Integrand size = 24, antiderivative size = 67 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=-\frac {1}{3 a^2 x^3}+\frac {2 b}{a^3 x}+\frac {b^2 x}{2 a^3 \left (a+b x^2\right )}+\frac {5 b^{3/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{7/2}} \] Output:
-1/3/a^2/x^3+2*b/a^3/x+1/2*b^2*x/a^3/(b*x^2+a)+5/2*b^(3/2)*arctan(b^(1/2)* x/a^(1/2))/a^(7/2)
Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=-\frac {1}{3 a^2 x^3}+\frac {2 b}{a^3 x}+\frac {b^2 x}{2 a^3 \left (a+b x^2\right )}+\frac {5 b^{3/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{7/2}} \] Input:
Integrate[1/(x^4*(a^2 + 2*a*b*x^2 + b^2*x^4)),x]
Output:
-1/3*1/(a^2*x^3) + (2*b)/(a^3*x) + (b^2*x)/(2*a^3*(a + b*x^2)) + (5*b^(3/2 )*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(7/2))
Time = 0.33 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.16, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1380, 27, 253, 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.
\(\displaystyle \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle b^2 \int \frac {1}{b^2 x^4 \left (b x^2+a\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {1}{x^4 \left (a+b x^2\right )^2}dx\) |
\(\Big \downarrow \) 253 |
\(\displaystyle \frac {5 \int \frac {1}{x^4 \left (b x^2+a\right )}dx}{2 a}+\frac {1}{2 a x^3 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {5 \left (-\frac {b \int \frac {1}{x^2 \left (b x^2+a\right )}dx}{a}-\frac {1}{3 a x^3}\right )}{2 a}+\frac {1}{2 a x^3 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {5 \left (-\frac {b \left (-\frac {b \int \frac {1}{b x^2+a}dx}{a}-\frac {1}{a x}\right )}{a}-\frac {1}{3 a x^3}\right )}{2 a}+\frac {1}{2 a x^3 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {5 \left (-\frac {b \left (-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2}}-\frac {1}{a x}\right )}{a}-\frac {1}{3 a x^3}\right )}{2 a}+\frac {1}{2 a x^3 \left (a+b x^2\right )}\) |
Input:
Int[1/(x^4*(a^2 + 2*a*b*x^2 + b^2*x^4)),x]
Output:
1/(2*a*x^3*(a + b*x^2)) + (5*(-1/3*1/(a*x^3) - (b*(-(1/(a*x)) - (Sqrt[b]*A rcTan[(Sqrt[b]*x)/Sqrt[a]])/a^(3/2)))/a))/(2*a)
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)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 2*a*(p + 1)) Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m }, x] && LtQ[p, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
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[1/c^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]
Time = 0.12 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.82
method | result | size |
default | \(\frac {b^{2} \left (\frac {x}{2 b \,x^{2}+2 a}+\frac {5 \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{3}}-\frac {1}{3 a^{2} x^{3}}+\frac {2 b}{a^{3} x}\) | \(55\) |
risch | \(\frac {\frac {5 b^{2} x^{4}}{2 a^{3}}+\frac {5 b \,x^{2}}{3 a^{2}}-\frac {1}{3 a}}{x^{3} \left (b \,x^{2}+a \right )}+\frac {5 \sqrt {-a b}\, b \ln \left (-b x -\sqrt {-a b}\right )}{4 a^{4}}-\frac {5 \sqrt {-a b}\, b \ln \left (-b x +\sqrt {-a b}\right )}{4 a^{4}}\) | \(91\) |
Input:
int(1/x^4/(b^2*x^4+2*a*b*x^2+a^2),x,method=_RETURNVERBOSE)
Output:
b^2/a^3*(1/2*x/(b*x^2+a)+5/2/(a*b)^(1/2)*arctan(b/(a*b)^(1/2)*x))-1/3/a^2/ x^3+2*b/a^3/x
