Integrand size = 30, antiderivative size = 121 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^4 \left (a+b x^2\right )^2} \, dx=-\frac {c}{3 a^2 x^3}+\frac {2 b c-a d}{a^3 x}+\frac {\left (\frac {b^2 c}{a^2}-\frac {b d}{a}+e-\frac {a f}{b}\right ) x}{2 a \left (a+b x^2\right )}+\frac {\left (5 b^3 c-3 a b^2 d+a^2 b e+a^3 f\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{7/2} b^{3/2}} \]
-1/3*c/a^2/x^3+(-a*d+2*b*c)/a^3/x+1/2*(b^2*c/a^2-b*d/a+e-a*f/b)*x/a/(b*x^2 +a)+1/2*(a^3*f+a^2*b*e-3*a*b^2*d+5*b^3*c)*arctan(x*b^(1/2)/a^(1/2))/a^(7/2 )/b^(3/2)
Time = 0.05 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.03 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^4 \left (a+b x^2\right )^2} \, dx=-\frac {c}{3 a^2 x^3}+\frac {2 b c-a d}{a^3 x}-\frac {\left (-b^3 c+a b^2 d-a^2 b e+a^3 f\right ) x}{2 a^3 b \left (a+b x^2\right )}+\frac {\left (5 b^3 c-3 a b^2 d+a^2 b e+a^3 f\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{7/2} b^{3/2}} \]
-1/3*c/(a^2*x^3) + (2*b*c - a*d)/(a^3*x) - ((-(b^3*c) + a*b^2*d - a^2*b*e + a^3*f)*x)/(2*a^3*b*(a + b*x^2)) + ((5*b^3*c - 3*a*b^2*d + a^2*b*e + a^3* f)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(7/2)*b^(3/2))
Time = 0.39 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2336, 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.
\(\displaystyle \int \frac {c+d x^2+e x^4+f x^6}{x^4 \left (a+b x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 2336 |
\(\displaystyle \frac {x \left (\frac {b^2 c}{a^2}-\frac {b d}{a}-\frac {a f}{b}+e\right )}{2 a \left (a+b x^2\right )}-\frac {\int -\frac {\left (\frac {c b^2}{a^2}-\frac {d b}{a}+e+\frac {a f}{b}\right ) x^4-2 \left (\frac {b c}{a}-d\right ) x^2+2 c}{x^4 \left (b x^2+a\right )}dx}{2 a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {\left (\frac {c b^2}{a^2}-\frac {d b}{a}+e+\frac {a f}{b}\right ) x^4-2 \left (\frac {b c}{a}-d\right ) x^2+2 c}{x^4 \left (b x^2+a\right )}dx}{2 a}+\frac {x \left (\frac {b^2 c}{a^2}-\frac {b d}{a}-\frac {a f}{b}+e\right )}{2 a \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 1584 |
\(\displaystyle \frac {\int \left (\frac {2 c}{a x^4}+\frac {f a^3+b e a^2-3 b^2 d a+5 b^3 c}{a^2 b \left (b x^2+a\right )}+\frac {2 (a d-2 b c)}{a^2 x^2}\right )dx}{2 a}+\frac {x \left (\frac {b^2 c}{a^2}-\frac {b d}{a}-\frac {a f}{b}+e\right )}{2 a \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x \left (\frac {b^2 c}{a^2}-\frac {b d}{a}-\frac {a f}{b}+e\right )}{2 a \left (a+b x^2\right )}+\frac {\frac {2 (2 b c-a d)}{a^2 x}+\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (a^3 f+a^2 b e-3 a b^2 d+5 b^3 c\right )}{a^{5/2} b^{3/2}}-\frac {2 c}{3 a x^3}}{2 a}\) |
(((b^2*c)/a^2 - (b*d)/a + e - (a*f)/b)*x)/(2*a*(a + b*x^2)) + ((-2*c)/(3*a *x^3) + (2*(2*b*c - a*d))/(a^2*x) + ((5*b^3*c - 3*a*b^2*d + a^2*b*e + a^3* f)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(a^(5/2)*b^(3/2)))/(2*a)
3.2.29.3.1 Defintions of rubi rules used
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]
