Integrand size = 20, antiderivative size = 97 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^3} \, dx=-\frac {A}{a^3 x}-\frac {(A b-a B) x}{4 a^2 \left (a+b x^2\right )^2}-\frac {(7 A b-3 a B) x}{8 a^3 \left (a+b x^2\right )}-\frac {3 (5 A b-a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {b}} \]
-A/a^3/x-1/4*(A*b-B*a)*x/a^2/(b*x^2+a)^2-1/8*(7*A*b-3*B*a)*x/a^3/(b*x^2+a) -3/8*(5*A*b-B*a)*arctan(x*b^(1/2)/a^(1/2))/a^(7/2)/b^(1/2)
Time = 0.04 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.99 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^3} \, dx=-\frac {A}{a^3 x}+\frac {(-A b+a B) x}{4 a^2 \left (a+b x^2\right )^2}+\frac {(-7 A b+3 a B) x}{8 a^3 \left (a+b x^2\right )}+\frac {3 (-5 A b+a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {b}} \]
-(A/(a^3*x)) + ((-(A*b) + a*B)*x)/(4*a^2*(a + b*x^2)^2) + ((-7*A*b + 3*a*B )*x)/(8*a^3*(a + b*x^2)) + (3*(-5*A*b + a*B)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/ (8*a^(7/2)*Sqrt[b])
Time = 0.27 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {361, 25, 27, 361, 25, 359, 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.
\(\displaystyle \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 361 |
\(\displaystyle -\frac {1}{4} \int -\frac {4 a A-3 (A b-a B) x^2}{a^2 x^2 \left (b x^2+a\right )^2}dx-\frac {x (A b-a B)}{4 a^2 \left (a+b x^2\right )^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{4} \int \frac {4 a A-3 (A b-a B) x^2}{a^2 x^2 \left (b x^2+a\right )^2}dx-\frac {x (A b-a B)}{4 a^2 \left (a+b x^2\right )^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {4 a A-3 (A b-a B) x^2}{x^2 \left (b x^2+a\right )^2}dx}{4 a^2}-\frac {x (A b-a B)}{4 a^2 \left (a+b x^2\right )^2}\) |
\(\Big \downarrow \) 361 |
\(\displaystyle \frac {-\frac {1}{2} \int -\frac {8 A-\left (\frac {7 A b}{a}-3 B\right ) x^2}{x^2 \left (b x^2+a\right )}dx-\frac {x (7 A b-3 a B)}{2 a \left (a+b x^2\right )}}{4 a^2}-\frac {x (A b-a B)}{4 a^2 \left (a+b x^2\right )^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {1}{2} \int \frac {8 A-\left (\frac {7 A b}{a}-3 B\right ) x^2}{x^2 \left (b x^2+a\right )}dx-\frac {x (7 A b-3 a B)}{2 a \left (a+b x^2\right )}}{4 a^2}-\frac {x (A b-a B)}{4 a^2 \left (a+b x^2\right )^2}\) |
\(\Big \downarrow \) 359 |
\(\displaystyle \frac {\frac {1}{2} \left (-\frac {3 (5 A b-a B) \int \frac {1}{b x^2+a}dx}{a}-\frac {8 A}{a x}\right )-\frac {x (7 A b-3 a B)}{2 a \left (a+b x^2\right )}}{4 a^2}-\frac {x (A b-a B)}{4 a^2 \left (a+b x^2\right )^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {1}{2} \left (-\frac {3 (5 A b-a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2} \sqrt {b}}-\frac {8 A}{a x}\right )-\frac {x (7 A b-3 a B)}{2 a \left (a+b x^2\right )}}{4 a^2}-\frac {x (A b-a B)}{4 a^2 \left (a+b x^2\right )^2}\) |
-1/4*((A*b - a*B)*x)/(a^2*(a + b*x^2)^2) + (-1/2*((7*A*b - 3*a*B)*x)/(a*(a + b*x^2)) + ((-8*A)/(a*x) - (3*(5*A*b - a*B)*ArcTan[(Sqrt[b]*x)/Sqrt[a]]) /(a^(3/2)*Sqrt[b]))/2)/(4*a^2)
