Integrand size = 30, antiderivative size = 153 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^2 \left (a+b x^2\right )^3} \, dx=-\frac {c}{a^3 x}-\frac {\left (\frac {b c}{a}-d+\frac {a e}{b}-\frac {a^2 f}{b^2}\right ) x}{4 a \left (a+b x^2\right )^2}-\frac {\left (7 b^3 c-3 a b^2 d-a^2 b e+5 a^3 f\right ) x}{8 a^3 b^2 \left (a+b x^2\right )}-\frac {\left (15 b^3 c-3 a b^2 d-a^2 b e-3 a^3 f\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} b^{5/2}} \]
-c/a^3/x-1/4*(b*c/a-d+a*e/b-a^2*f/b^2)*x/a/(b*x^2+a)^2-1/8*(5*a^3*f-a^2*b* e-3*a*b^2*d+7*b^3*c)*x/a^3/b^2/(b*x^2+a)-1/8*(-3*a^3*f-a^2*b*e-3*a*b^2*d+1 5*b^3*c)*arctan(x*b^(1/2)/a^(1/2))/a^(7/2)/b^(5/2)
Time = 0.08 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.01 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^2 \left (a+b x^2\right )^3} \, dx=-\frac {c}{a^3 x}+\frac {\left (-b^3 c+a b^2 d-a^2 b e+a^3 f\right ) x}{4 a^2 b^2 \left (a+b x^2\right )^2}-\frac {\left (7 b^3 c-3 a b^2 d-a^2 b e+5 a^3 f\right ) x}{8 a^3 b^2 \left (a+b x^2\right )}+\frac {\left (-15 b^3 c+3 a b^2 d+a^2 b e+3 a^3 f\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} b^{5/2}} \]
-(c/(a^3*x)) + ((-(b^3*c) + a*b^2*d - a^2*b*e + a^3*f)*x)/(4*a^2*b^2*(a + b*x^2)^2) - ((7*b^3*c - 3*a*b^2*d - a^2*b*e + 5*a^3*f)*x)/(8*a^3*b^2*(a + b*x^2)) + ((-15*b^3*c + 3*a*b^2*d + a^2*b*e + 3*a^3*f)*ArcTan[(Sqrt[b]*x)/ Sqrt[a]])/(8*a^(7/2)*b^(5/2))
Time = 0.42 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2336, 25, 1582, 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.
| \(\displaystyle \int \frac {c+d x^2+e x^4+f x^6}{x^2 \left (a+b x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 2336 |
| \(\displaystyle -\frac {\int -\frac {\frac {4 a f x^4}{b}-\left (\frac {f a^2}{b^2}-\frac {e a}{b}-3 d+\frac {3 b c}{a}\right ) x^2+4 c}{x^2 \left (b x^2+a\right )^2}dx}{4 a}-\frac {x \left (-\frac {a^2 f}{b^2}+\frac {b c}{a}+\frac {a e}{b}-d\right )}{4 a \left (a+b x^2\right )^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\frac {4 a f x^4}{b}-\left (\frac {f a^2}{b^2}-\frac {e a}{b}-3 d+\frac {3 b c}{a}\right ) x^2+4 c}{x^2 \left (b x^2+a\right )^2}dx}{4 a}-\frac {x \left (-\frac {a^2 f}{b^2}+\frac {b c}{a}+\frac {a e}{b}-d\right )}{4 a \left (a+b x^2\right )^2}\) |
| \(\Big \downarrow \) 1582 |
| \(\displaystyle \frac {-\frac {\int -\frac {8 a b^2 c-\left (-3 f a^3-b e a^2-3 b^2 d a+7 b^3 c\right ) x^2}{x^2 \left (b x^2+a\right )}dx}{2 a^2 b^2}-\frac {x \left (5 a^3 f-a^2 b e-3 a b^2 d+7 b^3 c\right )}{2 a^2 b^2 \left (a+b x^2\right )}}{4 a}-\frac {x \left (-\frac {a^2 f}{b^2}+\frac {b c}{a}+\frac {a e}{b}-d\right )}{4 a \left (a+b x^2\right )^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {8 a b^2 c-\left (-3 f a^3-b e a^2-3 b^2 d a+7 b^3 c\right ) x^2}{x^2 \left (b x^2+a\right )}dx}{2 a^2 b^2}-\frac {x \left (5 a^3 f-a^2 b e-3 a b^2 d+7 b^3 c\right )}{2 a^2 b^2 \left (a+b x^2\right )}}{4 a}-\frac {x \left (-\frac {a^2 f}{b^2}+\frac {b c}{a}+\frac {a e}{b}-d\right )}{4 a \left (a+b x^2\right )^2}\) |
