Integrand size = 20, antiderivative size = 87 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {(A b-2 a B) x}{b^3}+\frac {B x^3}{3 b^2}+\frac {a (A b-a B) x}{2 b^3 \left (a+b x^2\right )}-\frac {\sqrt {a} (3 A b-5 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 b^{7/2}} \]
(A*b-2*B*a)*x/b^3+1/3*B*x^3/b^2+1/2*a*(A*b-B*a)*x/b^3/(b*x^2+a)-1/2*(3*A*b -5*B*a)*arctan(x*b^(1/2)/a^(1/2))*a^(1/2)/b^(7/2)
Time = 0.05 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.02 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {(A b-2 a B) x}{b^3}+\frac {B x^3}{3 b^2}+\frac {\left (a A b-a^2 B\right ) x}{2 b^3 \left (a+b x^2\right )}+\frac {\sqrt {a} (-3 A b+5 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 b^{7/2}} \]
((A*b - 2*a*B)*x)/b^3 + (B*x^3)/(3*b^2) + ((a*A*b - a^2*B)*x)/(2*b^3*(a + b*x^2)) + (Sqrt[a]*(-3*A*b + 5*a*B)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*b^(7/2 ))
Time = 0.25 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.01, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {360, 1467, 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.
\(\displaystyle \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 360 |
\(\displaystyle \frac {a x (A b-a B)}{2 b^3 \left (a+b x^2\right )}-\frac {\int \frac {-2 b^2 B x^4-2 b (A b-a B) x^2+a (A b-a B)}{b x^2+a}dx}{2 b^3}\) |
\(\Big \downarrow \) 1467 |
\(\displaystyle \frac {a x (A b-a B)}{2 b^3 \left (a+b x^2\right )}-\frac {\int \left (-2 b B x^2-2 (A b-2 a B)+\frac {3 a A b-5 a^2 B}{b x^2+a}\right )dx}{2 b^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a x (A b-a B)}{2 b^3 \left (a+b x^2\right )}-\frac {\frac {\sqrt {a} (3 A b-5 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {b}}-2 x (A b-2 a B)-\frac {2}{3} b B x^3}{2 b^3}\) |
(a*(A*b - a*B)*x)/(2*b^3*(a + b*x^2)) - (-2*(A*b - 2*a*B)*x - (2*b*B*x^3)/ 3 + (Sqrt[a]*(3*A*b - 5*a*B)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/Sqrt[b])/(2*b^3)
3.1.76.3.1 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[(a + b*x^2)^(p + 1)*Expan dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & & (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]
Time = 2.66 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.86
method | result | size |
default | \(\frac {\frac {1}{3} b B \,x^{3}+A b x -2 B a x}{b^{3}}-\frac {a \left (\frac {\left (-\frac {A b}{2}+\frac {B a}{2}\right ) x}{b \,x^{2}+a}+\frac {\left (3 A b -5 B a \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{b^{3}}\) | \(75\) |
risch | \(\frac {B \,x^{3}}{3 b^{2}}+\frac {A x}{b^{2}}-\frac {2 B a x}{b^{3}}+\frac {\left (\frac {1}{2} a b A -\frac {1}{2} a^{2} B \right ) x}{b^{3} \left (b \,x^{2}+a \right )}+\frac {3 \sqrt {-a b}\, \ln \left (-\sqrt {-a b}\, x -a \right ) A}{4 b^{3}}-\frac {5 \sqrt {-a b}\, \ln \left (-\sqrt {-a b}\, x -a \right ) B a}{4 b^{4}}-\frac {3 \sqrt {-a b}\, \ln \left (\sqrt {-a b}\, x -a \right ) A}{4 b^{3}}+\frac {5 \sqrt {-a b}\, \ln \left (\sqrt {-a b}\, x -a \right ) B a}{4 b^{4}}\) | \(155\) |
1/b^3*(1/3*b*B*x^3+A*b*x-2*B*a*x)-a/b^3*((-1/2*A*b+1/2*B*a)*x/(b*x^2+a)+1/ 2*(3*A*b-5*B*a)/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))
