Integrand size = 28, antiderivative size = 95 \[ \int \frac {A+B x+C x^2+D x^3}{x \left (a+b x^2\right )^2} \, dx=\frac {A b-a C+(b B-a D) x}{2 a b \left (a+b x^2\right )}+\frac {(b B+a D) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{3/2} b^{3/2}}+\frac {A \log (x)}{a^2}-\frac {A \log \left (a+b x^2\right )}{2 a^2} \]
1/2*(A*b-C*a+(B*b-D*a)*x)/a/b/(b*x^2+a)+1/2*(B*b+D*a)*arctan(x*b^(1/2)/a^( 1/2))/a^(3/2)/b^(3/2)+A*ln(x)/a^2-1/2*A*ln(b*x^2+a)/a^2
Time = 0.05 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.89 \[ \int \frac {A+B x+C x^2+D x^3}{x \left (a+b x^2\right )^2} \, dx=\frac {\frac {a (A b+b B x-a (C+D x))}{b \left (a+b x^2\right )}+\frac {\sqrt {a} (b B+a D) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{3/2}}+2 A \log (x)-A \log \left (a+b x^2\right )}{2 a^2} \]
((a*(A*b + b*B*x - a*(C + D*x)))/(b*(a + b*x^2)) + (Sqrt[a]*(b*B + a*D)*Ar cTan[(Sqrt[b]*x)/Sqrt[a]])/b^(3/2) + 2*A*Log[x] - A*Log[a + b*x^2])/(2*a^2 )
Time = 0.32 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {2336, 25, 27, 523, 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.
\(\displaystyle \int \frac {A+B x+C x^2+D x^3}{x \left (a+b x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 2336 |
\(\displaystyle \frac {x (b B-a D)-a C+A b}{2 a b \left (a+b x^2\right )}-\frac {\int -\frac {2 A b+(b B+a D) x}{b x \left (b x^2+a\right )}dx}{2 a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {2 A b+(b B+a D) x}{b x \left (b x^2+a\right )}dx}{2 a}+\frac {x (b B-a D)-a C+A b}{2 a b \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {2 A b+(b B+a D) x}{x \left (b x^2+a\right )}dx}{2 a b}+\frac {x (b B-a D)-a C+A b}{2 a b \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 523 |
\(\displaystyle \frac {\int \left (\frac {2 A b}{a x}+\frac {D a^2+b B a-2 A b^2 x}{a \left (b x^2+a\right )}\right )dx}{2 a b}+\frac {x (b B-a D)-a C+A b}{2 a b \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {A b \log \left (a+b x^2\right )}{a}+\frac {2 A b \log (x)}{a}+\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) (a D+b B)}{\sqrt {a} \sqrt {b}}}{2 a b}+\frac {x (b B-a D)-a C+A b}{2 a b \left (a+b x^2\right )}\) |
(A*b - a*C + (b*B - a*D)*x)/(2*a*b*(a + b*x^2)) + (((b*B + a*D)*ArcTan[(Sq rt[b]*x)/Sqrt[a]])/(Sqrt[a]*Sqrt[b]) + (2*A*b*Log[x])/a - (A*b*Log[a + b*x ^2])/a)/(2*a*b)
3.1.99.3.1 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[((x_)^(m_.)*((c_) + (d_.)*(x_)))/((a_) + (b_.)*(x_)^2), x_Symbol] :> In t[ExpandIntegrand[x^m*((c + d*x)/(a + b*x^2)), x], x] /; FreeQ[{a, b, c, d} , x] && IntegerQ[m]
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 = 3.48 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.04
method | result | size |
default | \(\frac {A \ln \left (x \right )}{a^{2}}-\frac {\frac {-\frac {a \left (B b -D a \right ) x}{2 b}-\frac {a \left (A b -C a \right )}{2 b}}{b \,x^{2}+a}+\frac {b A \ln \left (b \,x^{2}+a \right )+\frac {\left (-a b B -D a^{2}\right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}}}{2 b}}{a^{2}}\) | \(99\) |
