Integrand size = 30, antiderivative size = 329 \[ \int \frac {d+e x^2+f x^4}{x^6 \left (a+b x^2+c x^4\right )} \, dx=-\frac {d}{5 a x^5}+\frac {b d-a e}{3 a^2 x^3}-\frac {b^2 d-a b e-a (c d-a f)}{a^3 x}-\frac {\sqrt {c} \left (b^2 d-a b e-a (c d-a f)+\frac {b^3 d-a b^2 e+2 a^2 c e-a b (3 c d-a f)}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} a^3 \sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\sqrt {c} \left (b^2 d-a b e-a (c d-a f)-\frac {b^3 d-a b^2 e+2 a^2 c e-a b (3 c d-a f)}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} a^3 \sqrt {b+\sqrt {b^2-4 a c}}} \]
-1/5*d/a/x^5+1/3*(-a*e+b*d)/a^2/x^3+(-b^2*d+a*b*e+a*(-a*f+c*d))/a^3/x-1/2* arctan(x*2^(1/2)*c^(1/2)/(b-(-4*a*c+b^2)^(1/2))^(1/2))*c^(1/2)*(b^2*d-a*b* e-a*(-a*f+c*d)+(b^3*d-a*b^2*e+2*a^2*c*e-a*b*(-a*f+3*c*d))/(-4*a*c+b^2)^(1/ 2))/a^3*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^2*d-a*b*e-a*(-a*f+c*d)+(-b^3*d+a* b^2*e-2*a^2*c*e+a*b*(-a*f+3*c*d))/(-4*a*c+b^2)^(1/2))/a^3*2^(1/2)/(b+(-4*a *c+b^2)^(1/2))^(1/2)
Time = 0.34 (sec) , antiderivative size = 394, normalized size of antiderivative = 1.20 \[ \int \frac {d+e x^2+f x^4}{x^6 \left (a+b x^2+c x^4\right )} \, dx=\frac {-\frac {6 a^2 d}{x^5}+\frac {10 a (b d-a e)}{x^3}+\frac {30 \left (-b^2 d+a b e+a (c d-a f)\right )}{x}-\frac {15 \sqrt {2} \sqrt {c} \left (b^3 d+b^2 \left (\sqrt {b^2-4 a c} d-a e\right )+a b \left (-3 c d-\sqrt {b^2-4 a c} e+a f\right )+a \left (-c \sqrt {b^2-4 a c} d+2 a c e+a \sqrt {b^2-4 a c} f\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {15 \sqrt {2} \sqrt {c} \left (b^3 d-b^2 \left (\sqrt {b^2-4 a c} d+a e\right )+a b \left (-3 c d+\sqrt {b^2-4 a c} e+a f\right )+a \left (c \sqrt {b^2-4 a c} d+2 a c e-a \sqrt {b^2-4 a c} f\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c} \sqrt {b+\sqrt {b^2-4 a c}}}}{30 a^3} \]
((-6*a^2*d)/x^5 + (10*a*(b*d - a*e))/x^3 + (30*(-(b^2*d) + a*b*e + a*(c*d - a*f)))/x - (15*Sqrt[2]*Sqrt[c]*(b^3*d + b^2*(Sqrt[b^2 - 4*a*c]*d - a*e) + a*b*(-3*c*d - Sqrt[b^2 - 4*a*c]*e + a*f) + a*(-(c*Sqrt[b^2 - 4*a*c]*d) + 2*a*c*e + a*Sqrt[b^2 - 4*a*c]*f))*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b - Sqr t[b^2 - 4*a*c]]])/(Sqrt[b^2 - 4*a*c]*Sqrt[b - Sqrt[b^2 - 4*a*c]]) + (15*Sq rt[2]*Sqrt[c]*(b^3*d - b^2*(Sqrt[b^2 - 4*a*c]*d + a*e) + a*b*(-3*c*d + Sqr t[b^2 - 4*a*c]*e + a*f) + a*(c*Sqrt[b^2 - 4*a*c]*d + 2*a*c*e - a*Sqrt[b^2 - 4*a*c]*f))*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(Sqr t[b^2 - 4*a*c]*Sqrt[b + Sqrt[b^2 - 4*a*c]]))/(30*a^3)
Time = 1.57 (sec) , antiderivative size = 329, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2195, 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.
