\(\int \frac {1}{x^4 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx\) [633]

Optimal result

Integrand size = 26, antiderivative size = 133 \[ \int \frac {1}{x^4 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {-a-b x^2}{3 a x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {b \left (a+b x^2\right )}{a^2 x \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {b^{3/2} \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

1/3*(-b*x^2-a)/a/x^3/((b*x^2+a)^2)^(1/2)+b*(b*x^2+a)/a^2/x/((b*x^2+a)^2)^(1/2)+b^(3/2)*(b*x^2+a)*arctan(x*b^(1
/2)/a^(1/2))/a^(5/2)/((b*x^2+a)^2)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1126, 331, 211} \[ \int \frac {1}{x^4 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {b \left (a+b x^2\right )}{a^2 x \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {a+b x^2}{3 a x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {b^{3/2} \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[In]

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

[Out]

-1/3*(a + b*x^2)/(a*x^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (b*(a + b*x^2))/(a^2*x*Sqrt[a^2 + 2*a*b*x^2 + b^2*x
^4]) + (b^(3/2)*(a + b*x^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(a^(5/2)*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^4 \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}{3 a x^3 \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}{x^2 \left (a b+b^2 x^2\right )} \, dx}{a \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = -\frac {a+b x^2}{3 a x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {b \left (a+b x^2\right )}{a^2 x \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (b^2 \left (a b+b^2 x^2\right )\right ) \int \frac {1}{a b+b^2 x^2} \, dx}{a^2 \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = -\frac {a+b x^2}{3 a x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {b \left (a+b x^2\right )}{a^2 x \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {b^{3/2} \left (a+b x^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.53 \[ \int \frac {1}{x^4 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=-\frac {\left (a+b x^2\right ) \left (\sqrt {a} \left (a-3 b x^2\right )-3 b^{3/2} x^3 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right )}{3 a^{5/2} x^3 \sqrt {\left (a+b x^2\right )^2}} \]

[In]

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

[Out]

-1/3*((a + b*x^2)*(Sqrt[a]*(a - 3*b*x^2) - 3*b^(3/2)*x^3*ArcTan[(Sqrt[b]*x)/Sqrt[a]]))/(a^(5/2)*x^3*Sqrt[(a +
b*x^2)^2])

Maple [A] (verified)

Time = 1.30 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.51

method	result	size
default	\(-\frac {\left (b \,x^{2}+a \right ) \left (-3 b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right ) x^{3}-3 \sqrt {a b}\, b \,x^{2}+\sqrt {a b}\, a \right )}{3 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{2} x^{3} \sqrt {a b}}\)	\(68\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (\frac {b \,x^{2}}{a^{2}}-\frac {1}{3 a}\right )}{\left (b \,x^{2}+a \right ) x^{3}}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \sqrt {-a b}\, b \ln \left (-b x -\sqrt {-a b}\right )}{2 \left (b \,x^{2}+a \right ) a^{3}}-\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \sqrt {-a b}\, b \ln \left (-b x +\sqrt {-a b}\right )}{2 \left (b \,x^{2}+a \right ) a^{3}}\)	\(130\)

[In]

int(1/x^4/((b*x^2+a)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.80 \[ \int \frac {1}{x^4 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\left [\frac {3 \, b x^{3} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} + 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right ) + 6 \, b x^{2} - 2 \, a}{6 \, a^{2} x^{3}}, \frac {3 \, b x^{3} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right ) + 3 \, b x^{2} - a}{3 \, a^{2} x^{3}}\right ] \]

[In]

integrate(1/x^4/((b*x^2+a)^2)^(1/2),x, algorithm="fricas")

[Out]

[1/6*(3*b*x^3*sqrt(-b/a)*log((b*x^2 + 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)) + 6*b*x^2 - 2*a)/(a^2*x^3), 1/3*(3*b*
x^3*sqrt(b/a)*arctan(x*sqrt(b/a)) + 3*b*x^2 - a)/(a^2*x^3)]

Sympy [F]

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

[In]

integrate(1/x**4/((b*x**2+a)**2)**(1/2),x)

[Out]

Integral(1/(x**4*sqrt((a + b*x**2)**2)), x)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.30 \[ \int \frac {1}{x^4 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2}} + \frac {3 \, b x^{2} - a}{3 \, a^{2} x^{3}} \]

[In]

integrate(1/x^4/((b*x^2+a)^2)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.38 \[ \int \frac {1}{x^4 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {1}{3} \, {\left (\frac {3 \, b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2}} + \frac {3 \, b x^{2} - a}{a^{2} x^{3}}\right )} \mathrm {sgn}\left (b x^{2} + a\right ) \]

[In]

integrate(1/x^4/((b*x^2+a)^2)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/(x^4*((a + b*x^2)^2)^(1/2)),x)

[Out]

int(1/(x^4*((a + b*x^2)^2)^(1/2)), x)