\(\int \frac {1}{x^3 (a x+b x^3+c x^5)^2} \, dx\) [43]

Optimal result

Integrand size = 20, antiderivative size = 219 \[ \int \frac {1}{x^3 \left (a x+b x^3+c x^5\right )^2} \, dx=-\frac {3 b^2-8 a c}{4 a^2 \left (b^2-4 a c\right ) x^4}+\frac {b \left (3 b^2-11 a c\right )}{2 a^3 \left (b^2-4 a c\right ) x^2}+\frac {b^2-2 a c+b c x^2}{2 a \left (b^2-4 a c\right ) x^4 \left (a+b x^2+c x^4\right )}+\frac {b \left (3 b^4-20 a b^2 c+30 a^2 c^2\right ) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 a^4 \left (b^2-4 a c\right )^{3/2}}+\frac {\left (3 b^2-2 a c\right ) \log (x)}{a^4}-\frac {\left (3 b^2-2 a c\right ) \log \left (a+b x^2+c x^4\right )}{4 a^4} \] Output:

-1/4*(-8*a*c+3*b^2)/a^2/(-4*a*c+b^2)/x^4+1/2*b*(-11*a*c+3*b^2)/a^3/(-4*a*c 
+b^2)/x^2+1/2*(b*c*x^2-2*a*c+b^2)/a/(-4*a*c+b^2)/x^4/(c*x^4+b*x^2+a)+1/2*b 
*(30*a^2*c^2-20*a*b^2*c+3*b^4)*arctanh((2*c*x^2+b)/(-4*a*c+b^2)^(1/2))/a^4 
/(-4*a*c+b^2)^(3/2)+(-2*a*c+3*b^2)*ln(x)/a^4-1/4*(-2*a*c+3*b^2)*ln(c*x^4+b 
*x^2+a)/a^4
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.50 \[ \int \frac {1}{x^3 \left (a x+b x^3+c x^5\right )^2} \, dx=\frac {-\frac {a^2}{x^4}+\frac {4 a b}{x^2}+\frac {2 a \left (b^4-4 a b^2 c+2 a^2 c^2+b^3 c x^2-3 a b c^2 x^2\right )}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+4 \left (3 b^2-2 a c\right ) \log (x)-\frac {\left (3 b^5-20 a b^3 c+30 a^2 b c^2+3 b^4 \sqrt {b^2-4 a c}-14 a b^2 c \sqrt {b^2-4 a c}+8 a^2 c^2 \sqrt {b^2-4 a c}\right ) \log \left (b-\sqrt {b^2-4 a c}+2 c x^2\right )}{\left (b^2-4 a c\right )^{3/2}}+\frac {\left (3 b^5-20 a b^3 c+30 a^2 b c^2-3 b^4 \sqrt {b^2-4 a c}+14 a b^2 c \sqrt {b^2-4 a c}-8 a^2 c^2 \sqrt {b^2-4 a c}\right ) \log \left (b+\sqrt {b^2-4 a c}+2 c x^2\right )}{\left (b^2-4 a c\right )^{3/2}}}{4 a^4} \] Input:

Integrate[1/(x^3*(a*x + b*x^3 + c*x^5)^2),x]
 

Output:

(-(a^2/x^4) + (4*a*b)/x^2 + (2*a*(b^4 - 4*a*b^2*c + 2*a^2*c^2 + b^3*c*x^2 
- 3*a*b*c^2*x^2))/((b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) + 4*(3*b^2 - 2*a*c)* 
Log[x] - ((3*b^5 - 20*a*b^3*c + 30*a^2*b*c^2 + 3*b^4*Sqrt[b^2 - 4*a*c] - 1 
4*a*b^2*c*Sqrt[b^2 - 4*a*c] + 8*a^2*c^2*Sqrt[b^2 - 4*a*c])*Log[b - Sqrt[b^ 
2 - 4*a*c] + 2*c*x^2])/(b^2 - 4*a*c)^(3/2) + ((3*b^5 - 20*a*b^3*c + 30*a^2 
*b*c^2 - 3*b^4*Sqrt[b^2 - 4*a*c] + 14*a*b^2*c*Sqrt[b^2 - 4*a*c] - 8*a^2*c^ 
2*Sqrt[b^2 - 4*a*c])*Log[b + Sqrt[b^2 - 4*a*c] + 2*c*x^2])/(b^2 - 4*a*c)^( 
3/2))/(4*a^4)
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {9, 1434, 1165, 25, 1200, 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.

