Integrand size = 22, antiderivative size = 87 \[ \int \frac {\left (a+b x^2\right )^2}{x^6 \left (c+d x^2\right )} \, dx=-\frac {a^2}{5 c x^5}-\frac {a (2 b c-a d)}{3 c^2 x^3}-\frac {(b c-a d)^2}{c^3 x}-\frac {\sqrt {d} (b c-a d)^2 \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{c^{7/2}} \] Output:
-1/5*a^2/c/x^5-1/3*a*(-a*d+2*b*c)/c^2/x^3-(-a*d+b*c)^2/c^3/x-d^(1/2)*(-a*d +b*c)^2*arctan(d^(1/2)*x/c^(1/2))/c^(7/2)
Time = 0.05 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+b x^2\right )^2}{x^6 \left (c+d x^2\right )} \, dx=-\frac {a^2}{5 c x^5}+\frac {a (-2 b c+a d)}{3 c^2 x^3}-\frac {(b c-a d)^2}{c^3 x}-\frac {\sqrt {d} (b c-a d)^2 \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{c^{7/2}} \] Input:
Integrate[(a + b*x^2)^2/(x^6*(c + d*x^2)),x]
Output:
-1/5*a^2/(c*x^5) + (a*(-2*b*c + a*d))/(3*c^2*x^3) - (b*c - a*d)^2/(c^3*x) - (Sqrt[d]*(b*c - a*d)^2*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/c^(7/2)
Time = 0.23 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {364, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^2}{x^6 \left (c+d x^2\right )} \, dx\) |
\(\Big \downarrow \) 364 |
\(\displaystyle \int \left (\frac {a^2}{c x^6}-\frac {d (b c-a d)^2}{c^3 \left (c+d x^2\right )}+\frac {(b c-a d)^2}{c^3 x^2}-\frac {a (a d-2 b c)}{c^2 x^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^2}{5 c x^5}-\frac {\sqrt {d} (b c-a d)^2 \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{c^{7/2}}-\frac {(b c-a d)^2}{c^3 x}-\frac {a (2 b c-a d)}{3 c^2 x^3}\) |
Input:
Int[(a + b*x^2)^2/(x^6*(c + d*x^2)),x]
Output:
-1/5*a^2/(c*x^5) - (a*(2*b*c - a*d))/(3*c^2*x^3) - (b*c - a*d)^2/(c^3*x) - (Sqrt[d]*(b*c - a*d)^2*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/c^(7/2)
Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*((a + b*x^2)^p/(c + d*x^2)), x], x ] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && (In tegerQ[m] || IGtQ[2*(m + 1), 0] || !RationalQ[m])
Time = 0.47 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.15
method | result | size |
default | \(-\frac {a^{2}}{5 c \,x^{5}}-\frac {a^{2} d^{2}-2 a b c d +b^{2} c^{2}}{c^{3} x}+\frac {a \left (a d -2 b c \right )}{3 c^{2} x^{3}}-\frac {d \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \arctan \left (\frac {x d}{\sqrt {c d}}\right )}{c^{3} \sqrt {c d}}\) | \(100\) |
risch | \(\frac {-\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) x^{4}}{c^{3}}+\frac {a \left (a d -2 b c \right ) x^{2}}{3 c^{2}}-\frac {a^{2}}{5 c}}{x^{5}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c^{7} \textit {\_Z}^{2}+a^{4} d^{5}-4 a^{3} b c \,d^{4}+6 a^{2} b^{2} c^{2} d^{3}-4 a \,b^{3} c^{3} d^{2}+b^{4} c^{4} d \right )}{\sum }\textit {\_R} \ln \left (\left (3 \textit {\_R}^{2} c^{7}+2 a^{4} d^{5}-8 a^{3} b c \,d^{4}+12 a^{2} b^{2} c^{2} d^{3}-8 a \,b^{3} c^{3} d^{2}+2 b^{4} c^{4} d \right ) x +\left (a^{2} c^{4} d^{2}-2 a b \,c^{5} d +b^{2} c^{6}\right ) \textit {\_R} \right )\right )}{2}\) | \(222\) |
Input:
int((b*x^2+a)^2/x^6/(d*x^2+c),x,method=_RETURNVERBOSE)
