### 3.197 $$\int \frac{x^3}{1+x^2} \, dx$$

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

[Out]

x^2/2 - Log[1 + x^2]/2

________________________________________________________________________________________

Rubi [A]  time = 0.0075449, antiderivative size = 18, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 11, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.182, Rules used = {266, 43} $\frac{x^2}{2}-\frac{1}{2} \log \left (x^2+1\right )$

Antiderivative was successfully veriﬁed.

[In]

Int[x^3/(1 + x^2),x]

[Out]

x^2/2 - Log[1 + x^2]/2

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0026922, size = 18, normalized size = 1. $\frac{x^2}{2}-\frac{1}{2} \log \left (x^2+1\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^3/(1 + x^2),x]

[Out]

x^2/2 - Log[1 + x^2]/2

________________________________________________________________________________________

Maple [A]  time = 0.001, size = 15, normalized size = 0.8 \begin{align*}{\frac{{x}^{2}}{2}}-{\frac{\ln \left ({x}^{2}+1 \right ) }{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 0.927049, size = 19, normalized size = 1.06 \begin{align*} \frac{1}{2} \, x^{2} - \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+1),x, algorithm="maxima")

[Out]

1/2*x^2 - 1/2*log(x^2 + 1)

________________________________________________________________________________________

Fricas [A]  time = 1.81593, size = 38, normalized size = 2.11 \begin{align*} \frac{1}{2} \, x^{2} - \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+1),x, algorithm="fricas")

[Out]

1/2*x^2 - 1/2*log(x^2 + 1)

________________________________________________________________________________________

Sympy [A]  time = 0.069481, size = 12, normalized size = 0.67 \begin{align*} \frac{x^{2}}{2} - \frac{\log{\left (x^{2} + 1 \right )}}{2} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2/2 - log(x**2 + 1)/2

________________________________________________________________________________________

Giac [A]  time = 1.05878, size = 19, normalized size = 1.06 \begin{align*} \frac{1}{2} \, x^{2} - \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+1),x, algorithm="giac")

[Out]

1/2*x^2 - 1/2*log(x^2 + 1)