Time = 0.29 (sec) , antiderivative size = 240, normalized size of antiderivative = 2.76 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\left [\frac {4 \, B b^{2} x^{5} - 4 \, {\left (5 \, B a b - 3 \, A b^{2}\right )} x^{3} - 3 \, {\left (5 \, B a^{2} - 3 \, A a b + {\left (5 \, B a b - 3 \, A b^{2}\right )} x^{2}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} - 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) - 6 \, {\left (5 \, B a^{2} - 3 \, A a b\right )} x}{12 \, {\left (b^{4} x^{2} + a b^{3}\right )}}, \frac {2 \, B b^{2} x^{5} - 2 \, {\left (5 \, B a b - 3 \, A b^{2}\right )} x^{3} + 3 \, {\left (5 \, B a^{2} - 3 \, A a b + {\left (5 \, B a b - 3 \, A b^{2}\right )} x^{2}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) - 3 \, {\left (5 \, B a^{2} - 3 \, A a b\right )} x}{6 \, {\left (b^{4} x^{2} + a b^{3}\right )}}\right ] \]
[1/12*(4*B*b^2*x^5 - 4*(5*B*a*b - 3*A*b^2)*x^3 - 3*(5*B*a^2 - 3*A*a*b + (5 *B*a*b - 3*A*b^2)*x^2)*sqrt(-a/b)*log((b*x^2 - 2*b*x*sqrt(-a/b) - a)/(b*x^ 2 + a)) - 6*(5*B*a^2 - 3*A*a*b)*x)/(b^4*x^2 + a*b^3), 1/6*(2*B*b^2*x^5 - 2 *(5*B*a*b - 3*A*b^2)*x^3 + 3*(5*B*a^2 - 3*A*a*b + (5*B*a*b - 3*A*b^2)*x^2) *sqrt(a/b)*arctan(b*x*sqrt(a/b)/a) - 3*(5*B*a^2 - 3*A*a*b)*x)/(b^4*x^2 + a *b^3)]
Time = 0.35 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.48 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {B x^{3}}{3 b^{2}} + x \left (\frac {A}{b^{2}} - \frac {2 B a}{b^{3}}\right ) + \frac {x \left (A a b - B a^{2}\right )}{2 a b^{3} + 2 b^{4} x^{2}} - \frac {\sqrt {- \frac {a}{b^{7}}} \left (- 3 A b + 5 B a\right ) \log {\left (- b^{3} \sqrt {- \frac {a}{b^{7}}} + x \right )}}{4} + \frac {\sqrt {- \frac {a}{b^{7}}} \left (- 3 A b + 5 B a\right ) \log {\left (b^{3} \sqrt {- \frac {a}{b^{7}}} + x \right )}}{4} \]
B*x**3/(3*b**2) + x*(A/b**2 - 2*B*a/b**3) + x*(A*a*b - B*a**2)/(2*a*b**3 + 2*b**4*x**2) - sqrt(-a/b**7)*(-3*A*b + 5*B*a)*log(-b**3*sqrt(-a/b**7) + x )/4 + sqrt(-a/b**7)*(-3*A*b + 5*B*a)*log(b**3*sqrt(-a/b**7) + x)/4
Time = 0.28 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.98 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=-\frac {{\left (B a^{2} - A a b\right )} x}{2 \, {\left (b^{4} x^{2} + a b^{3}\right )}} + \frac {{\left (5 \, B a^{2} - 3 \, A a b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{3}} + \frac {B b x^{3} - 3 \, {\left (2 \, B a - A b\right )} x}{3 \, b^{3}} \]
-1/2*(B*a^2 - A*a*b)*x/(b^4*x^2 + a*b^3) + 1/2*(5*B*a^2 - 3*A*a*b)*arctan( b*x/sqrt(a*b))/(sqrt(a*b)*b^3) + 1/3*(B*b*x^3 - 3*(2*B*a - A*b)*x)/b^3
Time = 0.28 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.01 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {{\left (5 \, B a^{2} - 3 \, A a b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{3}} - \frac {B a^{2} x - A a b x}{2 \, {\left (b x^{2} + a\right )} b^{3}} + \frac {B b^{4} x^{3} - 6 \, B a b^{3} x + 3 \, A b^{4} x}{3 \, b^{6}} \]
1/2*(5*B*a^2 - 3*A*a*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3) - 1/2*(B*a^2 *x - A*a*b*x)/((b*x^2 + a)*b^3) + 1/3*(B*b^4*x^3 - 6*B*a*b^3*x + 3*A*b^4*x )/b^6
Time = 0.07 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.20 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=x\,\left (\frac {A}{b^2}-\frac {2\,B\,a}{b^3}\right )-\frac {x\,\left (\frac {B\,a^2}{2}-\frac {A\,a\,b}{2}\right )}{b^4\,x^2+a\,b^3}+\frac {B\,x^3}{3\,b^2}+\frac {\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {b}\,x\,\left (3\,A\,b-5\,B\,a\right )}{5\,B\,a^2-3\,A\,a\,b}\right )\,\left (3\,A\,b-5\,B\,a\right )}{2\,b^{7/2}} \]