\(\int \frac {x^3}{(a x+b x^3+c x^5)^2} \, dx\) [40]

Optimal result

Integrand size = 20, antiderivative size = 74 \[ \int \frac {x^3}{\left (a x+b x^3+c x^5\right )^2} \, dx=-\frac {b+2 c x^2}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac {2 c \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}} \] Output:

-1/2*(2*c*x^2+b)/(-4*a*c+b^2)/(c*x^4+b*x^2+a)+2*c*arctanh((2*c*x^2+b)/(-4* 
a*c+b^2)^(1/2))/(-4*a*c+b^2)^(3/2)
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.07 \[ \int \frac {x^3}{\left (a x+b x^3+c x^5\right )^2} \, dx=-\frac {\frac {b+2 c x^2}{a+b x^2+c x^4}+\frac {4 c \arctan \left (\frac {b+2 c x^2}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}}{2 \left (b^2-4 a c\right )} \] Input:

Integrate[x^3/(a*x + b*x^3 + c*x^5)^2,x]
 

Output:

-1/2*((b + 2*c*x^2)/(a + b*x^2 + c*x^4) + (4*c*ArcTan[(b + 2*c*x^2)/Sqrt[- 
b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c])/(b^2 - 4*a*c)
 

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {9, 1432, 1086, 1083, 219}

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.

\(\displaystyle \int \frac {x^3}{\left (a x+b x^3+c x^5\right )^2} \, dx\)

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 1432

\(\displaystyle \frac {1}{2} \int \frac {1}{\left (c x^4+b x^2+a\right )^2}dx^2\)

\(\Big \downarrow \) 1086

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

\(\Big \downarrow \) 1083

\(\displaystyle \frac {1}{2} \left (\frac {4 c \int \frac {1}{-x^4+b^2-4 a c}d\left (2 c x^2+b\right )}{b^2-4 a c}-\frac {b+2 c x^2}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (\frac {4 c \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}-\frac {b+2 c x^2}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\)

Input:

Int[x^3/(a*x + b*x^3 + c*x^5)^2,x]
 

Output:

(-((b + 2*c*x^2)/((b^2 - 4*a*c)*(a + b*x^2 + c*x^4))) + (4*c*ArcTanh[(b + 
2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(b^2 - 4*a*c)^(3/2))/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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 1086

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) 
*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] - Simp[2*c*((2*p + 
 3)/((p + 1)*(b^2 - 4*a*c)))   Int[(a + b*x + c*x^2)^(p + 1), x], x] /; Fre 
eQ[{a, b, c}, x] && ILtQ[p, -1]
 

rule 1432

Int[(x_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[1/2 
 Subst[Int[(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x]
 

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.01

method	result	size
default	\(\frac {2 c \,x^{2}+b}{2 \left (4 a c -b^{2}\right ) \left (c \,x^{4}+b \,x^{2}+a \right )}+\frac {2 c \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {3}{2}}}\)	\(75\)
risch	\(\frac {\frac {c \,x^{2}}{4 a c -b^{2}}+\frac {b}{8 a c -2 b^{2}}}{c \,x^{4}+b \,x^{2}+a}+\frac {c \ln \left (\left (\left (-4 a c +b^{2}\right )^{\frac {3}{2}}+4 a b c -b^{3}\right ) x^{2}+8 a^{2} c -2 b^{2} a \right )}{\left (-4 a c +b^{2}\right )^{\frac {3}{2}}}-\frac {c \ln \left (\left (\left (-4 a c +b^{2}\right )^{\frac {3}{2}}-4 a b c +b^{3}\right ) x^{2}-8 a^{2} c +2 b^{2} a \right )}{\left (-4 a c +b^{2}\right )^{\frac {3}{2}}}\)	\(151\)

Input:

int(x^3/(c*x^5+b*x^3+a*x)^2,x,method=_RETURNVERBOSE)
 

Output:

1/2*(2*c*x^2+b)/(4*a*c-b^2)/(c*x^4+b*x^2+a)+2*c/(4*a*c-b^2)^(3/2)*arctan(( 
2*c*x^2+b)/(4*a*c-b^2)^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (68) = 136\).

