Integrand size = 27, antiderivative size = 348 \[ \int \frac {1}{x^4 \left (d+e x^2\right ) \left (a+b x^2+c x^4\right )} \, dx=-\frac {1}{3 a d x^3}+\frac {b d+a e}{a^2 d^2 x}+\frac {\sqrt {c} \left (b c d-b^2 e+a c e+\frac {b^2 c d-2 a c^2 d-b^3 e+3 a b c e}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} a^2 \sqrt {b-\sqrt {b^2-4 a c}} \left (c d^2-b d e+a e^2\right )}+\frac {\sqrt {c} \left (b c d-b^2 e+a c e-\frac {b^2 c d-2 a c^2 d-b^3 e+3 a b c e}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} a^2 \sqrt {b+\sqrt {b^2-4 a c}} \left (c d^2-b d e+a e^2\right )}+\frac {e^{7/2} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{d^{5/2} \left (c d^2-b d e+a e^2\right )} \]
-1/3/a/d/x^3+(a*e+b*d)/a^2/d^2/x+e^(7/2)*arctan(x*e^(1/2)/d^(1/2))/d^(5/2) /(a*e^2-b*d*e+c*d^2)+1/2*arctan(x*2^(1/2)*c^(1/2)/(b-(-4*a*c+b^2)^(1/2))^( 1/2))*c^(1/2)*(b*c*d-b^2*e+a*c*e+(3*a*b*c*e-2*a*c^2*d-b^3*e+b^2*c*d)/(-4*a *c+b^2)^(1/2))/a^2/(a*e^2-b*d*e+c*d^2)*2^(1/2)/(b-(-4*a*c+b^2)^(1/2))^(1/2 )+1/2*arctan(x*2^(1/2)*c^(1/2)/(b+(-4*a*c+b^2)^(1/2))^(1/2))*c^(1/2)*(b*c* d-b^2*e+a*c*e+(-3*a*b*c*e+2*a*c^2*d+b^3*e-b^2*c*d)/(-4*a*c+b^2)^(1/2))/a^2 /(a*e^2-b*d*e+c*d^2)*2^(1/2)/(b+(-4*a*c+b^2)^(1/2))^(1/2)
Time = 0.36 (sec) , antiderivative size = 410, normalized size of antiderivative = 1.18 \[ \int \frac {1}{x^4 \left (d+e x^2\right ) \left (a+b x^2+c x^4\right )} \, dx=-\frac {1}{3 a d x^3}+\frac {b d+a e}{a^2 d^2 x}+\frac {\sqrt {c} \left (-b^3 e+b c \left (\sqrt {b^2-4 a c} d+3 a e\right )+b^2 \left (c d-\sqrt {b^2-4 a c} e\right )+a c \left (-2 c d+\sqrt {b^2-4 a c} e\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} a^2 \sqrt {b^2-4 a c} \sqrt {b-\sqrt {b^2-4 a c}} \left (c d^2+e (-b d+a e)\right )}+\frac {\sqrt {c} \left (b^3 e+b c \left (\sqrt {b^2-4 a c} d-3 a e\right )-b^2 \left (c d+\sqrt {b^2-4 a c} e\right )+a c \left (2 c d+\sqrt {b^2-4 a c} e\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} a^2 \sqrt {b^2-4 a c} \sqrt {b+\sqrt {b^2-4 a c}} \left (c d^2+e (-b d+a e)\right )}+\frac {e^{7/2} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{d^{5/2} \left (c d^2-b d e+a e^2\right )} \]
-1/3*1/(a*d*x^3) + (b*d + a*e)/(a^2*d^2*x) + (Sqrt[c]*(-(b^3*e) + b*c*(Sqr t[b^2 - 4*a*c]*d + 3*a*e) + b^2*(c*d - Sqrt[b^2 - 4*a*c]*e) + a*c*(-2*c*d + Sqrt[b^2 - 4*a*c]*e))*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b - Sqrt[b^2 - 4*a *c]]])/(Sqrt[2]*a^2*Sqrt[b^2 - 4*a*c]*Sqrt[b - Sqrt[b^2 - 4*a*c]]*(c*d^2 + e*(-(b*d) + a*e))) + (Sqrt[c]*(b^3*e + b*c*(Sqrt[b^2 - 4*a*c]*d - 3*a*e) - b^2*(c*d + Sqrt[b^2 - 4*a*c]*e) + a*c*(2*c*d + Sqrt[b^2 - 4*a*c]*e))*Arc Tan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*a^2*Sqrt[b^ 2 - 4*a*c]*Sqrt[b + Sqrt[b^2 - 4*a*c]]*(c*d^2 + e*(-(b*d) + a*e))) + (e^(7 /2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(d^(5/2)*(c*d^2 - b*d*e + a*e^2))
Time = 1.28 (sec) , antiderivative size = 348, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {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.
