3.69 $$\int \frac{x^3 (1+x+x^2)}{(1+x^2)^2} \, dx$$

Optimal. Leaf size=43 $-\frac{x^3}{2 \left (x^2+1\right )}+\frac{x^2}{2}-\frac{1}{2} \log \left (x^2+1\right )+\frac{3 x}{2}-\frac{3}{2} \tan ^{-1}(x)$

[Out]

(3*x)/2 + x^2/2 - x^3/(2*(1 + x^2)) - (3*ArcTan[x])/2 - Log[1 + x^2]/2

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Rubi [A]  time = 0.0500244, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.294, Rules used = {1804, 801, 635, 203, 260} $-\frac{x^3}{2 \left (x^2+1\right )}+\frac{x^2}{2}-\frac{1}{2} \log \left (x^2+1\right )+\frac{3 x}{2}-\frac{3}{2} \tan ^{-1}(x)$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(3*x)/2 + x^2/2 - x^3/(2*(1 + x^2)) - (3*ArcTan[x])/2 - Log[1 + x^2]/2

Rule 1804

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x
^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x
], x, 1]}, Simp[((c*x)^m*(a + b*x^2)^(p + 1)*(a*g - b*f*x))/(2*a*b*(p + 1)), x] + Dist[c/(2*a*b*(p + 1)), Int[
(c*x)^(m - 1)*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*b*(p + 1)*x*Q - a*g*m + b*f*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && GtQ[m, 0]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

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 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]

Rubi steps

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

Mathematica [A]  time = 0.0178352, size = 29, normalized size = 0.67 $\frac{1}{2} \left (x \left (\frac{1}{x^2+1}+x+2\right )-\log \left (x^2+1\right )-3 \tan ^{-1}(x)\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(x*(2 + x + (1 + x^2)^(-1)) - 3*ArcTan[x] - Log[1 + x^2])/2

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Maple [A]  time = 0.046, size = 30, normalized size = 0.7 \begin{align*} x+{\frac{{x}^{2}}{2}}+{\frac{x}{2\,{x}^{2}+2}}-{\frac{\ln \left ({x}^{2}+1 \right ) }{2}}-{\frac{3\,\arctan \left ( x \right ) }{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x+1/2*x^2+1/2*x/(x^2+1)-1/2*ln(x^2+1)-3/2*arctan(x)

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Maxima [A]  time = 1.47725, size = 39, normalized size = 0.91 \begin{align*} \frac{1}{2} \, x^{2} + x + \frac{x}{2 \,{\left (x^{2} + 1\right )}} - \frac{3}{2} \, \arctan \left (x\right ) - \frac{1}{2} \, \log \left (x^{2} + 1\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2 + x + 1/2*x/(x^2 + 1) - 3/2*arctan(x) - 1/2*log(x^2 + 1)

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Fricas [A]  time = 0.974796, size = 122, normalized size = 2.84 \begin{align*} \frac{x^{4} + 2 \, x^{3} + x^{2} - 3 \,{\left (x^{2} + 1\right )} \arctan \left (x\right ) -{\left (x^{2} + 1\right )} \log \left (x^{2} + 1\right ) + 3 \, x}{2 \,{\left (x^{2} + 1\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.119279, size = 29, normalized size = 0.67 \begin{align*} \frac{x^{2}}{2} + x + \frac{x}{2 x^{2} + 2} - \frac{\log{\left (x^{2} + 1 \right )}}{2} - \frac{3 \operatorname{atan}{\left (x \right )}}{2} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2/2 + x + x/(2*x**2 + 2) - log(x**2 + 1)/2 - 3*atan(x)/2

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Giac [A]  time = 1.15523, size = 39, normalized size = 0.91 \begin{align*} \frac{1}{2} \, x^{2} + x + \frac{x}{2 \,{\left (x^{2} + 1\right )}} - \frac{3}{2} \, \arctan \left (x\right ) - \frac{1}{2} \, \log \left (x^{2} + 1\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2 + x + 1/2*x/(x^2 + 1) - 3/2*arctan(x) - 1/2*log(x^2 + 1)