Integrand size = 20, antiderivative size = 90 \[ \int \frac {A+B x^2}{x^4 \left (a+b x^2\right )^2} \, dx=-\frac {A}{3 a^2 x^3}+\frac {2 A b-a B}{a^3 x}+\frac {b (A b-a B) x}{2 a^3 \left (a+b x^2\right )}+\frac {\sqrt {b} (5 A b-3 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{7/2}} \]
-1/3*A/a^2/x^3+(2*A*b-B*a)/a^3/x+1/2*b*(A*b-B*a)*x/a^3/(b*x^2+a)+1/2*(5*A* b-3*B*a)*arctan(x*b^(1/2)/a^(1/2))*b^(1/2)/a^(7/2)
Time = 0.05 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00 \[ \int \frac {A+B x^2}{x^4 \left (a+b x^2\right )^2} \, dx=-\frac {A}{3 a^2 x^3}+\frac {2 A b-a B}{a^3 x}-\frac {b (-A b+a B) x}{2 a^3 \left (a+b x^2\right )}-\frac {\sqrt {b} (-5 A b+3 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{7/2}} \]
-1/3*A/(a^2*x^3) + (2*A*b - a*B)/(a^3*x) - (b*(-(A*b) + a*B)*x)/(2*a^3*(a + b*x^2)) - (Sqrt[b]*(-5*A*b + 3*a*B)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(7 /2))
Time = 0.29 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {361, 25, 1584, 2009}
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.
\(\displaystyle \int \frac {A+B x^2}{x^4 \left (a+b x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 361 |
\(\displaystyle \frac {b x (A b-a B)}{2 a^3 \left (a+b x^2\right )}-\frac {1}{2} b \int -\frac {\frac {(A b-a B) x^4}{a^3}-\frac {2 (A b-a B) x^2}{a^2 b}+\frac {2 A}{a b}}{x^4 \left (b x^2+a\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} b \int \frac {\frac {(A b-a B) x^4}{a^3}-\frac {2 (A b-a B) x^2}{a^2 b}+\frac {2 A}{a b}}{x^4 \left (b x^2+a\right )}dx+\frac {b x (A b-a B)}{2 a^3 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 1584 |
\(\displaystyle \frac {1}{2} b \int \left (\frac {2 A}{a^2 b x^4}+\frac {5 A b-3 a B}{a^3 \left (b x^2+a\right )}+\frac {2 (a B-2 A b)}{a^3 b x^2}\right )dx+\frac {b x (A b-a B)}{2 a^3 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b x (A b-a B)}{2 a^3 \left (a+b x^2\right )}+\frac {1}{2} b \left (\frac {(5 A b-3 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{7/2} \sqrt {b}}+\frac {2 (2 A b-a B)}{a^3 b x}-\frac {2 A}{3 a^2 b x^3}\right )\) |
(b*(A*b - a*B)*x)/(2*a^3*(a + b*x^2)) + (b*((-2*A)/(3*a^2*b*x^3) + (2*(2*A *b - a*B))/(a^3*b*x) + ((5*A*b - 3*a*B)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(a^(7 /2)*Sqrt[b])))/2
3.1.84.3.1 Defintions of rubi rules used
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : > Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1)) Int[x^m*(a + b*x^2)^(p + 1)*E xpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)] - ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ILtQ[m/ 2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* (a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[ b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]
Time = 2.48 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.87
method | result | size |
default | \(-\frac {A}{3 a^{2} x^{3}}-\frac {-2 A b +B a}{x \,a^{3}}+\frac {b \left (\frac {\left (\frac {A b}{2}-\frac {B a}{2}\right ) x}{b \,x^{2}+a}+\frac {\left (5 A b -3 B a \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{3}}\) | \(78\) |
risch | \(\frac {\frac {b \left (5 A b -3 B a \right ) x^{4}}{2 a^{3}}+\frac {\left (5 A b -3 B a \right ) x^{2}}{3 a^{2}}-\frac {A}{3 a}}{x^{3} \left (b \,x^{2}+a \right )}+\frac {5 \sqrt {-a b}\, \ln \left (-b x -\sqrt {-a b}\right ) A b}{4 a^{4}}-\frac {3 \sqrt {-a b}\, \ln \left (-b x -\sqrt {-a b}\right ) B}{4 a^{3}}-\frac {5 \sqrt {-a b}\, \ln \left (-b x +\sqrt {-a b}\right ) A b}{4 a^{4}}+\frac {3 \sqrt {-a b}\, \ln \left (-b x +\sqrt {-a b}\right ) B}{4 a^{3}}\) | \(159\) |
-1/3*A/a^2/x^3-(-2*A*b+B*a)/x/a^3+1/a^3*b*((1/2*A*b-1/2*B*a)*x/(b*x^2+a)+1 /2*(5*A*b-3*B*a)/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))
Time = 0.27 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.78 \[ \int \frac {A+B x^2}{x^4 \left (a+b x^2\right )^2} \, dx=\left [-\frac {6 \, {\left (3 \, B a b - 5 \, A b^{2}\right )} x^{4} + 4 \, A a^{2} + 4 \, {\left (3 \, B a^{2} - 5 \, A a b\right )} x^{2} + 3 \, {\left ({\left (3 \, B a b - 5 \, A b^{2}\right )} x^{5} + {\left (3 \, B a^{2} - 5 \, A a b\right )} x^{3}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} + 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right )}{12 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )}}, -\frac {3 \, {\left (3 \, B a b - 5 \, A b^{2}\right )} x^{4} + 2 \, A a^{2} + 2 \, {\left (3 \, B a^{2} - 5 \, A a b\right )} x^{2} + 3 \, {\left ({\left (3 \, B a b - 5 \, A b^{2}\right )} x^{5} + {\left (3 \, B a^{2} - 5 \, A a b\right )} x^{3}\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right )}{6 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )}}\right ] \]
[-1/12*(6*(3*B*a*b - 5*A*b^2)*x^4 + 4*A*a^2 + 4*(3*B*a^2 - 5*A*a*b)*x^2 + 3*((3*B*a*b - 5*A*b^2)*x^5 + (3*B*a^2 - 5*A*a*b)*x^3)*sqrt(-b/a)*log((b*x^ 2 + 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)))/(a^3*b*x^5 + a^4*x^3), -1/6*(3*(3* B*a*b - 5*A*b^2)*x^4 + 2*A*a^2 + 2*(3*B*a^2 - 5*A*a*b)*x^2 + 3*((3*B*a*b - 5*A*b^2)*x^5 + (3*B*a^2 - 5*A*a*b)*x^3)*sqrt(b/a)*arctan(x*sqrt(b/a)))/(a ^3*b*x^5 + a^4*x^3)]
Leaf count of result is larger than twice the leaf count of optimal. 184 vs. \(2 (82) = 164\).
