Integrand size = 24, antiderivative size = 95 \[ \int \frac {x^{10} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {B x}{c^3}-\frac {b (b B-A c) x}{4 c^3 \left (b+c x^2\right )^2}+\frac {(9 b B-5 A c) x}{8 c^3 \left (b+c x^2\right )}-\frac {3 (5 b B-A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 \sqrt {b} c^{7/2}} \] Output:
B*x/c^3-1/4*b*(-A*c+B*b)*x/c^3/(c*x^2+b)^2+1/8*(-5*A*c+9*B*b)*x/c^3/(c*x^2 +b)-3/8*(-A*c+5*B*b)*arctan(c^(1/2)*x/b^(1/2))/b^(1/2)/c^(7/2)
Time = 0.05 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.97 \[ \int \frac {x^{10} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {x \left (15 b^2 B+c^2 x^2 \left (-5 A+8 B x^2\right )+b \left (-3 A c+25 B c x^2\right )\right )}{8 c^3 \left (b+c x^2\right )^2}-\frac {3 (5 b B-A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 \sqrt {b} c^{7/2}} \] Input:
Integrate[(x^10*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]
Output:
(x*(15*b^2*B + c^2*x^2*(-5*A + 8*B*x^2) + b*(-3*A*c + 25*B*c*x^2)))/(8*c^3 *(b + c*x^2)^2) - (3*(5*b*B - A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(8*Sqrt[b] *c^(7/2))
Time = 0.45 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {9, 360, 25, 1471, 27, 299, 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.
\(\displaystyle \int \frac {x^{10} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {x^4 \left (A+B x^2\right )}{\left (b+c x^2\right )^3}dx\) |
\(\Big \downarrow \) 360 |
\(\displaystyle -\frac {\int -\frac {4 B c^2 x^4-4 c (b B-A c) x^2+b (b B-A c)}{\left (c x^2+b\right )^2}dx}{4 c^3}-\frac {b x (b B-A c)}{4 c^3 \left (b+c x^2\right )^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {4 B c^2 x^4-4 c (b B-A c) x^2+b (b B-A c)}{\left (c x^2+b\right )^2}dx}{4 c^3}-\frac {b x (b B-A c)}{4 c^3 \left (b+c x^2\right )^2}\) |
\(\Big \downarrow \) 1471 |
\(\displaystyle \frac {\frac {x (9 b B-5 A c)}{2 \left (b+c x^2\right )}-\frac {\int \frac {b \left (-8 B c x^2+7 b B-3 A c\right )}{c x^2+b}dx}{2 b}}{4 c^3}-\frac {b x (b B-A c)}{4 c^3 \left (b+c x^2\right )^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {x (9 b B-5 A c)}{2 \left (b+c x^2\right )}-\frac {1}{2} \int \frac {-8 B c x^2+7 b B-3 A c}{c x^2+b}dx}{4 c^3}-\frac {b x (b B-A c)}{4 c^3 \left (b+c x^2\right )^2}\) |
\(\Big \downarrow \) 299 |
\(\displaystyle \frac {\frac {1}{2} \left (8 B x-3 (5 b B-A c) \int \frac {1}{c x^2+b}dx\right )+\frac {x (9 b B-5 A c)}{2 \left (b+c x^2\right )}}{4 c^3}-\frac {b x (b B-A c)}{4 c^3 \left (b+c x^2\right )^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {1}{2} \left (8 B x-\frac {3 (5 b B-A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{\sqrt {b} \sqrt {c}}\right )+\frac {x (9 b B-5 A c)}{2 \left (b+c x^2\right )}}{4 c^3}-\frac {b x (b B-A c)}{4 c^3 \left (b+c x^2\right )^2}\) |
Input:
Int[(x^10*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]
Output:
-1/4*(b*(b*B - A*c)*x)/(c^3*(b + c*x^2)^2) + (((9*b*B - 5*A*c)*x)/(2*(b + c*x^2)) + (8*B*x - (3*(5*b*B - A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(Sqrt[b]* Sqrt[c]))/2)/(4*c^3)
