\(\int \frac {1}{x^2 (a+c x^4)^2} \, dx\) [84]

Optimal result

Integrand size = 13, antiderivative size = 162 \[ \int \frac {1}{x^2 \left (a+c x^4\right )^2} \, dx=-\frac {1}{a^2 x}-\frac {c x^3}{4 a^2 \left (a+c x^4\right )}+\frac {5 \sqrt [4]{c} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{9/4}}-\frac {5 \sqrt [4]{c} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{9/4}}+\frac {5 \sqrt [4]{c} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x}{\sqrt {a}+\sqrt {c} x^2}\right )}{8 \sqrt {2} a^{9/4}} \] Output:

-1/a^2/x-1/4*c*x^3/a^2/(c*x^4+a)-5/16*c^(1/4)*arctan(-1+2^(1/2)*c^(1/4)*x/ 
a^(1/4))*2^(1/2)/a^(9/4)-5/16*c^(1/4)*arctan(1+2^(1/2)*c^(1/4)*x/a^(1/4))* 
2^(1/2)/a^(9/4)+5/16*c^(1/4)*arctanh(2^(1/2)*a^(1/4)*c^(1/4)*x/(a^(1/2)+c^ 
(1/2)*x^2))*2^(1/2)/a^(9/4)
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.21 \[ \int \frac {1}{x^2 \left (a+c x^4\right )^2} \, dx=\frac {-\frac {32 \sqrt [4]{a}}{x}-\frac {8 \sqrt [4]{a} c x^3}{a+c x^4}+10 \sqrt {2} \sqrt [4]{c} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )-10 \sqrt {2} \sqrt [4]{c} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )-5 \sqrt {2} \sqrt [4]{c} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )+5 \sqrt {2} \sqrt [4]{c} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {c} x^2\right )}{32 a^{9/4}} \] Input:

Integrate[1/(x^2*(a + c*x^4)^2),x]
 

Output:

((-32*a^(1/4))/x - (8*a^(1/4)*c*x^3)/(a + c*x^4) + 10*Sqrt[2]*c^(1/4)*ArcT 
an[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)] - 10*Sqrt[2]*c^(1/4)*ArcTan[1 + (Sqrt[ 
2]*c^(1/4)*x)/a^(1/4)] - 5*Sqrt[2]*c^(1/4)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c 
^(1/4)*x + Sqrt[c]*x^2] + 5*Sqrt[2]*c^(1/4)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)* 
c^(1/4)*x + Sqrt[c]*x^2])/(32*a^(9/4))
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.49, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.769, Rules used = {819, 847, 826, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

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

\(\Big \downarrow \) 819

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

\(\Big \downarrow \) 847

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

\(\displaystyle \frac {5 \left (-\frac {c \left (\frac {\frac {\int \frac {1}{-\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )^2-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{c}}-\frac {\int \frac {1}{-\left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )^2-1}d\left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\int \frac {\sqrt {a}-\sqrt {c} x^2}{c x^4+a}dx}{2 \sqrt {c}}\right )}{a}-\frac {1}{a x}\right )}{4 a}+\frac {1}{4 a x \left (a+c x^4\right )}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {5 \left (-\frac {c \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\int \frac {\sqrt {a}-\sqrt {c} x^2}{c x^4+a}dx}{2 \sqrt {c}}\right )}{a}-\frac {1}{a x}\right )}{4 a}+\frac {1}{4 a x \left (a+c x^4\right )}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {5 \left (-\frac {c \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {-\frac {\int -\frac {\sqrt {2} \sqrt [4]{a}-2 \sqrt [4]{c} x}{\sqrt [4]{c} \left (x^2-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}+\frac {\sqrt {a}}{\sqrt {c}}\right )}dx}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{c}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} x+\sqrt [4]{a}\right )}{\sqrt [4]{c} \left (x^2+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}+\frac {\sqrt {a}}{\sqrt {c}}\right )}dx}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{c}}}{2 \sqrt {c}}\right )}{a}-\frac {1}{a x}\right )}{4 a}+\frac {1}{4 a x \left (a+c x^4\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {5 \left (-\frac {c \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{a}-2 \sqrt [4]{c} x}{\sqrt [4]{c} \left (x^2-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}+\frac {\sqrt {a}}{\sqrt {c}}\right )}dx}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{c}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} x+\sqrt [4]{a}\right )}{\sqrt [4]{c} \left (x^2+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}+\frac {\sqrt {a}}{\sqrt {c}}\right )}dx}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{c}}}{2 \sqrt {c}}\right )}{a}-\frac {1}{a x}\right )}{4 a}+\frac {1}{4 a x \left (a+c x^4\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {5 \left (-\frac {c \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{a}-2 \sqrt [4]{c} x}{x^2-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}+\frac {\sqrt {a}}{\sqrt {c}}}dx}{2 \sqrt {2} \sqrt [4]{a} \sqrt {c}}+\frac {\int \frac {\sqrt {2} \sqrt [4]{c} x+\sqrt [4]{a}}{x^2+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}+\frac {\sqrt {a}}{\sqrt {c}}}dx}{2 \sqrt [4]{a} \sqrt {c}}}{2 \sqrt {c}}\right )}{a}-\frac {1}{a x}\right )}{4 a}+\frac {1}{4 a x \left (a+c x^4\right )}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {5 \left (-\frac {c \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\frac {\log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{c}}-\frac {\log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{c}}}{2 \sqrt {c}}\right )}{a}-\frac {1}{a x}\right )}{4 a}+\frac {1}{4 a x \left (a+c x^4\right )}\)

