3.442 \(\int \frac {4 x^2+x^3}{x+x^3} \, dx\)

Optimal. Leaf size=14 \[ 2 \log \left (x^2+1\right )+x-\tan ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {1593, 1584, 774, 635, 203, 260} \[ 2 \log \left (x^2+1\right )+x-\tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(4*x^2 + x^3)/(x + x^3),x]

[Out]

x - ArcTan[x] + 2*Log[1 + x^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 774

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

Rule 1584

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

Rule 1593

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

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 14, normalized size = 1.00 \[ 2 \log \left (x^2+1\right )+x-\tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Integrate[(4*x^2 + x^3)/(x + x^3),x]

[Out]

x - ArcTan[x] + 2*Log[1 + x^2]

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fricas [A]  time = 0.53, size = 14, normalized size = 1.00 \[ x - \arctan \relax (x) + 2 \, \log \left (x^{2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - arctan(x) + 2*log(x^2 + 1)

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giac [A]  time = 0.25, size = 14, normalized size = 1.00 \[ x - \arctan \relax (x) + 2 \, \log \left (x^{2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - arctan(x) + 2*log(x^2 + 1)

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maple [A]  time = 0.00, size = 15, normalized size = 1.07 \[ x -\arctan \relax (x )+2 \ln \left (x^{2}+1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.56, size = 14, normalized size = 1.00 \[ x - \arctan \relax (x) + 2 \, \log \left (x^{2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - arctan(x) + 2*log(x^2 + 1)

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mupad [B]  time = 2.22, size = 14, normalized size = 1.00 \[ x+2\,\ln \left (x^2+1\right )-\mathrm {atan}\relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^2 + x^3)/(x + x^3),x)

[Out]

x + 2*log(x^2 + 1) - atan(x)

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sympy [A]  time = 0.10, size = 12, normalized size = 0.86 \[ x + 2 \log {\left (x^{2} + 1 \right )} - \operatorname {atan}{\relax (x )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 2*log(x**2 + 1) - atan(x)

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