Integrand size = 30, antiderivative size = 136 \[ \int \frac {x^2 \left (c+d x^2+e x^4+f x^6\right )}{a+b x^2} \, dx=\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x}{b^4}+\frac {\left (b^2 d-a b e+a^2 f\right ) x^3}{3 b^3}+\frac {(b e-a f) x^5}{5 b^2}+\frac {f x^7}{7 b}-\frac {\sqrt {a} \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{9/2}} \]
(-a^3*f+a^2*b*e-a*b^2*d+b^3*c)*x/b^4+1/3*(a^2*f-a*b*e+b^2*d)*x^3/b^3+1/5*( -a*f+b*e)*x^5/b^2+1/7*f*x^7/b-(-a^3*f+a^2*b*e-a*b^2*d+b^3*c)*arctan(x*b^(1 /2)/a^(1/2))*a^(1/2)/b^(9/2)
Time = 0.06 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.94 \[ \int \frac {x^2 \left (c+d x^2+e x^4+f x^6\right )}{a+b x^2} \, dx=\frac {x \left (-105 a^3 f+35 a^2 b \left (3 e+f x^2\right )-7 a b^2 \left (15 d+5 e x^2+3 f x^4\right )+b^3 \left (105 c+35 d x^2+21 e x^4+15 f x^6\right )\right )}{105 b^4}+\frac {\sqrt {a} \left (-b^3 c+a b^2 d-a^2 b e+a^3 f\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{9/2}} \]
(x*(-105*a^3*f + 35*a^2*b*(3*e + f*x^2) - 7*a*b^2*(15*d + 5*e*x^2 + 3*f*x^ 4) + b^3*(105*c + 35*d*x^2 + 21*e*x^4 + 15*f*x^6)))/(105*b^4) + (Sqrt[a]*( -(b^3*c) + a*b^2*d - a^2*b*e + a^3*f)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/b^(9/2)
Time = 0.32 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {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.
\(\displaystyle \int \frac {x^2 \left (c+d x^2+e x^4+f x^6\right )}{a+b x^2} \, dx\) |
\(\Big \downarrow \) 2333 |
\(\displaystyle \int \left (\frac {x^2 \left (a^2 f-a b e+b^2 d\right )}{b^3}+\frac {a^3 (-f)+a^2 b e-a b^2 d+b^3 c}{b^4}+\frac {a^4 f-a^3 b e+a^2 b^2 d-a b^3 c}{b^4 \left (a+b x^2\right )}+\frac {x^4 (b e-a f)}{b^2}+\frac {f x^6}{b}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x^3 \left (a^2 f-a b e+b^2 d\right )}{3 b^3}-\frac {\sqrt {a} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{b^{9/2}}+\frac {x \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{b^4}+\frac {x^5 (b e-a f)}{5 b^2}+\frac {f x^7}{7 b}\) |
((b^3*c - a*b^2*d + a^2*b*e - a^3*f)*x)/b^4 + ((b^2*d - a*b*e + a^2*f)*x^3 )/(3*b^3) + ((b*e - a*f)*x^5)/(5*b^2) + (f*x^7)/(7*b) - (Sqrt[a]*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/b^(9/2)
3.2.16.3.1 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]
Time = 3.49 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.01
method | result | size |
default | \(-\frac {-\frac {1}{7} f \,x^{7} b^{3}+\frac {1}{5} a \,b^{2} f \,x^{5}-\frac {1}{5} b^{3} e \,x^{5}-\frac {1}{3} a^{2} b f \,x^{3}+\frac {1}{3} a \,b^{2} e \,x^{3}-\frac {1}{3} b^{3} d \,x^{3}+f \,a^{3} x -a^{2} b e x +a \,b^{2} d x -b^{3} c x}{b^{4}}+\frac {a \left (f \,a^{3}-a^{2} b e +a \,b^{2} d -b^{3} c \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{b^{4} \sqrt {a b}}\) | \(137\) |
