Integrand size = 24, antiderivative size = 77 \[ \int \frac {x^6 \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {b (b B-A c) x}{c^3}-\frac {(b B-A c) x^3}{3 c^2}+\frac {B x^5}{5 c}-\frac {b^{3/2} (b B-A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{c^{7/2}} \] Output:
b*(-A*c+B*b)*x/c^3-1/3*(-A*c+B*b)*x^3/c^2+1/5*B*x^5/c-b^(3/2)*(-A*c+B*b)*a rctan(c^(1/2)*x/b^(1/2))/c^(7/2)
Time = 0.04 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00 \[ \int \frac {x^6 \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {b (b B-A c) x}{c^3}+\frac {(-b B+A c) x^3}{3 c^2}+\frac {B x^5}{5 c}-\frac {b^{3/2} (b B-A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{c^{7/2}} \] Input:
Integrate[(x^6*(A + B*x^2))/(b*x^2 + c*x^4),x]
Output:
(b*(b*B - A*c)*x)/c^3 + ((-(b*B) + A*c)*x^3)/(3*c^2) + (B*x^5)/(5*c) - (b^ (3/2)*(b*B - A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/c^(7/2)
Time = 0.35 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.87, 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.
\(\displaystyle \int \frac {x^6 \left (A+B x^2\right )}{b x^2+c x^4} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {x^4 \left (A+B x^2\right )}{b+c x^2}dx\) |
\(\Big \downarrow \) 363 |
\(\displaystyle \frac {B x^5}{5 c}-\frac {(b B-A c) \int \frac {x^4}{c x^2+b}dx}{c}\) |
\(\Big \downarrow \) 254 |
\(\displaystyle \frac {B x^5}{5 c}-\frac {(b B-A c) \int \left (\frac {b^2}{c^2 \left (c x^2+b\right )}-\frac {b}{c^2}+\frac {x^2}{c}\right )dx}{c}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {B x^5}{5 c}-\frac {(b B-A c) \left (\frac {b^{3/2} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{c^{5/2}}-\frac {b x}{c^2}+\frac {x^3}{3 c}\right )}{c}\) |
Input:
Int[(x^6*(A + B*x^2))/(b*x^2 + c*x^4),x]
Output:
(B*x^5)/(5*c) - ((b*B - A*c)*(-((b*x)/c^2) + x^3/(3*c) + (b^(3/2)*ArcTan[( Sqrt[c]*x)/Sqrt[b]])/c^(5/2)))/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 = 75, normalized size of antiderivative = 0.97
method | result | size |
default | \(-\frac {-\frac {1}{5} B \,c^{2} x^{5}-\frac {1}{3} A \,c^{2} x^{3}+\frac {1}{3} B b c \,x^{3}+A b c x -x B \,b^{2}}{c^{3}}+\frac {b^{2} \left (A c -B b \right ) \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{c^{3} \sqrt {b c}}\) | \(75\) |
risch | \(\frac {B \,x^{5}}{5 c}+\frac {A \,x^{3}}{3 c}-\frac {B b \,x^{3}}{3 c^{2}}-\frac {A b x}{c^{2}}+\frac {x B \,b^{2}}{c^{3}}+\frac {\sqrt {-b c}\, b \ln \left (-\sqrt {-b c}\, x +b \right ) A}{2 c^{3}}-\frac {\sqrt {-b c}\, b^{2} \ln \left (-\sqrt {-b c}\, x +b \right ) B}{2 c^{4}}-\frac {\sqrt {-b c}\, b \ln \left (\sqrt {-b c}\, x +b \right ) A}{2 c^{3}}+\frac {\sqrt {-b c}\, b^{2} \ln \left (\sqrt {-b c}\, x +b \right ) B}{2 c^{4}}\) | \(149\) |
Input:
int(x^6*(B*x^2+A)/(c*x^4+b*x^2),x,method=_RETURNVERBOSE)
Output:
-1/c^3*(-1/5*B*c^2*x^5-1/3*A*c^2*x^3+1/3*B*b*c*x^3+A*b*c*x-x*B*b^2)+b^2*(A *c-B*b)/c^3/(b*c)^(1/2)*arctan(c*x/(b*c)^(1/2))
Time = 0.08 (sec) , antiderivative size = 178, normalized size of antiderivative = 2.31 \[ \int \frac {x^6 \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\left [\frac {6 \, B c^{2} x^{5} - 10 \, {\left (B b c - A c^{2}\right )} x^{3} - 15 \, {\left (B b^{2} - A b c\right )} \sqrt {-\frac {b}{c}} \log \left (\frac {c x^{2} + 2 \, c x \sqrt {-\frac {b}{c}} - b}{c x^{2} + b}\right ) + 30 \, {\left (B b^{2} - A b c\right )} x}{30 \, c^{3}}, \frac {3 \, B c^{2} x^{5} - 5 \, {\left (B b c - A c^{2}\right )} x^{3} - 15 \, {\left (B b^{2} - A b c\right )} \sqrt {\frac {b}{c}} \arctan \left (\frac {c x \sqrt {\frac {b}{c}}}{b}\right ) + 15 \, {\left (B b^{2} - A b c\right )} x}{15 \, c^{3}}\right ] \] Input:
integrate(x^6*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="fricas")
Output:
[1/30*(6*B*c^2*x^5 - 10*(B*b*c - A*c^2)*x^3 - 15*(B*b^2 - A*b*c)*sqrt(-b/c )*log((c*x^2 + 2*c*x*sqrt(-b/c) - b)/(c*x^2 + b)) + 30*(B*b^2 - A*b*c)*x)/ c^3, 1/15*(3*B*c^2*x^5 - 5*(B*b*c - A*c^2)*x^3 - 15*(B*b^2 - A*b*c)*sqrt(b /c)*arctan(c*x*sqrt(b/c)/b) + 15*(B*b^2 - A*b*c)*x)/c^3]
Leaf count of result is larger than twice the leaf count of optimal. 153 vs. \(2 (68) = 136\).
