\(\int \frac {1}{x^5 (d+e x^4) (a+b x^4+c x^8)} \, dx\) [77]

Optimal result

Integrand size = 27, antiderivative size = 205 \[ \int \frac {1}{x^5 \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=-\frac {1}{4 a d x^4}-\frac {\left (b^2 c d-2 a c^2 d-b^3 e+3 a b c e\right ) \text {arctanh}\left (\frac {b+2 c x^4}{\sqrt {b^2-4 a c}}\right )}{4 a^2 \sqrt {b^2-4 a c} \left (c d^2-b d e+a e^2\right )}-\frac {(b d+a e) \log (x)}{a^2 d^2}+\frac {e^3 \log \left (d+e x^4\right )}{4 d^2 \left (c d^2-b d e+a e^2\right )}+\frac {\left (b c d-b^2 e+a c e\right ) \log \left (a+b x^4+c x^8\right )}{8 a^2 \left (c d^2-b d e+a e^2\right )} \] Output:

-1/4/a/d/x^4-1/4*(3*a*b*c*e-2*a*c^2*d-b^3*e+b^2*c*d)*arctanh((2*c*x^4+b)/( 
-4*a*c+b^2)^(1/2))/a^2/(-4*a*c+b^2)^(1/2)/(a*e^2-b*d*e+c*d^2)-(a*e+b*d)*ln 
(x)/a^2/d^2+1/4*e^3*ln(e*x^4+d)/d^2/(a*e^2-b*d*e+c*d^2)+1/8*(a*c*e-b^2*e+b 
*c*d)*ln(c*x^8+b*x^4+a)/a^2/(a*e^2-b*d*e+c*d^2)
 

Mathematica [C] (verified)

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

Time = 0.19 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.09 \[ \int \frac {1}{x^5 \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\frac {1}{4} \left (-\frac {1}{a d x^4}-\frac {4 (b d+a e) \log (x)}{a^2 d^2}+\frac {e^3 \log \left (d+e x^4\right )}{c d^4+d^2 e (-b d+a e)}-\frac {\text {RootSum}\left [a+b \text {$\#$1}^4+c \text {$\#$1}^8\&,\frac {-b^2 c d \log (x-\text {$\#$1})+a c^2 d \log (x-\text {$\#$1})+b^3 e \log (x-\text {$\#$1})-2 a b c e \log (x-\text {$\#$1})-b c^2 d \log (x-\text {$\#$1}) \text {$\#$1}^4+b^2 c e \log (x-\text {$\#$1}) \text {$\#$1}^4-a c^2 e \log (x-\text {$\#$1}) \text {$\#$1}^4}{b+2 c \text {$\#$1}^4}\&\right ]}{a^2 \left (c d^2+e (-b d+a e)\right )}\right ) \] Input:

Integrate[1/(x^5*(d + e*x^4)*(a + b*x^4 + c*x^8)),x]
 

Output:

(-(1/(a*d*x^4)) - (4*(b*d + a*e)*Log[x])/(a^2*d^2) + (e^3*Log[d + e*x^4])/ 
(c*d^4 + d^2*e*(-(b*d) + a*e)) - RootSum[a + b*#1^4 + c*#1^8 & , (-(b^2*c* 
d*Log[x - #1]) + a*c^2*d*Log[x - #1] + b^3*e*Log[x - #1] - 2*a*b*c*e*Log[x 
 - #1] - b*c^2*d*Log[x - #1]*#1^4 + b^2*c*e*Log[x - #1]*#1^4 - a*c^2*e*Log 
[x - #1]*#1^4)/(b + 2*c*#1^4) & ]/(a^2*(c*d^2 + e*(-(b*d) + a*e))))/4
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1802, 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.

\(\displaystyle \int \frac {1}{x^5 \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx\)

\(\Big \downarrow \) 1802

\(\displaystyle \frac {1}{4} \int \frac {1}{x^8 \left (e x^4+d\right ) \left (c x^8+b x^4+a\right )}dx^4\)

\(\Big \downarrow \) 1200

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

\(\Big \downarrow \) 2009

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

Input:

Int[1/(x^5*(d + e*x^4)*(a + b*x^4 + c*x^8)),x]
 

Output:

(-(1/(a*d*x^4)) - ((b^2*c*d - 2*a*c^2*d - b^3*e + 3*a*b*c*e)*ArcTanh[(b + 
2*c*x^4)/Sqrt[b^2 - 4*a*c]])/(a^2*Sqrt[b^2 - 4*a*c]*(c*d^2 - b*d*e + a*e^2 
)) - ((b*d + a*e)*Log[x^4])/(a^2*d^2) + (e^3*Log[d + e*x^4])/(d^2*(c*d^2 - 
 b*d*e + a*e^2)) + ((b*c*d - b^2*e + a*c*e)*Log[a + b*x^4 + c*x^8])/(2*a^2 
*(c*d^2 - b*d*e + a*e^2)))/4
 

