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

Optimal. Leaf size=92 \[ -\frac {a+b x^2}{a x \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\sqrt {b} \left (a+b x^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

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Rubi [A]  time = 0.03, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1112, 325, 205} \begin {gather*} -\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 {gather*}

Antiderivative was successfully verified.

[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 205

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

Rule 325

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 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^2 \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^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*}

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Mathematica [A]  time = 0.01, size = 56, normalized size = 0.61 \begin {gather*} -\frac {\left (a+b x^2\right ) \left (\sqrt {b} x \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )+\sqrt {a}\right )}{a^{3/2} x \sqrt {\left (a+b x^2\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[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]))

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IntegrateAlgebraic [A]  time = 8.78, size = 55, normalized size = 0.60 \begin {gather*} \frac {\left (a+b x^2\right ) \left (-\frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2}}-\frac {1}{a x}\right )}{\sqrt {\left (a+b x^2\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^2)*(-(1/(a*x)) - (Sqrt[b]*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/a^(3/2)))/Sqrt[(a + b*x^2)^2]

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fricas [A]  time = 1.56, size = 82, normalized size = 0.89 \begin {gather*} \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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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giac [A]  time = 0.15, size = 37, normalized size = 0.40 \begin {gather*} -{\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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.01, size = 50, normalized size = 0.54 \begin {gather*} -\frac {\left (b \,x^{2}+a \right ) \left (b x \arctan \left (\frac {b x}{\sqrt {a b}}\right )+\sqrt {a b}\right )}{\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \sqrt {a b}\, a x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.94, size = 29, normalized size = 0.32 \begin {gather*} -\frac {b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a} - \frac {1}{a x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^2\,\sqrt {{\left (b\,x^2+a\right )}^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.23, size = 65, normalized size = 0.71 \begin {gather*} \frac {\sqrt {- \frac {b}{a^{3}}} \log {\left (- \frac {a^{2} \sqrt {- \frac {b}{a^{3}}}}{b} + x \right )}}{2} - \frac {\sqrt {- \frac {b}{a^{3}}} \log {\left (\frac {a^{2} \sqrt {- \frac {b}{a^{3}}}}{b} + x \right )}}{2} - \frac {1}{a x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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