\(\displaystyle \int \frac {1}{x^3 \left (a x+b x^3+c x^5\right )^2} \, dx\)

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 1434

\(\displaystyle \frac {1}{2} \int \frac {1}{x^6 \left (c x^4+b x^2+a\right )^2}dx^2\)

\(\Big \downarrow \) 1165

\(\displaystyle \frac {1}{2} \left (\frac {-2 a c+b^2+b c x^2}{a x^4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\int -\frac {3 b^2+3 c x^2 b-8 a c}{x^6 \left (c x^4+b x^2+a\right )}dx^2}{a \left (b^2-4 a c\right )}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {3 b^2+3 c x^2 b-8 a c}{x^6 \left (c x^4+b x^2+a\right )}dx^2}{a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c x^2}{a x^4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\)

\(\Big \downarrow \) 1200

\(\displaystyle \frac {1}{2} \left (\frac {\int \left (\frac {3 b^2-8 a c}{a x^6}+\frac {-c \left (3 b^4-14 a c b^2+8 a^2 c^2\right ) x^2-b \left (3 b^4-17 a c b^2+19 a^2 c^2\right )}{a^3 \left (c x^4+b x^2+a\right )}+\frac {\left (b^2-4 a c\right ) \left (3 b^2-2 a c\right )}{a^3 x^2}+\frac {b \left (11 a c-3 b^2\right )}{a^2 x^4}\right )dx^2}{a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c x^2}{a x^4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (\frac {\frac {\log \left (x^2\right ) \left (b^2-4 a c\right ) \left (3 b^2-2 a c\right )}{a^3}-\frac {\left (b^2-4 a c\right ) \left (3 b^2-2 a c\right ) \log \left (a+b x^2+c x^4\right )}{2 a^3}+\frac {b \left (3 b^2-11 a c\right )}{a^2 x^2}+\frac {b \left (30 a^2 c^2-20 a b^2 c+3 b^4\right ) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{a^3 \sqrt {b^2-4 a c}}-\frac {3 b^2-8 a c}{2 a x^4}}{a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c x^2}{a x^4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\)

Input:

Int[1/(x^3*(a*x + b*x^3 + c*x^5)^2),x]
 

Output:

((b^2 - 2*a*c + b*c*x^2)/(a*(b^2 - 4*a*c)*x^4*(a + b*x^2 + c*x^4)) + (-1/2 
*(3*b^2 - 8*a*c)/(a*x^4) + (b*(3*b^2 - 11*a*c))/(a^2*x^2) + (b*(3*b^4 - 20 
*a*b^2*c + 30*a^2*c^2)*ArcTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(a^3*Sqrt 
[b^2 - 4*a*c]) + ((b^2 - 4*a*c)*(3*b^2 - 2*a*c)*Log[x^2])/a^3 - ((b^2 - 4* 
a*c)*(3*b^2 - 2*a*c)*Log[a + b*x^2 + c*x^4])/(2*a^3))/(a*(b^2 - 4*a*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 25

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

rule 1165

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(d + e*x)^(m + 1)*(b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e) 
*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^ 
2))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2))   Int[(d 
+ e*x)^m*Simp[b*c*d*e*(2*p - m + 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p 
+ 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x, x]*(a + 
 b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && LtQ[p, -1] 
 && IntQuadraticQ[a, b, c, d, e, m, p, x]
 

rule 1200

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* 
(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g* 
x)^n/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && In 
tegersQ[n]
 