Defintions of rubi rules used

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 1802

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + ( 
e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1 
)/n] - 1)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, 
c, d, e, m, n, p, q}, x] && EqQ[n2, 2*n] && IntegerQ[Simplify[(m + 1)/n]]
 

rule 2009

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

Maple [A] (verified)

Time = 1.81 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.04

method	result	size
default	\(\frac {\frac {\left (a \,c^{2} e -b^{2} c e +b \,c^{2} d \right ) \ln \left (c \,x^{8}+b \,x^{4}+a \right )}{4 c}+\frac {\left (2 a b c e -a \,c^{2} d -b^{3} e +b^{2} c d -\frac {\left (a \,c^{2} e -b^{2} c e +b \,c^{2} d \right ) b}{2 c}\right ) \arctan \left (\frac {2 c \,x^{4}+b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{2 a^{2} \left (a \,e^{2}-b d e +c \,d^{2}\right )}+\frac {e^{3} \ln \left (x^{4} e +d \right )}{4 d^{2} \left (a \,e^{2}-b d e +c \,d^{2}\right )}-\frac {1}{4 a d \,x^{4}}+\frac {\left (-a e -b d \right ) \ln \left (x \right )}{a^{2} d^{2}}\)	\(214\)
risch	\(-\frac {1}{4 a d \,x^{4}}-\frac {\ln \left (x \right ) e}{a \,d^{2}}-\frac {\ln \left (x \right ) b}{a^{2} d}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (4 a^{4} c \,e^{2}-a^{3} b^{2} e^{2}-4 a^{3} b c d e +4 a^{3} c^{2} d^{2}+a^{2} b^{3} d e -a^{2} b^{2} c \,d^{2}\right ) \textit {\_Z}^{2}+\left (-4 a^{2} c^{2} e +5 a \,b^{2} c e -4 d \,c^{2} b a -b^{4} e +d c \,b^{3}\right ) \textit {\_Z} +c^{3}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (-20 a^{5} c \,d^{2} e^{4}+5 a^{4} b^{2} d^{2} e^{4}+32 a^{4} b c \,d^{3} e^{3}-30 a^{4} c^{2} d^{4} e^{2}-8 a^{3} b^{3} d^{3} e^{3}-12 a^{3} b^{2} c \,d^{4} e^{2}+38 a^{3} b \,c^{2} d^{5} e -18 a^{3} c^{3} d^{6}+5 a^{2} b^{4} d^{4} e^{2}-10 a^{2} b^{3} c \,d^{5} e +5 a^{2} b^{2} c^{2} d^{6}\right ) \textit {\_R}^{3}+\left (-32 a^{3} b c d \,e^{4}+42 a^{3} c^{2} d^{2} e^{3}+8 a^{2} b^{3} d \,e^{4}-41 a^{2} b^{2} c \,d^{2} e^{3}+32 a^{2} b \,c^{2} d^{3} e^{2}+7 a^{2} c^{3} d^{4} e +8 a \,b^{4} d^{2} e^{3}-8 a \,b^{3} c \,d^{3} e^{2}-8 a \,b^{2} c^{2} d^{4} e +8 a b \,c^{3} d^{5}\right ) \textit {\_R}^{2}+\left (-8 a^{2} c^{2} e^{4}+16 a \,b^{2} c \,e^{4}-8 a b \,c^{2} d \,e^{3}-8 a \,c^{3} d^{2} e^{2}-4 b^{4} e^{4}-4 c^{4} d^{4}\right ) \textit {\_R} +4 c^{3} e^{3}\right ) x^{4}+\left (-4 a^{5} c \,d^{3} e^{3}+a^{4} b^{2} d^{3} e^{3}-3 a^{4} b c \,d^{4} e^{2}+4 a^{4} c^{2} d^{5} e +a^{3} b^{3} d^{4} e^{2}-2 a^{3} b^{2} c \,d^{5} e +a^{3} b \,c^{2} d^{6}\right ) \textit {\_R}^{3}+\left (-16 a^{4} c d \,e^{4}+4 a^{3} b^{2} d \,e^{4}-25 a^{3} b c \,d^{2} e^{3}+16 a^{3} c^{2} d^{3} e^{2}+7 a^{2} b^{3} d^{2} e^{3}-20 a^{2} b^{2} c \,d^{3} e^{2}+20 a^{2} b \,c^{2} d^{4} e -a^{2} c^{3} d^{5}+4 a \,b^{4} d^{3} e^{2}-8 a \,b^{3} c \,d^{4} e +4 a \,b^{2} c^{2} d^{5}\right ) \textit {\_R}^{2}+\left (12 a^{2} b c \,e^{4}-4 a \,b^{3} e^{4}+12 a \,b^{2} c d \,e^{3}-4 a \,c^{3} d^{3} e -4 b^{4} d \,e^{3}-4 b \,c^{3} d^{4}\right ) \textit {\_R} \right )\right )}{4}+\frac {e^{3} \ln \left (-x^{4} e -d \right )}{4 d^{2} \left (a \,e^{2}-b d e +c \,d^{2}\right )}\)	\(847\)