Time = 0.31 (sec) , antiderivative size = 184, normalized size of antiderivative = 2.04 \[ \int \frac {A+B x^2}{x^4 \left (a+b x^2\right )^2} \, dx=\frac {\sqrt {- \frac {b}{a^{7}}} \left (- 5 A b + 3 B a\right ) \log {\left (- \frac {a^{4} \sqrt {- \frac {b}{a^{7}}} \left (- 5 A b + 3 B a\right )}{- 5 A b^{2} + 3 B a b} + x \right )}}{4} - \frac {\sqrt {- \frac {b}{a^{7}}} \left (- 5 A b + 3 B a\right ) \log {\left (\frac {a^{4} \sqrt {- \frac {b}{a^{7}}} \left (- 5 A b + 3 B a\right )}{- 5 A b^{2} + 3 B a b} + x \right )}}{4} + \frac {- 2 A a^{2} + x^{4} \cdot \left (15 A b^{2} - 9 B a b\right ) + x^{2} \cdot \left (10 A a b - 6 B a^{2}\right )}{6 a^{4} x^{3} + 6 a^{3} b x^{5}} \]
sqrt(-b/a**7)*(-5*A*b + 3*B*a)*log(-a**4*sqrt(-b/a**7)*(-5*A*b + 3*B*a)/(- 5*A*b**2 + 3*B*a*b) + x)/4 - sqrt(-b/a**7)*(-5*A*b + 3*B*a)*log(a**4*sqrt( -b/a**7)*(-5*A*b + 3*B*a)/(-5*A*b**2 + 3*B*a*b) + x)/4 + (-2*A*a**2 + x**4 *(15*A*b**2 - 9*B*a*b) + x**2*(10*A*a*b - 6*B*a**2))/(6*a**4*x**3 + 6*a**3 *b*x**5)
Time = 0.28 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.03 \[ \int \frac {A+B x^2}{x^4 \left (a+b x^2\right )^2} \, dx=-\frac {3 \, {\left (3 \, B a b - 5 \, A b^{2}\right )} x^{4} + 2 \, A a^{2} + 2 \, {\left (3 \, B a^{2} - 5 \, A a b\right )} x^{2}}{6 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )}} - \frac {{\left (3 \, B a b - 5 \, A b^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{3}} \]
-1/6*(3*(3*B*a*b - 5*A*b^2)*x^4 + 2*A*a^2 + 2*(3*B*a^2 - 5*A*a*b)*x^2)/(a^ 3*b*x^5 + a^4*x^3) - 1/2*(3*B*a*b - 5*A*b^2)*arctan(b*x/sqrt(a*b))/(sqrt(a *b)*a^3)
Time = 0.28 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.94 \[ \int \frac {A+B x^2}{x^4 \left (a+b x^2\right )^2} \, dx=-\frac {{\left (3 \, B a b - 5 \, A b^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{3}} - \frac {B a b x - A b^{2} x}{2 \, {\left (b x^{2} + a\right )} a^{3}} - \frac {3 \, B a x^{2} - 6 \, A b x^{2} + A a}{3 \, a^{3} x^{3}} \]
-1/2*(3*B*a*b - 5*A*b^2)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^3) - 1/2*(B*a* b*x - A*b^2*x)/((b*x^2 + a)*a^3) - 1/3*(3*B*a*x^2 - 6*A*b*x^2 + A*a)/(a^3* x^3)
Time = 4.90 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.92 \[ \int \frac {A+B x^2}{x^4 \left (a+b x^2\right )^2} \, dx=\frac {\frac {x^2\,\left (5\,A\,b-3\,B\,a\right )}{3\,a^2}-\frac {A}{3\,a}+\frac {b\,x^4\,\left (5\,A\,b-3\,B\,a\right )}{2\,a^3}}{b\,x^5+a\,x^3}+\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (5\,A\,b-3\,B\,a\right )}{2\,a^{7/2}} \]