Output:
-1/5*a^2/c/x^5-(a^2*d^2-2*a*b*c*d+b^2*c^2)/c^3/x+1/3*a*(a*d-2*b*c)/c^2/x^3 -d*(a^2*d^2-2*a*b*c*d+b^2*c^2)/c^3/(c*d)^(1/2)*arctan(x*d/(c*d)^(1/2))
Time = 0.19 (sec) , antiderivative size = 236, normalized size of antiderivative = 2.71 \[ \int \frac {\left (a+b x^2\right )^2}{x^6 \left (c+d x^2\right )} \, dx=\left [\frac {15 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{5} \sqrt {-\frac {d}{c}} \log \left (\frac {d x^{2} - 2 \, c x \sqrt {-\frac {d}{c}} - c}{d x^{2} + c}\right ) - 30 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{4} - 6 \, a^{2} c^{2} - 10 \, {\left (2 \, a b c^{2} - a^{2} c d\right )} x^{2}}{30 \, c^{3} x^{5}}, -\frac {15 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{5} \sqrt {\frac {d}{c}} \arctan \left (x \sqrt {\frac {d}{c}}\right ) + 15 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{4} + 3 \, a^{2} c^{2} + 5 \, {\left (2 \, a b c^{2} - a^{2} c d\right )} x^{2}}{15 \, c^{3} x^{5}}\right ] \] Input:
integrate((b*x^2+a)^2/x^6/(d*x^2+c),x, algorithm="fricas")
Output:
[1/30*(15*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x^5*sqrt(-d/c)*log((d*x^2 - 2*c* x*sqrt(-d/c) - c)/(d*x^2 + c)) - 30*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x^4 - 6*a^2*c^2 - 10*(2*a*b*c^2 - a^2*c*d)*x^2)/(c^3*x^5), -1/15*(15*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x^5*sqrt(d/c)*arctan(x*sqrt(d/c)) + 15*(b^2*c^2 - 2*a *b*c*d + a^2*d^2)*x^4 + 3*a^2*c^2 + 5*(2*a*b*c^2 - a^2*c*d)*x^2)/(c^3*x^5) ]
Leaf count of result is larger than twice the leaf count of optimal. 207 vs. \(2 (75) = 150\).
Time = 0.38 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.38 \[ \int \frac {\left (a+b x^2\right )^2}{x^6 \left (c+d x^2\right )} \, dx=\frac {\sqrt {- \frac {d}{c^{7}}} \left (a d - b c\right )^{2} \log {\left (- \frac {c^{4} \sqrt {- \frac {d}{c^{7}}} \left (a d - b c\right )^{2}}{a^{2} d^{3} - 2 a b c d^{2} + b^{2} c^{2} d} + x \right )}}{2} - \frac {\sqrt {- \frac {d}{c^{7}}} \left (a d - b c\right )^{2} \log {\left (\frac {c^{4} \sqrt {- \frac {d}{c^{7}}} \left (a d - b c\right )^{2}}{a^{2} d^{3} - 2 a b c d^{2} + b^{2} c^{2} d} + x \right )}}{2} + \frac {- 3 a^{2} c^{2} + x^{4} \left (- 15 a^{2} d^{2} + 30 a b c d - 15 b^{2} c^{2}\right ) + x^{2} \cdot \left (5 a^{2} c d - 10 a b c^{2}\right )}{15 c^{3} x^{5}} \] Input:
integrate((b*x**2+a)**2/x**6/(d*x**2+c),x)
Output:
sqrt(-d/c**7)*(a*d - b*c)**2*log(-c**4*sqrt(-d/c**7)*(a*d - b*c)**2/(a**2* d**3 - 2*a*b*c*d**2 + b**2*c**2*d) + x)/2 - sqrt(-d/c**7)*(a*d - b*c)**2*l og(c**4*sqrt(-d/c**7)*(a*d - b*c)**2/(a**2*d**3 - 2*a*b*c*d**2 + b**2*c**2 *d) + x)/2 + (-3*a**2*c**2 + x**4*(-15*a**2*d**2 + 30*a*b*c*d - 15*b**2*c* *2) + x**2*(5*a**2*c*d - 10*a*b*c**2))/(15*c**3*x**5)
Time = 0.12 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.23 \[ \int \frac {\left (a+b x^2\right )^2}{x^6 \left (c+d x^2\right )} \, dx=-\frac {{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{\sqrt {c d} c^{3}} - \frac {15 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{4} + 3 \, a^{2} c^{2} + 5 \, {\left (2 \, a b c^{2} - a^{2} c d\right )} x^{2}}{15 \, c^{3} x^{5}} \] Input:
