\(\int \frac {x^8 (A+B x^2)}{(b x^2+c x^4)^3} \, dx\) [83]

Optimal result

Integrand size = 24, antiderivative size = 90 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {(b B-A c) x}{4 c^2 \left (b+c x^2\right )^2}-\frac {(5 b B-A c) x}{8 b c^2 \left (b+c x^2\right )}+\frac {(3 b B+A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{3/2} c^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.92 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {\frac {\sqrt {c} x \left (-3 b^2 B+A c^2 x^2-b c \left (A+5 B x^2\right )\right )}{b \left (b+c x^2\right )^2}+\frac {(3 b B+A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{b^{3/2}}}{8 c^{5/2}} \] Input:

Integrate[(x^8*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {9, 360, 298, 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.

\(\displaystyle \int \frac {x^8 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx\)

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 360

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

\(\Big \downarrow \) 298

\(\displaystyle \frac {x (b B-A c)}{4 c^2 \left (b+c x^2\right )^2}-\frac {\frac {x (5 b B-A c)}{2 b \left (b+c x^2\right )}-\frac {(A c+3 b B) \int \frac {1}{c x^2+b}dx}{2 b}}{4 c^2}\)

\(\Big \downarrow \) 218

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

Input:

Int[(x^8*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]
 

Output:

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

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 218

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]
 

rule 298

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( 
b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 
2*p + 3))/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, 
 c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
 

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])
 

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.84

method	result	size
default	\(\frac {\frac {\left (A c -5 B b \right ) x^{3}}{8 b c}-\frac {\left (A c +3 B b \right ) x}{8 c^{2}}}{\left (c \,x^{2}+b \right )^{2}}+\frac {\left (A c +3 B b \right ) \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 c^{2} b \sqrt {b c}}\)	\(76\)
risch	\(\frac {\frac {\left (A c -5 B b \right ) x^{3}}{8 b c}-\frac {\left (A c +3 B b \right ) x}{8 c^{2}}}{\left (c \,x^{2}+b \right )^{2}}-\frac {\ln \left (c x +\sqrt {-b c}\right ) A}{16 \sqrt {-b c}\, c b}-\frac {3 \ln \left (c x +\sqrt {-b c}\right ) B}{16 \sqrt {-b c}\, c^{2}}+\frac {\ln \left (-c x +\sqrt {-b c}\right ) A}{16 \sqrt {-b c}\, c b}+\frac {3 \ln \left (-c x +\sqrt {-b c}\right ) B}{16 \sqrt {-b c}\, c^{2}}\)	\(146\)

Input:

int(x^8*(B*x^2+A)/(c*x^4+b*x^2)^3,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 301, normalized size of antiderivative = 3.34 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\left [-\frac {2 \, {\left (5 \, B b^{2} c^{2} - A b c^{3}\right )} x^{3} + {\left ({\left (3 \, B b c^{2} + A c^{3}\right )} x^{4} + 3 \, B b^{3} + A b^{2} c + 2 \, {\left (3 \, B b^{2} c + A b c^{2}\right )} x^{2}\right )} \sqrt {-b c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-b c} x - b}{c x^{2} + b}\right ) + 2 \, {\left (3 \, B b^{3} c + A b^{2} c^{2}\right )} x}{16 \, {\left (b^{2} c^{5} x^{4} + 2 \, b^{3} c^{4} x^{2} + b^{4} c^{3}\right )}}, -\frac {{\left (5 \, B b^{2} c^{2} - A b c^{3}\right )} x^{3} - {\left ({\left (3 \, B b c^{2} + A c^{3}\right )} x^{4} + 3 \, B b^{3} + A b^{2} c + 2 \, {\left (3 \, B b^{2} c + A b c^{2}\right )} x^{2}\right )} \sqrt {b c} \arctan \left (\frac {\sqrt {b c} x}{b}\right ) + {\left (3 \, B b^{3} c + A b^{2} c^{2}\right )} x}{8 \, {\left (b^{2} c^{5} x^{4} + 2 \, b^{3} c^{4} x^{2} + b^{4} c^{3}\right )}}\right ] \] Input:

integrate(x^8*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="fricas")
 

Output:

[-1/16*(2*(5*B*b^2*c^2 - A*b*c^3)*x^3 + ((3*B*b*c^2 + A*c^3)*x^4 + 3*B*b^3 
 + A*b^2*c + 2*(3*B*b^2*c + A*b*c^2)*x^2)*sqrt(-b*c)*log((c*x^2 - 2*sqrt(- 
b*c)*x - b)/(c*x^2 + b)) + 2*(3*B*b^3*c + A*b^2*c^2)*x)/(b^2*c^5*x^4 + 2*b 
^3*c^4*x^2 + b^4*c^3), -1/8*((5*B*b^2*c^2 - A*b*c^3)*x^3 - ((3*B*b*c^2 + A 
*c^3)*x^4 + 3*B*b^3 + A*b^2*c + 2*(3*B*b^2*c + A*b*c^2)*x^2)*sqrt(b*c)*arc 
tan(sqrt(b*c)*x/b) + (3*B*b^3*c + A*b^2*c^2)*x)/(b^2*c^5*x^4 + 2*b^3*c^4*x 
^2 + b^4*c^3)]
 