| \(\Big \downarrow \) 359 |
| \(\displaystyle \frac {\frac {-\left (-3 a^3 f-a^2 b e-3 a b^2 d+15 b^3 c\right ) \int \frac {1}{b x^2+a}dx-\frac {8 b^2 c}{x}}{2 a^2 b^2}-\frac {x \left (5 a^3 f-a^2 b e-3 a b^2 d+7 b^3 c\right )}{2 a^2 b^2 \left (a+b x^2\right )}}{4 a}-\frac {x \left (-\frac {a^2 f}{b^2}+\frac {b c}{a}+\frac {a e}{b}-d\right )}{4 a \left (a+b x^2\right )^2}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {-\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (-3 a^3 f-a^2 b e-3 a b^2 d+15 b^3 c\right )}{\sqrt {a} \sqrt {b}}-\frac {8 b^2 c}{x}}{2 a^2 b^2}-\frac {x \left (5 a^3 f-a^2 b e-3 a b^2 d+7 b^3 c\right )}{2 a^2 b^2 \left (a+b x^2\right )}}{4 a}-\frac {x \left (-\frac {a^2 f}{b^2}+\frac {b c}{a}+\frac {a e}{b}-d\right )}{4 a \left (a+b x^2\right )^2}\) |
-1/4*(((b*c)/a - d + (a*e)/b - (a^2*f)/b^2)*x)/(a*(a + b*x^2)^2) + (-1/2*( (7*b^3*c - 3*a*b^2*d - a^2*b*e + 5*a^3*f)*x)/(a^2*b^2*(a + b*x^2)) + ((-8* b^2*c)/x - ((15*b^3*c - 3*a*b^2*d - a^2*b*e - 3*a^3*f)*ArcTan[(Sqrt[b]*x)/ Sqrt[a]])/(Sqrt[a]*Sqrt[b]))/(2*a^2*b^2))/(4*a)
3.2.38.3.1 Defintions of rubi rules used
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_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^ 4)^(p_.), x_Symbol] :> Simp[(-d)^(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*((d + e*x^2)^(q + 1)/(2*e^(2*p + m/2)*(q + 1))), x] + Simp[(-d)^(m/2 - 1)/(2*e^ (2*p)*(q + 1)) Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e *x^2))*(2*(-d)^(-m/2 + 1)*e^(2*p)*(q + 1)*(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2))], x], x], x] /; Fre eQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]
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 = 3.52 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.92
| method | result | size |
| default | \(-\frac {c}{a^{3} x}+\frac {\frac {-\frac {\left (5 f \,a^{3}-a^{2} b e -3 a \,b^{2} d +7 b^{3} c \right ) x^{3}}{8 b}-\frac {a \left (3 f \,a^{3}+a^{2} b e -5 a \,b^{2} d +9 b^{3} c \right ) x}{8 b^{2}}}{\left (b \,x^{2}+a \right )^{2}}+\frac {\left (3 f \,a^{3}+a^{2} b e +3 a \,b^{2} d -15 b^{3} c \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 b^{2} \sqrt {a b}}}{a^{3}}\) | \(140\) |
| risch | \(\frac {-\frac {\left (5 f \,a^{3}-a^{2} b e -3 a \,b^{2} d +15 b^{3} c \right ) x^{4}}{8 a^{3} b}-\frac {\left (3 f \,a^{3}+a^{2} b e -5 a \,b^{2} d +25 b^{3} c \right ) x^{2}}{8 a^{2} b^{2}}-\frac {c}{a}}{x \left (b \,x^{2}+a \right )^{2}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{7} b^{5} \textit {\_Z}^{2}+9 a^{6} f^{2}+6 a^{5} b e f +18 a^{4} b^{2} d f +a^{4} b^{2} e^{2}-90 a^{3} b^{3} c f +6 a^{3} b^{3} d e -30 a^{2} b^{4} c e +9 a^{2} b^{4} d^{2}-90 a c d \,b^{5}+225 c^{2} b^{6}\right )}{\sum }\textit {\_R} \ln \left (\left (3 \textit {\_R}^{2} a^{7} b^{5}+18 a^{6} f^{2}+12 a^{5} b e f +36 a^{4} b^{2} d f +2 a^{4} b^{2} e^{2}-180 a^{3} b^{3} c f +12 a^{3} b^{3} d e -60 a^{2} b^{4} c e +18 a^{2} b^{4} d^{2}-180 a c d \,b^{5}+450 c^{2} b^{6}\right ) x +\left (-3 a^{7} f \,b^{2}-e \,b^{3} a^{6}-3 b^{4} d \,a^{5}+15 b^{5} c \,a^{4}\right ) \textit {\_R} \right )\right )}{16}\) | \(358\) |
