### 3.2266 $$\int \frac{x^4}{4+4 x+x^2} \, dx$$

Optimal. Leaf size=29 $\frac{x^3}{3}-2 x^2+12 x-\frac{16}{x+2}-32 \log (x+2)$

[Out]

12*x - 2*x^2 + x^3/3 - 16/(2 + x) - 32*Log[2 + x]

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Rubi [A]  time = 0.0147971, antiderivative size = 29, 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} $\frac{x^3}{3}-2 x^2+12 x-\frac{16}{x+2}-32 \log (x+2)$

Antiderivative was successfully veriﬁed.

[In]

Int[x^4/(4 + 4*x + x^2),x]

[Out]

12*x - 2*x^2 + x^3/3 - 16/(2 + x) - 32*Log[2 + 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^4}{4+4 x+x^2} \, dx &=\int \frac{x^4}{(2+x)^2} \, dx\\ &=\int \left (12-4 x+x^2+\frac{16}{(2+x)^2}-\frac{32}{2+x}\right ) \, dx\\ &=12 x-2 x^2+\frac{x^3}{3}-\frac{16}{2+x}-32 \log (2+x)\\ \end{align*}

Mathematica [A]  time = 0.009439, size = 30, normalized size = 1.03 $\frac{1}{3} \left (x^3-6 x^2+36 x-\frac{48}{x+2}-96 \log (x+2)+104\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^4/(4 + 4*x + x^2),x]

[Out]

(104 + 36*x - 6*x^2 + x^3 - 48/(2 + x) - 96*Log[2 + x])/3

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Maple [A]  time = 0.044, size = 28, normalized size = 1. \begin{align*} 12\,x-2\,{x}^{2}+{\frac{{x}^{3}}{3}}-16\, \left ( 2+x \right ) ^{-1}-32\,\ln \left ( 2+x \right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

12*x-2*x^2+1/3*x^3-16/(2+x)-32*ln(2+x)

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Maxima [A]  time = 0.985296, size = 36, normalized size = 1.24 \begin{align*} \frac{1}{3} \, x^{3} - 2 \, x^{2} + 12 \, x - \frac{16}{x + 2} - 32 \, \log \left (x + 2\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3 - 2*x^2 + 12*x - 16/(x + 2) - 32*log(x + 2)

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Fricas [A]  time = 2.2301, size = 97, normalized size = 3.34 \begin{align*} \frac{x^{4} - 4 \, x^{3} + 24 \, x^{2} - 96 \,{\left (x + 2\right )} \log \left (x + 2\right ) + 72 \, x - 48}{3 \,{\left (x + 2\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(x^4 - 4*x^3 + 24*x^2 - 96*(x + 2)*log(x + 2) + 72*x - 48)/(x + 2)

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Sympy [A]  time = 0.08555, size = 24, normalized size = 0.83 \begin{align*} \frac{x^{3}}{3} - 2 x^{2} + 12 x - 32 \log{\left (x + 2 \right )} - \frac{16}{x + 2} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**3/3 - 2*x**2 + 12*x - 32*log(x + 2) - 16/(x + 2)

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Giac [A]  time = 1.11598, size = 38, normalized size = 1.31 \begin{align*} \frac{1}{3} \, x^{3} - 2 \, x^{2} + 12 \, x - \frac{16}{x + 2} - 32 \, \log \left ({\left | x + 2 \right |}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3 - 2*x^2 + 12*x - 16/(x + 2) - 32*log(abs(x + 2))