Sympy [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.72 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=- \frac {\sqrt {- \frac {1}{b^{3} c^{5}}} \left (A c + 3 B b\right ) \log {\left (- b^{2} c^{2} \sqrt {- \frac {1}{b^{3} c^{5}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{b^{3} c^{5}}} \left (A c + 3 B b\right ) \log {\left (b^{2} c^{2} \sqrt {- \frac {1}{b^{3} c^{5}}} + x \right )}}{16} + \frac {x^{3} \left (A c^{2} - 5 B b c\right ) + x \left (- A b c - 3 B b^{2}\right )}{8 b^{3} c^{2} + 16 b^{2} c^{3} x^{2} + 8 b c^{4} x^{4}} \] Input:

integrate(x**8*(B*x**2+A)/(c*x**4+b*x**2)**3,x)
 

Output:

-sqrt(-1/(b**3*c**5))*(A*c + 3*B*b)*log(-b**2*c**2*sqrt(-1/(b**3*c**5)) + 
x)/16 + sqrt(-1/(b**3*c**5))*(A*c + 3*B*b)*log(b**2*c**2*sqrt(-1/(b**3*c** 
5)) + x)/16 + (x**3*(A*c**2 - 5*B*b*c) + x*(-A*b*c - 3*B*b**2))/(8*b**3*c* 
*2 + 16*b**2*c**3*x**2 + 8*b*c**4*x**4)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {{\left (5 \, B b c - A c^{2}\right )} x^{3} + {\left (3 \, B b^{2} + A b c\right )} x}{8 \, {\left (b c^{4} x^{4} + 2 \, b^{2} c^{3} x^{2} + b^{3} c^{2}\right )}} + \frac {{\left (3 \, B b + A c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} b c^{2}} \] Input:

integrate(x^8*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.87 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {{\left (3 \, B b + A c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} b c^{2}} - \frac {5 \, B b c x^{3} - A c^{2} x^{3} + 3 \, B b^{2} x + A b c x}{8 \, {\left (c x^{2} + b\right )}^{2} b c^{2}} \] Input:

integrate(x^8*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="giac")
 

Output:

1/8*(3*B*b + A*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b*c^2) - 1/8*(5*B*b*c*x 
^3 - A*c^2*x^3 + 3*B*b^2*x + A*b*c*x)/((c*x^2 + b)^2*b*c^2)
 

Mupad [B] (verification not implemented)

Time = 9.33 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.91 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )\,\left (A\,c+3\,B\,b\right )}{8\,b^{3/2}\,c^{5/2}}-\frac {\frac {x\,\left (A\,c+3\,B\,b\right )}{8\,c^2}-\frac {x^3\,\left (A\,c-5\,B\,b\right )}{8\,b\,c}}{b^2+2\,b\,c\,x^2+c^2\,x^4} \] Input:

int((x^8*(A + B*x^2))/(b*x^2 + c*x^4)^3,x)
 

Output:

(atan((c^(1/2)*x)/b^(1/2))*(A*c + 3*B*b))/(8*b^(3/2)*c^(5/2)) - ((x*(A*c + 
 3*B*b))/(8*c^2) - (x^3*(A*c - 5*B*b))/(8*b*c))/(b^2 + c^2*x^4 + 2*b*c*x^2 
)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.34 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {\sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) a \,b^{2} c +2 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) a b \,c^{2} x^{2}+\sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) a \,c^{3} x^{4}+3 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{4}+6 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{3} c \,x^{2}+3 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{2} c^{2} x^{4}-a \,b^{2} c^{2} x +a b \,c^{3} x^{3}-3 b^{4} c x -5 b^{3} c^{2} x^{3}}{8 b^{2} c^{3} \left (c^{2} x^{4}+2 b c \,x^{2}+b^{2}\right )} \] Input:

int(x^8*(B*x^2+A)/(c*x^4+b*x^2)^3,x)
 

Output:

(sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*a*b**2*c + 2*sqrt(c)*sqrt(b 
)*atan((c*x)/(sqrt(c)*sqrt(b)))*a*b*c**2*x**2 + sqrt(c)*sqrt(b)*atan((c*x) 
/(sqrt(c)*sqrt(b)))*a*c**3*x**4 + 3*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sq 
rt(b)))*b**4 + 6*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*b**3*c*x**2 
 + 3*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*b**2*c**2*x**4 - a*b**2 
*c**2*x + a*b*c**3*x**3 - 3*b**4*c*x - 5*b**3*c**2*x**3)/(8*b**2*c**3*(b** 
2 + 2*b*c*x**2 + c**2*x**4))