{4} \right )}}{8} + \frac {3 \operatorname {atan}{\left (x + \frac {1}{2} \right )}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x**2+5*x+7)/(4*x**2+4*x+5),x)

[Out]

x + log(x**2 + x + 5/4)/8 + 3*atan(x + 1/2)/8

________________________________________________________________________________________