\(\int \frac {1}{x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx\) [632]

Optimal result

Integrand size = 26, antiderivative size = 92 \[ \int \frac {1}{x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=-\frac {a+b x^2}{a x \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\sqrt {b} \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1126, 331, 211} \[ \int \frac {1}{x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=-\frac {a+b x^2}{a x \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\sqrt {b} \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[In]

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

[Out]

-((a + b*x^2)/(a*x*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])) - (Sqrt[b]*(a + b*x^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(a^(3
/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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^2 \left (a b+b^2 x^2\right )} \, dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = -\frac {a+b x^2}{a x \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (b \left (a b+b^2 x^2\right )\right ) \int \frac {1}{a b+b^2 x^2} \, dx}{a \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = -\frac {a+b x^2}{a x \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\sqrt {b} \left (a+b x^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.61 \[ \int \frac {1}{x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=-\frac {\left (a+b x^2\right ) \left (\sqrt {a}+\sqrt {b} x \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right )}{a^{3/2} x \sqrt {\left (a+b x^2\right )^2}} \]

[In]

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

[Out]

-(((a + b*x^2)*(Sqrt[a] + Sqrt[b]*x*ArcTan[(Sqrt[b]*x)/Sqrt[a]]))/(a^(3/2)*x*Sqrt[(a + b*x^2)^2]))

Maple [A] (verified)

Time = 0.81 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.54

method	result	size
default	\(-\frac {\left (b \,x^{2}+a \right ) \left (b \arctan \left (\frac {b x}{\sqrt {a b}}\right ) x +\sqrt {a b}\right )}{\sqrt {\left (b \,x^{2}+a \right )^{2}}\, a \sqrt {a b}\, x}\)	\(50\)
risch	\(-\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}}{\left (b \,x^{2}+a \right ) a x}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \sqrt {-a b}\, \ln \left (-b x +\sqrt {-a b}\right )}{2 \left (b \,x^{2}+a \right ) a^{2}}-\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \sqrt {-a b}\, \ln \left (-b x -\sqrt {-a b}\right )}{2 \left (b \,x^{2}+a \right ) a^{2}}\)	\(118\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\left [\frac {x \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right ) - 2}{2 \, a x}, -\frac {x \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right ) + 1}{a x}\right ] \]

[In]

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

[Out]

[1/2*(x*sqrt(-b/a)*log((b*x^2 - 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)) - 2)/(a*x), -(x*sqrt(b/a)*arctan(x*sqrt(b/a
)) + 1)/(a*x)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.32 \[ \int \frac {1}{x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=-\frac {b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a} - \frac {1}{a x} \]

[In]

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

[Out]

-b*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a) - 1/(a*x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.40 \[ \int \frac {1}{x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=-{\left (\frac {b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a} + \frac {1}{a x}\right )} \mathrm {sgn}\left (b x^{2} + a\right ) \]

[In]

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

[Out]

-(b*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a) + 1/(a*x))*sgn(b*x^2 + a)

Mupad [F(-1)]

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

[In]

int(1/(x^2*((a + b*x^2)^2)^(1/2)),x)

[Out]

int(1/(x^2*((a + b*x^2)^2)^(1/2)), x)