Time = 0.08 (sec) , antiderivative size = 361, normalized size of antiderivative = 4.88 \[ \int \frac {x^3}{\left (a x+b x^3+c x^5\right )^2} \, dx=\left [-\frac {b^{3} - 4 \, a b c + 2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} x^{2} + 2 \, {\left (c^{2} x^{4} + b c x^{2} + a c\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{4} + 2 \, b c x^{2} + b^{2} - 2 \, a c - {\left (2 \, c x^{2} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{4} + b x^{2} + a}\right )}{2 \, {\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{4} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x^{2}\right )}}, -\frac {b^{3} - 4 \, a b c + 2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} x^{2} - 4 \, {\left (c^{2} x^{4} + b c x^{2} + a c\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {{\left (2 \, c x^{2} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right )}{2 \, {\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{4} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x^{2}\right )}}\right ] \] Input:

integrate(x^3/(c*x^5+b*x^3+a*x)^2,x, algorithm="fricas")
 

Output:

[-1/2*(b^3 - 4*a*b*c + 2*(b^2*c - 4*a*c^2)*x^2 + 2*(c^2*x^4 + b*c*x^2 + a* 
c)*sqrt(b^2 - 4*a*c)*log((2*c^2*x^4 + 2*b*c*x^2 + b^2 - 2*a*c - (2*c*x^2 + 
 b)*sqrt(b^2 - 4*a*c))/(c*x^4 + b*x^2 + a)))/(a*b^4 - 8*a^2*b^2*c + 16*a^3 
*c^2 + (b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*x^4 + (b^5 - 8*a*b^3*c + 16*a^2* 
b*c^2)*x^2), -1/2*(b^3 - 4*a*b*c + 2*(b^2*c - 4*a*c^2)*x^2 - 4*(c^2*x^4 + 
b*c*x^2 + a*c)*sqrt(-b^2 + 4*a*c)*arctan(-(2*c*x^2 + b)*sqrt(-b^2 + 4*a*c) 
/(b^2 - 4*a*c)))/(a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2 + (b^4*c - 8*a*b^2*c^2 
+ 16*a^2*c^3)*x^4 + (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*x^2)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 267 vs. \(2 (66) = 132\).

Time = 0.92 (sec) , antiderivative size = 267, normalized size of antiderivative = 3.61 \[ \int \frac {x^3}{\left (a x+b x^3+c x^5\right )^2} \, dx=- c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \log {\left (x^{2} + \frac {- 16 a^{2} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + 8 a b^{2} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} - b^{4} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + b c}{2 c^{2}} \right )} + c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \log {\left (x^{2} + \frac {16 a^{2} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} - 8 a b^{2} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + b^{4} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + b c}{2 c^{2}} \right )} + \frac {b + 2 c x^{2}}{8 a^{2} c - 2 a b^{2} + x^{4} \cdot \left (8 a c^{2} - 2 b^{2} c\right ) + x^{2} \cdot \left (8 a b c - 2 b^{3}\right )} \] Input:

integrate(x**3/(c*x**5+b*x**3+a*x)**2,x)
 

Output:

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

Maxima [F]

\[ \int \frac {x^3}{\left (a x+b x^3+c x^5\right )^2} \, dx=\int { \frac {x^{3}}{{\left (c x^{5} + b x^{3} + a x\right )}^{2}} \,d x } \] Input:

integrate(x^3/(c*x^5+b*x^3+a*x)^2,x, algorithm="maxima")
 

Output:

-2*c*integrate(x/(c*x^4 + b*x^2 + a), x)/(b^2 - 4*a*c) - 1/2*(2*c*x^2 + b) 
/((b^2*c - 4*a*c^2)*x^4 + a*b^2 - 4*a^2*c + (b^3 - 4*a*b*c)*x^2)
 