-c/a^3/x+1/a^3*((-1/8*(5*a^3*f-a^2*b*e-3*a*b^2*d+7*b^3*c)/b*x^3-1/8*a*(3*a ^3*f+a^2*b*e-5*a*b^2*d+9*b^3*c)/b^2*x)/(b*x^2+a)^2+1/8*(3*a^3*f+a^2*b*e+3* a*b^2*d-15*b^3*c)/b^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))
Time = 0.28 (sec) , antiderivative size = 517, normalized size of antiderivative = 3.38 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^2 \left (a+b x^2\right )^3} \, dx=\left [-\frac {16 \, a^{3} b^{3} c + 2 \, {\left (15 \, a b^{5} c - 3 \, a^{2} b^{4} d - a^{3} b^{3} e + 5 \, a^{4} b^{2} f\right )} x^{4} + 2 \, {\left (25 \, a^{2} b^{4} c - 5 \, a^{3} b^{3} d + a^{4} b^{2} e + 3 \, a^{5} b f\right )} x^{2} - {\left ({\left (15 \, b^{5} c - 3 \, a b^{4} d - a^{2} b^{3} e - 3 \, a^{3} b^{2} f\right )} x^{5} + 2 \, {\left (15 \, a b^{4} c - 3 \, a^{2} b^{3} d - a^{3} b^{2} e - 3 \, a^{4} b f\right )} x^{3} + {\left (15 \, a^{2} b^{3} c - 3 \, a^{3} b^{2} d - a^{4} b e - 3 \, a^{5} f\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{16 \, {\left (a^{4} b^{5} x^{5} + 2 \, a^{5} b^{4} x^{3} + a^{6} b^{3} x\right )}}, -\frac {8 \, a^{3} b^{3} c + {\left (15 \, a b^{5} c - 3 \, a^{2} b^{4} d - a^{3} b^{3} e + 5 \, a^{4} b^{2} f\right )} x^{4} + {\left (25 \, a^{2} b^{4} c - 5 \, a^{3} b^{3} d + a^{4} b^{2} e + 3 \, a^{5} b f\right )} x^{2} + {\left ({\left (15 \, b^{5} c - 3 \, a b^{4} d - a^{2} b^{3} e - 3 \, a^{3} b^{2} f\right )} x^{5} + 2 \, {\left (15 \, a b^{4} c - 3 \, a^{2} b^{3} d - a^{3} b^{2} e - 3 \, a^{4} b f\right )} x^{3} + {\left (15 \, a^{2} b^{3} c - 3 \, a^{3} b^{2} d - a^{4} b e - 3 \, a^{5} f\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{8 \, {\left (a^{4} b^{5} x^{5} + 2 \, a^{5} b^{4} x^{3} + a^{6} b^{3} x\right )}}\right ] \]
[-1/16*(16*a^3*b^3*c + 2*(15*a*b^5*c - 3*a^2*b^4*d - a^3*b^3*e + 5*a^4*b^2 *f)*x^4 + 2*(25*a^2*b^4*c - 5*a^3*b^3*d + a^4*b^2*e + 3*a^5*b*f)*x^2 - ((1 5*b^5*c - 3*a*b^4*d - a^2*b^3*e - 3*a^3*b^2*f)*x^5 + 2*(15*a*b^4*c - 3*a^2 *b^3*d - a^3*b^2*e - 3*a^4*b*f)*x^3 + (15*a^2*b^3*c - 3*a^3*b^2*d - a^4*b* e - 3*a^5*f)*x)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)))/ (a^4*b^5*x^5 + 2*a^5*b^4*x^3 + a^6*b^3*x), -1/8*(8*a^3*b^3*c + (15*a*b^5*c - 3*a^2*b^4*d - a^3*b^3*e + 5*a^4*b^2*f)*x^4 + (25*a^2*b^4*c - 5*a^3*b^3* d + a^4*b^2*e + 3*a^5*b*f)*x^2 + ((15*b^5*c - 3*a*b^4*d - a^2*b^3*e - 3*a^ 3*b^2*f)*x^5 + 2*(15*a*b^4*c - 3*a^2*b^3*d - a^3*b^2*e - 3*a^4*b*f)*x^3 + (15*a^2*b^3*c - 3*a^3*b^2*d - a^4*b*e - 3*a^5*f)*x)*sqrt(a*b)*arctan(sqrt( a*b)*x/a))/(a^4*b^5*x^5 + 2*a^5*b^4*x^3 + a^6*b^3*x)]