3.2.4.3.1 Defintions of rubi rules used
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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
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])
Time = 2.50 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.85
method | result | size |
default | \(-\frac {A}{a^{3} x}-\frac {\frac {\left (\frac {7}{8} b^{2} A -\frac {3}{8} a b B \right ) x^{3}+\frac {a \left (9 A b -5 B a \right ) x}{8}}{\left (b \,x^{2}+a \right )^{2}}+\frac {3 \left (5 A b -B a \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}}}{a^{3}}\) | \(82\) |
risch | \(\frac {-\frac {3 b \left (5 A b -B a \right ) x^{4}}{8 a^{3}}-\frac {5 \left (5 A b -B a \right ) x^{2}}{8 a^{2}}-\frac {A}{a}}{x \left (b \,x^{2}+a \right )^{2}}-\frac {15 \ln \left (-\sqrt {-a b}\, x -a \right ) A b}{16 \sqrt {-a b}\, a^{3}}+\frac {3 \ln \left (-\sqrt {-a b}\, x -a \right ) B}{16 \sqrt {-a b}\, a^{2}}+\frac {15 \ln \left (-\sqrt {-a b}\, x +a \right ) A b}{16 \sqrt {-a b}\, a^{3}}-\frac {3 \ln \left (-\sqrt {-a b}\, x +a \right ) B}{16 \sqrt {-a b}\, a^{2}}\) | \(159\) |
-A/a^3/x-1/a^3*(((7/8*b^2*A-3/8*a*b*B)*x^3+1/8*a*(9*A*b-5*B*a)*x)/(b*x^2+a )^2+3/8*(5*A*b-B*a)/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))
Time = 0.27 (sec) , antiderivative size = 324, normalized size of antiderivative = 3.34 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^3} \, dx=\left [-\frac {16 \, A a^{3} b - 6 \, {\left (B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{4} - 10 \, {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2} - 3 \, {\left ({\left (B a b^{2} - 5 \, A b^{3}\right )} x^{5} + 2 \, {\left (B a^{2} b - 5 \, A a b^{2}\right )} x^{3} + {\left (B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x^{2} + 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{16 \, {\left (a^{4} b^{3} x^{5} + 2 \, a^{5} b^{2} x^{3} + a^{6} b x\right )}}, -\frac {8 \, A a^{3} b - 3 \, {\left (B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{4} - 5 \, {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2} - 3 \, {\left ({\left (B a b^{2} - 5 \, A b^{3}\right )} x^{5} + 2 \, {\left (B a^{2} b - 5 \, A a b^{2}\right )} x^{3} + {\left (B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{8 \, {\left (a^{4} b^{3} x^{5} + 2 \, a^{5} b^{2} x^{3} + a^{6} b x\right )}}\right ] \]
[-1/16*(16*A*a^3*b - 6*(B*a^2*b^2 - 5*A*a*b^3)*x^4 - 10*(B*a^3*b - 5*A*a^2 *b^2)*x^2 - 3*((B*a*b^2 - 5*A*b^3)*x^5 + 2*(B*a^2*b - 5*A*a*b^2)*x^3 + (B* a^3 - 5*A*a^2*b)*x)*sqrt(-a*b)*log((b*x^2 + 2*sqrt(-a*b)*x - a)/(b*x^2 + a )))/(a^4*b^3*x^5 + 2*a^5*b^2*x^3 + a^6*b*x), -1/8*(8*A*a^3*b - 3*(B*a^2*b^ 2 - 5*A*a*b^3)*x^4 - 5*(B*a^3*b - 5*A*a^2*b^2)*x^2 - 3*((B*a*b^2 - 5*A*b^3 )*x^5 + 2*(B*a^2*b - 5*A*a*b^2)*x^3 + (B*a^3 - 5*A*a^2*b)*x)*sqrt(a*b)*arc tan(sqrt(a*b)*x/a))/(a^4*b^3*x^5 + 2*a^5*b^2*x^3 + a^6*b*x)]
Leaf count of result is larger than twice the leaf count of optimal. 194 vs. \(2 (94) = 188\).
