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]

((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])

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Rubi [A]  time = 0.0337835, antiderivative size = 80, normalized size of antiderivative = 1., 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 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]

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 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 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]

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*}

Mathematica [A]  time = 0.0112525, 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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Maple [A]  time = 0.21, size = 39, normalized size = 0.5 \begin{align*}{\frac{ \left ( b{x}^{2}+a \right ) \left ( 2\,\ln \left ( x \right ) -\ln \left ( b{x}^{2}+a \right ) \right ) }{2\,a}{\frac{1}{\sqrt{ \left ( b{x}^{2}+a \right ) ^{2}}}}} \end{align*}

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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.24601, size = 49, normalized size = 0.61 \begin{align*} -\frac{\log \left (b x^{2} + a\right ) - 2 \, \log \left (x\right )}{2 \, a} \end{align*}

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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Sympy [A]  time = 0.222443, size = 15, normalized size = 0.19 \begin{align*} \frac{\log{\left (x \right )}}{a} - \frac{\log{\left (\frac{a}{b} + x^{2} \right )}}{2 a} \end{align*}

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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Giac [A]  time = 1.11977, size = 45, normalized size = 0.56 \begin{align*} \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 ) \end{align*}

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)