Integrand size = 27, antiderivative size = 167 \[ \int \frac {1}{x \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\frac {\left (b c d-b^2 e+2 a c e\right ) \text {arctanh}\left (\frac {b+2 c x^4}{\sqrt {b^2-4 a c}}\right )}{4 a \sqrt {b^2-4 a c} \left (c d^2-b d e+a e^2\right )}+\frac {\log (x)}{a d}-\frac {e^2 \log \left (d+e x^4\right )}{4 d \left (c d^2-b d e+a e^2\right )}-\frac {(c d-b e) \log \left (a+b x^4+c x^8\right )}{8 a \left (c d^2-b d e+a e^2\right )} \] Output:
1/4*(2*a*c*e-b^2*e+b*c*d)*arctanh((2*c*x^4+b)/(-4*a*c+b^2)^(1/2))/a/(-4*a* c+b^2)^(1/2)/(a*e^2-b*d*e+c*d^2)+ln(x)/a/d-1/4*e^2*ln(e*x^4+d)/d/(a*e^2-b* d*e+c*d^2)-1/8*(-b*e+c*d)*ln(c*x^8+b*x^4+a)/a/(a*e^2-b*d*e+c*d^2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.10 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.98 \[ \int \frac {1}{x \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\frac {4 \left (c d^2-b d e+a e^2\right ) \log (x)-a e^2 \log \left (d+e x^4\right )+d \text {RootSum}\left [a+b \text {$\#$1}^4+c \text {$\#$1}^8\&,\frac {-b c d \log (x-\text {$\#$1})+b^2 e \log (x-\text {$\#$1})-a c e \log (x-\text {$\#$1})-c^2 d \log (x-\text {$\#$1}) \text {$\#$1}^4+b c e \log (x-\text {$\#$1}) \text {$\#$1}^4}{b+2 c \text {$\#$1}^4}\&\right ]}{4 a d \left (c d^2+e (-b d+a e)\right )} \] Input:
Integrate[1/(x*(d + e*x^4)*(a + b*x^4 + c*x^8)),x]
Output:
(4*(c*d^2 - b*d*e + a*e^2)*Log[x] - a*e^2*Log[d + e*x^4] + d*RootSum[a + b *#1^4 + c*#1^8 & , (-(b*c*d*Log[x - #1]) + b^2*e*Log[x - #1] - a*c*e*Log[x - #1] - c^2*d*Log[x - #1]*#1^4 + b*c*e*Log[x - #1]*#1^4)/(b + 2*c*#1^4) & ])/(4*a*d*(c*d^2 + e*(-(b*d) + a*e)))
Time = 0.46 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.01, 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.
\(\displaystyle \int \frac {1}{x \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx\) |
\(\Big \downarrow \) 1802 |
\(\displaystyle \frac {1}{4} \int \frac {1}{x^4 \left (e x^4+d\right ) \left (c x^8+b x^4+a\right )}dx^4\) |
\(\Big \downarrow \) 1200 |
\(\displaystyle \frac {1}{4} \int \left (-\frac {e^3}{d \left (c d^2-b e d+a e^2\right ) \left (e x^4+d\right )}+\frac {-c (c d-b e) x^4-b c d+b^2 e-a c e}{a \left (c d^2-b e d+a e^2\right ) \left (c x^8+b x^4+a\right )}+\frac {1}{a d x^4}\right )dx^4\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (\frac {\text {arctanh}\left (\frac {b+2 c x^4}{\sqrt {b^2-4 a c}}\right ) \left (2 a c e+b^2 (-e)+b c d\right )}{a \sqrt {b^2-4 a c} \left (a e^2-b d e+c d^2\right )}-\frac {e^2 \log \left (d+e x^4\right )}{d \left (a e^2-b d e+c d^2\right )}-\frac {(c d-b e) \log \left (a+b x^4+c x^8\right )}{2 a \left (a e^2-b d e+c d^2\right )}+\frac {\log \left (x^4\right )}{a d}\right )\) |
Input:
Int[1/(x*(d + e*x^4)*(a + b*x^4 + c*x^8)),x]
Output:
(((b*c*d - b^2*e + 2*a*c*e)*ArcTanh[(b + 2*c*x^4)/Sqrt[b^2 - 4*a*c]])/(a*S qrt[b^2 - 4*a*c]*(c*d^2 - b*d*e + a*e^2)) + Log[x^4]/(a*d) - (e^2*Log[d + e*x^4])/(d*(c*d^2 - b*d*e + a*e^2)) - ((c*d - b*e)*Log[a + b*x^4 + c*x^8]) /(2*a*(c*d^2 - b*d*e + a*e^2)))/4
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]
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]]
Time = 1.05 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.99
method | result | size |
default | \(-\frac {\frac {\left (-b c e +c^{2} d \right ) \ln \left (c \,x^{8}+b \,x^{4}+a \right )}{4 c}+\frac {\left (a c e -b^{2} e +c b d -\frac {\left (-b c e +c^{2} d \right ) b}{2 c}\right ) \arctan \left (\frac {2 c \,x^{4}+b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{2 \left (a \,e^{2}-b d e +c \,d^{2}\right ) a}-\frac {e^{2} \ln \left (x^{4} e +d \right )}{4 d \left (a \,e^{2}-b d e +c \,d^{2}\right )}+\frac {\ln \left (x \right )}{a d}\) | \(165\) |