\(\displaystyle \int \frac {d+e x^2+f x^4}{x^6 \left (a+b x^2+c x^4\right )} \, dx\) |
\(\Big \downarrow \) 2195 |
\(\displaystyle \int \left (\frac {-a b e-a (c d-a f)+b^2 d}{a^3 x^2}+\frac {a e-b d}{a^2 x^4}+\frac {-a^2 c e-c x^2 \left (-a b e-a (c d-a f)+b^2 d\right )+a b^2 e+a b (2 c d-a f)+b^3 (-d)}{a^3 \left (a+b x^2+c x^4\right )}+\frac {d}{a x^6}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-a b e-a (c d-a f)+b^2 d}{a^3 x}+\frac {b d-a e}{3 a^2 x^3}-\frac {\sqrt {c} \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right ) \left (\frac {2 a^2 c e-a b^2 e-a b (3 c d-a f)+b^3 d}{\sqrt {b^2-4 a c}}-a b e-a (c d-a f)+b^2 d\right )}{\sqrt {2} a^3 \sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\sqrt {c} \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {\sqrt {b^2-4 a c}+b}}\right ) \left (-\frac {2 a^2 c e-a b^2 e-a b (3 c d-a f)+b^3 d}{\sqrt {b^2-4 a c}}-a b e-a (c d-a f)+b^2 d\right )}{\sqrt {2} a^3 \sqrt {\sqrt {b^2-4 a c}+b}}-\frac {d}{5 a x^5}\) |
-1/5*d/(a*x^5) + (b*d - a*e)/(3*a^2*x^3) - (b^2*d - a*b*e - a*(c*d - a*f)) /(a^3*x) - (Sqrt[c]*(b^2*d - a*b*e - a*(c*d - a*f) + (b^3*d - a*b^2*e + 2* a^2*c*e - a*b*(3*c*d - a*f))/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x) /Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*a^3*Sqrt[b - Sqrt[b^2 - 4*a*c]]) - (Sqrt[c]*(b^2*d - a*b*e - a*(c*d - a*f) - (b^3*d - a*b^2*e + 2*a^2*c*e - a*b*(3*c*d - a*f))/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*a^3*Sqrt[b + Sqrt[b^2 - 4*a*c]])
3.1.60.3.1 Defintions of rubi rules used
Int[(Pq_)*((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_ Symbol] :> Int[ExpandIntegrand[(d*x)^m*Pq*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && PolyQ[Pq, x^2] && IGtQ[p, -2]
Time = 0.14 (sec) , antiderivative size = 360, normalized size of antiderivative = 1.09
method | result | size |
default | \(-\frac {d}{5 a \,x^{5}}-\frac {a e -b d}{3 a^{2} x^{3}}-\frac {f \,a^{2}-a b e -a c d +b^{2} d}{a^{3} x}+\frac {4 c \left (\frac {\left (-f \,a^{2} \sqrt {-4 a c +b^{2}}+a b e \sqrt {-4 a c +b^{2}}+a c d \sqrt {-4 a c +b^{2}}-b^{2} d \sqrt {-4 a c +b^{2}}+a^{2} b f +2 a^{2} c e -a \,b^{2} e -3 a b c d +b^{3} d \right ) \sqrt {2}\, \arctan \left (\frac {c x \sqrt {2}}{\sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 \sqrt {-4 a c +b^{2}}\, \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}-\frac {\left (-f \,a^{2} \sqrt {-4 a c +b^{2}}+a b e \sqrt {-4 a c +b^{2}}+a c d \sqrt {-4 a c +b^{2}}-b^{2} d \sqrt {-4 a c +b^{2}}-a^{2} b f -2 a^{2} c e +a \,b^{2} e +3 a b c d -b^{3} d \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {c x \sqrt {2}}{\sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 \sqrt {-4 a c +b^{2}}\, \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{a^{3}}\) | \(360\) |
risch | \(\text {Expression too large to display}\) | \(1569\) |