rule 1434

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

rule 2009

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

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.20

method	result	size
default	\(\frac {\frac {\frac {a c b \left (3 a c -b^{2}\right ) x^{2}}{4 a c -b^{2}}-\frac {a \left (2 a^{2} c^{2}-4 a \,b^{2} c +b^{4}\right )}{4 a c -b^{2}}}{c \,x^{4}+b \,x^{2}+a}+\frac {\frac {\left (8 a^{2} c^{3}-14 a \,b^{2} c^{2}+3 b^{4} c \right ) \ln \left (c \,x^{4}+b \,x^{2}+a \right )}{2 c}+\frac {2 \left (19 a^{2} b \,c^{2}-17 a \,b^{3} c +3 b^{5}-\frac {\left (8 a^{2} c^{3}-14 a \,b^{2} c^{2}+3 b^{4} c \right ) b}{2 c}\right ) \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{4 a c -b^{2}}}{2 a^{4}}-\frac {1}{4 a^{2} x^{4}}+\frac {\left (-2 a c +3 b^{2}\right ) \ln \left (x \right )}{a^{4}}+\frac {b}{a^{3} x^{2}}\)	\(263\)
risch	\(\frac {\frac {b c \left (11 a c -3 b^{2}\right ) x^{6}}{2 a^{3} \left (4 a c -b^{2}\right )}-\frac {\left (8 a^{2} c^{2}-25 a \,b^{2} c +6 b^{4}\right ) x^{4}}{4 a^{3} \left (4 a c -b^{2}\right )}+\frac {3 b \,x^{2}}{4 a^{2}}-\frac {1}{4 a}}{x^{4} \left (c \,x^{4}+b \,x^{2}+a \right )}-\frac {2 \ln \left (x \right ) c}{a^{3}}+\frac {3 \ln \left (x \right ) b^{2}}{a^{4}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (64 a^{7} c^{3}-48 a^{6} b^{2} c^{2}+12 a^{5} b^{4} c -a^{4} b^{6}\right ) \textit {\_Z}^{2}+\left (-128 c^{4} a^{4}+288 a^{3} b^{2} c^{3}-168 a^{2} b^{4} c^{2}+38 a \,b^{6} c -3 b^{8}\right ) \textit {\_Z} +64 a \,c^{5}-15 b^{2} c^{4}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (160 c^{3} a^{9}-128 c^{2} b^{2} a^{8}+34 a^{7} b^{4} c -3 a^{6} b^{6}\right ) \textit {\_R}^{2}+\left (-160 c^{4} a^{6}+276 c^{3} b^{2} a^{5}-107 a^{4} b^{4} c^{2}+12 a^{3} b^{6} c \right ) \textit {\_R} +242 a^{2} b^{2} c^{4}-132 a \,b^{4} c^{3}+18 b^{6} c^{2}\right ) x^{2}+\left (-16 c^{2} b \,a^{9}+8 a^{8} b^{3} c -a^{7} b^{5}\right ) \textit {\_R}^{2}+\left (-108 c^{3} b \,a^{6}+151 a^{5} b^{3} c^{2}-55 a^{4} b^{5} c +6 a^{3} b^{7}\right ) \textit {\_R} -176 a^{3} b \,c^{4}+356 a^{2} b^{3} c^{3}-150 a \,b^{5} c^{2}+18 b^{7} c \right )\right )}{2}\)	\(459\)

Input:

int(1/x^3/(c*x^5+b*x^3+a*x)^2,x,method=_RETURNVERBOSE)
 

Output:

1/2/a^4*((a*c*b*(3*a*c-b^2)/(4*a*c-b^2)*x^2-a*(2*a^2*c^2-4*a*b^2*c+b^4)/(4 
*a*c-b^2))/(c*x^4+b*x^2+a)+1/(4*a*c-b^2)*(1/2*(8*a^2*c^3-14*a*b^2*c^2+3*b^ 
4*c)/c*ln(c*x^4+b*x^2+a)+2*(19*a^2*b*c^2-17*a*b^3*c+3*b^5-1/2*(8*a^2*c^3-1 
4*a*b^2*c^2+3*b^4*c)*b/c)/(4*a*c-b^2)^(1/2)*arctan((2*c*x^2+b)/(4*a*c-b^2) 
^(1/2))))-1/4/a^2/x^4+(-2*a*c+3*b^2)*ln(x)/a^4+1/a^3*b/x^2
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 609 vs. \(2 (205) = 410\).

