\(\int \frac {x^4 (A+B x^2)}{(b x^2+c x^4)^3} \, dx\) [87]

Optimal result

Integrand size = 24, antiderivative size = 96 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {A}{b^3 x}+\frac {(b B-A c) x}{4 b^2 \left (b+c x^2\right )^2}+\frac {(3 b B-7 A c) x}{8 b^3 \left (b+c x^2\right )}+\frac {3 (b B-5 A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{7/2} \sqrt {c}} \] Output:

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {A}{b^3 x}+\frac {(b B-A c) x}{4 b^2 \left (b+c x^2\right )^2}+\frac {(3 b B-7 A c) x}{8 b^3 \left (b+c x^2\right )}+\frac {3 (b B-5 A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{7/2} \sqrt {c}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {9, 361, 25, 27, 361, 25, 359, 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.

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

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 361

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 361

\(\displaystyle \frac {\frac {x (3 b B-7 A c)}{2 b \left (b+c x^2\right )}-\frac {1}{2} \int -\frac {\left (3 B-\frac {7 A c}{b}\right ) x^2+8 A}{x^2 \left (c x^2+b\right )}dx}{4 b^2}+\frac {x (b B-A c)}{4 b^2 \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {1}{2} \int \frac {\left (3 B-\frac {7 A c}{b}\right ) x^2+8 A}{x^2 \left (c x^2+b\right )}dx+\frac {x (3 b B-7 A c)}{2 b \left (b+c x^2\right )}}{4 b^2}+\frac {x (b B-A c)}{4 b^2 \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 359

\(\displaystyle \frac {\frac {1}{2} \left (\frac {3 (b B-5 A c) \int \frac {1}{c x^2+b}dx}{b}-\frac {8 A}{b x}\right )+\frac {x (3 b B-7 A c)}{2 b \left (b+c x^2\right )}}{4 b^2}+\frac {x (b B-A c)}{4 b^2 \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {1}{2} \left (\frac {3 (b B-5 A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{b^{3/2} \sqrt {c}}-\frac {8 A}{b x}\right )+\frac {x (3 b B-7 A c)}{2 b \left (b+c x^2\right )}}{4 b^2}+\frac {x (b B-A c)}{4 b^2 \left (b+c x^2\right )^2}\)

Input:

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

Output:

((b*B - A*c)*x)/(4*b^2*(b + c*x^2)^2) + (((3*b*B - 7*A*c)*x)/(2*b*(b + c*x 
^2)) + ((-8*A)/(b*x) + (3*(b*B - 5*A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(b^(3 
/2)*Sqrt[c]))/2)/(4*b^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 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 359

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

rule 361

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[x^m*(a + b*x^2)^(p + 1)*E 
xpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c 
- a*d)*x^(-m + 2))/(a + b*x^2)] - ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], 
 x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ILtQ[m/ 
2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
 