-1/5*d/a/x^5-1/3*(a*e-b*d)/a^2/x^3-(a^2*f-a*b*e-a*c*d+b^2*d)/a^3/x+4/a^3*c *(1/8*(-f*a^2*(-4*a*c+b^2)^(1/2)+a*b*e*(-4*a*c+b^2)^(1/2)+a*c*d*(-4*a*c+b^ 2)^(1/2)-b^2*d*(-4*a*c+b^2)^(1/2)+a^2*b*f+2*a^2*c*e-a*b^2*e-3*a*b*c*d+b^3* 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*(-f*a^2*(-4*a*c+b^2)^(1/2)+a *b*e*(-4*a*c+b^2)^(1/2)+a*c*d*(-4*a*c+b^2)^(1/2)-b^2*d*(-4*a*c+b^2)^(1/2)- a^2*b*f-2*a^2*c*e+a*b^2*e+3*a*b*c*d-b^3*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)))
Leaf count of result is larger than twice the leaf count of optimal. 15830 vs. \(2 (289) = 578\).
Time = 45.94 (sec) , antiderivative size = 15830, normalized size of antiderivative = 48.12 \[ \int \frac {d+e x^2+f x^4}{x^6 \left (a+b x^2+c x^4\right )} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {d+e x^2+f x^4}{x^6 \left (a+b x^2+c x^4\right )} \, dx=\text {Timed out} \]
\[ \int \frac {d+e x^2+f x^4}{x^6 \left (a+b x^2+c x^4\right )} \, dx=\int { \frac {f x^{4} + e x^{2} + d}{{\left (c x^{4} + b x^{2} + a\right )} x^{6}} \,d x } \]
-integrate((a^2*b*f - (a*b*c*e - a^2*c*f - (b^2*c - a*c^2)*d)*x^2 + (b^3 - 2*a*b*c)*d - (a*b^2 - a^2*c)*e)/(c*x^4 + b*x^2 + a), x)/a^3 + 1/15*(15*(a *b*e - a^2*f - (b^2 - a*c)*d)*x^4 - 3*a^2*d + 5*(a*b*d - a^2*e)*x^2)/(a^3* x^5)
Leaf count of result is larger than twice the leaf count of optimal. 6710 vs. \(2 (289) = 578\).
Time = 1.35 (sec) , antiderivative size = 6710, normalized size of antiderivative = 20.40 \[ \int \frac {d+e x^2+f x^4}{x^6 \left (a+b x^2+c x^4\right )} \, dx=\text {Too large to display} \]
-1/8*((2*b^6*c^2 - 18*a*b^4*c^3 + 48*a^2*b^2*c^4 - 32*a^3*c^5 - sqrt(2)*sq rt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^6 + 9*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^4*c + 2*sqrt(2)*sqrt(b^2 - 4*a *c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^5*c - 24*sqrt(2)*sqrt(b^2 - 4*a*c)*s qrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b^2*c^2 - 10*sqrt(2)*sqrt(b^2 - 4*a*c)* sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^3*c^2 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt (b*c + sqrt(b^2 - 4*a*c)*c)*b^4*c^2 + 16*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b* c + sqrt(b^2 - 4*a*c)*c)*a^3*c^3 + 8*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b*c^3 + 5*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sq rt(b^2 - 4*a*c)*c)*a*b^2*c^3 - 4*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt (b^2 - 4*a*c)*c)*a^2*c^4 - 2*(b^2 - 4*a*c)*b^4*c^2 + 10*(b^2 - 4*a*c)*a*b^ 2*c^3 - 8*(b^2 - 4*a*c)*a^2*c^4)*a^2*d - (2*a*b^5*c^2 - 16*a^2*b^3*c^3 + 3 2*a^3*b*c^4 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a* b^5 + 8*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b^3* c + 2*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^4*c - 16*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^3*b*c^2 - 8 *sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b^2*c^2 - s qrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^3*c^2 + 4*sqr t(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b*c^3 - 2*(b^2 - 4*a*c)*a*b^3*c^2 + 8*(b^2 - 4*a*c)*a^2*b*c^3)*a^2*e + (2*a^2*b^4*c^2 ...
