\(\int \frac {1+x^2+x^3}{2 x^2+x^3+x^4} \, dx\) [314]

Optimal result

Integrand size = 23, antiderivative size = 46 \[ \int \frac {1+x^2+x^3}{2 x^2+x^3+x^4} \, dx=-\frac {1}{2 x}+\frac {\arctan \left (\frac {1+2 x}{\sqrt {7}}\right )}{4 \sqrt {7}}-\frac {\log (x)}{4}+\frac {5}{8} \log \left (2+x+x^2\right ) \]

[Out]

-1/2/x-1/4*ln(x)+5/8*ln(x^2+x+2)+1/28*arctan(1/7*(1+2*x)*7^(1/2))*7^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {1608, 1642, 648, 632, 210, 642} \[ \int \frac {1+x^2+x^3}{2 x^2+x^3+x^4} \, dx=\frac {\arctan \left (\frac {2 x+1}{\sqrt {7}}\right )}{4 \sqrt {7}}+\frac {5}{8} \log \left (x^2+x+2\right )-\frac {1}{2 x}-\frac {\log (x)}{4} \]

[In]

Int[(1 + x^2 + x^3)/(2*x^2 + x^3 + x^4),x]

[Out]

-1/2*1/x + ArcTan[(1 + 2*x)/Sqrt[7]]/(4*Sqrt[7]) - Log[x]/4 + (5*Log[2 + x + x^2])/8

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]

Rule 1608

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 1642

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1+x^2+x^3}{x^2 \left (2+x+x^2\right )} \, dx \\ & = \int \left (\frac {1}{2 x^2}-\frac {1}{4 x}+\frac {3+5 x}{4 \left (2+x+x^2\right )}\right ) \, dx \\ & = -\frac {1}{2 x}-\frac {\log (x)}{4}+\frac {1}{4} \int \frac {3+5 x}{2+x+x^2} \, dx \\ & = -\frac {1}{2 x}-\frac {\log (x)}{4}+\frac {1}{8} \int \frac {1}{2+x+x^2} \, dx+\frac {5}{8} \int \frac {1+2 x}{2+x+x^2} \, dx \\ & = -\frac {1}{2 x}-\frac {\log (x)}{4}+\frac {5}{8} \log \left (2+x+x^2\right )-\frac {1}{4} \text {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,1+2 x\right ) \\ & = -\frac {1}{2 x}+\frac {\tan ^{-1}\left (\frac {1+2 x}{\sqrt {7}}\right )}{4 \sqrt {7}}-\frac {\log (x)}{4}+\frac {5}{8} \log \left (2+x+x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00 \[ \int \frac {1+x^2+x^3}{2 x^2+x^3+x^4} \, dx=-\frac {1}{2 x}+\frac {\arctan \left (\frac {1+2 x}{\sqrt {7}}\right )}{4 \sqrt {7}}-\frac {\log (x)}{4}+\frac {5}{8} \log \left (2+x+x^2\right ) \]

[In]

Integrate[(1 + x^2 + x^3)/(2*x^2 + x^3 + x^4),x]

[Out]

-1/2*1/x + ArcTan[(1 + 2*x)/Sqrt[7]]/(4*Sqrt[7]) - Log[x]/4 + (5*Log[2 + x + x^2])/8

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78

method	result	size
default	\(-\frac {1}{2 x}-\frac {\ln \left (x \right )}{4}+\frac {5 \ln \left (x^{2}+x +2\right )}{8}+\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {7}}{7}\right ) \sqrt {7}}{28}\)	\(36\)
risch	\(-\frac {1}{2 x}+\frac {5 \ln \left (4 x^{2}+4 x +8\right )}{8}+\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {7}}{7}\right ) \sqrt {7}}{28}-\frac {\ln \left (x \right )}{4}\)	\(40\)

[In]

int((x^3+x^2+1)/(x^4+x^3+2*x^2),x,method=_RETURNVERBOSE)

[Out]

