3.627 \(\int \frac {1}{x \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx\)

Optimal. Leaf size=80 \[ \frac {\log (x) \left (a+b x^2\right )}{a \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1112, 266, 36, 29, 31} \[ \frac {\log (x) \left (a+b x^2\right )}{a \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^2)*Log[x])/(a*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - ((a + b*x^2)*Log[a + b*x^2])/(2*a*Sqrt[a^2 + 2*a*b*
x^2 + b^2*x^4])

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 266

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]]

Rule 1112

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 42, normalized size = 0.52 \[ \frac {\left (a+b x^2\right ) \left (2 \log (x)-\log \left (a+b x^2\right )\right )}{2 a \sqrt {\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.94, size = 18, normalized size = 0.22 \[ -\frac {\log \left (b x^{2} + a\right ) - 2 \, \log \relax (x)}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(log(b*x^2 + a) - 2*log(x))/a

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giac [A]  time = 0.15, size = 33, normalized size = 0.41 \[ \frac {1}{2} \, {\left (\frac {\log \left (x^{2}\right )}{a} - \frac {\log \left ({\left | b x^{2} + a \right |}\right )}{a}\right )} \mathrm {sgn}\left (b x^{2} + a\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 39, normalized size = 0.49 \[ \frac {\left (b \,x^{2}+a \right ) \left (2 \ln \relax (x )-\ln \left (b \,x^{2}+a \right )\right )}{2 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.28, size = 23, normalized size = 0.29 \[ -\frac {\log \left (b x^{2} + a\right )}{2 \, a} + \frac {\log \left (x^{2}\right )}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.45, size = 40, normalized size = 0.50 \[ -\frac {\ln \left (\sqrt {{\left (b\,x^2+a\right )}^2}\,\sqrt {a^2}+a^2+a\,b\,x^2\right )+\ln \left (\frac {1}{x^2}\right )}{2\,\sqrt {a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*((a + b*x^2)^2)^(1/2)),x)

[Out]

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

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sympy [A]  time = 0.26, size = 15, normalized size = 0.19 \[ \frac {\log {\relax (x )}}{a} - \frac {\log {\left (\frac {a}{b} + x^{2} \right )}}{2 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x)/a - log(a/b + x**2)/(2*a)

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