\(\int \frac {x^{12} (A+B x^2)}{(b x^2+c x^4)^3} \, dx\) [79]

Optimal result

Integrand size = 24, antiderivative size = 118 \[ \int \frac {x^{12} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {(3 b B-A c) x}{c^4}+\frac {B x^3}{3 c^3}+\frac {b^2 (b B-A c) x}{4 c^4 \left (b+c x^2\right )^2}-\frac {b (13 b B-9 A c) x}{8 c^4 \left (b+c x^2\right )}+\frac {5 \sqrt {b} (7 b B-3 A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 c^{9/2}} \] Output:

-(-A*c+3*B*b)*x/c^4+1/3*B*x^3/c^3+1/4*b^2*(-A*c+B*b)*x/c^4/(c*x^2+b)^2-1/8 
*b*(-9*A*c+13*B*b)*x/c^4/(c*x^2+b)+5/8*b^(1/2)*(-3*A*c+7*B*b)*arctan(c^(1/ 
2)*x/b^(1/2))/c^(9/2)
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.96 \[ \int \frac {x^{12} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {-105 b^3 B x+b c^2 x^3 \left (75 A-56 B x^2\right )+5 b^2 c x \left (9 A-35 B x^2\right )+8 c^3 x^5 \left (3 A+B x^2\right )}{24 c^4 \left (b+c x^2\right )^2}+\frac {5 \sqrt {b} (7 b B-3 A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 c^{9/2}} \] Input:

Integrate[(x^12*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]
 

Output:

(-105*b^3*B*x + b*c^2*x^3*(75*A - 56*B*x^2) + 5*b^2*c*x*(9*A - 35*B*x^2) + 
 8*c^3*x^5*(3*A + B*x^2))/(24*c^4*(b + c*x^2)^2) + (5*Sqrt[b]*(7*b*B - 3*A 
*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(8*c^(9/2))
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {9, 360, 2345, 1467, 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.

\(\displaystyle \int \frac {x^{12} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx\)

\(\Big \downarrow \) 9

\(\displaystyle \int \frac {x^6 \left (A+B x^2\right )}{\left (b+c x^2\right )^3}dx\)

\(\Big \downarrow \) 360

\(\displaystyle \frac {b^2 x (b B-A c)}{4 c^4 \left (b+c x^2\right )^2}-\frac {\int \frac {-4 B c^3 x^6+4 c^2 (b B-A c) x^4-4 b c (b B-A c) x^2+b^2 (b B-A c)}{\left (c x^2+b\right )^2}dx}{4 c^4}\)

\(\Big \downarrow \) 2345

\(\displaystyle \frac {b^2 x (b B-A c)}{4 c^4 \left (b+c x^2\right )^2}-\frac {\frac {b x (13 b B-9 A c)}{2 \left (b+c x^2\right )}-\frac {\int \frac {8 b B c^2 x^4-8 b c (2 b B-A c) x^2+b^2 (11 b B-7 A c)}{c x^2+b}dx}{2 b}}{4 c^4}\)

\(\Big \downarrow \) 1467

\(\displaystyle \frac {b^2 x (b B-A c)}{4 c^4 \left (b+c x^2\right )^2}-\frac {\frac {b x (13 b B-9 A c)}{2 \left (b+c x^2\right )}-\frac {\int \left (8 b B c x^2-8 b (3 b B-A c)+\frac {5 \left (7 b^3 B-3 A b^2 c\right )}{c x^2+b}\right )dx}{2 b}}{4 c^4}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b^2 x (b B-A c)}{4 c^4 \left (b+c x^2\right )^2}-\frac {\frac {b x (13 b B-9 A c)}{2 \left (b+c x^2\right )}-\frac {\frac {5 b^{3/2} (7 b B-3 A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{\sqrt {c}}-8 b x (3 b B-A c)+\frac {8}{3} b B c x^3}{2 b}}{4 c^4}\)

Input:

Int[(x^12*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]
 

Output:

(b^2*(b*B - A*c)*x)/(4*c^4*(b + c*x^2)^2) - ((b*(13*b*B - 9*A*c)*x)/(2*(b 
+ c*x^2)) - (-8*b*(3*b*B - A*c)*x + (8*b*B*c*x^3)/3 + (5*b^(3/2)*(7*b*B - 
3*A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/Sqrt[c])/(2*b))/(4*c^4)
 