risch | \(\frac {f \,x^{7}}{7 b}-\frac {a f \,x^{5}}{5 b^{2}}+\frac {e \,x^{5}}{5 b}+\frac {a^{2} f \,x^{3}}{3 b^{3}}-\frac {a e \,x^{3}}{3 b^{2}}+\frac {d \,x^{3}}{3 b}-\frac {f \,a^{3} x}{b^{4}}+\frac {a^{2} e x}{b^{3}}-\frac {a d x}{b^{2}}+\frac {c x}{b}+\frac {\sqrt {-a b}\, \ln \left (-\sqrt {-a b}\, x +a \right ) f \,a^{3}}{2 b^{5}}-\frac {\sqrt {-a b}\, \ln \left (-\sqrt {-a b}\, x +a \right ) a^{2} e}{2 b^{4}}+\frac {\sqrt {-a b}\, \ln \left (-\sqrt {-a b}\, x +a \right ) a d}{2 b^{3}}-\frac {\sqrt {-a b}\, \ln \left (-\sqrt {-a b}\, x +a \right ) c}{2 b^{2}}-\frac {\sqrt {-a b}\, \ln \left (\sqrt {-a b}\, x +a \right ) f \,a^{3}}{2 b^{5}}+\frac {\sqrt {-a b}\, \ln \left (\sqrt {-a b}\, x +a \right ) a^{2} e}{2 b^{4}}-\frac {\sqrt {-a b}\, \ln \left (\sqrt {-a b}\, x +a \right ) a d}{2 b^{3}}+\frac {\sqrt {-a b}\, \ln \left (\sqrt {-a b}\, x +a \right ) c}{2 b^{2}}\) | \(296\) |
-1/b^4*(-1/7*f*x^7*b^3+1/5*a*b^2*f*x^5-1/5*b^3*e*x^5-1/3*a^2*b*f*x^3+1/3*a *b^2*e*x^3-1/3*b^3*d*x^3+f*a^3*x-a^2*b*e*x+a*b^2*d*x-b^3*c*x)+a*(a^3*f-a^2 *b*e+a*b^2*d-b^3*c)/b^4/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))
Time = 0.29 (sec) , antiderivative size = 286, normalized size of antiderivative = 2.10 \[ \int \frac {x^2 \left (c+d x^2+e x^4+f x^6\right )}{a+b x^2} \, dx=\left [\frac {30 \, b^{3} f x^{7} + 42 \, {\left (b^{3} e - a b^{2} f\right )} x^{5} + 70 \, {\left (b^{3} d - a b^{2} e + a^{2} b f\right )} x^{3} - 105 \, {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} + 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) + 210 \, {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} x}{210 \, b^{4}}, \frac {15 \, b^{3} f x^{7} + 21 \, {\left (b^{3} e - a b^{2} f\right )} x^{5} + 35 \, {\left (b^{3} d - a b^{2} e + a^{2} b f\right )} x^{3} - 105 \, {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) + 105 \, {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} x}{105 \, b^{4}}\right ] \]
[1/210*(30*b^3*f*x^7 + 42*(b^3*e - a*b^2*f)*x^5 + 70*(b^3*d - a*b^2*e + a^ 2*b*f)*x^3 - 105*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*sqrt(-a/b)*log((b*x^2 + 2*b*x*sqrt(-a/b) - a)/(b*x^2 + a)) + 210*(b^3*c - a*b^2*d + a^2*b*e - a ^3*f)*x)/b^4, 1/105*(15*b^3*f*x^7 + 21*(b^3*e - a*b^2*f)*x^5 + 35*(b^3*d - a*b^2*e + a^2*b*f)*x^3 - 105*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*sqrt(a/b )*arctan(b*x*sqrt(a/b)/a) + 105*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*x)/b^4 ]
Time = 0.40 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.36 \[ \int \frac {x^2 \left (c+d x^2+e x^4+f x^6\right )}{a+b x^2} \, dx=x^{5} \left (- \frac {a f}{5 b^{2}} + \frac {e}{5 b}\right ) + x^{3} \left (\frac {a^{2} f}{3 b^{3}} - \frac {a e}{3 b^{2}} + \frac {d}{3 b}\right ) + x \left (- \frac {a^{3} f}{b^{4}} + \frac {a^{2} e}{b^{3}} - \frac {a d}{b^{2}} + \frac {c}{b}\right ) - \frac {\sqrt {- \frac {a}{b^{9}}} \left (a^{3} f - a^{2} b e + a b^{2} d - b^{3} c\right ) \log {\left (- b^{4} \sqrt {- \frac {a}{b^{9}}} + x \right )}}{2} + \frac {\sqrt {- \frac {a}{b^{9}}} \left (a^{3} f - a^{2} b e + a b^{2} d - b^{3} c\right ) \log {\left (b^{4} \sqrt {- \frac {a}{b^{9}}} + x \right )}}{2} + \frac {f x^{7}}{7 b} \]