risch | \(\frac {\ln \left (x \right )}{a d}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (4 a^{3} c \,e^{2}-a^{2} b^{2} e^{2}-4 a^{2} b c d e +4 a^{2} c^{2} d^{2}+a \,b^{3} d e -a \,b^{2} c \,d^{2}\right ) \textit {\_Z}^{2}+\left (-4 a b c e +4 a \,c^{2} d +b^{3} e -b^{2} c d \right ) \textit {\_Z} +c^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (-20 a^{3} c \,e^{4}+5 a^{2} b^{2} e^{4}+32 a^{2} b c d \,e^{3}-30 a^{2} c^{2} d^{2} e^{2}-8 a \,b^{3} d \,e^{3}-12 a \,b^{2} c \,d^{2} e^{2}+38 a b \,c^{2} d^{3} e -18 a \,c^{3} d^{4}+5 b^{4} d^{2} e^{2}-10 b^{3} c \,d^{3} e +5 b^{2} c^{2} d^{4}\right ) \textit {\_R}^{2}+\left (36 a b c \,e^{3}-42 a \,c^{2} d \,e^{2}-9 b^{3} e^{3}+9 b^{2} c d \,e^{2}+9 b \,c^{2} d^{2} e -9 c^{3} d^{3}\right ) \textit {\_R} -13 c^{2} e^{2}\right ) x^{4}+\left (-4 a^{3} c d \,e^{3}+a^{2} b^{2} d \,e^{3}-3 a^{2} b c \,d^{2} e^{2}+4 a^{2} c^{2} d^{3} e +a \,b^{3} d^{2} e^{2}-2 a \,b^{2} c \,d^{3} e +a b \,c^{2} d^{4}\right ) \textit {\_R}^{2}+\left (16 a^{2} c \,e^{3}-4 a \,b^{2} e^{3}+13 a b c d \,e^{2}-15 a \,c^{2} d^{2} e -4 b^{3} d \,e^{2}+8 b^{2} c \,d^{2} e -4 b \,c^{2} d^{3}\right ) \textit {\_R} -4 b c \,e^{2}-4 c^{2} d e \right )\right )}{4}-\frac {e^{2} \ln \left (-x^{4} e -d \right )}{4 d \left (a \,e^{2}-b d e +c \,d^{2}\right )}\) | \(511\) |
Input:
int(1/x/(e*x^4+d)/(c*x^8+b*x^4+a),x,method=_RETURNVERBOSE)
Output:
-1/2/(a*e^2-b*d*e+c*d^2)/a*(1/4*(-b*c*e+c^2*d)/c*ln(c*x^8+b*x^4+a)+(a*c*e- b^2*e+c*b*d-1/2*(-b*c*e+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^2*ln(e*x^4+d)/d/(a*e^2-b*d*e+c*d^2)+ln(x)/a/d
Timed out. \[ \int \frac {1}{x \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\text {Timed out} \] Input:
integrate(1/x/(e*x^4+d)/(c*x^8+b*x^4+a),x, algorithm="fricas")
Output:
Timed out
Timed out. \[ \int \frac {1}{x \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\text {Timed out} \] Input:
integrate(1/x/(e*x**4+d)/(c*x**8+b*x**4+a),x)
Output:
Timed out
Exception generated. \[ \int \frac {1}{x \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/x/(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
Time = 1.00 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.02 \[ \int \frac {1}{x \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=-\frac {e^{3} \log \left ({\left | e x^{4} + d \right |}\right )}{4 \, {\left (c d^{3} e - b d^{2} e^{2} + a d e^{3}\right )}} - \frac {{\left (c d - b e\right )} \log \left (c x^{8} + b x^{4} + a\right )}{8 \, {\left (a c d^{2} - a b d e + a^{2} e^{2}\right )}} - \frac {{\left (b c d - b^{2} e + 2 \, a c e\right )} \arctan \left (\frac {2 \, c x^{4} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{4 \, {\left (a c d^{2} - a b d e + a^{2} e^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {\log \left (x^{4}\right )}{4 \, a d} \] Input:
integrate(1/x/(e*x^4+d)/(c*x^8+b*x^4+a),x, algorithm="giac")
Output:
-1/4*e^3*log(abs(e*x^4 + d))/(c*d^3*e - b*d^2*e^2 + a*d*e^3) - 1/8*(c*d - b*e)*log(c*x^8 + b*x^4 + a)/(a*c*d^2 - a*b*d*e + a^2*e^2) - 1/4*(b*c*d - b ^2*e + 2*a*c*e)*arctan((2*c*x^4 + b)/sqrt(-b^2 + 4*a*c))/((a*c*d^2 - a*b*d *e + a^2*e^2)*sqrt(-b^2 + 4*a*c)) + 1/4*log(x^4)/(a*d)
Timed out. \[ \int \frac {1}{x \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\text {Hanged} \] Input:
int(1/(x*(d + e*x^4)*(a + b*x^4 + c*x^8)),x)
Output:
\text{Hanged}
\[ \int \frac {1}{x \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\int \frac {1}{x \left (e \,x^{4}+d \right ) \left (c \,x^{8}+b \,x^{4}+a \right )}d x \] Input:
int(1/x/(e*x^4+d)/(c*x^8+b*x^4+a),x)
Output:
int(1/x/(e*x^4+d)/(c*x^8+b*x^4+a),x)