-1/2/x-1/4*ln(x)+5/8*ln(x^2+x+2)+1/28*arctan(1/7*(1+2*x)*7^(1/2))*7^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.85 \[ \int \frac {1+x^2+x^3}{2 x^2+x^3+x^4} \, dx=\frac {2 \, \sqrt {7} x \arctan \left (\frac {1}{7} \, \sqrt {7} {\left (2 \, x + 1\right )}\right ) + 35 \, x \log \left (x^{2} + x + 2\right ) - 14 \, x \log \left (x\right ) - 28}{56 \, x} \]

[In]

integrate((x^3+x^2+1)/(x^4+x^3+2*x^2),x, algorithm="fricas")

[Out]

1/56*(2*sqrt(7)*x*arctan(1/7*sqrt(7)*(2*x + 1)) + 35*x*log(x^2 + x + 2) - 14*x*log(x) - 28)/x

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00 \[ \int \frac {1+x^2+x^3}{2 x^2+x^3+x^4} \, dx=- \frac {\log {\left (x \right )}}{4} + \frac {5 \log {\left (x^{2} + x + 2 \right )}}{8} + \frac {\sqrt {7} \operatorname {atan}{\left (\frac {2 \sqrt {7} x}{7} + \frac {\sqrt {7}}{7} \right )}}{28} - \frac {1}{2 x} \]

[In]

integrate((x**3+x**2+1)/(x**4+x**3+2*x**2),x)

[Out]

-log(x)/4 + 5*log(x**2 + x + 2)/8 + sqrt(7)*atan(2*sqrt(7)*x/7 + sqrt(7)/7)/28 - 1/(2*x)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.76 \[ \int \frac {1+x^2+x^3}{2 x^2+x^3+x^4} \, dx=\frac {1}{28} \, \sqrt {7} \arctan \left (\frac {1}{7} \, \sqrt {7} {\left (2 \, x + 1\right )}\right ) - \frac {1}{2 \, x} + \frac {5}{8} \, \log \left (x^{2} + x + 2\right ) - \frac {1}{4} \, \log \left (x\right ) \]

[In]

integrate((x^3+x^2+1)/(x^4+x^3+2*x^2),x, algorithm="maxima")

[Out]

1/28*sqrt(7)*arctan(1/7*sqrt(7)*(2*x + 1)) - 1/2/x + 5/8*log(x^2 + x + 2) - 1/4*log(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \frac {1+x^2+x^3}{2 x^2+x^3+x^4} \, dx=\frac {1}{28} \, \sqrt {7} \arctan \left (\frac {1}{7} \, \sqrt {7} {\left (2 \, x + 1\right )}\right ) - \frac {1}{2 \, x} + \frac {5}{8} \, \log \left (x^{2} + x + 2\right ) - \frac {1}{4} \, \log \left ({\left | x \right |}\right ) \]

[In]

integrate((x^3+x^2+1)/(x^4+x^3+2*x^2),x, algorithm="giac")

[Out]

1/28*sqrt(7)*arctan(1/7*sqrt(7)*(2*x + 1)) - 1/2/x + 5/8*log(x^2 + x + 2) - 1/4*log(abs(x))

Mupad [B] (verification not implemented)

Time = 9.43 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.07 \[ \int \frac {1+x^2+x^3}{2 x^2+x^3+x^4} \, dx=-\frac {\ln \left (x\right )}{4}-\ln \left (x+\frac {1}{2}-\frac {\sqrt {7}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {5}{8}+\frac {\sqrt {7}\,1{}\mathrm {i}}{56}\right )+\ln \left (x+\frac {1}{2}+\frac {\sqrt {7}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {5}{8}+\frac {\sqrt {7}\,1{}\mathrm {i}}{56}\right )-\frac {1}{2\,x} \]

[In]

int((x^2 + x^3 + 1)/(2*x^2 + x^3 + x^4),x)

[Out]

log(x + (7^(1/2)*1i)/2 + 1/2)*((7^(1/2)*1i)/56 + 5/8) - log(x - (7^(1/2)*1i)/2 + 1/2)*((7^(1/2)*1i)/56 - 5/8)
- log(x)/4 - 1/(2*x)