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.85

method	result	size
default	\(-\frac {A}{b^{3} x}-\frac {\frac {\left (\frac {7}{8} A \,c^{2}-\frac {3}{8} B b c \right ) x^{3}+\frac {b \left (9 A c -5 B b \right ) x}{8}}{\left (c \,x^{2}+b \right )^{2}}+\frac {3 \left (5 A c -B b \right ) \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \sqrt {b c}}}{b^{3}}\)	\(82\)
risch	\(\frac {-\frac {3 c \left (5 A c -B b \right ) x^{4}}{8 b^{3}}-\frac {5 \left (5 A c -B b \right ) x^{2}}{8 b^{2}}-\frac {A}{b}}{\left (c \,x^{2}+b \right )^{2} x}-\frac {15 \ln \left (-\sqrt {-b c}\, x -b \right ) A c}{16 \sqrt {-b c}\, b^{3}}+\frac {3 \ln \left (-\sqrt {-b c}\, x -b \right ) B}{16 \sqrt {-b c}\, b^{2}}+\frac {15 \ln \left (-\sqrt {-b c}\, x +b \right ) A c}{16 \sqrt {-b c}\, b^{3}}-\frac {3 \ln \left (-\sqrt {-b c}\, x +b \right ) B}{16 \sqrt {-b c}\, b^{2}}\)	\(159\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 324, normalized size of antiderivative = 3.38 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\left [-\frac {16 \, A b^{3} c - 6 \, {\left (B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{4} - 10 \, {\left (B b^{3} c - 5 \, A b^{2} c^{2}\right )} x^{2} - 3 \, {\left ({\left (B b c^{2} - 5 \, A c^{3}\right )} x^{5} + 2 \, {\left (B b^{2} c - 5 \, A b c^{2}\right )} x^{3} + {\left (B b^{3} - 5 \, A b^{2} c\right )} x\right )} \sqrt {-b c} \log \left (\frac {c x^{2} + 2 \, \sqrt {-b c} x - b}{c x^{2} + b}\right )}{16 \, {\left (b^{4} c^{3} x^{5} + 2 \, b^{5} c^{2} x^{3} + b^{6} c x\right )}}, -\frac {8 \, A b^{3} c - 3 \, {\left (B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{4} - 5 \, {\left (B b^{3} c - 5 \, A b^{2} c^{2}\right )} x^{2} - 3 \, {\left ({\left (B b c^{2} - 5 \, A c^{3}\right )} x^{5} + 2 \, {\left (B b^{2} c - 5 \, A b c^{2}\right )} x^{3} + {\left (B b^{3} - 5 \, A b^{2} c\right )} x\right )} \sqrt {b c} \arctan \left (\frac {\sqrt {b c} x}{b}\right )}{8 \, {\left (b^{4} c^{3} x^{5} + 2 \, b^{5} c^{2} x^{3} + b^{6} c x\right )}}\right ] \] Input:

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

Output:

[-1/16*(16*A*b^3*c - 6*(B*b^2*c^2 - 5*A*b*c^3)*x^4 - 10*(B*b^3*c - 5*A*b^2 
*c^2)*x^2 - 3*((B*b*c^2 - 5*A*c^3)*x^5 + 2*(B*b^2*c - 5*A*b*c^2)*x^3 + (B* 
b^3 - 5*A*b^2*c)*x)*sqrt(-b*c)*log((c*x^2 + 2*sqrt(-b*c)*x - b)/(c*x^2 + b 
)))/(b^4*c^3*x^5 + 2*b^5*c^2*x^3 + b^6*c*x), -1/8*(8*A*b^3*c - 3*(B*b^2*c^ 
2 - 5*A*b*c^3)*x^4 - 5*(B*b^3*c - 5*A*b^2*c^2)*x^2 - 3*((B*b*c^2 - 5*A*c^3 
)*x^5 + 2*(B*b^2*c - 5*A*b*c^2)*x^3 + (B*b^3 - 5*A*b^2*c)*x)*sqrt(b*c)*arc 
tan(sqrt(b*c)*x/b))/(b^4*c^3*x^5 + 2*b^5*c^2*x^3 + b^6*c*x)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 194 vs. \(2 (92) = 184\).

Time = 0.35 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.02 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=- \frac {3 \sqrt {- \frac {1}{b^{7} c}} \left (- 5 A c + B b\right ) \log {\left (- \frac {3 b^{4} \sqrt {- \frac {1}{b^{7} c}} \left (- 5 A c + B b\right )}{- 15 A c + 3 B b} + x \right )}}{16} + \frac {3 \sqrt {- \frac {1}{b^{7} c}} \left (- 5 A c + B b\right ) \log {\left (\frac {3 b^{4} \sqrt {- \frac {1}{b^{7} c}} \left (- 5 A c + B b\right )}{- 15 A c + 3 B b} + x \right )}}{16} + \frac {- 8 A b^{2} + x^{4} \left (- 15 A c^{2} + 3 B b c\right ) + x^{2} \left (- 25 A b c + 5 B b^{2}\right )}{8 b^{5} x + 16 b^{4} c x^{3} + 8 b^{3} c^{2} x^{5}} \] Input:

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

Output:

-3*sqrt(-1/(b**7*c))*(-5*A*c + B*b)*log(-3*b**4*sqrt(-1/(b**7*c))*(-5*A*c 
+ B*b)/(-15*A*c + 3*B*b) + x)/16 + 3*sqrt(-1/(b**7*c))*(-5*A*c + B*b)*log( 
3*b**4*sqrt(-1/(b**7*c))*(-5*A*c + B*b)/(-15*A*c + 3*B*b) + x)/16 + (-8*A* 
b**2 + x**4*(-15*A*c**2 + 3*B*b*c) + x**2*(-25*A*b*c + 5*B*b**2))/(8*b**5* 
x + 16*b**4*c*x**3 + 8*b**3*c**2*x**5)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {3 \, {\left (B b c - 5 \, A c^{2}\right )} x^{4} - 8 \, A b^{2} + 5 \, {\left (B b^{2} - 5 \, A b c\right )} x^{2}}{8 \, {\left (b^{3} c^{2} x^{5} + 2 \, b^{4} c x^{3} + b^{5} x\right )}} + \frac {3 \, {\left (B b - 5 \, A c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} b^{3}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.85 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {3 \, {\left (B b - 5 \, A c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} b^{3}} - \frac {A}{b^{3} x} + \frac {3 \, B b c x^{3} - 7 \, A c^{2} x^{3} + 5 \, B b^{2} x - 9 \, A b c x}{8 \, {\left (c x^{2} + b\right )}^{2} b^{3}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 9.26 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.18 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {\frac {A}{b}+\frac {5\,x^2\,\left (5\,A\,c-B\,b\right )}{8\,b^2}+\frac {3\,c\,x^4\,\left (5\,A\,c-B\,b\right )}{8\,b^3}}{b^2\,x+2\,b\,c\,x^3+c^2\,x^5}-\frac {3\,\mathrm {atan}\left (\frac {3\,\sqrt {c}\,x\,\left (5\,A\,c-B\,b\right )}{\sqrt {b}\,\left (15\,A\,c-3\,B\,b\right )}\right )\,\left (5\,A\,c-B\,b\right )}{8\,b^{7/2}\,\sqrt {c}} \] Input:

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

Output:

- (A/b + (5*x^2*(5*A*c - B*b))/(8*b^2) + (3*c*x^4*(5*A*c - B*b))/(8*b^3))/ 
(b^2*x + c^2*x^5 + 2*b*c*x^3) - (3*atan((3*c^(1/2)*x*(5*A*c - B*b))/(b^(1/ 
2)*(15*A*c - 3*B*b)))*(5*A*c - B*b))/(8*b^(7/2)*c^(1/2))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 230, normalized size of antiderivative = 2.40 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {-15 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) a \,b^{2} c x -30 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) a b \,c^{2} x^{3}-15 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) a \,c^{3} x^{5}+3 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{4} x +6 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{3} c \,x^{3}+3 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{2} c^{2} x^{5}-8 a \,b^{3} c -25 a \,b^{2} c^{2} x^{2}-15 a b \,c^{3} x^{4}+5 b^{4} c \,x^{2}+3 b^{3} c^{2} x^{4}}{8 b^{4} c x \left (c^{2} x^{4}+2 b c \,x^{2}+b^{2}\right )} \] Input:

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

Output:

( - 15*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*a*b**2*c*x - 30*sqrt( 
c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*a*b*c**2*x**3 - 15*sqrt(c)*sqrt(b 
)*atan((c*x)/(sqrt(c)*sqrt(b)))*a*c**3*x**5 + 3*sqrt(c)*sqrt(b)*atan((c*x) 
/(sqrt(c)*sqrt(b)))*b**4*x + 6*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b) 
))*b**3*c*x**3 + 3*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*b**2*c**2 
*x**5 - 8*a*b**3*c - 25*a*b**2*c**2*x**2 - 15*a*b*c**3*x**4 + 5*b**4*c*x** 
2 + 3*b**3*c**2*x**4)/(8*b**4*c*x*(b**2 + 2*b*c*x**2 + c**2*x**4))