\(\displaystyle \int \frac {1}{x^4 \left (d+e x^2\right ) \left (a+b x^2+c x^4\right )} \, dx\) |
\(\Big \downarrow \) 1610 |
\(\displaystyle \int \left (\frac {c x^2 \left (a c e+b^2 (-e)+b c d\right )+2 a b c e-a c^2 d+b^3 (-e)+b^2 c d}{a^2 \left (a+b x^2+c x^4\right ) \left (a e^2-b d e+c d^2\right )}+\frac {-a e-b d}{a^2 d^2 x^2}+\frac {e^4}{d^2 \left (d+e x^2\right ) \left (a e^2-b d e+c d^2\right )}+\frac {1}{a d x^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {c} \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right ) \left (\frac {3 a b c e-2 a c^2 d+b^3 (-e)+b^2 c d}{\sqrt {b^2-4 a c}}+a c e+b^2 (-e)+b c d\right )}{\sqrt {2} a^2 \sqrt {b-\sqrt {b^2-4 a c}} \left (a e^2-b d e+c d^2\right )}+\frac {\sqrt {c} \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {\sqrt {b^2-4 a c}+b}}\right ) \left (-\frac {3 a b c e-2 a c^2 d+b^3 (-e)+b^2 c d}{\sqrt {b^2-4 a c}}+a c e+b^2 (-e)+b c d\right )}{\sqrt {2} a^2 \sqrt {\sqrt {b^2-4 a c}+b} \left (a e^2-b d e+c d^2\right )}+\frac {a e+b d}{a^2 d^2 x}+\frac {e^{7/2} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{d^{5/2} \left (a e^2-b d e+c d^2\right )}-\frac {1}{3 a d x^3}\) |
-1/3*1/(a*d*x^3) + (b*d + a*e)/(a^2*d^2*x) + (Sqrt[c]*(b*c*d - b^2*e + a*c *e + (b^2*c*d - 2*a*c^2*d - b^3*e + 3*a*b*c*e)/Sqrt[b^2 - 4*a*c])*ArcTan[( Sqrt[2]*Sqrt[c]*x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*a^2*Sqrt[b - Sqr t[b^2 - 4*a*c]]*(c*d^2 - b*d*e + a*e^2)) + (Sqrt[c]*(b*c*d - b^2*e + a*c*e - (b^2*c*d - 2*a*c^2*d - b^3*e + 3*a*b*c*e)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sq rt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*a^2*Sqrt[b + Sqrt[ b^2 - 4*a*c]]*(c*d^2 - b*d*e + a*e^2)) + (e^(7/2)*ArcTan[(Sqrt[e]*x)/Sqrt[ d]])/(d^(5/2)*(c*d^2 - b*d*e + a*e^2))
3.4.9.3.1 Defintions of rubi rules used
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]
Time = 0.49 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.00
method | result | size |
default | \(-\frac {1}{3 a d \,x^{3}}-\frac {-a e -b d}{x \,a^{2} d^{2}}+\frac {4 c \left (\frac {\left (a c e \sqrt {-4 a c +b^{2}}-b^{2} e \sqrt {-4 a c +b^{2}}+b c d \sqrt {-4 a c +b^{2}}-3 a b c e +2 a \,c^{2} d +b^{3} e -b^{2} c d \right ) \sqrt {2}\, \arctan \left (\frac {c x \sqrt {2}}{\sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 \sqrt {-4 a c +b^{2}}\, \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}-\frac {\left (a c e \sqrt {-4 a c +b^{2}}-b^{2} e \sqrt {-4 a c +b^{2}}+b c d \sqrt {-4 a c +b^{2}}+3 a b c e -2 a \,c^{2} d -b^{3} e +b^{2} c d \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {c x \sqrt {2}}{\sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 \sqrt {-4 a c +b^{2}}\, \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) a^{2}}+\frac {e^{4} \arctan \left (\frac {e x}{\sqrt {e d}}\right )}{d^{2} \left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {e d}}\) | \(349\) |
risch | \(\text {Expression too large to display}\) | \(5012\) |
