3.179 \(\int \frac {4+3 x^4}{x^2 (1+x^2)^3} \, dx\)

Optimal. Leaf size=36 \[ -\frac {25 x}{8 \left (x^2+1\right )}-\frac {7 x}{4 \left (x^2+1\right )^2}-\frac {4}{x}-\frac {57}{8} \tan ^{-1}(x) \]

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Rubi [A]  time = 0.02, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1260, 456, 453, 203} \[ -\frac {25 x}{8 \left (x^2+1\right )}-\frac {7 x}{4 \left (x^2+1\right )^2}-\frac {4}{x}-\frac {57}{8} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(4 + 3*x^4)/(x^2*(1 + x^2)^3),x]

[Out]

-4/x - (7*x)/(4*(1 + x^2)^2) - (25*x)/(8*(1 + x^2)) - (57*ArcTan[x])/8

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 453

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

Rule 456

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)]
- ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &
& ILtQ[m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1260

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((-d)^(m/2 - 1)*(c*d^2
+ a*e^2)^p*x*(d + e*x^2)^(q + 1))/(2*e^(2*p + m/2)*(q + 1)), x] + Dist[(-d)^(m/2 - 1)/(2*e^(2*p)*(q + 1)), In
t[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*(-d)^(-(m/2) + 1)*e^(2*p)*(q + 1)*(a + c*x^4)^p - ((c*d^2
+ a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2)))/(d + e*x^2)], x], x], x] /; FreeQ[{a, c, d, e}, x] && IGtQ[
p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]

Rubi steps

\begin {align*} \int \frac {4+3 x^4}{x^2 \left (1+x^2\right )^3} \, dx &=-\frac {7 x}{4 \left (1+x^2\right )^2}-\frac {1}{4} \int \frac {-16+9 x^2}{x^2 \left (1+x^2\right )^2} \, dx\\ &=-\frac {7 x}{4 \left (1+x^2\right )^2}-\frac {25 x}{8 \left (1+x^2\right )}+\frac {1}{8} \int \frac {32-25 x^2}{x^2 \left (1+x^2\right )} \, dx\\ &=-\frac {4}{x}-\frac {7 x}{4 \left (1+x^2\right )^2}-\frac {25 x}{8 \left (1+x^2\right )}-\frac {57}{8} \int \frac {1}{1+x^2} \, dx\\ &=-\frac {4}{x}-\frac {7 x}{4 \left (1+x^2\right )^2}-\frac {25 x}{8 \left (1+x^2\right )}-\frac {57}{8} \tan ^{-1}(x)\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 33, normalized size = 0.92 \[ -\frac {57 x^4+103 x^2+32}{8 x \left (x^2+1\right )^2}-\frac {57}{8} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Integrate[(4 + 3*x^4)/(x^2*(1 + x^2)^3),x]

[Out]

-1/8*(32 + 103*x^2 + 57*x^4)/(x*(1 + x^2)^2) - (57*ArcTan[x])/8

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IntegrateAlgebraic [A]  time = 0.00, size = 33, normalized size = 0.92 \[ \frac {-57 x^4-103 x^2-32}{8 x \left (x^2+1\right )^2}-\frac {57}{8} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(4 + 3*x^4)/(x^2*(1 + x^2)^3),x]

[Out]

(-32 - 103*x^2 - 57*x^4)/(8*x*(1 + x^2)^2) - (57*ArcTan[x])/8

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fricas [A]  time = 0.70, size = 40, normalized size = 1.11 \[ -\frac {57 \, x^{4} + 103 \, x^{2} + 57 \, {\left (x^{5} + 2 \, x^{3} + x\right )} \arctan \relax (x) + 32}{8 \, {\left (x^{5} + 2 \, x^{3} + x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^4+4)/x^2/(x^2+1)^3,x, algorithm="fricas")

[Out]

-1/8*(57*x^4 + 103*x^2 + 57*(x^5 + 2*x^3 + x)*arctan(x) + 32)/(x^5 + 2*x^3 + x)

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giac [A]  time = 0.92, size = 28, normalized size = 0.78 \[ -\frac {25 \, x^{3} + 39 \, x}{8 \, {\left (x^{2} + 1\right )}^{2}} - \frac {4}{x} - \frac {57}{8} \, \arctan \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^4+4)/x^2/(x^2+1)^3,x, algorithm="giac")

[Out]

-1/8*(25*x^3 + 39*x)/(x^2 + 1)^2 - 4/x - 57/8*arctan(x)

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maple [A]  time = 0.00, size = 29, normalized size = 0.81

method	result	size
default	\(-\frac {4}{x}-\frac {\frac {25}{8} x^{3}+\frac {39}{8} x}{\left (x^{2}+1\right )^{2}}-\frac {57 \arctan \relax (x )}{8}\)	\(29\)
risch	\(\frac {-\frac {57}{8} x^{4}-\frac {103}{8} x^{2}-4}{\left (x^{2}+1\right )^{2} x}-\frac {57 \arctan \relax (x )}{8}\)	\(29\)
meijerg	\(-\frac {15 x^{4}+25 x^{2}+8}{2 x \left (x^{2}+1\right )^{2}}-\frac {57 \arctan \relax (x )}{8}-\frac {x \left (-3 x^{2}+3\right )}{8 \left (x^{2}+1\right )^{2}}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^4+4)/x^2/(x^2+1)^3,x,method=_RETURNVERBOSE)

[Out]

-4/x-(25/8*x^3+39/8*x)/(x^2+1)^2-57/8*arctan(x)

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maxima [A]  time = 1.42, size = 31, normalized size = 0.86 \[ -\frac {57 \, x^{4} + 103 \, x^{2} + 32}{8 \, {\left (x^{5} + 2 \, x^{3} + x\right )}} - \frac {57}{8} \, \arctan \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^4+4)/x^2/(x^2+1)^3,x, algorithm="maxima")

[Out]

-1/8*(57*x^4 + 103*x^2 + 32)/(x^5 + 2*x^3 + x) - 57/8*arctan(x)

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mupad [B]  time = 0.00, size = 29, normalized size = 0.81 \[ -\frac {57\,\mathrm {atan}\relax (x)}{8}-\frac {\frac {57\,x^4}{8}+\frac {103\,x^2}{8}+4}{x\,{\left (x^2+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^4 + 4)/(x^2*(x^2 + 1)^3),x)

[Out]

- (57*atan(x))/8 - ((103*x^2)/8 + (57*x^4)/8 + 4)/(x*(x^2 + 1)^2)

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sympy [A]  time = 0.15, size = 32, normalized size = 0.89 \[ \frac {- 57 x^{4} - 103 x^{2} - 32}{8 x^{5} + 16 x^{3} + 8 x} - \frac {57 \operatorname {atan}{\relax (x )}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x**4+4)/x**2/(x**2+1)**3,x)

[Out]

(-57*x**4 - 103*x**2 - 32)/(8*x**5 + 16*x**3 + 8*x) - 57*atan(x)/8

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