Input:

Int[1/(x^2*(a + c*x^4)^2),x]
 

Output:

1/(4*a*x*(a + c*x^4)) + (5*(-(1/(a*x)) - (c*((-(ArcTan[1 - (Sqrt[2]*c^(1/4 
)*x)/a^(1/4)]/(Sqrt[2]*a^(1/4)*c^(1/4))) + ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/ 
a^(1/4)]/(Sqrt[2]*a^(1/4)*c^(1/4)))/(2*Sqrt[c]) - (-1/2*Log[Sqrt[a] - Sqrt 
[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2]/(Sqrt[2]*a^(1/4)*c^(1/4)) + Log[Sqrt[ 
a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2]/(2*Sqrt[2]*a^(1/4)*c^(1/4))) 
/(2*Sqrt[c])))/a))/(4*a)
 

Defintions of rubi rules used

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 217

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

rule 819

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-( 
c*x)^(m + 1))*((a + b*x^n)^(p + 1)/(a*c*n*(p + 1))), x] + Simp[(m + n*(p + 
1) + 1)/(a*n*(p + 1))   Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a 
, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p 
, x]
 

rule 826

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 
2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s)   Int[(r + s*x^2)/(a + b*x^ 
4), x], x] - Simp[1/(2*s)   Int[(r - s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{ 
a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] 
 && AtomQ[SplitProduct[SumBaseQ, b]]))
 

rule 847

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x 
)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + n*(p + 1) 
+ 1)/(a*c^n*(m + 1)))   Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a 
, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p 
, x]
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.48 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.43

method	result	size
risch	\(\frac {-\frac {5 c \,x^{4}}{4 a^{2}}-\frac {1}{a}}{x \left (c \,x^{4}+a \right )}+\frac {5 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{9} \textit {\_Z}^{4}+c \right )}{\sum }\textit {\_R} \ln \left (\left (5 \textit {\_R}^{4} a^{9}+4 c \right ) x +a^{7} \textit {\_R}^{3}\right )\right )}{16}\)	\(70\)
default	\(-\frac {1}{a^{2} x}-\frac {c \left (\frac {x^{3}}{4 c \,x^{4}+4 a}+\frac {5 \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}{x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{32 c \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{a^{2}}\)	\(132\)

Input:

int(1/x^2/(c*x^4+a)^2,x,method=_RETURNVERBOSE)
 

Output:

(-5/4/a^2*c*x^4-1/a)/x/(c*x^4+a)+5/16*sum(_R*ln((5*_R^4*a^9+4*c)*x+a^7*_R^ 
3),_R=RootOf(_Z^4*a^9+c))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.26 \[ \int \frac {1}{x^2 \left (a+c x^4\right )^2} \, dx=-\frac {20 \, c x^{4} + 5 \, {\left (a^{2} c x^{5} + a^{3} x\right )} \left (-\frac {c}{a^{9}}\right )^{\frac {1}{4}} \log \left (125 \, a^{7} \left (-\frac {c}{a^{9}}\right )^{\frac {3}{4}} + 125 \, c x\right ) + 5 \, {\left (-i \, a^{2} c x^{5} - i \, a^{3} x\right )} \left (-\frac {c}{a^{9}}\right )^{\frac {1}{4}} \log \left (125 i \, a^{7} \left (-\frac {c}{a^{9}}\right )^{\frac {3}{4}} + 125 \, c x\right ) + 5 \, {\left (i \, a^{2} c x^{5} + i \, a^{3} x\right )} \left (-\frac {c}{a^{9}}\right )^{\frac {1}{4}} \log \left (-125 i \, a^{7} \left (-\frac {c}{a^{9}}\right )^{\frac {3}{4}} + 125 \, c x\right ) - 5 \, {\left (a^{2} c x^{5} + a^{3} x\right )} \left (-\frac {c}{a^{9}}\right )^{\frac {1}{4}} \log \left (-125 \, a^{7} \left (-\frac {c}{a^{9}}\right )^{\frac {3}{4}} + 125 \, c x\right ) + 16 \, a}{16 \, {\left (a^{2} c x^{5} + a^{3} x\right )}} \] Input:

integrate(1/x^2/(c*x^4+a)^2,x, algorithm="fricas")
 

