3.5.81 \(\int \frac {1}{x (a^2+2 a b x^2+b^2 x^4)} \, dx\) [481]

Optimal. Leaf size=38 \[ \frac {1}{2 a \left (a+b x^2\right )}+\frac {\log (x)}{a^2}-\frac {\log \left (a+b x^2\right )}{2 a^2} \]

[Out]

1/2/a/(b*x^2+a)+ln(x)/a^2-1/2*ln(b*x^2+a)/a^2

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Rubi [A]
time = 0.03, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {28, 272, 46} \begin {gather*} -\frac {\log \left (a+b x^2\right )}{2 a^2}+\frac {\log (x)}{a^2}+\frac {1}{2 a \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*(a^2 + 2*a*b*x^2 + b^2*x^4)),x]

[Out]

1/(2*a*(a + b*x^2)) + Log[x]/a^2 - Log[a + b*x^2]/(2*a^2)

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 272

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

Rubi steps

\begin {align*} \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx &=b^2 \int \frac {1}{x \left (a b+b^2 x^2\right )^2} \, dx\\ &=\frac {1}{2} b^2 \text {Subst}\left (\int \frac {1}{x \left (a b+b^2 x\right )^2} \, dx,x,x^2\right )\\ &=\frac {1}{2} b^2 \text {Subst}\left (\int \left (\frac {1}{a^2 b^2 x}-\frac {1}{a b (a+b x)^2}-\frac {1}{a^2 b (a+b x)}\right ) \, dx,x,x^2\right )\\ &=\frac {1}{2 a \left (a+b x^2\right )}+\frac {\log (x)}{a^2}-\frac {\log \left (a+b x^2\right )}{2 a^2}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 33, normalized size = 0.87 \begin {gather*} \frac {\frac {a}{a+b x^2}+2 \log (x)-\log \left (a+b x^2\right )}{2 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a/(a + b*x^2) + 2*Log[x] - Log[a + b*x^2])/(2*a^2)

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Maple [A]
time = 0.03, size = 42, normalized size = 1.11

method	result	size
risch	\(\frac {1}{2 a \left (b \,x^{2}+a \right )}+\frac {\ln \left (x \right )}{a^{2}}-\frac {\ln \left (b \,x^{2}+a \right )}{2 a^{2}}\)	\(35\)
norman	\(-\frac {b \,x^{2}}{2 a^{2} \left (b \,x^{2}+a \right )}+\frac {\ln \left (x \right )}{a^{2}}-\frac {\ln \left (b \,x^{2}+a \right )}{2 a^{2}}\)	\(39\)
default	\(-\frac {b \left (-\frac {a}{b \left (b \,x^{2}+a \right )}+\frac {\ln \left (b \,x^{2}+a \right )}{b}\right )}{2 a^{2}}+\frac {\ln \left (x \right )}{a^{2}}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b/a^2*(-a/b/(b*x^2+a)+1/b*ln(b*x^2+a))+ln(x)/a^2

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Maxima [A]
time = 0.27, size = 37, normalized size = 0.97 \begin {gather*} \frac {1}{2 \, {\left (a b x^{2} + a^{2}\right )}} - \frac {\log \left (b x^{2} + a\right )}{2 \, a^{2}} + \frac {\log \left (x^{2}\right )}{2 \, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/(a*b*x^2 + a^2) - 1/2*log(b*x^2 + a)/a^2 + 1/2*log(x^2)/a^2

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Fricas [A]
time = 0.35, size = 47, normalized size = 1.24 \begin {gather*} -\frac {{\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right ) - 2 \, {\left (b x^{2} + a\right )} \log \left (x\right ) - a}{2 \, {\left (a^{2} b x^{2} + a^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.13, size = 34, normalized size = 0.89 \begin {gather*} \frac {1}{2 a^{2} + 2 a b x^{2}} + \frac {\log {\left (x \right )}}{a^{2}} - \frac {\log {\left (\frac {a}{b} + x^{2} \right )}}{2 a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b**2*x**4+2*a*b*x**2+a**2),x)

[Out]

1/(2*a**2 + 2*a*b*x**2) + log(x)/a**2 - log(a/b + x**2)/(2*a**2)

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Giac [A]
time = 3.56, size = 47, normalized size = 1.24 \begin {gather*} \frac {\log \left (x^{2}\right )}{2 \, a^{2}} - \frac {\log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{2}} + \frac {b x^{2} + 2 \, a}{2 \, {\left (b x^{2} + a\right )} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(x^2)/a^2 - 1/2*log(abs(b*x^2 + a))/a^2 + 1/2*(b*x^2 + 2*a)/((b*x^2 + a)*a^2)

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Mupad [B]
time = 4.39, size = 34, normalized size = 0.89 \begin {gather*} \frac {\ln \left (x\right )}{a^2}+\frac {1}{2\,a\,\left (b\,x^2+a\right )}-\frac {\ln \left (b\,x^2+a\right )}{2\,a^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a^2 + b^2*x^4 + 2*a*b*x^2)),x)

[Out]

log(x)/a^2 + 1/(2*a*(a + b*x^2)) - log(a + b*x^2)/(2*a^2)

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