Integrand size = 28, antiderivative size = 110 \[ \int \frac {A+B x+C x^2+D x^3}{x^2 \left (a+b x^2\right )^2} \, dx=-\frac {A}{a^2 x}+\frac {b B-a D-b \left (\frac {A b}{a}-C\right ) x}{2 a b \left (a+b x^2\right )}-\frac {(3 A b-a C) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{5/2} \sqrt {b}}+\frac {B \log (x)}{a^2}-\frac {B \log \left (a+b x^2\right )}{2 a^2} \]
-A/a^2/x+1/2*(B*b-D*a-b*(A*b/a-C)*x)/a/b/(b*x^2+a)+B*ln(x)/a^2-1/2*B*ln(b* x^2+a)/a^2-1/2*(3*A*b-C*a)*arctan(x*b^(1/2)/a^(1/2))/a^(5/2)/b^(1/2)
Time = 0.04 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00 \[ \int \frac {A+B x+C x^2+D x^3}{x^2 \left (a+b x^2\right )^2} \, dx=-\frac {A}{a^2 x}+\frac {a b B-a^2 D-A b^2 x+a b C x}{2 a^2 b \left (a+b x^2\right )}+\frac {(-3 A b+a C) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{5/2} \sqrt {b}}+\frac {B \log (x)}{a^2}-\frac {B \log \left (a+b x^2\right )}{2 a^2} \]
-(A/(a^2*x)) + (a*b*B - a^2*D - A*b^2*x + a*b*C*x)/(2*a^2*b*(a + b*x^2)) + ((-3*A*b + a*C)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(5/2)*Sqrt[b]) + (B*Log [x])/a^2 - (B*Log[a + b*x^2])/(2*a^2)
Time = 0.40 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2336, 25, 2333, 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.
\(\displaystyle \int \frac {A+B x+C x^2+D x^3}{x^2 \left (a+b x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 2336 |
\(\displaystyle \frac {-b x \left (\frac {A b}{a}-C\right )-a D+b B}{2 a b \left (a+b x^2\right )}-\frac {\int -\frac {-\left (\left (\frac {A b}{a}-C\right ) x^2\right )+2 B x+2 A}{x^2 \left (b x^2+a\right )}dx}{2 a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {-\left (\left (\frac {A b}{a}-C\right ) x^2\right )+2 B x+2 A}{x^2 \left (b x^2+a\right )}dx}{2 a}+\frac {-b x \left (\frac {A b}{a}-C\right )-a D+b B}{2 a b \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 2333 |
\(\displaystyle \frac {\int \left (\frac {2 A}{a x^2}+\frac {2 B}{a x}+\frac {-3 A b-2 B x b+a C}{a \left (b x^2+a\right )}\right )dx}{2 a}+\frac {-b x \left (\frac {A b}{a}-C\right )-a D+b B}{2 a b \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {(3 A b-a C) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2} \sqrt {b}}-\frac {2 A}{a x}-\frac {B \log \left (a+b x^2\right )}{a}+\frac {2 B \log (x)}{a}}{2 a}+\frac {-b x \left (\frac {A b}{a}-C\right )-a D+b B}{2 a b \left (a+b x^2\right )}\) |
(b*B - a*D - b*((A*b)/a - C)*x)/(2*a*b*(a + b*x^2)) + ((-2*A)/(a*x) - ((3* A*b - a*C)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(a^(3/2)*Sqrt[b]) + (2*B*Log[x])/a - (B*Log[a + b*x^2])/a)/(2*a)
3.1.100.3.1 Defintions of rubi rules used
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ {Q = PolynomialQuotient[(c*x)^m*Pq, a + b*x^2, x], f = Coeff[PolynomialRema inder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[(c*x) ^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a* b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[(c*x)^m*(a + b*x^2)^(p + 1)*Ex pandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x], x]] /; F reeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]
Time = 3.55 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.87
method | result | size |
default | \(-\frac {A}{a^{2} x}+\frac {B \ln \left (x \right )}{a^{2}}-\frac {\frac {\left (\frac {A b}{2}-\frac {C a}{2}\right ) x -\frac {a \left (B b -D a \right )}{2 b}}{b \,x^{2}+a}+\frac {B \ln \left (b \,x^{2}+a \right )}{2}+\frac {\left (3 A b -C a \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}}}{a^{2}}\) | \(96\) |
-A/a^2/x+B*ln(x)/a^2-1/a^2*(((1/2*A*b-1/2*C*a)*x-1/2*a*(B*b-D*a)/b)/(b*x^2 +a)+1/2*B*ln(b*x^2+a)+1/2*(3*A*b-C*a)/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))
