3.3.67 \(\int \frac {3 x-4 x^2+3 x^3}{1+x^2} \, dx\) [267]

Optimal. Leaf size=15 \[ -4 x+\frac {3 x^2}{2}+4 \tan ^{-1}(x) \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1608, 1816, 209} \begin {gather*} 4 \text {ArcTan}(x)+\frac {3 x^2}{2}-4 x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-4*x + (3*x^2)/2 + 4*ArcTan[x]

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

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Mathematica [A]
time = 0.00, size = 15, normalized size = 1.00 \begin {gather*} -4 x+\frac {3 x^2}{2}+4 \tan ^{-1}(x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-4*x + (3*x^2)/2 + 4*ArcTan[x]

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Maple [A]
time = 0.16, size = 14, normalized size = 0.93

method	result	size
default	\(-4 x +\frac {3 x^{2}}{2}+4 \arctan \left (x \right )\)	\(14\)
meijerg	\(-4 x +\frac {3 x^{2}}{2}+4 \arctan \left (x \right )\)	\(14\)
risch	\(-4 x +\frac {3 x^{2}}{2}+4 \arctan \left (x \right )\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.47, size = 13, normalized size = 0.87 \begin {gather*} \frac {3}{2} \, x^{2} - 4 \, x + 4 \, \arctan \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*x^2 - 4*x + 4*arctan(x)

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Fricas [A]
time = 0.37, size = 13, normalized size = 0.87 \begin {gather*} \frac {3}{2} \, x^{2} - 4 \, x + 4 \, \arctan \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*x^2 - 4*x + 4*arctan(x)

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Sympy [A]
time = 0.02, size = 14, normalized size = 0.93 \begin {gather*} \frac {3 x^{2}}{2} - 4 x + 4 \operatorname {atan}{\left (x \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*x**2/2 - 4*x + 4*atan(x)

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Giac [A]
time = 2.93, size = 13, normalized size = 0.87 \begin {gather*} \frac {3}{2} \, x^{2} - 4 \, x + 4 \, \arctan \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*x^2 - 4*x + 4*arctan(x)

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Mupad [B]
time = 2.13, size = 13, normalized size = 0.87 \begin {gather*} 4\,\mathrm {atan}\left (x\right )-4\,x+\frac {3\,x^2}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*atan(x) - 4*x + (3*x^2)/2

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