3.446 \(\int \frac {1-5 x^2}{x^3 (1+x^2)} \, dx\)

Optimal. Leaf size=20 \[ -\frac {1}{2 x^2}+3 \log \left (x^2+1\right )-6 \log (x) \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {446, 77} \[ -\frac {1}{2 x^2}+3 \log \left (x^2+1\right )-6 \log (x) \]

Antiderivative was successfully verified.

[In]

Int[(1 - 5*x^2)/(x^3*(1 + x^2)),x]

[Out]

-1/(2*x^2) - 6*Log[x] + 3*Log[1 + x^2]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {1-5 x^2}{x^3 \left (1+x^2\right )} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1-5 x}{x^2 (1+x)} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {1}{x^2}-\frac {6}{x}+\frac {6}{1+x}\right ) \, dx,x,x^2\right )\\ &=-\frac {1}{2 x^2}-6 \log (x)+3 \log \left (1+x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 20, normalized size = 1.00 \[ -\frac {1}{2 x^2}+3 \log \left (x^2+1\right )-6 \log (x) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - 5*x^2)/(x^3*(1 + x^2)),x]

[Out]

-1/2*1/x^2 - 6*Log[x] + 3*Log[1 + x^2]

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fricas [A]  time = 0.67, size = 25, normalized size = 1.25 \[ \frac {6 \, x^{2} \log \left (x^{2} + 1\right ) - 12 \, x^{2} \log \relax (x) - 1}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*x^2+1)/x^3/(x^2+1),x, algorithm="fricas")

[Out]

1/2*(6*x^2*log(x^2 + 1) - 12*x^2*log(x) - 1)/x^2

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giac [A]  time = 0.38, size = 27, normalized size = 1.35 \[ \frac {6 \, x^{2} - 1}{2 \, x^{2}} + 3 \, \log \left (x^{2} + 1\right ) - 3 \, \log \left (x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*x^2+1)/x^3/(x^2+1),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 19, normalized size = 0.95 \[ -6 \ln \relax (x )+3 \ln \left (x^{2}+1\right )-\frac {1}{2 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-5*x^2+1)/x^3/(x^2+1),x)

[Out]

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

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maxima [A]  time = 0.72, size = 20, normalized size = 1.00 \[ -\frac {1}{2 \, x^{2}} + 3 \, \log \left (x^{2} + 1\right ) - 3 \, \log \left (x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*x^2+1)/x^3/(x^2+1),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.04, size = 18, normalized size = 0.90 \[ 3\,\ln \left (x^2+1\right )-6\,\ln \relax (x)-\frac {1}{2\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*x^2 - 1)/(x^3*(x^2 + 1)),x)

[Out]

3*log(x^2 + 1) - 6*log(x) - 1/(2*x^2)

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sympy [A]  time = 0.11, size = 19, normalized size = 0.95 \[ - 6 \log {\relax (x )} + 3 \log {\left (x^{2} + 1 \right )} - \frac {1}{2 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*x**2+1)/x**3/(x**2+1),x)

[Out]

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

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