Time = 11.08 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.63 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^2 \left (a+b x^2\right )^3} \, dx=- \frac {\sqrt {- \frac {1}{a^{7} b^{5}}} \cdot \left (3 a^{3} f + a^{2} b e + 3 a b^{2} d - 15 b^{3} c\right ) \log {\left (- a^{4} b^{2} \sqrt {- \frac {1}{a^{7} b^{5}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{a^{7} b^{5}}} \cdot \left (3 a^{3} f + a^{2} b e + 3 a b^{2} d - 15 b^{3} c\right ) \log {\left (a^{4} b^{2} \sqrt {- \frac {1}{a^{7} b^{5}}} + x \right )}}{16} + \frac {- 8 a^{2} b^{2} c + x^{4} \left (- 5 a^{3} b f + a^{2} b^{2} e + 3 a b^{3} d - 15 b^{4} c\right ) + x^{2} \left (- 3 a^{4} f - a^{3} b e + 5 a^{2} b^{2} d - 25 a b^{3} c\right )}{8 a^{5} b^{2} x + 16 a^{4} b^{3} x^{3} + 8 a^{3} b^{4} x^{5}} \]
-sqrt(-1/(a**7*b**5))*(3*a**3*f + a**2*b*e + 3*a*b**2*d - 15*b**3*c)*log(- a**4*b**2*sqrt(-1/(a**7*b**5)) + x)/16 + sqrt(-1/(a**7*b**5))*(3*a**3*f + a**2*b*e + 3*a*b**2*d - 15*b**3*c)*log(a**4*b**2*sqrt(-1/(a**7*b**5)) + x) /16 + (-8*a**2*b**2*c + x**4*(-5*a**3*b*f + a**2*b**2*e + 3*a*b**3*d - 15* b**4*c) + x**2*(-3*a**4*f - a**3*b*e + 5*a**2*b**2*d - 25*a*b**3*c))/(8*a* *5*b**2*x + 16*a**4*b**3*x**3 + 8*a**3*b**4*x**5)
Time = 0.29 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.05 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^2 \left (a+b x^2\right )^3} \, dx=-\frac {8 \, a^{2} b^{2} c + {\left (15 \, b^{4} c - 3 \, a b^{3} d - a^{2} b^{2} e + 5 \, a^{3} b f\right )} x^{4} + {\left (25 \, a b^{3} c - 5 \, a^{2} b^{2} d + a^{3} b e + 3 \, a^{4} f\right )} x^{2}}{8 \, {\left (a^{3} b^{4} x^{5} + 2 \, a^{4} b^{3} x^{3} + a^{5} b^{2} x\right )}} - \frac {{\left (15 \, b^{3} c - 3 \, a b^{2} d - a^{2} b e - 3 \, a^{3} f\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3} b^{2}} \]
-1/8*(8*a^2*b^2*c + (15*b^4*c - 3*a*b^3*d - a^2*b^2*e + 5*a^3*b*f)*x^4 + ( 25*a*b^3*c - 5*a^2*b^2*d + a^3*b*e + 3*a^4*f)*x^2)/(a^3*b^4*x^5 + 2*a^4*b^ 3*x^3 + a^5*b^2*x) - 1/8*(15*b^3*c - 3*a*b^2*d - a^2*b*e - 3*a^3*f)*arctan (b*x/sqrt(a*b))/(sqrt(a*b)*a^3*b^2)
Time = 0.29 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.98 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^2 \left (a+b x^2\right )^3} \, dx=-\frac {c}{a^{3} x} - \frac {{\left (15 \, b^{3} c - 3 \, a b^{2} d - a^{2} b e - 3 \, a^{3} f\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3} b^{2}} - \frac {7 \, b^{4} c x^{3} - 3 \, a b^{3} d x^{3} - a^{2} b^{2} e x^{3} + 5 \, a^{3} b f x^{3} + 9 \, a b^{3} c x - 5 \, a^{2} b^{2} d x + a^{3} b e x + 3 \, a^{4} f x}{8 \, {\left (b x^{2} + a\right )}^{2} a^{3} b^{2}} \]
-c/(a^3*x) - 1/8*(15*b^3*c - 3*a*b^2*d - a^2*b*e - 3*a^3*f)*arctan(b*x/sqr t(a*b))/(sqrt(a*b)*a^3*b^2) - 1/8*(7*b^4*c*x^3 - 3*a*b^3*d*x^3 - a^2*b^2*e *x^3 + 5*a^3*b*f*x^3 + 9*a*b^3*c*x - 5*a^2*b^2*d*x + a^3*b*e*x + 3*a^4*f*x )/((b*x^2 + a)^2*a^3*b^2)
Time = 0.17 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.97 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^2 \left (a+b x^2\right )^3} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (3\,f\,a^3+e\,a^2\,b+3\,d\,a\,b^2-15\,c\,b^3\right )}{8\,a^{7/2}\,b^{5/2}}-\frac {\frac {c}{a}+\frac {x^4\,\left (5\,f\,a^3-e\,a^2\,b-3\,d\,a\,b^2+15\,c\,b^3\right )}{8\,a^3\,b}+\frac {x^2\,\left (3\,f\,a^3+e\,a^2\,b-5\,d\,a\,b^2+25\,c\,b^3\right )}{8\,a^2\,b^2}}{a^2\,x+2\,a\,b\,x^3+b^2\,x^5} \]