Defintions of rubi rules used

rule 9

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]
 

rule 360

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])
 

rule 1467

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
 x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, 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] && IGtQ[q, -2]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2345

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot 
ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 
 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b 
*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   In 
t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] 
/; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
 

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.81

method	result	size
default	\(\frac {\frac {1}{3} B c \,x^{3}+A c x -3 B b x}{c^{4}}-\frac {b \left (\frac {\left (-\frac {9}{8} A \,c^{2}+\frac {13}{8} B b c \right ) x^{3}-\frac {b \left (7 A c -11 B b \right ) x}{8}}{\left (c \,x^{2}+b \right )^{2}}+\frac {5 \left (3 A c -7 B b \right ) \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \sqrt {b c}}\right )}{c^{4}}\)	\(95\)
risch	\(\frac {B \,x^{3}}{3 c^{3}}+\frac {A x}{c^{3}}-\frac {3 B b x}{c^{4}}+\frac {\left (\frac {9}{8} A b \,c^{2}-\frac {13}{8} B \,b^{2} c \right ) x^{3}+\frac {b^{2} \left (7 A c -11 B b \right ) x}{8}}{c^{4} \left (c \,x^{2}+b \right )^{2}}+\frac {15 \sqrt {-b c}\, \ln \left (-\sqrt {-b c}\, x -b \right ) A}{16 c^{4}}-\frac {35 \sqrt {-b c}\, \ln \left (-\sqrt {-b c}\, x -b \right ) B b}{16 c^{5}}-\frac {15 \sqrt {-b c}\, \ln \left (\sqrt {-b c}\, x -b \right ) A}{16 c^{4}}+\frac {35 \sqrt {-b c}\, \ln \left (\sqrt {-b c}\, x -b \right ) B b}{16 c^{5}}\)	\(177\)

Input:

int(x^12*(B*x^2+A)/(c*x^4+b*x^2)^3,x,method=_RETURNVERBOSE)
 

Output:

1/c^4*(1/3*B*c*x^3+A*c*x-3*B*b*x)-b/c^4*(((-9/8*A*c^2+13/8*B*b*c)*x^3-1/8* 
b*(7*A*c-11*B*b)*x)/(c*x^2+b)^2+5/8*(3*A*c-7*B*b)/(b*c)^(1/2)*arctan(c*x/( 
b*c)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 358, normalized size of antiderivative = 3.03 \[ \int \frac {x^{12} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\left [\frac {16 \, B c^{3} x^{7} - 16 \, {\left (7 \, B b c^{2} - 3 \, A c^{3}\right )} x^{5} - 50 \, {\left (7 \, B b^{2} c - 3 \, A b c^{2}\right )} x^{3} - 15 \, {\left ({\left (7 \, B b c^{2} - 3 \, A c^{3}\right )} x^{4} + 7 \, B b^{3} - 3 \, A b^{2} c + 2 \, {\left (7 \, B b^{2} c - 3 \, A b c^{2}\right )} x^{2}\right )} \sqrt {-\frac {b}{c}} \log \left (\frac {c x^{2} - 2 \, c x \sqrt {-\frac {b}{c}} - b}{c x^{2} + b}\right ) - 30 \, {\left (7 \, B b^{3} - 3 \, A b^{2} c\right )} x}{48 \, {\left (c^{6} x^{4} + 2 \, b c^{5} x^{2} + b^{2} c^{4}\right )}}, \frac {8 \, B c^{3} x^{7} - 8 \, {\left (7 \, B b c^{2} - 3 \, A c^{3}\right )} x^{5} - 25 \, {\left (7 \, B b^{2} c - 3 \, A b c^{2}\right )} x^{3} + 15 \, {\left ({\left (7 \, B b c^{2} - 3 \, A c^{3}\right )} x^{4} + 7 \, B b^{3} - 3 \, A b^{2} c + 2 \, {\left (7 \, B b^{2} c - 3 \, A b c^{2}\right )} x^{2}\right )} \sqrt {\frac {b}{c}} \arctan \left (\frac {c x \sqrt {\frac {b}{c}}}{b}\right ) - 15 \, {\left (7 \, B b^{3} - 3 \, A b^{2} c\right )} x}{24 \, {\left (c^{6} x^{4} + 2 \, b c^{5} x^{2} + b^{2} c^{4}\right )}}\right ] \] Input:

integrate(x^12*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="fricas")
 

Output:

[1/48*(16*B*c^3*x^7 - 16*(7*B*b*c^2 - 3*A*c^3)*x^5 - 50*(7*B*b^2*c - 3*A*b 
*c^2)*x^3 - 15*((7*B*b*c^2 - 3*A*c^3)*x^4 + 7*B*b^3 - 3*A*b^2*c + 2*(7*B*b 
^2*c - 3*A*b*c^2)*x^2)*sqrt(-b/c)*log((c*x^2 - 2*c*x*sqrt(-b/c) - b)/(c*x^ 
2 + b)) - 30*(7*B*b^3 - 3*A*b^2*c)*x)/(c^6*x^4 + 2*b*c^5*x^2 + b^2*c^4), 1 
/24*(8*B*c^3*x^7 - 8*(7*B*b*c^2 - 3*A*c^3)*x^5 - 25*(7*B*b^2*c - 3*A*b*c^2 
)*x^3 + 15*((7*B*b*c^2 - 3*A*c^3)*x^4 + 7*B*b^3 - 3*A*b^2*c + 2*(7*B*b^2*c 
 - 3*A*b*c^2)*x^2)*sqrt(b/c)*arctan(c*x*sqrt(b/c)/b) - 15*(7*B*b^3 - 3*A*b 
^2*c)*x)/(c^6*x^4 + 2*b*c^5*x^2 + b^2*c^4)]
 

Sympy [A] (verification not implemented)

Time = 0.70 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.81 \[ \int \frac {x^{12} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {B x^{3}}{3 c^{3}} + x \left (\frac {A}{c^{3}} - \frac {3 B b}{c^{4}}\right ) - \frac {5 \sqrt {- \frac {b}{c^{9}}} \left (- 3 A c + 7 B b\right ) \log {\left (- \frac {5 c^{4} \sqrt {- \frac {b}{c^{9}}} \left (- 3 A c + 7 B b\right )}{- 15 A c + 35 B b} + x \right )}}{16} + \frac {5 \sqrt {- \frac {b}{c^{9}}} \left (- 3 A c + 7 B b\right ) \log {\left (\frac {5 c^{4} \sqrt {- \frac {b}{c^{9}}} \left (- 3 A c + 7 B b\right )}{- 15 A c + 35 B b} + x \right )}}{16} + \frac {x^{3} \cdot \left (9 A b c^{2} - 13 B b^{2} c\right ) + x \left (7 A b^{2} c - 11 B b^{3}\right )}{8 b^{2} c^{4} + 16 b c^{5} x^{2} + 8 c^{6} x^{4}} \] Input:

integrate(x**12*(B*x**2+A)/(c*x**4+b*x**2)**3,x)
 

Output:

B*x**3/(3*c**3) + x*(A/c**3 - 3*B*b/c**4) - 5*sqrt(-b/c**9)*(-3*A*c + 7*B* 
b)*log(-5*c**4*sqrt(-b/c**9)*(-3*A*c + 7*B*b)/(-15*A*c + 35*B*b) + x)/16 + 
 5*sqrt(-b/c**9)*(-3*A*c + 7*B*b)*log(5*c**4*sqrt(-b/c**9)*(-3*A*c + 7*B*b 
)/(-15*A*c + 35*B*b) + x)/16 + (x**3*(9*A*b*c**2 - 13*B*b**2*c) + x*(7*A*b 
**2*c - 11*B*b**3))/(8*b**2*c**4 + 16*b*c**5*x**2 + 8*c**6*x**4)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.02 \[ \int \frac {x^{12} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {{\left (13 \, B b^{2} c - 9 \, A b c^{2}\right )} x^{3} + {\left (11 \, B b^{3} - 7 \, A b^{2} c\right )} x}{8 \, {\left (c^{6} x^{4} + 2 \, b c^{5} x^{2} + b^{2} c^{4}\right )}} + \frac {5 \, {\left (7 \, B b^{2} - 3 \, A b c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} c^{4}} + \frac {B c x^{3} - 3 \, {\left (3 \, B b - A c\right )} x}{3 \, c^{4}} \] Input:

integrate(x^12*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="maxima")
 

Output:

-1/8*((13*B*b^2*c - 9*A*b*c^2)*x^3 + (11*B*b^3 - 7*A*b^2*c)*x)/(c^6*x^4 + 
2*b*c^5*x^2 + b^2*c^4) + 5/8*(7*B*b^2 - 3*A*b*c)*arctan(c*x/sqrt(b*c))/(sq 
rt(b*c)*c^4) + 1/3*(B*c*x^3 - 3*(3*B*b - A*c)*x)/c^4
 

