Integrand size = 27, antiderivative size = 268 \[ \int \frac {1}{x^9 \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=-\frac {1}{8 a d x^8}+\frac {b d+a e}{4 a^2 d^2 x^4}+\frac {\left (b^3 c d-3 a b c^2 d-b^4 e+4 a b^2 c e-2 a^2 c^2 e\right ) \text {arctanh}\left (\frac {b+2 c x^4}{\sqrt {b^2-4 a c}}\right )}{4 a^3 \sqrt {b^2-4 a c} \left (c d^2-b d e+a e^2\right )}+\frac {\left (b^2 d^2+a b d e-a \left (c d^2-a e^2\right )\right ) \log (x)}{a^3 d^3}-\frac {e^4 \log \left (d+e x^4\right )}{4 d^3 \left (c d^2-b d e+a e^2\right )}-\frac {\left (b^2 c d-a c^2 d-b^3 e+2 a b c e\right ) \log \left (a+b x^4+c x^8\right )}{8 a^3 \left (c d^2-b d e+a e^2\right )} \] Output:
-1/8/a/d/x^8+1/4*(a*e+b*d)/a^2/d^2/x^4+1/4*(-2*a^2*c^2*e+4*a*b^2*c*e-3*a*b *c^2*d-b^4*e+b^3*c*d)*arctanh((2*c*x^4+b)/(-4*a*c+b^2)^(1/2))/a^3/(-4*a*c+ b^2)^(1/2)/(a*e^2-b*d*e+c*d^2)+(b^2*d^2+a*b*d*e-a*(-a*e^2+c*d^2))*ln(x)/a^ 3/d^3-1/4*e^4*ln(e*x^4+d)/d^3/(a*e^2-b*d*e+c*d^2)-1/8*(2*a*b*c*e-a*c^2*d-b ^3*e+b^2*c*d)*ln(c*x^8+b*x^4+a)/a^3/(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 = 0.26 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.13 \[ \int \frac {1}{x^9 \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\frac {1}{8} \left (-\frac {1}{a d x^8}+\frac {2 (b d+a e)}{a^2 d^2 x^4}+\frac {8 \left (b^2 d^2+a b d e+a \left (-c d^2+a e^2\right )\right ) \log (x)}{a^3 d^3}-\frac {2 e^4 \log \left (d+e x^4\right )}{c d^5+d^3 e (-b d+a e)}+\frac {2 \text {RootSum}\left [a+b \text {$\#$1}^4+c \text {$\#$1}^8\&,\frac {-b^3 c d \log (x-\text {$\#$1})+2 a b c^2 d \log (x-\text {$\#$1})+b^4 e \log (x-\text {$\#$1})-3 a b^2 c e \log (x-\text {$\#$1})+a^2 c^2 e \log (x-\text {$\#$1})-b^2 c^2 d \log (x-\text {$\#$1}) \text {$\#$1}^4+a c^3 d \log (x-\text {$\#$1}) \text {$\#$1}^4+b^3 c e \log (x-\text {$\#$1}) \text {$\#$1}^4-2 a b c^2 e \log (x-\text {$\#$1}) \text {$\#$1}^4}{b+2 c \text {$\#$1}^4}\&\right ]}{a^3 \left (c d^2+e (-b d+a e)\right )}\right ) \] Input:
Integrate[1/(x^9*(d + e*x^4)*(a + b*x^4 + c*x^8)),x]
Output:
(-(1/(a*d*x^8)) + (2*(b*d + a*e))/(a^2*d^2*x^4) + (8*(b^2*d^2 + a*b*d*e + a*(-(c*d^2) + a*e^2))*Log[x])/(a^3*d^3) - (2*e^4*Log[d + e*x^4])/(c*d^5 + d^3*e*(-(b*d) + a*e)) + (2*RootSum[a + b*#1^4 + c*#1^8 & , (-(b^3*c*d*Log[ x - #1]) + 2*a*b*c^2*d*Log[x - #1] + b^4*e*Log[x - #1] - 3*a*b^2*c*e*Log[x - #1] + a^2*c^2*e*Log[x - #1] - b^2*c^2*d*Log[x - #1]*#1^4 + a*c^3*d*Log[ x - #1]*#1^4 + b^3*c*e*Log[x - #1]*#1^4 - 2*a*b*c^2*e*Log[x - #1]*#1^4)/(b + 2*c*#1^4) & ])/(a^3*(c*d^2 + e*(-(b*d) + a*e))))/8
Time = 0.68 (sec) , antiderivative size = 266, normalized size of antiderivative = 0.99, 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.