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 = 3.46 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.96
method | result | size |
default | \(-\frac {c}{3 a^{2} x^{3}}-\frac {a d -2 b c}{a^{3} x}+\frac {-\frac {\left (f \,a^{3}-a^{2} b e +a \,b^{2} d -b^{3} c \right ) x}{2 b \left (b \,x^{2}+a \right )}+\frac {\left (f \,a^{3}+a^{2} b e -3 a \,b^{2} d +5 b^{3} c \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 b \sqrt {a b}}}{a^{3}}\) | \(116\) |
risch | \(\frac {-\frac {\left (f \,a^{3}-a^{2} b e +3 a \,b^{2} d -5 b^{3} c \right ) x^{4}}{2 a^{3} b}-\frac {\left (3 a d -5 b c \right ) x^{2}}{3 a^{2}}-\frac {c}{3 a}}{x^{3} \left (b \,x^{2}+a \right )}-\frac {\ln \left (-\sqrt {-a b}\, x +a \right ) f}{4 \sqrt {-a b}\, b}-\frac {\ln \left (-\sqrt {-a b}\, x +a \right ) e}{4 \sqrt {-a b}\, a}+\frac {3 b \ln \left (-\sqrt {-a b}\, x +a \right ) d}{4 \sqrt {-a b}\, a^{2}}-\frac {5 b^{2} \ln \left (-\sqrt {-a b}\, x +a \right ) c}{4 \sqrt {-a b}\, a^{3}}+\frac {\ln \left (-\sqrt {-a b}\, x -a \right ) f}{4 \sqrt {-a b}\, b}+\frac {\ln \left (-\sqrt {-a b}\, x -a \right ) e}{4 \sqrt {-a b}\, a}-\frac {3 b \ln \left (-\sqrt {-a b}\, x -a \right ) d}{4 \sqrt {-a b}\, a^{2}}+\frac {5 b^{2} \ln \left (-\sqrt {-a b}\, x -a \right ) c}{4 \sqrt {-a b}\, a^{3}}\) | \(284\) |
-1/3*c/a^2/x^3-(a*d-2*b*c)/a^3/x+1/a^3*(-1/2*(a^3*f-a^2*b*e+a*b^2*d-b^3*c) /b*x/(b*x^2+a)+1/2*(a^3*f+a^2*b*e-3*a*b^2*d+5*b^3*c)/b/(a*b)^(1/2)*arctan( b*x/(a*b)^(1/2)))
Time = 0.28 (sec) , antiderivative size = 378, normalized size of antiderivative = 3.12 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^4 \left (a+b x^2\right )^2} \, dx=\left [-\frac {4 \, a^{3} b^{2} c - 6 \, {\left (5 \, a b^{4} c - 3 \, a^{2} b^{3} d + a^{3} b^{2} e - a^{4} b f\right )} x^{4} - 4 \, {\left (5 \, a^{2} b^{3} c - 3 \, a^{3} b^{2} d\right )} x^{2} + 3 \, {\left ({\left (5 \, b^{4} c - 3 \, a b^{3} d + a^{2} b^{2} e + a^{3} b f\right )} x^{5} + {\left (5 \, a b^{3} c - 3 \, a^{2} b^{2} d + a^{3} b e + a^{4} f\right )} x^{3}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{12 \, {\left (a^{4} b^{3} x^{5} + a^{5} b^{2} x^{3}\right )}}, -\frac {2 \, a^{3} b^{2} c - 3 \, {\left (5 \, a b^{4} c - 3 \, a^{2} b^{3} d + a^{3} b^{2} e - a^{4} b f\right )} x^{4} - 2 \, {\left (5 \, a^{2} b^{3} c - 3 \, a^{3} b^{2} d\right )} x^{2} - 3 \, {\left ({\left (5 \, b^{4} c - 3 \, a b^{3} d + a^{2} b^{2} e + a^{3} b f\right )} x^{5} + {\left (5 \, a b^{3} c - 3 \, a^{2} b^{2} d + a^{3} b e + a^{4} f\right )} x^{3}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{6 \, {\left (a^{4} b^{3} x^{5} + a^{5} b^{2} x^{3}\right )}}\right ] \]
[-1/12*(4*a^3*b^2*c - 6*(5*a*b^4*c - 3*a^2*b^3*d + a^3*b^2*e - a^4*b*f)*x^ 4 - 4*(5*a^2*b^3*c - 3*a^3*b^2*d)*x^2 + 3*((5*b^4*c - 3*a*b^3*d + a^2*b^2* e + a^3*b*f)*x^5 + (5*a*b^3*c - 3*a^2*b^2*d + a^3*b*e + a^4*f)*x^3)*sqrt(- a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)))/(a^4*b^3*x^5 + a^5*b^2 *x^3), -1/6*(2*a^3*b^2*c - 3*(5*a*b^4*c - 3*a^2*b^3*d + a^3*b^2*e - a^4*b* f)*x^4 - 2*(5*a^2*b^3*c - 3*a^3*b^2*d)*x^2 - 3*((5*b^4*c - 3*a*b^3*d + a^2 *b^2*e + a^3*b*f)*x^5 + (5*a*b^3*c - 3*a^2*b^2*d + a^3*b*e + a^4*f)*x^3)*s qrt(a*b)*arctan(sqrt(a*b)*x/a))/(a^4*b^3*x^5 + a^5*b^2*x^3)]