x**5*(-a*f/(5*b**2) + e/(5*b)) + x**3*(a**2*f/(3*b**3) - a*e/(3*b**2) + d/ (3*b)) + x*(-a**3*f/b**4 + a**2*e/b**3 - a*d/b**2 + c/b) - sqrt(-a/b**9)*( a**3*f - a**2*b*e + a*b**2*d - b**3*c)*log(-b**4*sqrt(-a/b**9) + x)/2 + sq rt(-a/b**9)*(a**3*f - a**2*b*e + a*b**2*d - b**3*c)*log(b**4*sqrt(-a/b**9) + x)/2 + f*x**7/(7*b)
Time = 0.29 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.98 \[ \int \frac {x^2 \left (c+d x^2+e x^4+f x^6\right )}{a+b x^2} \, dx=-\frac {{\left (a b^{3} c - a^{2} b^{2} d + a^{3} b e - a^{4} f\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b^{4}} + \frac {15 \, b^{3} f x^{7} + 21 \, {\left (b^{3} e - a b^{2} f\right )} x^{5} + 35 \, {\left (b^{3} d - a b^{2} e + a^{2} b f\right )} x^{3} + 105 \, {\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} x}{105 \, b^{4}} \]
-(a*b^3*c - a^2*b^2*d + a^3*b*e - a^4*f)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)* b^4) + 1/105*(15*b^3*f*x^7 + 21*(b^3*e - a*b^2*f)*x^5 + 35*(b^3*d - a*b^2* e + a^2*b*f)*x^3 + 105*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*x)/b^4
Time = 0.28 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.09 \[ \int \frac {x^2 \left (c+d x^2+e x^4+f x^6\right )}{a+b x^2} \, dx=-\frac {{\left (a b^{3} c - a^{2} b^{2} d + a^{3} b e - a^{4} f\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b^{4}} + \frac {15 \, b^{6} f x^{7} + 21 \, b^{6} e x^{5} - 21 \, a b^{5} f x^{5} + 35 \, b^{6} d x^{3} - 35 \, a b^{5} e x^{3} + 35 \, a^{2} b^{4} f x^{3} + 105 \, b^{6} c x - 105 \, a b^{5} d x + 105 \, a^{2} b^{4} e x - 105 \, a^{3} b^{3} f x}{105 \, b^{7}} \]
-(a*b^3*c - a^2*b^2*d + a^3*b*e - a^4*f)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)* b^4) + 1/105*(15*b^6*f*x^7 + 21*b^6*e*x^5 - 21*a*b^5*f*x^5 + 35*b^6*d*x^3 - 35*a*b^5*e*x^3 + 35*a^2*b^4*f*x^3 + 105*b^6*c*x - 105*a*b^5*d*x + 105*a^ 2*b^4*e*x - 105*a^3*b^3*f*x)/b^7
Time = 5.58 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.42 \[ \int \frac {x^2 \left (c+d x^2+e x^4+f x^6\right )}{a+b x^2} \, dx=x^5\,\left (\frac {e}{5\,b}-\frac {a\,f}{5\,b^2}\right )+x^3\,\left (\frac {d}{3\,b}-\frac {a\,\left (\frac {e}{b}-\frac {a\,f}{b^2}\right )}{3\,b}\right )+x\,\left (\frac {c}{b}-\frac {a\,\left (\frac {d}{b}-\frac {a\,\left (\frac {e}{b}-\frac {a\,f}{b^2}\right )}{b}\right )}{b}\right )+\frac {f\,x^7}{7\,b}+\frac {\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {b}\,x\,\left (-f\,a^3+e\,a^2\,b-d\,a\,b^2+c\,b^3\right )}{f\,a^4-e\,a^3\,b+d\,a^2\,b^2-c\,a\,b^3}\right )\,\left (-f\,a^3+e\,a^2\,b-d\,a\,b^2+c\,b^3\right )}{b^{9/2}} \]
x^5*(e/(5*b) - (a*f)/(5*b^2)) + x^3*(d/(3*b) - (a*(e/b - (a*f)/b^2))/(3*b) ) + x*(c/b - (a*(d/b - (a*(e/b - (a*f)/b^2))/b))/b) + (f*x^7)/(7*b) + (a^( 1/2)*atan((a^(1/2)*b^(1/2)*x*(b^3*c - a^3*f - a*b^2*d + a^2*b*e))/(a^4*f + a^2*b^2*d - a*b^3*c - a^3*b*e))*(b^3*c - a^3*f - a*b^2*d + a^2*b*e))/b^(9 /2)