Giac [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.11 \[ \int \frac {x^3}{\left (a x+b x^3+c x^5\right )^2} \, dx=-\frac {2 \, c \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{2} - 4 \, a c\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {2 \, c x^{2} + b}{2 \, {\left (c x^{4} + b x^{2} + a\right )} {\left (b^{2} - 4 \, a c\right )}} \] Input:

integrate(x^3/(c*x^5+b*x^3+a*x)^2,x, algorithm="giac")
 

Output:

-2*c*arctan((2*c*x^2 + b)/sqrt(-b^2 + 4*a*c))/((b^2 - 4*a*c)*sqrt(-b^2 + 4 
*a*c)) - 1/2*(2*c*x^2 + b)/((c*x^4 + b*x^2 + a)*(b^2 - 4*a*c))
 

Mupad [B] (verification not implemented)

Time = 12.22 (sec) , antiderivative size = 172, normalized size of antiderivative = 2.32 \[ \int \frac {x^3}{\left (a x+b x^3+c x^5\right )^2} \, dx=\frac {\frac {b}{2\,\left (4\,a\,c-b^2\right )}+\frac {c\,x^2}{4\,a\,c-b^2}}{c\,x^4+b\,x^2+a}-\frac {2\,c\,\mathrm {atan}\left (\frac {b^3-4\,a\,b\,c}{{\left (4\,a\,c-b^2\right )}^{3/2}}-\frac {x^2\,{\left (4\,a\,c-b^2\right )}^4\,\left (\frac {4\,c^4}{a\,{\left (4\,a\,c-b^2\right )}^{7/2}}+\frac {4\,c^2\,\left (b^3\,c^2-4\,a\,b\,c^3\right )\,\left (b^3-4\,a\,b\,c\right )}{a\,{\left (4\,a\,c-b^2\right )}^{13/2}}\right )}{8\,c^4}\right )}{{\left (4\,a\,c-b^2\right )}^{3/2}} \] Input:

int(x^3/(a*x + b*x^3 + c*x^5)^2,x)
 

Output:

(b/(2*(4*a*c - b^2)) + (c*x^2)/(4*a*c - b^2))/(a + b*x^2 + c*x^4) - (2*c*a 
tan((b^3 - 4*a*b*c)/(4*a*c - b^2)^(3/2) - (x^2*(4*a*c - b^2)^4*((4*c^4)/(a 
*(4*a*c - b^2)^(7/2)) + (4*c^2*(b^3*c^2 - 4*a*b*c^3)*(b^3 - 4*a*b*c))/(a*( 
4*a*c - b^2)^(13/2))))/(8*c^4)))/(4*a*c - b^2)^(3/2)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 477, normalized size of antiderivative = 6.45 \[ \int \frac {x^3}{\left (a x+b x^3+c x^5\right )^2} \, dx=\frac {-4 \sqrt {2 \sqrt {c}\, \sqrt {a}+b}\, \sqrt {2 \sqrt {c}\, \sqrt {a}-b}\, \mathit {atan} \left (\frac {\sqrt {2 \sqrt {c}\, \sqrt {a}-b}-2 \sqrt {c}\, x}{\sqrt {2 \sqrt {c}\, \sqrt {a}+b}}\right ) a b c -4 \sqrt {2 \sqrt {c}\, \sqrt {a}+b}\, \sqrt {2 \sqrt {c}\, \sqrt {a}-b}\, \mathit {atan} \left (\frac {\sqrt {2 \sqrt {c}\, \sqrt {a}-b}-2 \sqrt {c}\, x}{\sqrt {2 \sqrt {c}\, \sqrt {a}+b}}\right ) b^{2} c \,x^{2}-4 \sqrt {2 \sqrt {c}\, \sqrt {a}+b}\, \sqrt {2 \sqrt {c}\, \sqrt {a}-b}\, \mathit {atan} \left (\frac {\sqrt {2 \sqrt {c}\, \sqrt {a}-b}-2 \sqrt {c}\, x}{\sqrt {2 \sqrt {c}\, \sqrt {a}+b}}\right ) b \,c^{2} x^{4}-4 \sqrt {2 \sqrt {c}\, \sqrt {a}+b}\, \sqrt {2 \sqrt {c}\, \sqrt {a}-b}\, \mathit {atan} \left (\frac {\sqrt {2 \sqrt {c}\, \sqrt {a}-b}+2 \sqrt {c}\, x}{\sqrt {2 \sqrt {c}\, \sqrt {a}+b}}\right ) a b c -4 \sqrt {2 \sqrt {c}\, \sqrt {a}+b}\, \sqrt {2 \sqrt {c}\, \sqrt {a}-b}\, \mathit {atan} \left (\frac {\sqrt {2 \sqrt {c}\, \sqrt {a}-b}+2 \sqrt {c}\, x}{\sqrt {2 \sqrt {c}\, \sqrt {a}+b}}\right ) b^{2} c \,x^{2}-4 \sqrt {2 \sqrt {c}\, \sqrt {a}+b}\, \sqrt {2 \sqrt {c}\, \sqrt {a}-b}\, \mathit {atan} \left (\frac {\sqrt {2 \sqrt {c}\, \sqrt {a}-b}+2 \sqrt {c}\, x}{\sqrt {2 \sqrt {c}\, \sqrt {a}+b}}\right ) b \,c^{2} x^{4}-8 a^{2} c^{2}+6 a \,b^{2} c -8 a \,c^{3} x^{4}-b^{4}+2 b^{2} c^{2} x^{4}}{2 b \left (16 a^{2} c^{3} x^{4}-8 a \,b^{2} c^{2} x^{4}+b^{4} c \,x^{4}+16 a^{2} b \,c^{2} x^{2}-8 a \,b^{3} c \,x^{2}+b^{5} x^{2}+16 a^{3} c^{2}-8 a^{2} b^{2} c +a \,b^{4}\right )} \] Input:

int(x^3/(c*x^5+b*x^3+a*x)^2,x)
 

Output:

( - 4*sqrt(2*sqrt(c)*sqrt(a) + b)*sqrt(2*sqrt(c)*sqrt(a) - b)*atan((sqrt(2 
*sqrt(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*a*b*c - 
4*sqrt(2*sqrt(c)*sqrt(a) + b)*sqrt(2*sqrt(c)*sqrt(a) - b)*atan((sqrt(2*sqr 
t(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*b**2*c*x**2 
- 4*sqrt(2*sqrt(c)*sqrt(a) + b)*sqrt(2*sqrt(c)*sqrt(a) - b)*atan((sqrt(2*s 
qrt(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*b*c**2*x** 
4 - 4*sqrt(2*sqrt(c)*sqrt(a) + b)*sqrt(2*sqrt(c)*sqrt(a) - b)*atan((sqrt(2 
*sqrt(c)*sqrt(a) - b) + 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*a*b*c - 
4*sqrt(2*sqrt(c)*sqrt(a) + b)*sqrt(2*sqrt(c)*sqrt(a) - b)*atan((sqrt(2*sqr 
t(c)*sqrt(a) - b) + 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*b**2*c*x**2 
- 4*sqrt(2*sqrt(c)*sqrt(a) + b)*sqrt(2*sqrt(c)*sqrt(a) - b)*atan((sqrt(2*s 
qrt(c)*sqrt(a) - b) + 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*b*c**2*x** 
4 - 8*a**2*c**2 + 6*a*b**2*c - 8*a*c**3*x**4 - b**4 + 2*b**2*c**2*x**4)/(2 
*b*(16*a**3*c**2 - 8*a**2*b**2*c + 16*a**2*b*c**2*x**2 + 16*a**2*c**3*x**4 
 + a*b**4 - 8*a*b**3*c*x**2 - 8*a*b**2*c**2*x**4 + b**5*x**2 + b**4*c*x**4 
))