### 3.2264 $$\int \frac{x^3}{1+2 x+x^2} \, dx$$

Optimal. Leaf size=22 $\frac{x^2}{2}-2 x+\frac{1}{x+1}+3 \log (x+1)$

[Out]

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

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Rubi [A]  time = 0.0100248, antiderivative size = 22, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 14, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.143, Rules used = {27, 43} $\frac{x^2}{2}-2 x+\frac{1}{x+1}+3 \log (x+1)$

Antiderivative was successfully veriﬁed.

[In]

Int[x^3/(1 + 2*x + x^2),x]

[Out]

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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

Mathematica [A]  time = 0.008752, size = 26, normalized size = 1.18 $\frac{1}{2} (x+1)^2-3 (x+1)+\frac{1}{x+1}+3 \log (x+1)$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^3/(1 + 2*x + x^2),x]

[Out]

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

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Maple [A]  time = 0.043, size = 21, normalized size = 1. \begin{align*} -2\,x+{\frac{{x}^{2}}{2}}+ \left ( 1+x \right ) ^{-1}+3\,\ln \left ( 1+x \right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.957131, size = 27, normalized size = 1.23 \begin{align*} \frac{1}{2} \, x^{2} - 2 \, x + \frac{1}{x + 1} + 3 \, \log \left (x + 1\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.24834, size = 81, normalized size = 3.68 \begin{align*} \frac{x^{3} - 3 \, x^{2} + 6 \,{\left (x + 1\right )} \log \left (x + 1\right ) - 4 \, x + 2}{2 \,{\left (x + 1\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(x^3 - 3*x^2 + 6*(x + 1)*log(x + 1) - 4*x + 2)/(x + 1)

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Sympy [A]  time = 0.079324, size = 19, normalized size = 0.86 \begin{align*} \frac{x^{2}}{2} - 2 x + 3 \log{\left (x + 1 \right )} + \frac{1}{x + 1} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.12364, size = 28, normalized size = 1.27 \begin{align*} \frac{1}{2} \, x^{2} - 2 \, x + \frac{1}{x + 1} + 3 \, \log \left ({\left | x + 1 \right |}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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