Time = 0.08 (sec) , antiderivative size = 172, normalized size of antiderivative = 2.57 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=\left [\frac {30 \, b^{2} x^{4} + 20 \, a b x^{2} + 15 \, {\left (b^{2} x^{5} + a b x^{3}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} + 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right ) - 4 \, a^{2}}{12 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )}}, \frac {15 \, b^{2} x^{4} + 10 \, a b x^{2} + 15 \, {\left (b^{2} x^{5} + a b x^{3}\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right ) - 2 \, a^{2}}{6 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )}}\right ] \] Input:
integrate(1/x^4/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="fricas")
Output:
[1/12*(30*b^2*x^4 + 20*a*b*x^2 + 15*(b^2*x^5 + a*b*x^3)*sqrt(-b/a)*log((b* x^2 + 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)) - 4*a^2)/(a^3*b*x^5 + a^4*x^3), 1 /6*(15*b^2*x^4 + 10*a*b*x^2 + 15*(b^2*x^5 + a*b*x^3)*sqrt(b/a)*arctan(x*sq rt(b/a)) - 2*a^2)/(a^3*b*x^5 + a^4*x^3)]
Time = 0.16 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.70 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=- \frac {5 \sqrt {- \frac {b^{3}}{a^{7}}} \log {\left (- \frac {a^{4} \sqrt {- \frac {b^{3}}{a^{7}}}}{b^{2}} + x \right )}}{4} + \frac {5 \sqrt {- \frac {b^{3}}{a^{7}}} \log {\left (\frac {a^{4} \sqrt {- \frac {b^{3}}{a^{7}}}}{b^{2}} + x \right )}}{4} + \frac {- 2 a^{2} + 10 a b x^{2} + 15 b^{2} x^{4}}{6 a^{4} x^{3} + 6 a^{3} b x^{5}} \] Input:
integrate(1/x**4/(b**2*x**4+2*a*b*x**2+a**2),x)
Output:
-5*sqrt(-b**3/a**7)*log(-a**4*sqrt(-b**3/a**7)/b**2 + x)/4 + 5*sqrt(-b**3/ a**7)*log(a**4*sqrt(-b**3/a**7)/b**2 + x)/4 + (-2*a**2 + 10*a*b*x**2 + 15* b**2*x**4)/(6*a**4*x**3 + 6*a**3*b*x**5)
Time = 0.11 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.96 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=\frac {15 \, b^{2} x^{4} + 10 \, a b x^{2} - 2 \, a^{2}}{6 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )}} + \frac {5 \, b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{3}} \] Input:
integrate(1/x^4/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="maxima")
Output:
1/6*(15*b^2*x^4 + 10*a*b*x^2 - 2*a^2)/(a^3*b*x^5 + a^4*x^3) + 5/2*b^2*arct an(b*x/sqrt(a*b))/(sqrt(a*b)*a^3)
Time = 0.12 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=\frac {5 \, b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{3}} + \frac {b^{2} x}{2 \, {\left (b x^{2} + a\right )} a^{3}} + \frac {6 \, b x^{2} - a}{3 \, a^{3} x^{3}} \] Input:
integrate(1/x^4/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="giac")
Output:
5/2*b^2*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^3) + 1/2*b^2*x/((b*x^2 + a)*a^3 ) + 1/3*(6*b*x^2 - a)/(a^3*x^3)
Time = 18.40 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.87 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=\frac {\frac {5\,b\,x^2}{3\,a^2}-\frac {1}{3\,a}+\frac {5\,b^2\,x^4}{2\,a^3}}{b\,x^5+a\,x^3}+\frac {5\,b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{2\,a^{7/2}} \] Input:
int(1/(x^4*(a^2 + b^2*x^4 + 2*a*b*x^2)),x)
Output:
((5*b*x^2)/(3*a^2) - 1/(3*a) + (5*b^2*x^4)/(2*a^3))/(a*x^3 + b*x^5) + (5*b ^(3/2)*atan((b^(1/2)*x)/a^(1/2)))/(2*a^(7/2))
Time = 0.16 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.31 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=\frac {15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a b \,x^{3}+15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{2} x^{5}-2 a^{3}+10 a^{2} b \,x^{2}+15 a \,b^{2} x^{4}}{6 a^{4} x^{3} \left (b \,x^{2}+a \right )} \] Input:
int(1/x^4/(b^2*x^4+2*a*b*x^2+a^2),x)
Output:
(15*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b*x**3 + 15*sqrt(b)*sq rt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*b**2*x**5 - 2*a**3 + 10*a**2*b*x**2 + 15*a*b**2*x**4)/(6*a**4*x**3*(a + b*x**2))