Integrand size = 24, antiderivative size = 98 \[ \int \frac {x^8 \left (A+B x^2\right )}{b x^2+c x^4} \, dx=-\frac {b^2 (b B-A c) x}{c^4}+\frac {b (b B-A c) x^3}{3 c^3}-\frac {(b B-A c) x^5}{5 c^2}+\frac {B x^7}{7 c}+\frac {b^{5/2} (b B-A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{c^{9/2}} \] Output:
-b^2*(-A*c+B*b)*x/c^4+1/3*b*(-A*c+B*b)*x^3/c^3-1/5*(-A*c+B*b)*x^5/c^2+1/7* B*x^7/c+b^(5/2)*(-A*c+B*b)*arctan(c^(1/2)*x/b^(1/2))/c^(9/2)
Time = 0.05 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00 \[ \int \frac {x^8 \left (A+B x^2\right )}{b x^2+c x^4} \, dx=-\frac {b^2 (b B-A c) x}{c^4}+\frac {b (b B-A c) x^3}{3 c^3}+\frac {(-b B+A c) x^5}{5 c^2}+\frac {B x^7}{7 c}+\frac {b^{5/2} (b B-A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{c^{9/2}} \] Input:
Integrate[(x^8*(A + B*x^2))/(b*x^2 + c*x^4),x]
Output:
-((b^2*(b*B - A*c)*x)/c^4) + (b*(b*B - A*c)*x^3)/(3*c^3) + ((-(b*B) + A*c) *x^5)/(5*c^2) + (B*x^7)/(7*c) + (b^(5/2)*(b*B - A*c)*ArcTan[(Sqrt[c]*x)/Sq rt[b]])/c^(9/2)
Time = 0.37 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.82, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {9, 363, 254, 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.
\(\displaystyle \int \frac {x^8 \left (A+B x^2\right )}{b x^2+c x^4} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {x^6 \left (A+B x^2\right )}{b+c x^2}dx\) |
\(\Big \downarrow \) 363 |
\(\displaystyle \frac {B x^7}{7 c}-\frac {(b B-A c) \int \frac {x^6}{c x^2+b}dx}{c}\) |
\(\Big \downarrow \) 254 |
\(\displaystyle \frac {B x^7}{7 c}-\frac {(b B-A c) \int \left (\frac {x^4}{c}-\frac {b x^2}{c^2}-\frac {b^3}{c^3 \left (c x^2+b\right )}+\frac {b^2}{c^3}\right )dx}{c}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {B x^7}{7 c}-\frac {(b B-A c) \left (-\frac {b^{5/2} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{c^{7/2}}+\frac {b^2 x}{c^3}-\frac {b x^3}{3 c^2}+\frac {x^5}{5 c}\right )}{c}\) |
Input:
Int[(x^8*(A + B*x^2))/(b*x^2 + c*x^4),x]
Output:
(B*x^7)/(7*c) - ((b*B - A*c)*((b^2*x)/c^3 - (b*x^3)/(3*c^2) + x^5/(5*c) - (b^(5/2)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/c^(7/2)))/c
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3)) Int[(e*x)^ m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d , 0] && NeQ[m + 2*p + 3, 0]
Time = 0.41 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.01
method | result | size |
default | \(\frac {\frac {1}{7} B \,c^{3} x^{7}+\frac {1}{5} A \,c^{3} x^{5}-\frac {1}{5} B b \,c^{2} x^{5}-\frac {1}{3} A b \,c^{2} x^{3}+\frac {1}{3} B \,b^{2} c \,x^{3}+A \,b^{2} c x -x B \,b^{3}}{c^{4}}-\frac {b^{3} \left (A c -B b \right ) \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{c^{4} \sqrt {b c}}\) | \(99\) |
risch | \(\frac {B \,x^{7}}{7 c}+\frac {A \,x^{5}}{5 c}-\frac {B b \,x^{5}}{5 c^{2}}-\frac {A b \,x^{3}}{3 c^{2}}+\frac {B \,b^{2} x^{3}}{3 c^{3}}+\frac {A \,b^{2} x}{c^{3}}-\frac {x B \,b^{3}}{c^{4}}+\frac {\sqrt {-b c}\, b^{2} \ln \left (-\sqrt {-b c}\, x -b \right ) A}{2 c^{4}}-\frac {\sqrt {-b c}\, b^{3} \ln \left (-\sqrt {-b c}\, x -b \right ) B}{2 c^{5}}-\frac {\sqrt {-b c}\, b^{2} \ln \left (\sqrt {-b c}\, x -b \right ) A}{2 c^{4}}+\frac {\sqrt {-b c}\, b^{3} \ln \left (\sqrt {-b c}\, x -b \right ) B}{2 c^{5}}\) | \(185\) |