Time = 0.24 (sec) , antiderivative size = 1242, normalized size of antiderivative = 5.67 \[ \int \frac {1}{x^3 \left (a x+b x^3+c x^5\right )^2} \, dx=\text {Too large to display} \] Input:

integrate(1/x^3/(c*x^5+b*x^3+a*x)^2,x, algorithm="fricas")
 

Output:

[-1/4*(a^3*b^4 - 8*a^4*b^2*c + 16*a^5*c^2 - 2*(3*a*b^5*c - 23*a^2*b^3*c^2 
+ 44*a^3*b*c^3)*x^6 - (6*a*b^6 - 49*a^2*b^4*c + 108*a^3*b^2*c^2 - 32*a^4*c 
^3)*x^4 - 3*(a^2*b^5 - 8*a^3*b^3*c + 16*a^4*b*c^2)*x^2 + ((3*b^5*c - 20*a* 
b^3*c^2 + 30*a^2*b*c^3)*x^8 + (3*b^6 - 20*a*b^4*c + 30*a^2*b^2*c^2)*x^6 + 
(3*a*b^5 - 20*a^2*b^3*c + 30*a^3*b*c^2)*x^4)*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)) + ((3*b^6*c - 26*a*b^4*c^2 + 64*a^2*b^2*c^3 - 32*a^3*c^4)*x^8 
+ (3*b^7 - 26*a*b^5*c + 64*a^2*b^3*c^2 - 32*a^3*b*c^3)*x^6 + (3*a*b^6 - 26 
*a^2*b^4*c + 64*a^3*b^2*c^2 - 32*a^4*c^3)*x^4)*log(c*x^4 + b*x^2 + a) - 4* 
((3*b^6*c - 26*a*b^4*c^2 + 64*a^2*b^2*c^3 - 32*a^3*c^4)*x^8 + (3*b^7 - 26* 
a*b^5*c + 64*a^2*b^3*c^2 - 32*a^3*b*c^3)*x^6 + (3*a*b^6 - 26*a^2*b^4*c + 6 
4*a^3*b^2*c^2 - 32*a^4*c^3)*x^4)*log(x))/((a^4*b^4*c - 8*a^5*b^2*c^2 + 16* 
a^6*c^3)*x^8 + (a^4*b^5 - 8*a^5*b^3*c + 16*a^6*b*c^2)*x^6 + (a^5*b^4 - 8*a 
^6*b^2*c + 16*a^7*c^2)*x^4), -1/4*(a^3*b^4 - 8*a^4*b^2*c + 16*a^5*c^2 - 2* 
(3*a*b^5*c - 23*a^2*b^3*c^2 + 44*a^3*b*c^3)*x^6 - (6*a*b^6 - 49*a^2*b^4*c 
+ 108*a^3*b^2*c^2 - 32*a^4*c^3)*x^4 - 3*(a^2*b^5 - 8*a^3*b^3*c + 16*a^4*b* 
c^2)*x^2 - 2*((3*b^5*c - 20*a*b^3*c^2 + 30*a^2*b*c^3)*x^8 + (3*b^6 - 20*a* 
b^4*c + 30*a^2*b^2*c^2)*x^6 + (3*a*b^5 - 20*a^2*b^3*c + 30*a^3*b*c^2)*x^4) 
*sqrt(-b^2 + 4*a*c)*arctan(-(2*c*x^2 + b)*sqrt(-b^2 + 4*a*c)/(b^2 - 4*a*c) 
) + ((3*b^6*c - 26*a*b^4*c^2 + 64*a^2*b^2*c^3 - 32*a^3*c^4)*x^8 + (3*b^...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{x^3 \left (a x+b x^3+c x^5\right )^2} \, dx=\text {Timed out} \] Input:

integrate(1/x**3/(c*x**5+b*x**3+a*x)**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {1}{x^3 \left (a x+b x^3+c x^5\right )^2} \, dx=\int { \frac {1}{{\left (c x^{5} + b x^{3} + a x\right )}^{2} x^{3}} \,d x } \] Input:

integrate(1/x^3/(c*x^5+b*x^3+a*x)^2,x, algorithm="maxima")
 

Output:

1/4*(2*(3*b^3*c - 11*a*b*c^2)*x^6 + (6*b^4 - 25*a*b^2*c + 8*a^2*c^2)*x^4 - 
 a^2*b^2 + 4*a^3*c + 3*(a*b^3 - 4*a^2*b*c)*x^2)/((a^3*b^2*c - 4*a^4*c^2)*x 
^8 + (a^3*b^3 - 4*a^4*b*c)*x^6 + (a^4*b^2 - 4*a^5*c)*x^4) - integrate(((3* 
b^4*c - 14*a*b^2*c^2 + 8*a^2*c^3)*x^3 + (3*b^5 - 17*a*b^3*c + 19*a^2*b*c^2 
)*x)/(c*x^4 + b*x^2 + a), x)/(a^4*b^2 - 4*a^5*c) + (3*b^2 - 2*a*c)*log(x)/ 
a^4
 

Giac [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.25 \[ \int \frac {1}{x^3 \left (a x+b x^3+c x^5\right )^2} \, dx=-\frac {{\left (3 \, b^{5} - 20 \, a b^{3} c + 30 \, a^{2} b c^{2}\right )} \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{2 \, {\left (a^{4} b^{2} - 4 \, a^{5} c\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {3 \, b^{4} c x^{4} - 14 \, a b^{2} c^{2} x^{4} + 8 \, a^{2} c^{3} x^{4} + 3 \, b^{5} x^{2} - 12 \, a b^{3} c x^{2} + 2 \, a^{2} b c^{2} x^{2} + 5 \, a b^{4} - 22 \, a^{2} b^{2} c + 12 \, a^{3} c^{2}}{4 \, {\left (a^{4} b^{2} - 4 \, a^{5} c\right )} {\left (c x^{4} + b x^{2} + a\right )}} - \frac {{\left (3 \, b^{2} - 2 \, a c\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, a^{4}} + \frac {{\left (3 \, b^{2} - 2 \, a c\right )} \log \left (x^{2}\right )}{2 \, a^{4}} - \frac {9 \, b^{2} x^{4} - 6 \, a c x^{4} - 4 \, a b x^{2} + a^{2}}{4 \, a^{4} x^{4}} \] Input:

integrate(1/x^3/(c*x^5+b*x^3+a*x)^2,x, algorithm="giac")
 

Output:

-1/2*(3*b^5 - 20*a*b^3*c + 30*a^2*b*c^2)*arctan((2*c*x^2 + b)/sqrt(-b^2 + 
4*a*c))/((a^4*b^2 - 4*a^5*c)*sqrt(-b^2 + 4*a*c)) + 1/4*(3*b^4*c*x^4 - 14*a 
*b^2*c^2*x^4 + 8*a^2*c^3*x^4 + 3*b^5*x^2 - 12*a*b^3*c*x^2 + 2*a^2*b*c^2*x^ 
2 + 5*a*b^4 - 22*a^2*b^2*c + 12*a^3*c^2)/((a^4*b^2 - 4*a^5*c)*(c*x^4 + b*x 
^2 + a)) - 1/4*(3*b^2 - 2*a*c)*log(c*x^4 + b*x^2 + a)/a^4 + 1/2*(3*b^2 - 2 
*a*c)*log(x^2)/a^4 - 1/4*(9*b^2*x^4 - 6*a*c*x^4 - 4*a*b*x^2 + a^2)/(a^4*x^ 
4)
 

Mupad [B] (verification not implemented)

Time = 15.83 (sec) , antiderivative size = 5999, normalized size of antiderivative = 27.39 \[ \int \frac {1}{x^3 \left (a x+b x^3+c x^5\right )^2} \, dx=\text {Too large to display} \] Input:

int(1/(x^3*(a*x + b*x^3 + c*x^5)^2),x)
 

Output:

(b*atan((x^2*((((((b*((2240*a^10*b*c^7 - 6*a^6*b^9*c^3 + 40*a^7*b^7*c^4 + 
108*a^8*b^5*c^5 - 1248*a^9*b^3*c^6)/(a^9*b^6 - 64*a^12*c^3 - 12*a^10*b^4*c 
 + 48*a^11*b^2*c^2) - ((2560*a^13*b*c^6 + 12*a^9*b^9*c^2 - 184*a^10*b^7*c^ 
3 + 1056*a^11*b^5*c^4 - 2688*a^12*b^3*c^5)*(6*b^8 + 256*a^4*c^4 + 336*a^2* 
b^4*c^2 - 576*a^3*b^2*c^3 - 76*a*b^6*c))/(2*(a^9*b^6 - 64*a^12*c^3 - 12*a^ 
10*b^4*c + 48*a^11*b^2*c^2)*(4*a^4*b^6 - 256*a^7*c^3 - 48*a^5*b^4*c + 192* 
a^6*b^2*c^2)))*(3*b^4 + 30*a^2*c^2 - 20*a*b^2*c))/(4*a^4*(4*a*c - b^2)^(3/ 
2)) - (b*(3*b^4 + 30*a^2*c^2 - 20*a*b^2*c)*(2560*a^13*b*c^6 + 12*a^9*b^9*c 
^2 - 184*a^10*b^7*c^3 + 1056*a^11*b^5*c^4 - 2688*a^12*b^3*c^5)*(6*b^8 + 25 
6*a^4*c^4 + 336*a^2*b^4*c^2 - 576*a^3*b^2*c^3 - 76*a*b^6*c))/(8*a^4*(4*a*c 
 - b^2)^(3/2)*(a^9*b^6 - 64*a^12*c^3 - 12*a^10*b^4*c + 48*a^11*b^2*c^2)*(4 
*a^4*b^6 - 256*a^7*c^3 - 48*a^5*b^4*c + 192*a^6*b^2*c^2)))*(6*b^8 + 256*a^ 
4*c^4 + 336*a^2*b^4*c^2 - 576*a^3*b^2*c^3 - 76*a*b^6*c))/(2*(4*a^4*b^6 - 2 
56*a^7*c^3 - 48*a^5*b^4*c + 192*a^6*b^2*c^2)) + (b*((1760*a^7*b*c^8 + 54*a 
^3*b^9*c^4 - 657*a^4*b^7*c^5 + 2775*a^5*b^5*c^6 - 4484*a^6*b^3*c^7)/(a^9*b 
^6 - 64*a^12*c^3 - 12*a^10*b^4*c + 48*a^11*b^2*c^2) + (((2240*a^10*b*c^7 - 
 6*a^6*b^9*c^3 + 40*a^7*b^7*c^4 + 108*a^8*b^5*c^5 - 1248*a^9*b^3*c^6)/(a^9 
*b^6 - 64*a^12*c^3 - 12*a^10*b^4*c + 48*a^11*b^2*c^2) - ((2560*a^13*b*c^6 
+ 12*a^9*b^9*c^2 - 184*a^10*b^7*c^3 + 1056*a^11*b^5*c^4 - 2688*a^12*b^3*c^ 
5)*(6*b^8 + 256*a^4*c^4 + 336*a^2*b^4*c^2 - 576*a^3*b^2*c^3 - 76*a*b^6*...
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 2329, normalized size of antiderivative = 10.63 \[ \int \frac {1}{x^3 \left (a x+b x^3+c x^5\right )^2} \, dx =\text {Too large to display} \] Input:

int(1/x^3/(c*x^5+b*x^3+a*x)^2,x)
 

Output:

( - 60*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**3*b* 
c**2*x**4 + 40*sqrt(2*sqrt(c)*sqrt(a) + b)*sqrt(2*sqrt(c)*sqrt(a) - b)*ata 
n((sqrt(2*sqrt(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b)) 
*a**2*b**3*c*x**4 - 60*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**2*b**2*c**2*x**6 - 60*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*sq 
rt(c)*sqrt(a) + b))*a**2*b*c**3*x**8 - 6*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)/sq 
rt(2*sqrt(c)*sqrt(a) + b))*a*b**5*x**4 + 40*sqrt(2*sqrt(c)*sqrt(a) + b)*sq 
rt(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**4*c*x**6 + 40*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**3*c**2*x**8 - 6*sqrt(2*sqrt(c)*sqr 
t(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))*b**6*x**6 - 6*sqrt(2*sqrt(c)*sqr 
t(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))*b**5*c*x**8 - 60*sqrt(2*sqrt(c)* 
sqrt(a) + b)*sqrt(2*sqrt(c)*sqrt(a) - b)*atan((sqrt(2*sqrt(c)*sqrt(a) -...