Integrand size = 27, antiderivative size = 312 \[ \int \frac {1}{x^3 \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=-\frac {1}{2 a d x^2}-\frac {\sqrt {c} \left (c d-b e+\frac {b c d-b^2 e+2 a c e}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{2 \sqrt {2} a \sqrt {b-\sqrt {b^2-4 a c}} \left (c d^2-b d e+a e^2\right )}-\frac {\sqrt {c} \left (c d-b e-\frac {b c d-b^2 e+2 a c e}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{2 \sqrt {2} a \sqrt {b+\sqrt {b^2-4 a c}} \left (c d^2-b d e+a e^2\right )}-\frac {e^{5/2} \arctan \left (\frac {\sqrt {e} x^2}{\sqrt {d}}\right )}{2 d^{3/2} \left (c d^2-b d e+a e^2\right )} \] Output:
-1/2/a/d/x^2-1/4*c^(1/2)*(c*d-b*e+(2*a*c*e-b^2*e+b*c*d)/(-4*a*c+b^2)^(1/2) )*arctan(2^(1/2)*c^(1/2)*x^2/(b-(-4*a*c+b^2)^(1/2))^(1/2))*2^(1/2)/a/(b-(- 4*a*c+b^2)^(1/2))^(1/2)/(a*e^2-b*d*e+c*d^2)-1/4*c^(1/2)*(c*d-b*e-(2*a*c*e- b^2*e+b*c*d)/(-4*a*c+b^2)^(1/2))*arctan(2^(1/2)*c^(1/2)*x^2/(b+(-4*a*c+b^2 )^(1/2))^(1/2))*2^(1/2)/a/(b+(-4*a*c+b^2)^(1/2))^(1/2)/(a*e^2-b*d*e+c*d^2) -1/2*e^(5/2)*arctan(e^(1/2)*x^2/d^(1/2))/d^(3/2)/(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 = 0.28 (sec) , antiderivative size = 226, normalized size of antiderivative = 0.72 \[ \int \frac {1}{x^3 \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\frac {2 \left (-c d+b e-\frac {a e^2}{d}+\frac {a e^{5/2} x^2 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{e} x}{\sqrt [4]{d}}\right )}{d^{3/2}}+\frac {a e^{5/2} x^2 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{e} x}{\sqrt [4]{d}}\right )}{d^{3/2}}\right )+x^2 \text {RootSum}\left [a+b \text {$\#$1}^4+c \text {$\#$1}^8\&,\frac {-b c d \log (x-\text {$\#$1})+b^2 e \log (x-\text {$\#$1})-a c e \log (x-\text {$\#$1})-c^2 d \log (x-\text {$\#$1}) \text {$\#$1}^4+b c e \log (x-\text {$\#$1}) \text {$\#$1}^4}{b \text {$\#$1}^2+2 c \text {$\#$1}^6}\&\right ]}{4 a \left (c d^2+e (-b d+a e)\right ) x^2} \] Input:
Integrate[1/(x^3*(d + e*x^4)*(a + b*x^4 + c*x^8)),x]
Output:
(2*(-(c*d) + b*e - (a*e^2)/d + (a*e^(5/2)*x^2*ArcTan[1 - (Sqrt[2]*e^(1/4)* x)/d^(1/4)])/d^(3/2) + (a*e^(5/2)*x^2*ArcTan[1 + (Sqrt[2]*e^(1/4)*x)/d^(1/ 4)])/d^(3/2)) + x^2*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*L og[x - #1]*#1^4)/(b*#1^2 + 2*c*#1^6) & ])/(4*a*(c*d^2 + e*(-(b*d) + a*e))* x^2)
Time = 0.96 (sec) , antiderivative size = 308, 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 = {1814, 1610, 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.