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[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : > Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1)) Int[(a + b*x^2)^(p + 1)*Expan dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & & (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, d + e*x^2 , x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], x , 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Simp[1/(2*d*(q + 1)) Int[(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^ 2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]
Time = 0.40 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.81
method | result | size |
default | \(\frac {B x}{c^{3}}+\frac {\frac {\left (-\frac {5}{8} A \,c^{2}+\frac {9}{8} B b c \right ) x^{3}-\frac {b \left (3 A c -7 B b \right ) x}{8}}{\left (c \,x^{2}+b \right )^{2}}+\frac {3 \left (A c -5 B b \right ) \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \sqrt {b c}}}{c^{3}}\) | \(77\) |
risch | \(\frac {B x}{c^{3}}+\frac {\left (-\frac {5}{8} A \,c^{2}+\frac {9}{8} B b c \right ) x^{3}-\frac {b \left (3 A c -7 B b \right ) x}{8}}{c^{3} \left (c \,x^{2}+b \right )^{2}}-\frac {3 \ln \left (c x +\sqrt {-b c}\right ) A}{16 c^{2} \sqrt {-b c}}+\frac {15 \ln \left (c x +\sqrt {-b c}\right ) B b}{16 c^{3} \sqrt {-b c}}+\frac {3 \ln \left (-c x +\sqrt {-b c}\right ) A}{16 c^{2} \sqrt {-b c}}-\frac {15 \ln \left (-c x +\sqrt {-b c}\right ) B b}{16 c^{3} \sqrt {-b c}}\) | \(147\) |
Input:
int(x^10*(B*x^2+A)/(c*x^4+b*x^2)^3,x,method=_RETURNVERBOSE)
Output:
B*x/c^3+1/c^3*(((-5/8*A*c^2+9/8*B*b*c)*x^3-1/8*b*(3*A*c-7*B*b)*x)/(c*x^2+b )^2+3/8*(A*c-5*B*b)/(b*c)^(1/2)*arctan(c*x/(b*c)^(1/2)))
Time = 0.10 (sec) , antiderivative size = 328, normalized size of antiderivative = 3.45 \[ \int \frac {x^{10} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\left [\frac {16 \, B b c^{3} x^{5} + 10 \, {\left (5 \, B b^{2} c^{2} - A b c^{3}\right )} x^{3} + 3 \, {\left ({\left (5 \, B b c^{2} - A c^{3}\right )} x^{4} + 5 \, B b^{3} - A b^{2} c + 2 \, {\left (5 \, B b^{2} c - A b c^{2}\right )} x^{2}\right )} \sqrt {-b c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-b c} x - b}{c x^{2} + b}\right ) + 6 \, {\left (5 \, B b^{3} c - A b^{2} c^{2}\right )} x}{16 \, {\left (b c^{6} x^{4} + 2 \, b^{2} c^{5} x^{2} + b^{3} c^{4}\right )}}, \frac {8 \, B b c^{3} x^{5} + 5 \, {\left (5 \, B b^{2} c^{2} - A b c^{3}\right )} x^{3} - 3 \, {\left ({\left (5 \, B b c^{2} - A c^{3}\right )} x^{4} + 5 \, B b^{3} - A b^{2} c + 2 \, {\left (5 \, B b^{2} c - A b c^{2}\right )} x^{2}\right )} \sqrt {b c} \arctan \left (\frac {\sqrt {b c} x}{b}\right ) + 3 \, {\left (5 \, B b^{3} c - A b^{2} c^{2}\right )} x}{8 \, {\left (b c^{6} x^{4} + 2 \, b^{2} c^{5} x^{2} + b^{3} c^{4}\right )}}\right ] \] Input:
integrate(x^10*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="fricas")
Output:
[1/16*(16*B*b*c^3*x^5 + 10*(5*B*b^2*c^2 - A*b*c^3)*x^3 + 3*((5*B*b*c^2 - A *c^3)*x^4 + 5*B*b^3 - A*b^2*c + 2*(5*B*b^2*c - A*b*c^2)*x^2)*sqrt(-b*c)*lo g((c*x^2 - 2*sqrt(-b*c)*x - b)/(c*x^2 + b)) + 6*(5*B*b^3*c - A*b^2*c^2)*x) /(b*c^6*x^4 + 2*b^2*c^5*x^2 + b^3*c^4), 1/8*(8*B*b*c^3*x^5 + 5*(5*B*b^2*c^ 2 - A*b*c^3)*x^3 - 3*((5*B*b*c^2 - A*c^3)*x^4 + 5*B*b^3 - A*b^2*c + 2*(5*B *b^2*c - A*b*c^2)*x^2)*sqrt(b*c)*arctan(sqrt(b*c)*x/b) + 3*(5*B*b^3*c - A* b^2*c^2)*x)/(b*c^6*x^4 + 2*b^2*c^5*x^2 + b^3*c^4)]
Leaf count of result is larger than twice the leaf count of optimal. 194 vs. \(2 (92) = 184\).