Time = 0.38 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.00 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^3} \, dx=- \frac {3 \sqrt {- \frac {1}{a^{7} b}} \left (- 5 A b + B a\right ) \log {\left (- \frac {3 a^{4} \sqrt {- \frac {1}{a^{7} b}} \left (- 5 A b + B a\right )}{- 15 A b + 3 B a} + x \right )}}{16} + \frac {3 \sqrt {- \frac {1}{a^{7} b}} \left (- 5 A b + B a\right ) \log {\left (\frac {3 a^{4} \sqrt {- \frac {1}{a^{7} b}} \left (- 5 A b + B a\right )}{- 15 A b + 3 B a} + x \right )}}{16} + \frac {- 8 A a^{2} + x^{4} \left (- 15 A b^{2} + 3 B a b\right ) + x^{2} \left (- 25 A a b + 5 B a^{2}\right )}{8 a^{5} x + 16 a^{4} b x^{3} + 8 a^{3} b^{2} x^{5}} \]
-3*sqrt(-1/(a**7*b))*(-5*A*b + B*a)*log(-3*a**4*sqrt(-1/(a**7*b))*(-5*A*b + B*a)/(-15*A*b + 3*B*a) + x)/16 + 3*sqrt(-1/(a**7*b))*(-5*A*b + B*a)*log( 3*a**4*sqrt(-1/(a**7*b))*(-5*A*b + B*a)/(-15*A*b + 3*B*a) + x)/16 + (-8*A* a**2 + x**4*(-15*A*b**2 + 3*B*a*b) + x**2*(-25*A*a*b + 5*B*a**2))/(8*a**5* x + 16*a**4*b*x**3 + 8*a**3*b**2*x**5)
Time = 0.29 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.99 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^3} \, dx=\frac {3 \, {\left (B a b - 5 \, A b^{2}\right )} x^{4} - 8 \, A a^{2} + 5 \, {\left (B a^{2} - 5 \, A a b\right )} x^{2}}{8 \, {\left (a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x\right )}} + \frac {3 \, {\left (B a - 5 \, A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3}} \]
1/8*(3*(B*a*b - 5*A*b^2)*x^4 - 8*A*a^2 + 5*(B*a^2 - 5*A*a*b)*x^2)/(a^3*b^2 *x^5 + 2*a^4*b*x^3 + a^5*x) + 3/8*(B*a - 5*A*b)*arctan(b*x/sqrt(a*b))/(sqr t(a*b)*a^3)
Time = 0.28 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.85 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^3} \, dx=\frac {3 \, {\left (B a - 5 \, A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3}} - \frac {A}{a^{3} x} + \frac {3 \, B a b x^{3} - 7 \, A b^{2} x^{3} + 5 \, B a^{2} x - 9 \, A a b x}{8 \, {\left (b x^{2} + a\right )}^{2} a^{3}} \]
3/8*(B*a - 5*A*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^3) - A/(a^3*x) + 1/8* (3*B*a*b*x^3 - 7*A*b^2*x^3 + 5*B*a^2*x - 9*A*a*b*x)/((b*x^2 + a)^2*a^3)
Time = 5.11 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.16 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^3} \, dx=-\frac {\frac {A}{a}+\frac {5\,x^2\,\left (5\,A\,b-B\,a\right )}{8\,a^2}+\frac {3\,b\,x^4\,\left (5\,A\,b-B\,a\right )}{8\,a^3}}{a^2\,x+2\,a\,b\,x^3+b^2\,x^5}-\frac {3\,\mathrm {atan}\left (\frac {3\,\sqrt {b}\,x\,\left (5\,A\,b-B\,a\right )}{\sqrt {a}\,\left (15\,A\,b-3\,B\,a\right )}\right )\,\left (5\,A\,b-B\,a\right )}{8\,a^{7/2}\,\sqrt {b}} \]