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.94 \[ \int \frac {x^{12} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {5 \, {\left (7 \, B b^{2} - 3 \, A b c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} c^{4}} - \frac {13 \, B b^{2} c x^{3} - 9 \, A b c^{2} x^{3} + 11 \, B b^{3} x - 7 \, A b^{2} c x}{8 \, {\left (c x^{2} + b\right )}^{2} c^{4}} + \frac {B c^{6} x^{3} - 9 \, B b c^{5} x + 3 \, A c^{6} x}{3 \, c^{9}} \] Input:

integrate(x^12*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="giac")
 

Output:

5/8*(7*B*b^2 - 3*A*b*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*c^4) - 1/8*(13*B* 
b^2*c*x^3 - 9*A*b*c^2*x^3 + 11*B*b^3*x - 7*A*b^2*c*x)/((c*x^2 + b)^2*c^4) 
+ 1/3*(B*c^6*x^3 - 9*B*b*c^5*x + 3*A*c^6*x)/c^9
 

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.17 \[ \int \frac {x^{12} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {x^3\,\left (\frac {9\,A\,b\,c^2}{8}-\frac {13\,B\,b^2\,c}{8}\right )-x\,\left (\frac {11\,B\,b^3}{8}-\frac {7\,A\,b^2\,c}{8}\right )}{b^2\,c^4+2\,b\,c^5\,x^2+c^6\,x^4}+x\,\left (\frac {A}{c^3}-\frac {3\,B\,b}{c^4}\right )+\frac {B\,x^3}{3\,c^3}+\frac {5\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c}\,x\,\left (3\,A\,c-7\,B\,b\right )}{7\,B\,b^2-3\,A\,b\,c}\right )\,\left (3\,A\,c-7\,B\,b\right )}{8\,c^{9/2}} \] Input:

int((x^12*(A + B*x^2))/(b*x^2 + c*x^4)^3,x)
 

Output:

(x^3*((9*A*b*c^2)/8 - (13*B*b^2*c)/8) - x*((11*B*b^3)/8 - (7*A*b^2*c)/8))/ 
(b^2*c^4 + c^6*x^4 + 2*b*c^5*x^2) + x*(A/c^3 - (3*B*b)/c^4) + (B*x^3)/(3*c 
^3) + (5*b^(1/2)*atan((b^(1/2)*c^(1/2)*x*(3*A*c - 7*B*b))/(7*B*b^2 - 3*A*b 
*c))*(3*A*c - 7*B*b))/(8*c^(9/2))
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 240, normalized size of antiderivative = 2.03 \[ \int \frac {x^{12} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {-45 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) a \,b^{2} c -90 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) a b \,c^{2} x^{2}-45 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) a \,c^{3} x^{4}+105 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{4}+210 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{3} c \,x^{2}+105 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{2} c^{2} x^{4}+45 a \,b^{2} c^{2} x +75 a b \,c^{3} x^{3}+24 a \,c^{4} x^{5}-105 b^{4} c x -175 b^{3} c^{2} x^{3}-56 b^{2} c^{3} x^{5}+8 b \,c^{4} x^{7}}{24 c^{5} \left (c^{2} x^{4}+2 b c \,x^{2}+b^{2}\right )} \] Input:

int(x^12*(B*x^2+A)/(c*x^4+b*x^2)^3,x)
 

Output:

( - 45*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*a*b**2*c - 90*sqrt(c) 
*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*a*b*c**2*x**2 - 45*sqrt(c)*sqrt(b)* 
atan((c*x)/(sqrt(c)*sqrt(b)))*a*c**3*x**4 + 105*sqrt(c)*sqrt(b)*atan((c*x) 
/(sqrt(c)*sqrt(b)))*b**4 + 210*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b) 
))*b**3*c*x**2 + 105*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*b**2*c* 
*2*x**4 + 45*a*b**2*c**2*x + 75*a*b*c**3*x**3 + 24*a*c**4*x**5 - 105*b**4* 
c*x - 175*b**3*c**2*x**3 - 56*b**2*c**3*x**5 + 8*b*c**4*x**7)/(24*c**5*(b* 
*2 + 2*b*c*x**2 + c**2*x**4))