Time = 5.30 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.75 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^4 \left (a+b x^2\right )^2} \, dx=- \frac {\sqrt {- \frac {1}{a^{7} b^{3}}} \left (a^{3} f + a^{2} b e - 3 a b^{2} d + 5 b^{3} c\right ) \log {\left (- a^{4} b \sqrt {- \frac {1}{a^{7} b^{3}}} + x \right )}}{4} + \frac {\sqrt {- \frac {1}{a^{7} b^{3}}} \left (a^{3} f + a^{2} b e - 3 a b^{2} d + 5 b^{3} c\right ) \log {\left (a^{4} b \sqrt {- \frac {1}{a^{7} b^{3}}} + x \right )}}{4} + \frac {- 2 a^{2} b c + x^{4} \left (- 3 a^{3} f + 3 a^{2} b e - 9 a b^{2} d + 15 b^{3} c\right ) + x^{2} \left (- 6 a^{2} b d + 10 a b^{2} c\right )}{6 a^{4} b x^{3} + 6 a^{3} b^{2} x^{5}} \]
-sqrt(-1/(a**7*b**3))*(a**3*f + a**2*b*e - 3*a*b**2*d + 5*b**3*c)*log(-a** 4*b*sqrt(-1/(a**7*b**3)) + x)/4 + sqrt(-1/(a**7*b**3))*(a**3*f + a**2*b*e - 3*a*b**2*d + 5*b**3*c)*log(a**4*b*sqrt(-1/(a**7*b**3)) + x)/4 + (-2*a**2 *b*c + x**4*(-3*a**3*f + 3*a**2*b*e - 9*a*b**2*d + 15*b**3*c) + x**2*(-6*a **2*b*d + 10*a*b**2*c))/(6*a**4*b*x**3 + 6*a**3*b**2*x**5)
Time = 0.28 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.07 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^4 \left (a+b x^2\right )^2} \, dx=\frac {3 \, {\left (5 \, b^{3} c - 3 \, a b^{2} d + a^{2} b e - a^{3} f\right )} x^{4} - 2 \, a^{2} b c + 2 \, {\left (5 \, a b^{2} c - 3 \, a^{2} b d\right )} x^{2}}{6 \, {\left (a^{3} b^{2} x^{5} + a^{4} b x^{3}\right )}} + \frac {{\left (5 \, b^{3} c - 3 \, a b^{2} d + a^{2} b e + a^{3} f\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{3} b} \]
1/6*(3*(5*b^3*c - 3*a*b^2*d + a^2*b*e - a^3*f)*x^4 - 2*a^2*b*c + 2*(5*a*b^ 2*c - 3*a^2*b*d)*x^2)/(a^3*b^2*x^5 + a^4*b*x^3) + 1/2*(5*b^3*c - 3*a*b^2*d + a^2*b*e + a^3*f)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^3*b)
Time = 0.29 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^4 \left (a+b x^2\right )^2} \, dx=\frac {{\left (5 \, b^{3} c - 3 \, a b^{2} d + a^{2} b e + a^{3} f\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{3} b} + \frac {b^{3} c x - a b^{2} d x + a^{2} b e x - a^{3} f x}{2 \, {\left (b x^{2} + a\right )} a^{3} b} + \frac {6 \, b c x^{2} - 3 \, a d x^{2} - a c}{3 \, a^{3} x^{3}} \]
1/2*(5*b^3*c - 3*a*b^2*d + a^2*b*e + a^3*f)*arctan(b*x/sqrt(a*b))/(sqrt(a* b)*a^3*b) + 1/2*(b^3*c*x - a*b^2*d*x + a^2*b*e*x - a^3*f*x)/((b*x^2 + a)*a ^3*b) + 1/3*(6*b*c*x^2 - 3*a*d*x^2 - a*c)/(a^3*x^3)
Time = 0.15 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.98 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^4 \left (a+b x^2\right )^2} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (f\,a^3+e\,a^2\,b-3\,d\,a\,b^2+5\,c\,b^3\right )}{2\,a^{7/2}\,b^{3/2}}-\frac {\frac {c}{3\,a}+\frac {x^2\,\left (3\,a\,d-5\,b\,c\right )}{3\,a^2}-\frac {x^4\,\left (-f\,a^3+e\,a^2\,b-3\,d\,a\,b^2+5\,c\,b^3\right )}{2\,a^3\,b}}{b\,x^5+a\,x^3} \]