3.2.35 \(\int \frac {1-x^3}{x (1+x^2)} \, dx\) [135]

Optimal. Leaf size=18 \[ -x+\tan ^{-1}(x)+\log (x)-\frac {1}{2} \log \left (1+x^2\right ) \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1816, 649, 209, 266} \begin {gather*} -\frac {1}{2} \log \left (x^2+1\right )-x+\log (x)+\tan ^{-1}(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - x^3)/(x*(1 + x^2)),x]

[Out]

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

Rule 209

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

Rule 266

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 649

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 1816

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

Rubi steps

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

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Mathematica [A]
time = 0.00, size = 18, normalized size = 1.00 \begin {gather*} -x+\tan ^{-1}(x)+\log (x)-\frac {1}{2} \log \left (1+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - x^3)/(x*(1 + x^2)),x]

[Out]

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

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Mathics [A]
time = 1.68, size = 16, normalized size = 0.89 \begin {gather*} -x+\text {ArcTan}\left [x\right ]+\text {Log}\left [x\right ]-\frac {\text {Log}\left [1+x^2\right ]}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[(1 - x^3)/(x*(x^2 + 1)),x]')

[Out]

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

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Maple [A]
time = 0.08, size = 17, normalized size = 0.94

method	result	size
default	\(-x +\arctan \left (x \right )+\ln \left (x \right )-\frac {\ln \left (x^{2}+1\right )}{2}\)	\(17\)
meijerg	\(-x +\arctan \left (x \right )+\ln \left (x \right )-\frac {\ln \left (x^{2}+1\right )}{2}\)	\(17\)
risch	\(-x +\arctan \left (x \right )+\ln \left (x \right )-\frac {\ln \left (x^{2}+1\right )}{2}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.35, size = 16, normalized size = 0.89 \begin {gather*} -x + \arctan \left (x\right ) - \frac {1}{2} \, \log \left (x^{2} + 1\right ) + \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 16, normalized size = 0.89 \begin {gather*} -x + \arctan \left (x\right ) - \frac {1}{2} \, \log \left (x^{2} + 1\right ) + \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.06, size = 15, normalized size = 0.83 \begin {gather*} - x + \log {\left (x \right )} - \frac {\log {\left (x^{2} + 1 \right )}}{2} + \operatorname {atan}{\left (x \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.00, size = 18, normalized size = 1.00 \begin {gather*} \ln \left |x\right |-\frac {\ln \left (x^{2}+1\right )}{2}+\arctan x-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^3+1)/x/(x^2+1),x)

[Out]

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

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Mupad [B]
time = 0.04, size = 24, normalized size = 1.33 \begin {gather*} \ln \left (x\right )-x+\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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^3 - 1)/(x*(x^2 + 1)),x)

[Out]

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

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