\(\int \frac {1}{x \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx\) [627]

Optimal result

Integrand size = 26, antiderivative size = 80 \[ \int \frac {1}{x \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {\left (a+b x^2\right ) \log (x)}{a \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[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)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1126, 272, 36, 29, 31} \[ \int \frac {1}{x \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {\log (x) \left (a+b x^2\right )}{a \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[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 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]]

Rule 1126

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*} \text {integral}& = \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 ) \text {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 ) \text {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 ) \text {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] (verified)

Time = 0.16 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.68 \[ \int \frac {1}{x \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {-2 a \log \left (x^2\right )+\left (a-\sqrt {a^2}\right ) \log \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )+a \log \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )+\sqrt {a^2} \log \left (a \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )\right )}{4 a \sqrt {a^2}} \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.14 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.36

method	result	size
pseudoelliptic	\(-\frac {\left (-\ln \left (x^{2}\right )+\ln \left (b \,x^{2}+a \right )\right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{2 a}\)	\(29\)
default	\(\frac {\left (b \,x^{2}+a \right ) \left (2 \ln \left (x \right )-\ln \left (b \,x^{2}+a \right )\right )}{2 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a}\)	\(39\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \ln \left (x \right )}{\left (b \,x^{2}+a \right ) a}-\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \ln \left (b \,x^{2}+a \right )}{2 \left (b \,x^{2}+a \right ) a}\)	\(61\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.22 \[ \int \frac {1}{x \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=-\frac {\log \left (b x^{2} + a\right ) - 2 \, \log \left (x\right )}{2 \, a} \]

[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

Sympy [F]

\[ \int \frac {1}{x \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\int \frac {1}{x \sqrt {\left (a + b x^{2}\right )^{2}}}\, dx \]

[In]

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

[Out]

Integral(1/(x*sqrt((a + b*x**2)**2)), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.29 \[ \int \frac {1}{x \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=-\frac {\log \left (b x^{2} + a\right )}{2 \, a} + \frac {\log \left (x^{2}\right )}{2 \, a} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.41 \[ \int \frac {1}{x \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {1}{2} \, {\left (\frac {\log \left (x^{2}\right )}{a} - \frac {\log \left ({\left | b x^{2} + a \right |}\right )}{a}\right )} \mathrm {sgn}\left (b x^{2} + a\right ) \]

[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)

Mupad [B] (verification not implemented)

Time = 13.96 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.50 \[ \int \frac {1}{x \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=-\frac {\ln \left (\sqrt {{\left (b\,x^2+a\right )}^2}\,\sqrt {a^2}+a^2+a\,b\,x^2\right )+\ln \left (\frac {1}{x^2}\right )}{2\,\sqrt {a^2}} \]

[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))