Input:
int(x^8*(B*x^2+A)/(c*x^4+b*x^2),x,method=_RETURNVERBOSE)
Output:
1/c^4*(1/7*B*c^3*x^7+1/5*A*c^3*x^5-1/5*B*b*c^2*x^5-1/3*A*b*c^2*x^3+1/3*B*b ^2*c*x^3+A*b^2*c*x-x*B*b^3)-b^3*(A*c-B*b)/c^4/(b*c)^(1/2)*arctan(c*x/(b*c) ^(1/2))
Time = 0.08 (sec) , antiderivative size = 228, normalized size of antiderivative = 2.33 \[ \int \frac {x^8 \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\left [\frac {30 \, B c^{3} x^{7} - 42 \, {\left (B b c^{2} - A c^{3}\right )} x^{5} + 70 \, {\left (B b^{2} c - A b c^{2}\right )} x^{3} - 105 \, {\left (B b^{3} - A b^{2} c\right )} \sqrt {-\frac {b}{c}} \log \left (\frac {c x^{2} - 2 \, c x \sqrt {-\frac {b}{c}} - b}{c x^{2} + b}\right ) - 210 \, {\left (B b^{3} - A b^{2} c\right )} x}{210 \, c^{4}}, \frac {15 \, B c^{3} x^{7} - 21 \, {\left (B b c^{2} - A c^{3}\right )} x^{5} + 35 \, {\left (B b^{2} c - A b c^{2}\right )} x^{3} + 105 \, {\left (B b^{3} - A b^{2} c\right )} \sqrt {\frac {b}{c}} \arctan \left (\frac {c x \sqrt {\frac {b}{c}}}{b}\right ) - 105 \, {\left (B b^{3} - A b^{2} c\right )} x}{105 \, c^{4}}\right ] \] Input:
integrate(x^8*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="fricas")
Output:
[1/210*(30*B*c^3*x^7 - 42*(B*b*c^2 - A*c^3)*x^5 + 70*(B*b^2*c - A*b*c^2)*x ^3 - 105*(B*b^3 - A*b^2*c)*sqrt(-b/c)*log((c*x^2 - 2*c*x*sqrt(-b/c) - b)/( c*x^2 + b)) - 210*(B*b^3 - A*b^2*c)*x)/c^4, 1/105*(15*B*c^3*x^7 - 21*(B*b* c^2 - A*c^3)*x^5 + 35*(B*b^2*c - A*b*c^2)*x^3 + 105*(B*b^3 - A*b^2*c)*sqrt (b/c)*arctan(c*x*sqrt(b/c)/b) - 105*(B*b^3 - A*b^2*c)*x)/c^4]
Leaf count of result is larger than twice the leaf count of optimal. 180 vs. \(2 (87) = 174\).
Time = 0.21 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.84 \[ \int \frac {x^8 \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {B x^{7}}{7 c} + x^{5} \left (\frac {A}{5 c} - \frac {B b}{5 c^{2}}\right ) + x^{3} \left (- \frac {A b}{3 c^{2}} + \frac {B b^{2}}{3 c^{3}}\right ) + x \left (\frac {A b^{2}}{c^{3}} - \frac {B b^{3}}{c^{4}}\right ) - \frac {\sqrt {- \frac {b^{5}}{c^{9}}} \left (- A c + B b\right ) \log {\left (- \frac {c^{4} \sqrt {- \frac {b^{5}}{c^{9}}} \left (- A c + B b\right )}{- A b^{2} c + B b^{3}} + x \right )}}{2} + \frac {\sqrt {- \frac {b^{5}}{c^{9}}} \left (- A c + B b\right ) \log {\left (\frac {c^{4} \sqrt {- \frac {b^{5}}{c^{9}}} \left (- A c + B b\right )}{- A b^{2} c + B b^{3}} + x \right )}}{2} \] Input:
integrate(x**8*(B*x**2+A)/(c*x**4+b*x**2),x)
Output:
B*x**7/(7*c) + x**5*(A/(5*c) - B*b/(5*c**2)) + x**3*(-A*b/(3*c**2) + B*b** 2/(3*c**3)) + x*(A*b**2/c**3 - B*b**3/c**4) - sqrt(-b**5/c**9)*(-A*c + B*b )*log(-c**4*sqrt(-b**5/c**9)*(-A*c + B*b)/(-A*b**2*c + B*b**3) + x)/2 + sq rt(-b**5/c**9)*(-A*c + B*b)*log(c**4*sqrt(-b**5/c**9)*(-A*c + B*b)/(-A*b** 2*c + B*b**3) + x)/2