Time = 0.56 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.04 \[ \int \frac {x^{10} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {B x}{c^{3}} + \frac {3 \sqrt {- \frac {1}{b c^{7}}} \left (- A c + 5 B b\right ) \log {\left (- \frac {3 b c^{3} \sqrt {- \frac {1}{b c^{7}}} \left (- A c + 5 B b\right )}{- 3 A c + 15 B b} + x \right )}}{16} - \frac {3 \sqrt {- \frac {1}{b c^{7}}} \left (- A c + 5 B b\right ) \log {\left (\frac {3 b c^{3} \sqrt {- \frac {1}{b c^{7}}} \left (- A c + 5 B b\right )}{- 3 A c + 15 B b} + x \right )}}{16} + \frac {x^{3} \left (- 5 A c^{2} + 9 B b c\right ) + x \left (- 3 A b c + 7 B b^{2}\right )}{8 b^{2} c^{3} + 16 b c^{4} x^{2} + 8 c^{5} x^{4}} \] Input:
integrate(x**10*(B*x**2+A)/(c*x**4+b*x**2)**3,x)
Output:
B*x/c**3 + 3*sqrt(-1/(b*c**7))*(-A*c + 5*B*b)*log(-3*b*c**3*sqrt(-1/(b*c** 7))*(-A*c + 5*B*b)/(-3*A*c + 15*B*b) + x)/16 - 3*sqrt(-1/(b*c**7))*(-A*c + 5*B*b)*log(3*b*c**3*sqrt(-1/(b*c**7))*(-A*c + 5*B*b)/(-3*A*c + 15*B*b) + x)/16 + (x**3*(-5*A*c**2 + 9*B*b*c) + x*(-3*A*b*c + 7*B*b**2))/(8*b**2*c** 3 + 16*b*c**4*x**2 + 8*c**5*x**4)
Time = 0.12 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.99 \[ \int \frac {x^{10} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {{\left (9 \, B b c - 5 \, A c^{2}\right )} x^{3} + {\left (7 \, B b^{2} - 3 \, A b c\right )} x}{8 \, {\left (c^{5} x^{4} + 2 \, b c^{4} x^{2} + b^{2} c^{3}\right )}} + \frac {B x}{c^{3}} - \frac {3 \, {\left (5 \, B b - A c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} c^{3}} \] Input:
integrate(x^10*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="maxima")
Output:
1/8*((9*B*b*c - 5*A*c^2)*x^3 + (7*B*b^2 - 3*A*b*c)*x)/(c^5*x^4 + 2*b*c^4*x ^2 + b^2*c^3) + B*x/c^3 - 3/8*(5*B*b - A*c)*arctan(c*x/sqrt(b*c))/(sqrt(b* c)*c^3)
Time = 0.23 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.84 \[ \int \frac {x^{10} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {B x}{c^{3}} - \frac {3 \, {\left (5 \, B b - A c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} c^{3}} + \frac {9 \, B b c x^{3} - 5 \, A c^{2} x^{3} + 7 \, B b^{2} x - 3 \, A b c x}{8 \, {\left (c x^{2} + b\right )}^{2} c^{3}} \] Input:
integrate(x^10*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="giac")
Output:
B*x/c^3 - 3/8*(5*B*b - A*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*c^3) + 1/8*(9 *B*b*c*x^3 - 5*A*c^2*x^3 + 7*B*b^2*x - 3*A*b*c*x)/((c*x^2 + b)^2*c^3)
Time = 8.84 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.97 \[ \int \frac {x^{10} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {B\,x}{c^3}-\frac {x^3\,\left (\frac {5\,A\,c^2}{8}-\frac {9\,B\,b\,c}{8}\right )-x\,\left (\frac {7\,B\,b^2}{8}-\frac {3\,A\,b\,c}{8}\right )}{b^2\,c^3+2\,b\,c^4\,x^2+c^5\,x^4}+\frac {3\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )\,\left (A\,c-5\,B\,b\right )}{8\,\sqrt {b}\,c^{7/2}} \] Input:
int((x^10*(A + B*x^2))/(b*x^2 + c*x^4)^3,x)
Output:
(B*x)/c^3 - (x^3*((5*A*c^2)/8 - (9*B*b*c)/8) - x*((7*B*b^2)/8 - (3*A*b*c)/ 8))/(b^2*c^3 + c^5*x^4 + 2*b*c^4*x^2) + (3*atan((c^(1/2)*x)/b^(1/2))*(A*c - 5*B*b))/(8*b^(1/2)*c^(7/2))
Time = 0.19 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.37 \[ \int \frac {x^{10} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {3 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) a \,b^{2} c +6 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) a b \,c^{2} x^{2}+3 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) a \,c^{3} x^{4}-15 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{4}-30 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{3} c \,x^{2}-15 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{2} c^{2} x^{4}-3 a \,b^{2} c^{2} x -5 a b \,c^{3} x^{3}+15 b^{4} c x +25 b^{3} c^{2} x^{3}+8 b^{2} c^{3} x^{5}}{8 b \,c^{4} \left (c^{2} x^{4}+2 b c \,x^{2}+b^{2}\right )} \] Input:
int(x^10*(B*x^2+A)/(c*x^4+b*x^2)^3,x)
Output:
(3*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*a*b**2*c + 6*sqrt(c)*sqrt (b)*atan((c*x)/(sqrt(c)*sqrt(b)))*a*b*c**2*x**2 + 3*sqrt(c)*sqrt(b)*atan(( c*x)/(sqrt(c)*sqrt(b)))*a*c**3*x**4 - 15*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt( c)*sqrt(b)))*b**4 - 30*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*b**3* c*x**2 - 15*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*b**2*c**2*x**4 - 3*a*b**2*c**2*x - 5*a*b*c**3*x**3 + 15*b**4*c*x + 25*b**3*c**2*x**3 + 8*b **2*c**3*x**5)/(8*b*c**4*(b**2 + 2*b*c*x**2 + c**2*x**4))