-1/3/a/d/x^3-(-a*e-b*d)/x/a^2/d^2+4/(a*e^2-b*d*e+c*d^2)/a^2*c*(1/8*(a*c*e* (-4*a*c+b^2)^(1/2)-b^2*e*(-4*a*c+b^2)^(1/2)+b*c*d*(-4*a*c+b^2)^(1/2)-3*a*b *c*e+2*a*c^2*d+b^3*e-b^2*c*d)/(-4*a*c+b^2)^(1/2)*2^(1/2)/((b+(-4*a*c+b^2)^ (1/2))*c)^(1/2)*arctan(c*x*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2))-1/8*( a*c*e*(-4*a*c+b^2)^(1/2)-b^2*e*(-4*a*c+b^2)^(1/2)+b*c*d*(-4*a*c+b^2)^(1/2) +3*a*b*c*e-2*a*c^2*d-b^3*e+b^2*c*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^(1/2)/((-b+(-4*a*c+b^2)^(1/2))*c)^(1/ 2)))+1/d^2*e^4/(a*e^2-b*d*e+c*d^2)/(e*d)^(1/2)*arctan(e*x/(e*d)^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 12482 vs. \(2 (304) = 608\).
Time = 293.42 (sec) , antiderivative size = 24988, normalized size of antiderivative = 71.80 \[ \int \frac {1}{x^4 \left (d+e x^2\right ) \left (a+b x^2+c x^4\right )} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {1}{x^4 \left (d+e x^2\right ) \left (a+b x^2+c x^4\right )} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {1}{x^4 \left (d+e x^2\right ) \left (a+b x^2+c x^4\right )} \, dx=\text {Exception raised: ValueError} \]
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. 12281 vs. \(2 (304) = 608\).
Time = 2.65 (sec) , antiderivative size = 12281, normalized size of antiderivative = 35.29 \[ \int \frac {1}{x^4 \left (d+e x^2\right ) \left (a+b x^2+c x^4\right )} \, dx=\text {Too large to display} \]
e^4*arctan(e*x/sqrt(d*e))/((c*d^4 - b*d^3*e + a*d^2*e^2)*sqrt(d*e)) - 1/8* ((2*a^4*b^5*c^5 - 12*a^5*b^3*c^6 + 16*a^6*b*c^7 - sqrt(2)*sqrt(b^2 - 4*a*c )*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^4*b^5*c^3 + 6*sqrt(2)*sqrt(b^2 - 4*a*c )*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^5*b^3*c^4 + 2*sqrt(2)*sqrt(b^2 - 4*a*c )*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^4*b^4*c^4 - 8*sqrt(2)*sqrt(b^2 - 4*a*c )*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^6*b*c^5 - 4*sqrt(2)*sqrt(b^2 - 4*a*c)* sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^5*b^2*c^5 - sqrt(2)*sqrt(b^2 - 4*a*c)*sq rt(b*c + sqrt(b^2 - 4*a*c)*c)*a^4*b^3*c^5 + 2*sqrt(2)*sqrt(b^2 - 4*a*c)*sq rt(b*c + sqrt(b^2 - 4*a*c)*c)*a^5*b*c^6 - 2*(b^2 - 4*a*c)*a^4*b^3*c^5 + 4* (b^2 - 4*a*c)*a^5*b*c^6)*d^5 - (6*a^4*b^6*c^4 - 38*a^5*b^4*c^5 + 56*a^6*b^ 2*c^6 - 3*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^4*b^ 6*c^2 + 19*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^5*b ^4*c^3 + 6*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^4*b ^5*c^3 - 28*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^6* b^2*c^4 - 14*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^5 *b^3*c^4 - 3*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^4 *b^4*c^4 + 7*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^5 *b^2*c^5 - 6*(b^2 - 4*a*c)*a^4*b^4*c^4 + 14*(b^2 - 4*a*c)*a^5*b^2*c^5)*d^4 *e + (6*a^4*b^7*c^3 - 36*a^5*b^5*c^4 + 40*a^6*b^3*c^5 + 32*a^7*b*c^6 - 3*s qrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^4*b^7*c + 18...