Time = 11.57 (sec) , antiderivative size = 23019, normalized size of antiderivative = 69.97 \[ \int \frac {d+e x^2+f x^4}{x^6 \left (a+b x^2+c x^4\right )} \, dx=\text {Too large to display} \]
atan(((x*(4*a^13*c^5*e^2 - 4*a^12*c^6*d^2 - 4*a^14*c^4*f^2 + 2*a^9*b^6*c^3 *d^2 - 12*a^10*b^4*c^4*d^2 + 18*a^11*b^2*c^5*d^2 + 2*a^11*b^4*c^3*e^2 - 8* a^12*b^2*c^4*e^2 + 2*a^13*b^2*c^3*f^2 + 8*a^13*c^5*d*f - 20*a^12*b*c^5*d*e + 12*a^13*b*c^4*e*f - 4*a^10*b^5*c^3*d*e + 20*a^11*b^3*c^4*d*e + 4*a^11*b ^4*c^3*d*f - 16*a^12*b^2*c^4*d*f - 4*a^12*b^3*c^3*e*f) - (-(b^9*d^2 + a^2* b^7*e^2 + b^6*d^2*(-(4*a*c - b^2)^3)^(1/2) + a^4*b^5*f^2 + 28*a^4*b*c^4*d^ 2 - 9*a^3*b^5*c*e^2 - 20*a^5*b*c^3*e^2 - 7*a^5*b^3*c*f^2 + 12*a^6*b*c^2*f^ 2 - a^5*c*f^2*(-(4*a*c - b^2)^3)^(1/2) - 2*a*b^8*d*e + 42*a^2*b^5*c^2*d^2 - 63*a^3*b^3*c^3*d^2 + a^2*b^4*e^2*(-(4*a*c - b^2)^3)^(1/2) - a^3*c^3*d^2* (-(4*a*c - b^2)^3)^(1/2) + 25*a^4*b^3*c^2*e^2 + a^4*b^2*f^2*(-(4*a*c - b^2 )^3)^(1/2) + a^4*c^2*e^2*(-(4*a*c - b^2)^3)^(1/2) - 11*a*b^7*c*d^2 + 2*a^2 *b^7*d*f - 16*a^5*c^4*d*e - 2*a^3*b^6*e*f + 16*a^6*c^3*e*f - 2*a*b^5*d*e*( -(4*a*c - b^2)^3)^(1/2) + 20*a^2*b^6*c*d*e - 18*a^3*b^5*c*d*f - 40*a^5*b*c ^3*d*f + 16*a^4*b^4*c*e*f + 6*a^2*b^2*c^2*d^2*(-(4*a*c - b^2)^3)^(1/2) - 5 *a*b^4*c*d^2*(-(4*a*c - b^2)^3)^(1/2) - 66*a^3*b^4*c^2*d*e + 76*a^4*b^2*c^ 3*d*e + 2*a^2*b^4*d*f*(-(4*a*c - b^2)^3)^(1/2) + 50*a^4*b^3*c^2*d*f - 2*a^ 3*b^3*e*f*(-(4*a*c - b^2)^3)^(1/2) + 2*a^4*c^2*d*f*(-(4*a*c - b^2)^3)^(1/2 ) - 36*a^5*b^2*c^2*e*f - 3*a^3*b^2*c*e^2*(-(4*a*c - b^2)^3)^(1/2) + 4*a^4* b*c*e*f*(-(4*a*c - b^2)^3)^(1/2) + 8*a^2*b^3*c*d*e*(-(4*a*c - b^2)^3)^(1/2 ) - 6*a^3*b*c^2*d*e*(-(4*a*c - b^2)^3)^(1/2) - 6*a^3*b^2*c*d*f*(-(4*a*c...