\(\displaystyle \int \frac {1}{x^3 \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx\) |
\(\Big \downarrow \) 1814 |
\(\displaystyle \frac {1}{2} \int \frac {1}{x^4 \left (e x^4+d\right ) \left (c x^8+b x^4+a\right )}dx^2\) |
\(\Big \downarrow \) 1610 |
\(\displaystyle \frac {1}{2} \int \left (-\frac {e^3}{d \left (c d^2-b e d+a e^2\right ) \left (e x^4+d\right )}+\frac {-c (c d-b e) x^4-b c d+b^2 e-a c e}{a \left (c d^2-b e d+a e^2\right ) \left (c x^8+b x^4+a\right )}+\frac {1}{a d x^4}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {\sqrt {c} \arctan \left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {b-\sqrt {b^2-4 a c}}}\right ) \left (\frac {2 a c e+b^2 (-e)+b c d}{\sqrt {b^2-4 a c}}-b e+c d\right )}{\sqrt {2} a \sqrt {b-\sqrt {b^2-4 a c}} \left (a e^2-b d e+c d^2\right )}-\frac {\sqrt {c} \arctan \left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {\sqrt {b^2-4 a c}+b}}\right ) \left (-\frac {2 a c e+b^2 (-e)+b c d}{\sqrt {b^2-4 a c}}-b e+c d\right )}{\sqrt {2} a \sqrt {\sqrt {b^2-4 a c}+b} \left (a e^2-b d e+c d^2\right )}-\frac {e^{5/2} \arctan \left (\frac {\sqrt {e} x^2}{\sqrt {d}}\right )}{d^{3/2} \left (a e^2-b d e+c d^2\right )}-\frac {1}{a d x^2}\right )\) |
Input:
Int[1/(x^3*(d + e*x^4)*(a + b*x^4 + c*x^8)),x]
Output:
(-(1/(a*d*x^2)) - (Sqrt[c]*(c*d - b*e + (b*c*d - b^2*e + 2*a*c*e)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x^2)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(Sqr t[2]*a*Sqrt[b - Sqrt[b^2 - 4*a*c]]*(c*d^2 - b*d*e + a*e^2)) - (Sqrt[c]*(c* d - b*e - (b*c*d - b^2*e + 2*a*c*e)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqr t[c]*x^2)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*a*Sqrt[b + Sqrt[b^2 - 4*a *c]]*(c*d^2 - b*d*e + a*e^2)) - (e^(5/2)*ArcTan[(Sqrt[e]*x^2)/Sqrt[d]])/(d ^(3/2)*(c*d^2 - b*d*e + a*e^2)))/2
Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.))/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*((d + e*x^2)^q/(a + b*x^2 + c*x^4)), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b^2 - 4 *a*c, 0] && IntegerQ[q] && IntegerQ[m]
Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_)*((d_) + (e _.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Sub st[Int[x^((m + 1)/k - 1)*(d + e*x^(n/k))^q*(a + b*x^(n/k) + c*x^(2*(n/k)))^ p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, c, d, e, p, q}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && IntegerQ[m]
Time = 0.43 (sec) , antiderivative size = 282, normalized size of antiderivative = 0.90
method | result | size |
default | \(-\frac {1}{2 d a \,x^{2}}-\frac {2 c \left (-\frac {\left (-e b \sqrt {-4 a c +b^{2}}+c d \sqrt {-4 a c +b^{2}}+2 a c e -b^{2} e +c b d \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {c \,x^{2} \sqrt {2}}{\sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 \sqrt {-4 a c +b^{2}}\, \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}+\frac {\left (-e b \sqrt {-4 a c +b^{2}}+c d \sqrt {-4 a c +b^{2}}-2 a c e +b^{2} e -c b d \right ) \sqrt {2}\, \arctan \left (\frac {c \,x^{2} \sqrt {2}}{\sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 \sqrt {-4 a c +b^{2}}\, \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) a}-\frac {e^{3} \arctan \left (\frac {e \,x^{2}}{\sqrt {d e}}\right )}{2 d \left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {d e}}\) | \(282\) |
risch | \(\text {Expression too large to display}\) | \(4232\) |
Input:
int(1/x^3/(e*x^4+d)/(c*x^8+b*x^4+a),x,method=_RETURNVERBOSE)
Output:
-1/2/d/a/x^2-2/(a*e^2-b*d*e+c*d^2)/a*c*(-1/8*(-e*b*(-4*a*c+b^2)^(1/2)+c*d* (-4*a*c+b^2)^(1/2)+2*a*c*e-b^2*e+c*b*d)/(-4*a*c+b^2)^(1/2)*2^(1/2)/((-b+(- 4*a*c+b^2)^(1/2))*c)^(1/2)*arctanh(c*x^2*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))* c)^(1/2))+1/8*(-e*b*(-4*a*c+b^2)^(1/2)+c*d*(-4*a*c+b^2)^(1/2)-2*a*c*e+b^2* e-c*b*d)/(-4*a*c+b^2)^(1/2)*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arcta n(c*x^2*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2)))-1/2*e^3/d/(a*e^2-b*d*e+ c*d^2)/(d*e)^(1/2)*arctan(e*x^2/(d*e)^(1/2))
Timed out. \[ \int \frac {1}{x^3 \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\text {Timed out} \] Input:
integrate(1/x^3/(e*x^4+d)/(c*x^8+b*x^4+a),x, algorithm="fricas")
Output:
Timed out
Timed out. \[ \int \frac {1}{x^3 \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\text {Timed out} \] Input:
integrate(1/x**3/(e*x**4+d)/(c*x**8+b*x**4+a),x)
Output:
Timed out
Exception generated. \[ \int \frac {1}{x^3 \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/x^3/(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(e>0)', see `assume?` for more de tails)Is e
Leaf count of result is larger than twice the leaf count of optimal. 7626 vs. \(2 (260) = 520\).
Time = 2.50 (sec) , antiderivative size = 7626, normalized size of antiderivative = 24.44 \[ \int \frac {1}{x^3 \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\text {Too large to display} \] Input:
integrate(1/x^3/(e*x^4+d)/(c*x^8+b*x^4+a),x, algorithm="giac")
Output:
-1/2*e^3*arctan(e*x^2/sqrt(d*e))/((c*d^3 - b*d^2*e + a*d*e^2)*sqrt(d*e)) - 1/8*((2*a*b^3*c^5 - 8*a^2*b*c^6 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sq rt(b^2 - 4*a*c)*c)*a*b^3*c^3 + 4*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt (b^2 - 4*a*c)*c)*a^2*b*c^4 + 2*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b ^2 - 4*a*c)*c)*a*b^2*c^4 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b*c^5 - 2*(b^2 - 4*a*c)*a*b*c^5)*d^3*x^4 - 2*(2*a*b^4*c^4 - 8 *a^2*b^2*c^5 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a *b^4*c^2 + 4*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2 *b^2*c^3 + 2*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b ^3*c^3 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^2*c ^4 - 2*(b^2 - 4*a*c)*a*b^2*c^4)*d^2*e*x^4 + (2*a*b^5*c^3 - 6*a^2*b^3*c^4 - 8*a^3*b*c^5 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a *b^5*c + 3*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b ^3*c^2 + 2*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^4 *c^2 + 4*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^3*b*c ^3 + 2*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b^2*c ^3 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^3*c^3 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b*c^4 - 2*( b^2 - 4*a*c)*a*b^3*c^3 - 2*(b^2 - 4*a*c)*a^2*b*c^4)*d*e^2*x^4 - (2*a^2*b^4 *c^3 - 8*a^3*b^2*c^4 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - ...
Timed out. \[ \int \frac {1}{x^3 \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\text {Hanged} \] Input:
int(1/(x^3*(d + e*x^4)*(a + b*x^4 + c*x^8)),x)
Output:
\text{Hanged}
\[ \int \frac {1}{x^3 \left (d+e x^4\right ) \left (a+b x^4+c x^8\right )} \, dx=\int \frac {1}{x^{3} \left (e \,x^{4}+d \right ) \left (c \,x^{8}+b \,x^{4}+a \right )}d x \] Input:
int(1/x^3/(e*x^4+d)/(c*x^8+b*x^4+a),x)
Output:
int(1/x^3/(e*x^4+d)/(c*x^8+b*x^4+a),x)