Time = 0.30 (sec) , antiderivative size = 336, normalized size of antiderivative = 3.05 \[ \int \frac {A+B x+C x^2+D x^3}{x^2 \left (a+b x^2\right )^2} \, dx=\left [-\frac {4 \, A a^{2} b - 2 \, {\left (C a^{2} b - 3 \, A a b^{2}\right )} x^{2} - {\left ({\left (C a b - 3 \, A b^{2}\right )} x^{3} + {\left (C a^{2} - 3 \, A a b\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x^{2} + 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) + 2 \, {\left (D a^{3} - B a^{2} b\right )} x + 2 \, {\left (B a b^{2} x^{3} + B a^{2} b x\right )} \log \left (b x^{2} + a\right ) - 4 \, {\left (B a b^{2} x^{3} + B a^{2} b x\right )} \log \left (x\right )}{4 \, {\left (a^{3} b^{2} x^{3} + a^{4} b x\right )}}, -\frac {2 \, A a^{2} b - {\left (C a^{2} b - 3 \, A a b^{2}\right )} x^{2} - {\left ({\left (C a b - 3 \, A b^{2}\right )} x^{3} + {\left (C a^{2} - 3 \, A a b\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + {\left (D a^{3} - B a^{2} b\right )} x + {\left (B a b^{2} x^{3} + B a^{2} b x\right )} \log \left (b x^{2} + a\right ) - 2 \, {\left (B a b^{2} x^{3} + B a^{2} b x\right )} \log \left (x\right )}{2 \, {\left (a^{3} b^{2} x^{3} + a^{4} b x\right )}}\right ] \]
[-1/4*(4*A*a^2*b - 2*(C*a^2*b - 3*A*a*b^2)*x^2 - ((C*a*b - 3*A*b^2)*x^3 + (C*a^2 - 3*A*a*b)*x)*sqrt(-a*b)*log((b*x^2 + 2*sqrt(-a*b)*x - a)/(b*x^2 + a)) + 2*(D*a^3 - B*a^2*b)*x + 2*(B*a*b^2*x^3 + B*a^2*b*x)*log(b*x^2 + a) - 4*(B*a*b^2*x^3 + B*a^2*b*x)*log(x))/(a^3*b^2*x^3 + a^4*b*x), -1/2*(2*A*a^ 2*b - (C*a^2*b - 3*A*a*b^2)*x^2 - ((C*a*b - 3*A*b^2)*x^3 + (C*a^2 - 3*A*a* b)*x)*sqrt(a*b)*arctan(sqrt(a*b)*x/a) + (D*a^3 - B*a^2*b)*x + (B*a*b^2*x^3 + B*a^2*b*x)*log(b*x^2 + a) - 2*(B*a*b^2*x^3 + B*a^2*b*x)*log(x))/(a^3*b^ 2*x^3 + a^4*b*x)]
Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{x^2 \left (a+b x^2\right )^2} \, dx=\text {Timed out} \]
Time = 0.29 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.95 \[ \int \frac {A+B x+C x^2+D x^3}{x^2 \left (a+b x^2\right )^2} \, dx=-\frac {2 \, A a b - {\left (C a b - 3 \, A b^{2}\right )} x^{2} + {\left (D a^{2} - B a b\right )} x}{2 \, {\left (a^{2} b^{2} x^{3} + a^{3} b x\right )}} - \frac {B \log \left (b x^{2} + a\right )}{2 \, a^{2}} + \frac {B \log \left (x\right )}{a^{2}} + \frac {{\left (C a - 3 \, A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{2}} \]
-1/2*(2*A*a*b - (C*a*b - 3*A*b^2)*x^2 + (D*a^2 - B*a*b)*x)/(a^2*b^2*x^3 + a^3*b*x) - 1/2*B*log(b*x^2 + a)/a^2 + B*log(x)/a^2 + 1/2*(C*a - 3*A*b)*arc tan(b*x/sqrt(a*b))/(sqrt(a*b)*a^2)
Time = 0.28 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.94 \[ \int \frac {A+B x+C x^2+D x^3}{x^2 \left (a+b x^2\right )^2} \, dx=-\frac {B \log \left (b x^{2} + a\right )}{2 \, a^{2}} + \frac {B \log \left ({\left | x \right |}\right )}{a^{2}} + \frac {{\left (C a - 3 \, A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{2}} + \frac {C a b x^{2} - 3 \, A b^{2} x^{2} - D a^{2} x + B a b x - 2 \, A a b}{2 \, {\left (b x^{3} + a x\right )} a^{2} b} \]
-1/2*B*log(b*x^2 + a)/a^2 + B*log(abs(x))/a^2 + 1/2*(C*a - 3*A*b)*arctan(b *x/sqrt(a*b))/(sqrt(a*b)*a^2) + 1/2*(C*a*b*x^2 - 3*A*b^2*x^2 - D*a^2*x + B *a*b*x - 2*A*a*b)/((b*x^3 + a*x)*a^2*b)
Time = 6.19 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.21 \[ \int \frac {A+B x+C x^2+D x^3}{x^2 \left (a+b x^2\right )^2} \, dx=\frac {B}{2\,a\,\left (b\,x^2+a\right )}-\frac {\frac {A}{a}+\frac {3\,A\,b\,x^2}{2\,a^2}}{b\,x^3+a\,x}-\frac {B\,\ln \left (b\,x^2+a\right )}{2\,a^2}+\frac {B\,\ln \left (x\right )}{a^2}-\frac {D}{2\,b\,\left (b\,x^2+a\right )}+\frac {C\,x}{2\,a\,\left (b\,x^2+a\right )}-\frac {3\,A\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{2\,a^{5/2}}+\frac {C\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{2\,a^{3/2}\,\sqrt {b}} \]