3.2.17 \(\int \frac {1+x^2}{(16+x^2)^3} \, dx\)

Optimal. Leaf size=35 \[ \frac {19 x}{2048 \left (x^2+16\right )}-\frac {15 x}{64 \left (x^2+16\right )^2}+\frac {19 \tan ^{-1}\left (\frac {x}{4}\right )}{8192} \]

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Rubi [A]  time = 0.01, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {385, 199, 203} \begin {gather*} \frac {19 x}{2048 \left (x^2+16\right )}-\frac {15 x}{64 \left (x^2+16\right )^2}+\frac {19 \tan ^{-1}\left (\frac {x}{4}\right )}{8192} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + x^2)/(16 + x^2)^3,x]

[Out]

(-15*x)/(64*(16 + x^2)^2) + (19*x)/(2048*(16 + x^2)) + (19*ArcTan[x/4])/8192

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rubi steps

\begin {align*} \int \frac {1+x^2}{\left (16+x^2\right )^3} \, dx &=-\frac {15 x}{64 \left (16+x^2\right )^2}+\frac {19}{64} \int \frac {1}{\left (16+x^2\right )^2} \, dx\\ &=-\frac {15 x}{64 \left (16+x^2\right )^2}+\frac {19 x}{2048 \left (16+x^2\right )}+\frac {19 \int \frac {1}{16+x^2} \, dx}{2048}\\ &=-\frac {15 x}{64 \left (16+x^2\right )^2}+\frac {19 x}{2048 \left (16+x^2\right )}+\frac {19 \tan ^{-1}\left (\frac {x}{4}\right )}{8192}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 35, normalized size = 1.00 \begin {gather*} \frac {19 x}{2048 \left (x^2+16\right )}-\frac {15 x}{64 \left (x^2+16\right )^2}+\frac {19 \tan ^{-1}\left (\frac {x}{4}\right )}{8192} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + x^2)/(16 + x^2)^3,x]

[Out]

(-15*x)/(64*(16 + x^2)^2) + (19*x)/(2048*(16 + x^2)) + (19*ArcTan[x/4])/8192

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1+x^2}{\left (16+x^2\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(1 + x^2)/(16 + x^2)^3,x]

[Out]

IntegrateAlgebraic[(1 + x^2)/(16 + x^2)^3, x]

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fricas [A]  time = 0.48, size = 39, normalized size = 1.11 \begin {gather*} \frac {76 \, x^{3} + 19 \, {\left (x^{4} + 32 \, x^{2} + 256\right )} \arctan \left (\frac {1}{4} \, x\right ) - 704 \, x}{8192 \, {\left (x^{4} + 32 \, x^{2} + 256\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+1)/(x^2+16)^3,x, algorithm="fricas")

[Out]

1/8192*(76*x^3 + 19*(x^4 + 32*x^2 + 256)*arctan(1/4*x) - 704*x)/(x^4 + 32*x^2 + 256)

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giac [A]  time = 0.41, size = 25, normalized size = 0.71 \begin {gather*} \frac {19 \, x^{3} - 176 \, x}{2048 \, {\left (x^{2} + 16\right )}^{2}} + \frac {19}{8192} \, \arctan \left (\frac {1}{4} \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+1)/(x^2+16)^3,x, algorithm="giac")

[Out]

1/2048*(19*x^3 - 176*x)/(x^2 + 16)^2 + 19/8192*arctan(1/4*x)

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maple [A]  time = 0.01, size = 25, normalized size = 0.71 \begin {gather*} \frac {19 \arctan \left (\frac {x}{4}\right )}{8192}+\frac {\frac {19}{2048} x^{3}-\frac {11}{128} x}{\left (x^{2}+16\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2+1)/(x^2+16)^3,x)

[Out]

(19/2048*x^3-11/128*x)/(x^2+16)^2+19/8192*arctan(1/4*x)

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maxima [A]  time = 2.47, size = 30, normalized size = 0.86 \begin {gather*} \frac {19 \, x^{3} - 176 \, x}{2048 \, {\left (x^{4} + 32 \, x^{2} + 256\right )}} + \frac {19}{8192} \, \arctan \left (\frac {1}{4} \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+1)/(x^2+16)^3,x, algorithm="maxima")

[Out]

1/2048*(19*x^3 - 176*x)/(x^4 + 32*x^2 + 256) + 19/8192*arctan(1/4*x)

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mupad [B]  time = 0.08, size = 30, normalized size = 0.86 \begin {gather*} \frac {19\,\mathrm {atan}\left (\frac {x}{4}\right )}{8192}-\frac {\frac {11\,x}{128}-\frac {19\,x^3}{2048}}{x^4+32\,x^2+256} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 + 1)/(x^2 + 16)^3,x)

[Out]

(19*atan(x/4))/8192 - ((11*x)/128 - (19*x^3)/2048)/(32*x^2 + x^4 + 256)

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sympy [A]  time = 0.13, size = 27, normalized size = 0.77 \begin {gather*} \frac {19 x^{3} - 176 x}{2048 x^{4} + 65536 x^{2} + 524288} + \frac {19 \operatorname {atan}{\left (\frac {x}{4} \right )}}{8192} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2+1)/(x**2+16)**3,x)

[Out]

(19*x**3 - 176*x)/(2048*x**4 + 65536*x**2 + 524288) + 19*atan(x/4)/8192

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