Time = 0.12 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.02 \[ \int \frac {x^8 \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {{\left (B b^{4} - A b^{3} c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c} c^{4}} + \frac {15 \, B c^{3} x^{7} - 21 \, {\left (B b c^{2} - A c^{3}\right )} x^{5} + 35 \, {\left (B b^{2} c - A b c^{2}\right )} x^{3} - 105 \, {\left (B b^{3} - A b^{2} c\right )} x}{105 \, c^{4}} \] Input:
integrate(x^8*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="maxima")
Output:
(B*b^4 - A*b^3*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*c^4) + 1/105*(15*B*c^3* x^7 - 21*(B*b*c^2 - A*c^3)*x^5 + 35*(B*b^2*c - A*b*c^2)*x^3 - 105*(B*b^3 - A*b^2*c)*x)/c^4
Time = 0.20 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.10 \[ \int \frac {x^8 \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {{\left (B b^{4} - A b^{3} c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c} c^{4}} + \frac {15 \, B c^{6} x^{7} - 21 \, B b c^{5} x^{5} + 21 \, A c^{6} x^{5} + 35 \, B b^{2} c^{4} x^{3} - 35 \, A b c^{5} x^{3} - 105 \, B b^{3} c^{3} x + 105 \, A b^{2} c^{4} x}{105 \, c^{7}} \] Input:
integrate(x^8*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="giac")
Output:
(B*b^4 - A*b^3*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*c^4) + 1/105*(15*B*c^6* x^7 - 21*B*b*c^5*x^5 + 21*A*c^6*x^5 + 35*B*b^2*c^4*x^3 - 35*A*b*c^5*x^3 - 105*B*b^3*c^3*x + 105*A*b^2*c^4*x)/c^7
Time = 0.05 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.20 \[ \int \frac {x^8 \left (A+B x^2\right )}{b x^2+c x^4} \, dx=x^5\,\left (\frac {A}{5\,c}-\frac {B\,b}{5\,c^2}\right )+\frac {B\,x^7}{7\,c}+\frac {b^{5/2}\,\mathrm {atan}\left (\frac {b^{5/2}\,\sqrt {c}\,x\,\left (A\,c-B\,b\right )}{B\,b^4-A\,b^3\,c}\right )\,\left (A\,c-B\,b\right )}{c^{9/2}}-\frac {b\,x^3\,\left (\frac {A}{c}-\frac {B\,b}{c^2}\right )}{3\,c}+\frac {b^2\,x\,\left (\frac {A}{c}-\frac {B\,b}{c^2}\right )}{c^2} \] Input:
int((x^8*(A + B*x^2))/(b*x^2 + c*x^4),x)
Output:
x^5*(A/(5*c) - (B*b)/(5*c^2)) + (B*x^7)/(7*c) + (b^(5/2)*atan((b^(5/2)*c^( 1/2)*x*(A*c - B*b))/(B*b^4 - A*b^3*c))*(A*c - B*b))/c^(9/2) - (b*x^3*(A/c - (B*b)/c^2))/(3*c) + (b^2*x*(A/c - (B*b)/c^2))/c^2
Time = 0.22 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.19 \[ \int \frac {x^8 \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {-105 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) a \,b^{2} c +105 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{4}+105 a \,b^{2} c^{2} x -35 a b \,c^{3} x^{3}+21 a \,c^{4} x^{5}-105 b^{4} c x +35 b^{3} c^{2} x^{3}-21 b^{2} c^{3} x^{5}+15 b \,c^{4} x^{7}}{105 c^{5}} \] Input:
int(x^8*(B*x^2+A)/(c*x^4+b*x^2),x)
Output:
( - 105*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*a*b**2*c + 105*sqrt( c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*b**4 + 105*a*b**2*c**2*x - 35*a*b *c**3*x**3 + 21*a*c**4*x**5 - 105*b**4*c*x + 35*b**3*c**2*x**3 - 21*b**2*c **3*x**5 + 15*b*c**4*x**7)/(105*c**5)