3.69.37 $\int \frac{-112-96x^2+2x^3+16x^4}{x^2+x^4}dx$

Optimal. Leaf size=22 $$\log \left(e^{\frac{16(7+x)(x+x^2)}{x^2}}(1+x^2)\right)$$

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Rubi [A]  time = 0.05, antiderivative size = 15, normalized size of antiderivative = 0.68, number of steps used = 4, number of rules used = 3, integrand size = 27, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.111, Rules used = {1593, 1802, 260} $$\begin{array}{c}\log (x^2+1)+16x+\frac{112}{x}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(-112 - 96*x^2 + 2*x^3 + 16*x^4)/(x^2 + x^4),x]

[Out]

112/x + 16*x + Log[1 + x^2]

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 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]

Rule 1802

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{-112-96x^2+2x^3+16x^4}{x^2(1+x^2)}dx\\ & =\int (16-\frac{112}{x^2}+\frac{2x}{1+x^2})dx\\ & =\frac{112}{x}+16x+2\int \frac{x}{1+x^2}dx\\ & =\frac{112}{x}+16x+\log (1+x^2)\end{array}\end{array}$$

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Mathematica [A]  time = 0.01, size = 21, normalized size = 0.95 $$\begin{array}{c}2(\frac{56}{x}+8x+\frac{1}{2}\log (1+x^2))\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-112 - 96*x^2 + 2*x^3 + 16*x^4)/(x^2 + x^4),x]

[Out]

2*(56/x + 8*x + Log[1 + x^2]/2)

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fricas [A]  time = 1.02, size = 19, normalized size = 0.86 $$\begin{array}{c}\frac{16x^2+x\log (x^2+1)+112}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x^4+2*x^3-96*x^2-112)/(x^4+x^2),x, algorithm="fricas")

[Out]

(16*x^2 + x*log(x^2 + 1) + 112)/x

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giac [A]  time = 0.15, size = 15, normalized size = 0.68 $$\begin{array}{c}16x+\frac{112}{x}+\log (x^2+1)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x^4+2*x^3-96*x^2-112)/(x^4+x^2),x, algorithm="giac")

[Out]

16*x + 112/x + log(x^2 + 1)

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maple [A]  time = 0.08, size = 16, normalized size = 0.73

method	result	size
default	$16x+\ln (x^2+1)+\frac{112}{x}$	$16$
meijerg	$16x+\ln (x^2+1)+\frac{112}{x}$	$16$
risch	$16x+\ln (x^2+1)+\frac{112}{x}$	$16$
norman	$\frac{16x^2+112}{x}+\ln (x^2+1)$	$19$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x^4+2*x^3-96*x^2-112)/(x^4+x^2),x,method=_RETURNVERBOSE)

[Out]

16*x+ln(x^2+1)+112/x

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maxima [A]  time = 0.44, size = 15, normalized size = 0.68 $$\begin{array}{c}16x+\frac{112}{x}+\log (x^2+1)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x^4+2*x^3-96*x^2-112)/(x^4+x^2),x, algorithm="maxima")

[Out]

16*x + 112/x + log(x^2 + 1)

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mupad [B]  time = 0.04, size = 15, normalized size = 0.68 $$\begin{array}{c}16x+\ln (x^2+1)+\frac{112}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(96*x^2 - 2*x^3 - 16*x^4 + 112)/(x^2 + x^4),x)

[Out]

16*x + log(x^2 + 1) + 112/x

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sympy [A]  time = 0.12, size = 12, normalized size = 0.55 $$\begin{array}{c}16x+\log (x^2+1)+\frac{112}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x**4+2*x**3-96*x**2-112)/(x**4+x**2),x)

[Out]

16*x + log(x**2 + 1) + 112/x

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