Output:

-1/16*(20*c*x^4 + 5*(a^2*c*x^5 + a^3*x)*(-c/a^9)^(1/4)*log(125*a^7*(-c/a^9 
)^(3/4) + 125*c*x) + 5*(-I*a^2*c*x^5 - I*a^3*x)*(-c/a^9)^(1/4)*log(125*I*a 
^7*(-c/a^9)^(3/4) + 125*c*x) + 5*(I*a^2*c*x^5 + I*a^3*x)*(-c/a^9)^(1/4)*lo 
g(-125*I*a^7*(-c/a^9)^(3/4) + 125*c*x) - 5*(a^2*c*x^5 + a^3*x)*(-c/a^9)^(1 
/4)*log(-125*a^7*(-c/a^9)^(3/4) + 125*c*x) + 16*a)/(a^2*c*x^5 + a^3*x)
 

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.35 \[ \int \frac {1}{x^2 \left (a+c x^4\right )^2} \, dx=\frac {- 4 a - 5 c x^{4}}{4 a^{3} x + 4 a^{2} c x^{5}} + \operatorname {RootSum} {\left (65536 t^{4} a^{9} + 625 c, \left ( t \mapsto t \log {\left (- \frac {4096 t^{3} a^{7}}{125 c} + x \right )} \right )\right )} \] Input:

integrate(1/x**2/(c*x**4+a)**2,x)
                                                                                    
                                                                                    
 

Output:

(-4*a - 5*c*x**4)/(4*a**3*x + 4*a**2*c*x**5) + RootSum(65536*_t**4*a**9 + 
625*c, Lambda(_t, _t*log(-4096*_t**3*a**7/(125*c) + x)))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.25 \[ \int \frac {1}{x^2 \left (a+c x^4\right )^2} \, dx=-\frac {5 \, c x^{4} + 4 \, a}{4 \, {\left (a^{2} c x^{5} + a^{3} x\right )}} - \frac {5 \, c {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{\sqrt {\sqrt {a} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{\sqrt {\sqrt {a} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {c} x^{2} + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {1}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {c} x^{2} - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {1}{4}} c^{\frac {3}{4}}}\right )}}{32 \, a^{2}} \] Input:

integrate(1/x^2/(c*x^4+a)^2,x, algorithm="maxima")
 

Output:

-1/4*(5*c*x^4 + 4*a)/(a^2*c*x^5 + a^3*x) - 5/32*c*(2*sqrt(2)*arctan(1/2*sq 
rt(2)*(2*sqrt(c)*x + sqrt(2)*a^(1/4)*c^(1/4))/sqrt(sqrt(a)*sqrt(c)))/(sqrt 
(sqrt(a)*sqrt(c))*sqrt(c)) + 2*sqrt(2)*arctan(1/2*sqrt(2)*(2*sqrt(c)*x - s 
qrt(2)*a^(1/4)*c^(1/4))/sqrt(sqrt(a)*sqrt(c)))/(sqrt(sqrt(a)*sqrt(c))*sqrt 
(c)) - sqrt(2)*log(sqrt(c)*x^2 + sqrt(2)*a^(1/4)*c^(1/4)*x + sqrt(a))/(a^( 
1/4)*c^(3/4)) + sqrt(2)*log(sqrt(c)*x^2 - sqrt(2)*a^(1/4)*c^(1/4)*x + sqrt 
(a))/(a^(1/4)*c^(3/4)))/a^2
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.27 \[ \int \frac {1}{x^2 \left (a+c x^4\right )^2} \, dx=-\frac {5 \, c x^{4} + 4 \, a}{4 \, {\left (c x^{5} + a x\right )} a^{2}} - \frac {5 \, \sqrt {2} \left (a c^{3}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{16 \, a^{3} c^{2}} - \frac {5 \, \sqrt {2} \left (a c^{3}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{16 \, a^{3} c^{2}} + \frac {5 \, \sqrt {2} \left (a c^{3}\right )^{\frac {3}{4}} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{32 \, a^{3} c^{2}} - \frac {5 \, \sqrt {2} \left (a c^{3}\right )^{\frac {3}{4}} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{32 \, a^{3} c^{2}} \] Input:

integrate(1/x^2/(c*x^4+a)^2,x, algorithm="giac")
 

Output:

-1/4*(5*c*x^4 + 4*a)/((c*x^5 + a*x)*a^2) - 5/16*sqrt(2)*(a*c^3)^(3/4)*arct 
an(1/2*sqrt(2)*(2*x + sqrt(2)*(a/c)^(1/4))/(a/c)^(1/4))/(a^3*c^2) - 5/16*s 
qrt(2)*(a*c^3)^(3/4)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)*(a/c)^(1/4))/(a/c)^ 
(1/4))/(a^3*c^2) + 5/32*sqrt(2)*(a*c^3)^(3/4)*log(x^2 + sqrt(2)*x*(a/c)^(1 
/4) + sqrt(a/c))/(a^3*c^2) - 5/32*sqrt(2)*(a*c^3)^(3/4)*log(x^2 - sqrt(2)* 
x*(a/c)^(1/4) + sqrt(a/c))/(a^3*c^2)
 

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.43 \[ \int \frac {1}{x^2 \left (a+c x^4\right )^2} \, dx=\frac {5\,{\left (-c\right )}^{1/4}\,\mathrm {atanh}\left (\frac {{\left (-c\right )}^{1/4}\,x}{a^{1/4}}\right )}{8\,a^{9/4}}-\frac {5\,{\left (-c\right )}^{1/4}\,\mathrm {atan}\left (\frac {{\left (-c\right )}^{1/4}\,x}{a^{1/4}}\right )}{8\,a^{9/4}}-\frac {\frac {1}{a}+\frac {5\,c\,x^4}{4\,a^2}}{c\,x^5+a\,x} \] Input:

int(1/(x^2*(a + c*x^4)^2),x)
 

Output:

(5*(-c)^(1/4)*atanh(((-c)^(1/4)*x)/a^(1/4)))/(8*a^(9/4)) - (5*(-c)^(1/4)*a 
tan(((-c)^(1/4)*x)/a^(1/4)))/(8*a^(9/4)) - (1/a + (5*c*x^4)/(4*a^2))/(a*x 
+ c*x^5)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.95 \[ \int \frac {1}{x^2 \left (a+c x^4\right )^2} \, dx=\frac {10 c^{\frac {1}{4}} a^{\frac {7}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}-2 \sqrt {c}\, x}{c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) x +10 c^{\frac {5}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}-2 \sqrt {c}\, x}{c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) x^{5}-10 c^{\frac {1}{4}} a^{\frac {7}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+2 \sqrt {c}\, x}{c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) x -10 c^{\frac {5}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+2 \sqrt {c}\, x}{c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) x^{5}-5 c^{\frac {1}{4}} a^{\frac {7}{4}} \sqrt {2}\, \mathrm {log}\left (-c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {a}+\sqrt {c}\, x^{2}\right ) x -5 c^{\frac {5}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (-c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {a}+\sqrt {c}\, x^{2}\right ) x^{5}+5 c^{\frac {1}{4}} a^{\frac {7}{4}} \sqrt {2}\, \mathrm {log}\left (c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {a}+\sqrt {c}\, x^{2}\right ) x +5 c^{\frac {5}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (c^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {a}+\sqrt {c}\, x^{2}\right ) x^{5}-32 a^{2}-40 a c \,x^{4}}{32 a^{3} x \left (c \,x^{4}+a \right )} \] Input:

int(1/x^2/(c*x^4+a)^2,x)
 

Output:

(10*c**(1/4)*a**(3/4)*sqrt(2)*atan((c**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(c)* 
x)/(c**(1/4)*a**(1/4)*sqrt(2)))*a*x + 10*c**(1/4)*a**(3/4)*sqrt(2)*atan((c 
**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(c)*x)/(c**(1/4)*a**(1/4)*sqrt(2)))*c*x** 
5 - 10*c**(1/4)*a**(3/4)*sqrt(2)*atan((c**(1/4)*a**(1/4)*sqrt(2) + 2*sqrt( 
c)*x)/(c**(1/4)*a**(1/4)*sqrt(2)))*a*x - 10*c**(1/4)*a**(3/4)*sqrt(2)*atan 
((c**(1/4)*a**(1/4)*sqrt(2) + 2*sqrt(c)*x)/(c**(1/4)*a**(1/4)*sqrt(2)))*c* 
x**5 - 5*c**(1/4)*a**(3/4)*sqrt(2)*log( - c**(1/4)*a**(1/4)*sqrt(2)*x + sq 
rt(a) + sqrt(c)*x**2)*a*x - 5*c**(1/4)*a**(3/4)*sqrt(2)*log( - c**(1/4)*a* 
*(1/4)*sqrt(2)*x + sqrt(a) + sqrt(c)*x**2)*c*x**5 + 5*c**(1/4)*a**(3/4)*sq 
rt(2)*log(c**(1/4)*a**(1/4)*sqrt(2)*x + sqrt(a) + sqrt(c)*x**2)*a*x + 5*c* 
*(1/4)*a**(3/4)*sqrt(2)*log(c**(1/4)*a**(1/4)*sqrt(2)*x + sqrt(a) + sqrt(c 
)*x**2)*c*x**5 - 32*a**2 - 40*a*c*x**4)/(32*a**3*x*(a + c*x**4))