Integrand size = 26, antiderivative size = 133 \[ \int \frac {1}{x^4 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {-a-b x^2}{3 a x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {b \left (a+b x^2\right )}{a^2 x \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {b^{3/2} \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \] Output:
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(b^(1/2)*x/a^(1/2))/a^(5/2)/((b*x^2+a)^2)^(1/ 2)
Time = 1.02 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.53 \[ \int \frac {1}{x^4 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=-\frac {\left (a+b x^2\right ) \left (\sqrt {a} \left (a-3 b x^2\right )-3 b^{3/2} x^3 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right )}{3 a^{5/2} x^3 \sqrt {\left (a+b x^2\right )^2}} \] Input:
Integrate[1/(x^4*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]),x]
Output:
-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])
Time = 0.33 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.61, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1384, 27, 264, 264, 218}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{x^4 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {b \left (a+b x^2\right ) \int \frac {1}{b x^4 \left (b x^2+a\right )}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (a+b x^2\right ) \int \frac {1}{x^4 \left (b x^2+a\right )}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {\left (a+b x^2\right ) \left (-\frac {b \int \frac {1}{x^2 \left (b x^2+a\right )}dx}{a}-\frac {1}{3 a x^3}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {\left (a+b x^2\right ) \left (-\frac {b \left (-\frac {b \int \frac {1}{b x^2+a}dx}{a}-\frac {1}{a x}\right )}{a}-\frac {1}{3 a x^3}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\left (a+b x^2\right ) \left (-\frac {b \left (-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2}}-\frac {1}{a x}\right )}{a}-\frac {1}{3 a x^3}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
Input:
Int[1/(x^4*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]),x]
Output:
((a + b*x^2)*(-1/3*1/(a*x^3) - (b*(-(1/(a*x)) - (Sqrt[b]*ArcTan[(Sqrt[b]*x )/Sqrt[a]])/a^(3/2)))/a))/Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> S imp[(a + b*x^n + c*x^(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*Frac Part[p])) Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && NeQ[u, x^(n - 1)] && NeQ[u, x^(2*n - 1)] && !(EqQ[p, 1/2] && EqQ[u, x^(-2*n - 1)])
Time = 1.54 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.51
method | result | size |
default | \(-\frac {\left (b \,x^{2}+a \right ) \left (-3 b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right ) x^{3}-3 \sqrt {a b}\, b \,x^{2}+\sqrt {a b}\, a \right )}{3 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{2} \sqrt {a b}\, x^{3}}\) | \(68\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (\frac {b \,x^{2}}{a^{2}}-\frac {1}{3 a}\right )}{\left (b \,x^{2}+a \right ) x^{3}}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \sqrt {-a b}\, b \ln \left (-b x -\sqrt {-a b}\right )}{2 \left (b \,x^{2}+a \right ) a^{3}}-\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \sqrt {-a b}\, b \ln \left (-b x +\sqrt {-a b}\right )}{2 \left (b \,x^{2}+a \right ) a^{3}}\) | \(130\) |
Input:
int(1/x^4/((b*x^2+a)^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/3*(b*x^2+a)*(-3*b^2*arctan(b/(a*b)^(1/2)*x)*x^3-3*(a*b)^(1/2)*b*x^2+(a* b)^(1/2)*a)/((b*x^2+a)^2)^(1/2)/a^2/(a*b)^(1/2)/x^3
Time = 0.08 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.80 \[ \int \frac {1}{x^4 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\left [\frac {3 \, b x^{3} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} + 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right ) + 6 \, b x^{2} - 2 \, a}{6 \, a^{2} x^{3}}, \frac {3 \, b x^{3} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right ) + 3 \, b x^{2} - a}{3 \, a^{2} x^{3}}\right ] \] Input:
integrate(1/x^4/((b*x^2+a)^2)^(1/2),x, algorithm="fricas")
Output:
[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)]
\[ \int \frac {1}{x^4 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\int \frac {1}{x^{4} \sqrt {\left (a + b x^{2}\right )^{2}}}\, dx \] Input:
integrate(1/x**4/((b*x**2+a)**2)**(1/2),x)
Output:
Integral(1/(x**4*sqrt((a + b*x**2)**2)), x)
Time = 0.10 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.30 \[ \int \frac {1}{x^4 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2}} + \frac {3 \, b x^{2} - a}{3 \, a^{2} x^{3}} \] Input:
integrate(1/x^4/((b*x^2+a)^2)^(1/2),x, algorithm="maxima")
Output:
b^2*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^2) + 1/3*(3*b*x^2 - a)/(a^2*x^3)
Time = 0.12 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.38 \[ \int \frac {1}{x^4 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {1}{3} \, {\left (\frac {3 \, b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2}} + \frac {3 \, b x^{2} - a}{a^{2} x^{3}}\right )} \mathrm {sgn}\left (b x^{2} + a\right ) \] Input:
integrate(1/x^4/((b*x^2+a)^2)^(1/2),x, algorithm="giac")
Output:
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)
Timed out. \[ \int \frac {1}{x^4 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\int \frac {1}{x^4\,\sqrt {{\left (b\,x^2+a\right )}^2}} \,d x \] Input:
int(1/(x^4*((a + b*x^2)^2)^(1/2)),x)
Output:
int(1/(x^4*((a + b*x^2)^2)^(1/2)), x)
Time = 0.17 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.32 \[ \int \frac {1}{x^4 \sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b \,x^{3}-a^{2}+3 a b \,x^{2}}{3 a^{3} x^{3}} \] Input:
int(1/x^4/((b*x^2+a)^2)^(1/2),x)
Output:
(3*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*b*x**3 - a**2 + 3*a*b*x** 2)/(3*a**3*x**3)