Time = 0.19 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.99 \[ \int \frac {x^6 \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {B x^{5}}{5 c} + x^{3} \left (\frac {A}{3 c} - \frac {B b}{3 c^{2}}\right ) + x \left (- \frac {A b}{c^{2}} + \frac {B b^{2}}{c^{3}}\right ) + \frac {\sqrt {- \frac {b^{3}}{c^{7}}} \left (- A c + B b\right ) \log {\left (- \frac {c^{3} \sqrt {- \frac {b^{3}}{c^{7}}} \left (- A c + B b\right )}{- A b c + B b^{2}} + x \right )}}{2} - \frac {\sqrt {- \frac {b^{3}}{c^{7}}} \left (- A c + B b\right ) \log {\left (\frac {c^{3} \sqrt {- \frac {b^{3}}{c^{7}}} \left (- A c + B b\right )}{- A b c + B b^{2}} + x \right )}}{2} \] Input:
integrate(x**6*(B*x**2+A)/(c*x**4+b*x**2),x)
Output:
B*x**5/(5*c) + x**3*(A/(3*c) - B*b/(3*c**2)) + x*(-A*b/c**2 + B*b**2/c**3) + sqrt(-b**3/c**7)*(-A*c + B*b)*log(-c**3*sqrt(-b**3/c**7)*(-A*c + B*b)/( -A*b*c + B*b**2) + x)/2 - sqrt(-b**3/c**7)*(-A*c + B*b)*log(c**3*sqrt(-b** 3/c**7)*(-A*c + B*b)/(-A*b*c + B*b**2) + x)/2
Time = 0.11 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.01 \[ \int \frac {x^6 \left (A+B x^2\right )}{b x^2+c x^4} \, dx=-\frac {{\left (B b^{3} - A b^{2} c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c} c^{3}} + \frac {3 \, B c^{2} x^{5} - 5 \, {\left (B b c - A c^{2}\right )} x^{3} + 15 \, {\left (B b^{2} - A b c\right )} x}{15 \, c^{3}} \] Input:
integrate(x^6*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="maxima")
Output:
-(B*b^3 - A*b^2*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*c^3) + 1/15*(3*B*c^2*x ^5 - 5*(B*b*c - A*c^2)*x^3 + 15*(B*b^2 - A*b*c)*x)/c^3
Time = 0.17 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.10 \[ \int \frac {x^6 \left (A+B x^2\right )}{b x^2+c x^4} \, dx=-\frac {{\left (B b^{3} - A b^{2} c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c} c^{3}} + \frac {3 \, B c^{4} x^{5} - 5 \, B b c^{3} x^{3} + 5 \, A c^{4} x^{3} + 15 \, B b^{2} c^{2} x - 15 \, A b c^{3} x}{15 \, c^{5}} \] Input:
integrate(x^6*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="giac")
Output:
-(B*b^3 - A*b^2*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*c^3) + 1/15*(3*B*c^4*x ^5 - 5*B*b*c^3*x^3 + 5*A*c^4*x^3 + 15*B*b^2*c^2*x - 15*A*b*c^3*x)/c^5
Time = 8.88 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.25 \[ \int \frac {x^6 \left (A+B x^2\right )}{b x^2+c x^4} \, dx=x^3\,\left (\frac {A}{3\,c}-\frac {B\,b}{3\,c^2}\right )+\frac {B\,x^5}{5\,c}-\frac {b^{3/2}\,\mathrm {atan}\left (\frac {b^{3/2}\,\sqrt {c}\,x\,\left (A\,c-B\,b\right )}{B\,b^3-A\,b^2\,c}\right )\,\left (A\,c-B\,b\right )}{c^{7/2}}-\frac {b\,x\,\left (\frac {A}{c}-\frac {B\,b}{c^2}\right )}{c} \] Input:
int((x^6*(A + B*x^2))/(b*x^2 + c*x^4),x)
Output:
x^3*(A/(3*c) - (B*b)/(3*c^2)) + (B*x^5)/(5*c) - (b^(3/2)*atan((b^(3/2)*c^( 1/2)*x*(A*c - B*b))/(B*b^3 - A*b^2*c))*(A*c - B*b))/c^(7/2) - (b*x*(A/c - (B*b)/c^2))/c
Time = 0.24 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.19 \[ \int \frac {x^6 \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {15 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) a b c -15 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{3}-15 a b \,c^{2} x +5 a \,c^{3} x^{3}+15 b^{3} c x -5 b^{2} c^{2} x^{3}+3 b \,c^{3} x^{5}}{15 c^{4}} \] Input:
int(x^6*(B*x^2+A)/(c*x^4+b*x^2),x)
Output:
(15*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*a*b*c - 15*sqrt(c)*sqrt( b)*atan((c*x)/(sqrt(c)*sqrt(b)))*b**3 - 15*a*b*c**2*x + 5*a*c**3*x**3 + 15 *b**3*c*x - 5*b**2*c**2*x**3 + 3*b*c**3*x**5)/(15*c**4)