Input:

int(1/x^5/(e*x^4+d)/(c*x^8+b*x^4+a),x,method=_RETURNVERBOSE)
 

Output:

1/2/a^2/(a*e^2-b*d*e+c*d^2)*(1/4*(a*c^2*e-b^2*c*e+b*c^2*d)/c*ln(c*x^8+b*x^ 
4+a)+(2*a*b*c*e-a*c^2*d-b^3*e+b^2*c*d-1/2*(a*c^2*e-b^2*c*e+b*c^2*d)*b/c)/( 
4*a*c-b^2)^(1/2)*arctan((2*c*x^4+b)/(4*a*c-b^2)^(1/2)))+1/4*e^3*ln(e*x^4+d 
)/d^2/(a*e^2-b*d*e+c*d^2)-1/4/a/d/x^4+1/a^2/d^2*(-a*e-b*d)*ln(x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{x^5 \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\text {Timed out} \] Input:

integrate(1/x^5/(e*x^4+d)/(c*x^8+b*x^4+a),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{x^5 \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\text {Timed out} \] Input:

integrate(1/x**5/(e*x**4+d)/(c*x**8+b*x**4+a),x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{x^5 \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/x^5/(e*x^4+d)/(c*x^8+b*x^4+a),x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for 
 more deta
 

Giac [A] (verification not implemented)

Time = 1.04 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.13 \[ \int \frac {1}{x^5 \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\frac {e^{4} \log \left ({\left | e x^{4} + d \right |}\right )}{4 \, {\left (c d^{4} e - b d^{3} e^{2} + a d^{2} e^{3}\right )}} + \frac {{\left (b c d - b^{2} e + a c e\right )} \log \left (c x^{8} + b x^{4} + a\right )}{8 \, {\left (a^{2} c d^{2} - a^{2} b d e + a^{3} e^{2}\right )}} + \frac {{\left (b^{2} c d - 2 \, a c^{2} d - b^{3} e + 3 \, a b c e\right )} \arctan \left (\frac {2 \, c x^{4} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{4 \, {\left (a^{2} c d^{2} - a^{2} b d e + a^{3} e^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {{\left (b d + a e\right )} \log \left (x^{4}\right )}{4 \, a^{2} d^{2}} + \frac {b d x^{4} + a e x^{4} - a d}{4 \, a^{2} d^{2} x^{4}} \] Input:

integrate(1/x^5/(e*x^4+d)/(c*x^8+b*x^4+a),x, algorithm="giac")
 

Output:

1/4*e^4*log(abs(e*x^4 + d))/(c*d^4*e - b*d^3*e^2 + a*d^2*e^3) + 1/8*(b*c*d 
 - b^2*e + a*c*e)*log(c*x^8 + b*x^4 + a)/(a^2*c*d^2 - a^2*b*d*e + a^3*e^2) 
 + 1/4*(b^2*c*d - 2*a*c^2*d - b^3*e + 3*a*b*c*e)*arctan((2*c*x^4 + b)/sqrt 
(-b^2 + 4*a*c))/((a^2*c*d^2 - a^2*b*d*e + a^3*e^2)*sqrt(-b^2 + 4*a*c)) - 1 
/4*(b*d + a*e)*log(x^4)/(a^2*d^2) + 1/4*(b*d*x^4 + a*e*x^4 - a*d)/(a^2*d^2 
*x^4)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^5 \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\text {Hanged} \] Input:

int(1/(x^5*(d + e*x^4)*(a + b*x^4 + c*x^8)),x)
 

Output:

\text{Hanged}
 

Reduce [F]

\[ \int \frac {1}{x^5 \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\int \frac {1}{x^{5} \left (e \,x^{4}+d \right ) \left (c \,x^{8}+b \,x^{4}+a \right )}d x \] Input:

int(1/x^5/(e*x^4+d)/(c*x^8+b*x^4+a),x)
                                                                                    
                                                                                    
 

Output:

int(1/x^5/(e*x^4+d)/(c*x^8+b*x^4+a),x)