\(\displaystyle \int \frac {1}{x^9 \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx\) |
\(\Big \downarrow \) 1802 |
\(\displaystyle \frac {1}{4} \int \frac {1}{x^{12} \left (e x^4+d\right ) \left (c x^8+b x^4+a\right )}dx^4\) |
\(\Big \downarrow \) 1200 |
\(\displaystyle \frac {1}{4} \int \left (-\frac {e^5}{d^3 \left (c d^2-b e d+a e^2\right ) \left (e x^4+d\right )}+\frac {e b^4-c d b^3-3 a c e b^2+2 a c^2 d b-c \left (-e b^3+c d b^2+2 a c e b-a c^2 d\right ) x^4+a^2 c^2 e}{a^3 \left (c d^2-b e d+a e^2\right ) \left (c x^8+b x^4+a\right )}+\frac {b^2 d^2+a b e d-a \left (c d^2-a e^2\right )}{a^3 d^3 x^4}+\frac {-b d-a e}{a^2 d^2 x^8}+\frac {1}{a d x^{12}}\right )dx^4\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (\frac {\log \left (x^4\right ) \left (a b d e-a \left (c d^2-a e^2\right )+b^2 d^2\right )}{a^3 d^3}-\frac {\left (2 a b c e-a c^2 d+b^3 (-e)+b^2 c d\right ) \log \left (a+b x^4+c x^8\right )}{2 a^3 \left (a e^2-b d e+c d^2\right )}+\frac {a e+b d}{a^2 d^2 x^4}+\frac {\text {arctanh}\left (\frac {b+2 c x^4}{\sqrt {b^2-4 a c}}\right ) \left (-2 a^2 c^2 e+4 a b^2 c e-3 a b c^2 d+b^4 (-e)+b^3 c d\right )}{a^3 \sqrt {b^2-4 a c} \left (a e^2-b d e+c d^2\right )}-\frac {e^4 \log \left (d+e x^4\right )}{d^3 \left (a e^2-b d e+c d^2\right )}-\frac {1}{2 a d x^8}\right )\) |
Input:
Int[1/(x^9*(d + e*x^4)*(a + b*x^4 + c*x^8)),x]
Output:
(-1/2*1/(a*d*x^8) + (b*d + a*e)/(a^2*d^2*x^4) + ((b^3*c*d - 3*a*b*c^2*d - b^4*e + 4*a*b^2*c*e - 2*a^2*c^2*e)*ArcTanh[(b + 2*c*x^4)/Sqrt[b^2 - 4*a*c] ])/(a^3*Sqrt[b^2 - 4*a*c]*(c*d^2 - b*d*e + a*e^2)) + ((b^2*d^2 + a*b*d*e - a*(c*d^2 - a*e^2))*Log[x^4])/(a^3*d^3) - (e^4*Log[d + e*x^4])/(d^3*(c*d^2 - b*d*e + a*e^2)) - ((b^2*c*d - a*c^2*d - b^3*e + 2*a*b*c*e)*Log[a + b*x^ 4 + c*x^8])/(2*a^3*(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 = 3.24 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.06
method | result | size |
default | \(\frac {\frac {\left (-2 a b \,c^{2} e +a \,c^{3} d +b^{3} c e -b^{2} c^{2} d \right ) \ln \left (c \,x^{8}+b \,x^{4}+a \right )}{4 c}+\frac {\left (a^{2} c^{2} e -3 a \,b^{2} c e +2 d \,c^{2} b a +b^{4} e -d c \,b^{3}-\frac {\left (-2 a b \,c^{2} e +a \,c^{3} d +b^{3} c e -b^{2} c^{2} d \right ) b}{2 c}\right ) \arctan \left (\frac {2 c \,x^{4}+b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{2 \left (a \,e^{2}-b d e +c \,d^{2}\right ) a^{3}}-\frac {e^{4} \ln \left (x^{4} e +d \right )}{4 d^{3} \left (a \,e^{2}-b d e +c \,d^{2}\right )}-\frac {1}{8 a d \,x^{8}}-\frac {-a e -b d}{4 a^{2} d^{2} x^{4}}+\frac {\left (a^{2} e^{2}+a b d e -a c \,d^{2}+b^{2} d^{2}\right ) \ln \left (x \right )}{d^{3} a^{3}}\) | \(283\) |
risch | \(\text {Expression too large to display}\) | \(1325\) |
Input:
int(1/x^9/(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^3*(1/4*(-2*a*b*c^2*e+a*c^3*d+b^3*c*e-b^2*c^2*d)/ c*ln(c*x^8+b*x^4+a)+(a^2*c^2*e-3*a*b^2*c*e+2*d*c^2*b*a+b^4*e-d*c*b^3-1/2*( -2*a*b*c^2*e+a*c^3*d+b^3*c*e-b^2*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^4*ln(e*x^4+d)/d^3/(a*e^2-b*d*e+c*d^2)-1/ 8/a/d/x^8-1/4*(-a*e-b*d)/a^2/d^2/x^4+(a^2*e^2+a*b*d*e-a*c*d^2+b^2*d^2)/d^3 /a^3*ln(x)
Timed out. \[ \int \frac {1}{x^9 \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\text {Timed out} \] Input:
integrate(1/x^9/(e*x^4+d)/(c*x^8+b*x^4+a),x, algorithm="fricas")
Output:
Timed out
Timed out. \[ \int \frac {1}{x^9 \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\text {Timed out} \] Input:
integrate(1/x**9/(e*x**4+d)/(c*x**8+b*x**4+a),x)
Output:
Timed out
Exception generated. \[ \int \frac {1}{x^9 \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/x^9/(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 = 1.06 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.22 \[ \int \frac {1}{x^9 \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=-\frac {e^{5} \log \left ({\left | e x^{4} + d \right |}\right )}{4 \, {\left (c d^{5} e - b d^{4} e^{2} + a d^{3} e^{3}\right )}} - \frac {{\left (b^{2} c d - a c^{2} d - b^{3} e + 2 \, a b c e\right )} \log \left (c x^{8} + b x^{4} + a\right )}{8 \, {\left (a^{3} c d^{2} - a^{3} b d e + a^{4} e^{2}\right )}} - \frac {{\left (b^{3} c d - 3 \, a b c^{2} d - b^{4} e + 4 \, a b^{2} c e - 2 \, a^{2} c^{2} e\right )} \arctan \left (\frac {2 \, c x^{4} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{4 \, {\left (a^{3} c d^{2} - a^{3} b d e + a^{4} e^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {{\left (b^{2} d^{2} - a c d^{2} + a b d e + a^{2} e^{2}\right )} \log \left (x^{4}\right )}{4 \, a^{3} d^{3}} - \frac {3 \, b^{2} d^{2} x^{8} - 3 \, a c d^{2} x^{8} + 3 \, a b d e x^{8} + 3 \, a^{2} e^{2} x^{8} - 2 \, a b d^{2} x^{4} - 2 \, a^{2} d e x^{4} + a^{2} d^{2}}{8 \, a^{3} d^{3} x^{8}} \] Input:
integrate(1/x^9/(e*x^4+d)/(c*x^8+b*x^4+a),x, algorithm="giac")
Output:
-1/4*e^5*log(abs(e*x^4 + d))/(c*d^5*e - b*d^4*e^2 + a*d^3*e^3) - 1/8*(b^2* c*d - a*c^2*d - b^3*e + 2*a*b*c*e)*log(c*x^8 + b*x^4 + a)/(a^3*c*d^2 - a^3 *b*d*e + a^4*e^2) - 1/4*(b^3*c*d - 3*a*b*c^2*d - b^4*e + 4*a*b^2*c*e - 2*a ^2*c^2*e)*arctan((2*c*x^4 + b)/sqrt(-b^2 + 4*a*c))/((a^3*c*d^2 - a^3*b*d*e + a^4*e^2)*sqrt(-b^2 + 4*a*c)) + 1/4*(b^2*d^2 - a*c*d^2 + a*b*d*e + a^2*e ^2)*log(x^4)/(a^3*d^3) - 1/8*(3*b^2*d^2*x^8 - 3*a*c*d^2*x^8 + 3*a*b*d*e*x^ 8 + 3*a^2*e^2*x^8 - 2*a*b*d^2*x^4 - 2*a^2*d*e*x^4 + a^2*d^2)/(a^3*d^3*x^8)
Timed out. \[ \int \frac {1}{x^9 \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\text {Hanged} \] Input:
int(1/(x^9*(d + e*x^4)*(a + b*x^4 + c*x^8)),x)
Output:
\text{Hanged}
\[ \int \frac {1}{x^9 \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\int \frac {1}{x^{9} \left (e \,x^{4}+d \right ) \left (c \,x^{8}+b \,x^{4}+a \right )}d x \] Input:
int(1/x^9/(e*x^4+d)/(c*x^8+b*x^4+a),x)
Output:
int(1/x^9/(e*x^4+d)/(c*x^8+b*x^4+a),x)