Time = 11.88 (sec) , antiderivative size = 42882, normalized size of antiderivative = 123.22 \[ \int \frac {1}{x^4 \left (d+e x^2\right ) \left (a+b x^2+c x^4\right )} \, dx=\text {Too large to display} \]
(log(c^9*d^27*e^6 - b^9*d^18*e^15 + 2*a*c^8*d^25*e^8 - 2*b*c^8*d^26*e^7 + 2*b^8*c*d^19*e^14 + a^5*b^4*d^13*e^20 + a^2*c^7*d^23*e^10 + 16*a^4*c^5*d^1 9*e^14 + 16*a^7*c^2*d^13*e^20 + b^2*c^7*d^25*e^8 - b^7*c^2*d^20*e^13 - 25* a^2*b^3*c^4*d^20*e^13 + 66*a^2*b^4*c^3*d^19*e^14 - 42*a^2*b^5*c^2*d^18*e^1 5 - 76*a^3*b^2*c^4*d^19*e^14 + 63*a^3*b^3*c^3*d^18*e^15 - a^5*b^4*e^3*x*(- d^5*e^7)^(5/2) + a^2*c^7*d^15*x*(-d^5*e^7)^(3/2) - 16*a^7*c^2*e^3*x*(-d^5* e^7)^(5/2) - b^9*d^10*e^5*x*(-d^5*e^7)^(3/2) - c^9*d^24*e^3*x*(-d^5*e^7)^( 1/2) - 2*a*b*c^7*d^24*e^9 + 11*a*b^7*c*d^18*e^15 + 9*a*b^5*c^3*d^20*e^13 - 20*a*b^6*c^2*d^19*e^14 + 20*a^3*b*c^5*d^20*e^13 - 28*a^4*b*c^4*d^18*e^15 - 8*a^6*b^2*c*d^13*e^20 + 16*a^4*c^5*d^11*e^4*x*(-d^5*e^7)^(3/2) - b^7*c^2 *d^12*e^3*x*(-d^5*e^7)^(3/2) - b^2*c^7*d^22*e^5*x*(-d^5*e^7)^(1/2) + 8*a^6 *b^2*c*e^3*x*(-d^5*e^7)^(5/2) - 2*a*c^8*d^22*e^5*x*(-d^5*e^7)^(1/2) + 2*b^ 8*c*d^11*e^4*x*(-d^5*e^7)^(3/2) + 2*b*c^8*d^23*e^4*x*(-d^5*e^7)^(1/2) + 11 *a*b^7*c*d^10*e^5*x*(-d^5*e^7)^(3/2) + 2*a*b*c^7*d^21*e^6*x*(-d^5*e^7)^(1/ 2) + 9*a*b^5*c^3*d^12*e^3*x*(-d^5*e^7)^(3/2) - 20*a*b^6*c^2*d^11*e^4*x*(-d ^5*e^7)^(3/2) + 20*a^3*b*c^5*d^12*e^3*x*(-d^5*e^7)^(3/2) - 28*a^4*b*c^4*d^ 10*e^5*x*(-d^5*e^7)^(3/2) - 25*a^2*b^3*c^4*d^12*e^3*x*(-d^5*e^7)^(3/2) + 6 6*a^2*b^4*c^3*d^11*e^4*x*(-d^5*e^7)^(3/2) - 42*a^2*b^5*c^2*d^10*e^5*x*(-d^ 5*e^7)^(3/2) - 76*a^3*b^2*c^4*d^11*e^4*x*(-d^5*e^7)^(3/2) + 63*a^3*b^3*c^3 *d^10*e^5*x*(-d^5*e^7)^(3/2))*(-d^5*e^7)^(1/2))/(2*c*d^7 + 2*a*d^5*e^2 ...