3.107 \(\int \frac {1+x^4}{-1+x-x^2+x^3} \, dx\)

Optimal. Leaf size=29 \[ \frac {x^2}{2}-\frac {1}{2} \log \left (x^2+1\right )+x+\log (1-x)-\tan ^{-1}(x) \]

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Rubi [A]  time = 0.03, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2074, 635, 203, 260} \[ \frac {x^2}{2}-\frac {1}{2} \log \left (x^2+1\right )+x+\log (1-x)-\tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(1 + x^4)/(-1 + x - x^2 + x^3),x]

[Out]

x + x^2/2 - ArcTan[x] + Log[1 - x] - Log[1 + x^2]/2

Rule 203

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

Rule 260

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

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps

\begin {align*} \int \frac {1+x^4}{-1+x-x^2+x^3} \, dx &=\int \left (1+\frac {1}{-1+x}+x+\frac {-1-x}{1+x^2}\right ) \, dx\\ &=x+\frac {x^2}{2}+\log (1-x)+\int \frac {-1-x}{1+x^2} \, dx\\ &=x+\frac {x^2}{2}+\log (1-x)-\int \frac {1}{1+x^2} \, dx-\int \frac {x}{1+x^2} \, dx\\ &=x+\frac {x^2}{2}-\tan ^{-1}(x)+\log (1-x)-\frac {1}{2} \log \left (1+x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 29, normalized size = 1.00 \[ \frac {x^2}{2}-\frac {1}{2} \log \left (x^2+1\right )+x+\log (1-x)-\tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

x + x^2/2 - ArcTan[x] + Log[1 - x] - Log[1 + x^2]/2

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IntegrateAlgebraic [A]  time = 0.01, size = 27, normalized size = 0.93 \[ -\frac {1}{2} \log \left (x^2+1\right )+\frac {1}{2} x (x+2)+\log (x-1)-\tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(1 + x^4)/(-1 + x - x^2 + x^3),x]

[Out]

(x*(2 + x))/2 - ArcTan[x] + Log[-1 + x] - Log[1 + x^2]/2

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fricas [A]  time = 0.91, size = 23, normalized size = 0.79 \[ \frac {1}{2} \, x^{2} + x - \arctan \relax (x) - \frac {1}{2} \, \log \left (x^{2} + 1\right ) + \log \left (x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2 + x - arctan(x) - 1/2*log(x^2 + 1) + log(x - 1)

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giac [A]  time = 1.06, size = 24, normalized size = 0.83 \[ \frac {1}{2} \, x^{2} + x - \arctan \relax (x) - \frac {1}{2} \, \log \left (x^{2} + 1\right ) + \log \left ({\left | x - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2 + x - arctan(x) - 1/2*log(x^2 + 1) + log(abs(x - 1))

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maple [A]  time = 0.03, size = 24, normalized size = 0.83

method	result	size
default	\(x +\frac {x^{2}}{2}-\frac {\ln \left (x^{2}+1\right )}{2}-\arctan \relax (x )+\ln \left (-1+x \right )\)	\(24\)
risch	\(x +\frac {x^{2}}{2}-\frac {\ln \left (x^{2}+1\right )}{2}-\arctan \relax (x )+\ln \left (-1+x \right )\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x+1/2*x^2-1/2*ln(x^2+1)-arctan(x)+ln(-1+x)

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maxima [A]  time = 0.96, size = 23, normalized size = 0.79 \[ \frac {1}{2} \, x^{2} + x - \arctan \relax (x) - \frac {1}{2} \, \log \left (x^{2} + 1\right ) + \log \left (x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2 + x - arctan(x) - 1/2*log(x^2 + 1) + log(x - 1)

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mupad [B]  time = 0.05, size = 29, normalized size = 1.00 \[ x+\ln \left (x-1\right )+\frac {x^2}{2}+\ln \left (x-\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )+\ln \left (x+1{}\mathrm {i}\right )\,\left (-\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4 + 1)/(x - x^2 + x^3 - 1),x)

[Out]

x + log(x - 1) - log(x - 1i)*(1/2 - 1i/2) - log(x + 1i)*(1/2 + 1i/2) + x^2/2

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sympy [A]  time = 0.13, size = 22, normalized size = 0.76 \[ \frac {x^{2}}{2} + x + \log {\left (x - 1 \right )} - \frac {\log {\left (x^{2} + 1 \right )}}{2} - \operatorname {atan}{\relax (x )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2/2 + x + log(x - 1) - log(x**2 + 1)/2 - atan(x)

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