integrate((b*x^2+a)^2/x^6/(d*x^2+c),x, algorithm="maxima")
Output:
-(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*arctan(d*x/sqrt(c*d))/(sqrt(c*d)*c^3) - 1/15*(15*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x^4 + 3*a^2*c^2 + 5*(2*a*b*c^2 - a^2*c*d)*x^2)/(c^3*x^5)
Time = 0.13 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.29 \[ \int \frac {\left (a+b x^2\right )^2}{x^6 \left (c+d x^2\right )} \, dx=-\frac {{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{\sqrt {c d} c^{3}} - \frac {15 \, b^{2} c^{2} x^{4} - 30 \, a b c d x^{4} + 15 \, a^{2} d^{2} x^{4} + 10 \, a b c^{2} x^{2} - 5 \, a^{2} c d x^{2} + 3 \, a^{2} c^{2}}{15 \, c^{3} x^{5}} \] Input:
integrate((b*x^2+a)^2/x^6/(d*x^2+c),x, algorithm="giac")
Output:
-(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*arctan(d*x/sqrt(c*d))/(sqrt(c*d)*c^3) - 1/15*(15*b^2*c^2*x^4 - 30*a*b*c*d*x^4 + 15*a^2*d^2*x^4 + 10*a*b*c^2*x^2 - 5*a^2*c*d*x^2 + 3*a^2*c^2)/(c^3*x^5)
Time = 0.49 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.48 \[ \int \frac {\left (a+b x^2\right )^2}{x^6 \left (c+d x^2\right )} \, dx=\frac {a^2\,d}{3\,c^2\,x^3}-\frac {b^2}{c\,x}-\frac {a^2}{5\,c\,x^5}-\frac {a^2\,d^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {d}\,x}{\sqrt {c}}\right )}{c^{7/2}}-\frac {b^2\,\sqrt {d}\,\mathrm {atan}\left (\frac {\sqrt {d}\,x}{\sqrt {c}}\right )}{c^{3/2}}-\frac {a^2\,d^2}{c^3\,x}-\frac {2\,a\,b}{3\,c\,x^3}+\frac {2\,a\,b\,d}{c^2\,x}+\frac {2\,a\,b\,d^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {d}\,x}{\sqrt {c}}\right )}{c^{5/2}} \] Input:
int((a + b*x^2)^2/(x^6*(c + d*x^2)),x)
Output:
(a^2*d)/(3*c^2*x^3) - b^2/(c*x) - a^2/(5*c*x^5) - (a^2*d^(5/2)*atan((d^(1/ 2)*x)/c^(1/2)))/c^(7/2) - (b^2*d^(1/2)*atan((d^(1/2)*x)/c^(1/2)))/c^(3/2) - (a^2*d^2)/(c^3*x) - (2*a*b)/(3*c*x^3) + (2*a*b*d)/(c^2*x) + (2*a*b*d^(3/ 2)*atan((d^(1/2)*x)/c^(1/2)))/c^(5/2)
Time = 0.19 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.75 \[ \int \frac {\left (a+b x^2\right )^2}{x^6 \left (c+d x^2\right )} \, dx=\frac {-15 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a^{2} d^{2} x^{5}+30 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a b c d \,x^{5}-15 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) b^{2} c^{2} x^{5}-3 a^{2} c^{3}+5 a^{2} c^{2} d \,x^{2}-15 a^{2} c \,d^{2} x^{4}-10 a b \,c^{3} x^{2}+30 a b \,c^{2} d \,x^{4}-15 b^{2} c^{3} x^{4}}{15 c^{4} x^{5}} \] Input:
int((b*x^2+a)^2/x^6/(d*x^2+c),x)
Output:
( - 15*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a**2*d**2*x**5 + 30*s qrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a*b*c*d*x**5 - 15*sqrt(d)*sqr t(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*b**2*c**2*x**5 - 3*a**2*c**3 + 5*a**2*c **2*d*x**2 - 15*a**2*c*d**2*x**4 - 10*a*b*c**3*x**2 + 30*a*b*c**2*d*x**4 - 15*b**2*c**3*x**4)/(15*c**4*x**5)