A*ln(x)/a^2-1/a^2*((-1/2*a*(B*b-D*a)/b*x-1/2*a*(A*b-C*a)/b)/(b*x^2+a)+1/2/ b*(b*A*ln(b*x^2+a)+(-B*a*b-D*a^2)/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))))
Time = 0.30 (sec) , antiderivative size = 296, normalized size of antiderivative = 3.12 \[ \int \frac {A+B x+C x^2+D x^3}{x \left (a+b x^2\right )^2} \, dx=\left [-\frac {2 \, C a^{2} b - 2 \, A a b^{2} + {\left (D a^{2} + B a b + {\left (D a b + B b^{2}\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) + 2 \, {\left (D a^{2} b - B a b^{2}\right )} x + 2 \, {\left (A b^{3} x^{2} + A a b^{2}\right )} \log \left (b x^{2} + a\right ) - 4 \, {\left (A b^{3} x^{2} + A a b^{2}\right )} \log \left (x\right )}{4 \, {\left (a^{2} b^{3} x^{2} + a^{3} b^{2}\right )}}, -\frac {C a^{2} b - A a b^{2} - {\left (D a^{2} + B a b + {\left (D a b + B b^{2}\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + {\left (D a^{2} b - B a b^{2}\right )} x + {\left (A b^{3} x^{2} + A a b^{2}\right )} \log \left (b x^{2} + a\right ) - 2 \, {\left (A b^{3} x^{2} + A a b^{2}\right )} \log \left (x\right )}{2 \, {\left (a^{2} b^{3} x^{2} + a^{3} b^{2}\right )}}\right ] \]
[-1/4*(2*C*a^2*b - 2*A*a*b^2 + (D*a^2 + B*a*b + (D*a*b + B*b^2)*x^2)*sqrt( -a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)) + 2*(D*a^2*b - B*a*b^2 )*x + 2*(A*b^3*x^2 + A*a*b^2)*log(b*x^2 + a) - 4*(A*b^3*x^2 + A*a*b^2)*log (x))/(a^2*b^3*x^2 + a^3*b^2), -1/2*(C*a^2*b - A*a*b^2 - (D*a^2 + B*a*b + ( D*a*b + B*b^2)*x^2)*sqrt(a*b)*arctan(sqrt(a*b)*x/a) + (D*a^2*b - B*a*b^2)* x + (A*b^3*x^2 + A*a*b^2)*log(b*x^2 + a) - 2*(A*b^3*x^2 + A*a*b^2)*log(x)) /(a^2*b^3*x^2 + a^3*b^2)]
Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{x \left (a+b x^2\right )^2} \, dx=\text {Timed out} \]
Time = 0.28 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.92 \[ \int \frac {A+B x+C x^2+D x^3}{x \left (a+b x^2\right )^2} \, dx=-\frac {C a - A b + {\left (D a - B b\right )} x}{2 \, {\left (a b^{2} x^{2} + a^{2} b\right )}} - \frac {A \log \left (b x^{2} + a\right )}{2 \, a^{2}} + \frac {A \log \left (x\right )}{a^{2}} + \frac {{\left (D a + B b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a b} \]
-1/2*(C*a - A*b + (D*a - B*b)*x)/(a*b^2*x^2 + a^2*b) - 1/2*A*log(b*x^2 + a )/a^2 + A*log(x)/a^2 + 1/2*(D*a + B*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a* b)
Time = 0.30 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.98 \[ \int \frac {A+B x+C x^2+D x^3}{x \left (a+b x^2\right )^2} \, dx=-\frac {A \log \left (b x^{2} + a\right )}{2 \, a^{2}} + \frac {A \log \left ({\left | x \right |}\right )}{a^{2}} + \frac {{\left (D a + B b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a b} - \frac {C a^{2} - A a b + {\left (D a^{2} - B a b\right )} x}{2 \, {\left (b x^{2} + a\right )} a^{2} b} \]
-1/2*A*log(b*x^2 + a)/a^2 + A*log(abs(x))/a^2 + 1/2*(D*a + B*b)*arctan(b*x /sqrt(a*b))/(sqrt(a*b)*a*b) - 1/2*(C*a^2 - A*a*b + (D*a^2 - B*a*b)*x)/((b* x^2 + a)*a^2*b)
Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{x \left (a+b x^2\right )^2} \, dx=\int \frac {A+B\,x+C\,x^2+x^3\,D}{x\,{\left (b\,x^2+a\right )}^2} \,d x \]