Integrand size = 30, antiderivative size = 174 \[ \int \frac {d+e x^2+f x^4}{x^5 \left (a+b x^2+c x^4\right )} \, dx=-\frac {d}{4 a x^4}+\frac {b d-a e}{2 a^2 x^2}+\frac {\left (b^3 d-a b^2 e+2 a^2 c e-a b (3 c d-a f)\right ) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 a^3 \sqrt {b^2-4 a c}}+\frac {\left (b^2 d-a b e-a (c d-a f)\right ) \log (x)}{a^3}-\frac {\left (b^2 d-a b e-a (c d-a f)\right ) \log \left (a+b x^2+c x^4\right )}{4 a^3} \]
-1/4*d/a/x^4+1/2*(-a*e+b*d)/a^2/x^2+(b^2*d-a*b*e-a*(-a*f+c*d))*ln(x)/a^3-1 /4*(b^2*d-a*b*e-a*(-a*f+c*d))*ln(c*x^4+b*x^2+a)/a^3+1/2*(b^3*d-a*b^2*e+2*a ^2*c*e-a*b*(-a*f+3*c*d))*arctanh((2*c*x^2+b)/(-4*a*c+b^2)^(1/2))/a^3/(-4*a *c+b^2)^(1/2)
Time = 0.22 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.80 \[ \int \frac {d+e x^2+f x^4}{x^5 \left (a+b x^2+c x^4\right )} \, dx=-\frac {\frac {a^2 d}{x^4}+\frac {2 a (-b d+a e)}{x^2}-4 \left (b^2 d-a b e+a (-c d+a f)\right ) \log (x)+\frac {\left (b^3 d+b^2 \left (\sqrt {b^2-4 a c} d-a e\right )+a b \left (-3 c d-\sqrt {b^2-4 a c} e+a f\right )+a \left (-c \sqrt {b^2-4 a c} d+2 a c e+a \sqrt {b^2-4 a c} f\right )\right ) \log \left (b-\sqrt {b^2-4 a c}+2 c x^2\right )}{\sqrt {b^2-4 a c}}+\frac {\left (-b^3 d+b^2 \left (\sqrt {b^2-4 a c} d+a e\right )-a b \left (-3 c d+\sqrt {b^2-4 a c} e+a f\right )+a \left (-c \left (\sqrt {b^2-4 a c} d+2 a e\right )+a \sqrt {b^2-4 a c} f\right )\right ) \log \left (b+\sqrt {b^2-4 a c}+2 c x^2\right )}{\sqrt {b^2-4 a c}}}{4 a^3} \]
-1/4*((a^2*d)/x^4 + (2*a*(-(b*d) + a*e))/x^2 - 4*(b^2*d - a*b*e + a*(-(c*d ) + a*f))*Log[x] + ((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))*Log[b - Sqrt[b^2 - 4*a*c] + 2*c*x^2])/Sqrt[b^2 - 4*a *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*e)) + a*Sqrt[b^2 - 4*a*c]*f))*Log[b + Sqrt[b^2 - 4*a*c] + 2*c*x^2])/Sqrt[b^2 - 4*a*c])/a^3
Time = 0.53 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2194, 2159, 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.
| \(\displaystyle \int \frac {d+e x^2+f x^4}{x^5 \left (a+b x^2+c x^4\right )} \, dx\) |
| \(\Big \downarrow \) 2194 |
| \(\displaystyle \frac {1}{2} \int \frac {f x^4+e x^2+d}{x^6 \left (c x^4+b x^2+a\right )}dx^2\) |
| \(\Big \downarrow \) 2159 |
| \(\displaystyle \frac {1}{2} \int \left (\frac {d}{a x^6}+\frac {-d b^3+a e b^2+a (2 c d-a f) b-c \left (d b^2-a e b-a (c d-a f)\right ) x^2-a^2 c e}{a^3 \left (c x^4+b x^2+a\right )}+\frac {d b^2-a e b-a (c d-a f)}{a^3 x^2}+\frac {a e-b d}{a^2 x^4}\right )dx^2\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} \left (\frac {\log \left (x^2\right ) \left (-a b e-a (c d-a f)+b^2 d\right )}{a^3}-\frac {\log \left (a+b x^2+c x^4\right ) \left (-a b e-a (c d-a f)+b^2 d\right )}{2 a^3}+\frac {b d-a e}{a^2 x^2}+\frac {\text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right ) \left (2 a^2 c e-a b^2 e-a b (3 c d-a f)+b^3 d\right )}{a^3 \sqrt {b^2-4 a c}}-\frac {d}{2 a x^4}\right )\) |
(-1/2*d/(a*x^4) + (b*d - a*e)/(a^2*x^2) + ((b^3*d - a*b^2*e + 2*a^2*c*e - a*b*(3*c*d - a*f))*ArcTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(a^3*Sqrt[b^2 - 4*a*c]) + ((b^2*d - a*b*e - a*(c*d - a*f))*Log[x^2])/a^3 - ((b^2*d - a* b*e - a*(c*d - a*f))*Log[a + b*x^2 + c*x^4])/(2*a^3))/2
3.1.53.3.1 Defintions of rubi rules used
Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x ], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] : > Simp[1/2 Subst[Int[x^((m - 1)/2)*SubstFor[x^2, Pq, x]*(a + b*x + c*x^2) ^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x^2] && IntegerQ [(m - 1)/2]
Time = 0.12 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.17
| method | result | size |
| default | \(-\frac {d}{4 a \,x^{4}}-\frac {a e -b d}{2 a^{2} x^{2}}+\frac {\left (f \,a^{2}-a b e -a c d +b^{2} d \right ) \ln \left (x \right )}{a^{3}}-\frac {\frac {\left (a^{2} c f -a b c e -a \,c^{2} d +b^{2} c d \right ) \ln \left (c \,x^{4}+b \,x^{2}+a \right )}{2 c}+\frac {2 \left (a^{2} b f +a^{2} c e -a \,b^{2} e -2 a b c d +b^{3} d -\frac {\left (a^{2} c f -a b c e -a \,c^{2} d +b^{2} c d \right ) b}{2 c}\right ) \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{2 a^{3}}\) | \(203\) |
| risch | \(\frac {-\frac {\left (a e -b d \right ) x^{2}}{2 a^{2}}-\frac {d}{4 a}}{x^{4}}+\frac {\ln \left (x \right ) f}{a}-\frac {\ln \left (x \right ) b e}{a^{2}}-\frac {\ln \left (x \right ) c d}{a^{2}}+\frac {\ln \left (x \right ) b^{2} d}{a^{3}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (4 c \,a^{4}-a^{3} b^{2}\right ) \textit {\_Z}^{2}+\left (4 a^{3} c f -a^{2} b^{2} f -4 a^{2} b c e -4 a^{2} c^{2} d +a \,b^{3} e +5 a \,b^{2} c d -d \,b^{4}\right ) \textit {\_Z} +c \,f^{2} a^{2}-a b c e f -2 c^{2} d f a +a \,c^{2} e^{2}+b^{2} c d f -e d \,c^{2} b +c^{3} d^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (10 c \,a^{5}-3 b^{2} a^{4}\right ) \textit {\_R}^{2}+\left (5 a^{4} c f -4 e c b \,a^{3}-5 d \,c^{2} a^{3}+4 d c \,b^{2} a^{2}\right ) \textit {\_R} +2 a^{2} c^{2} e^{2}-4 a b \,c^{2} d e +2 d^{2} c^{2} b^{2}\right ) x^{2}-a^{5} b \,\textit {\_R}^{2}+\left (2 a^{4} b f +a^{4} c e -2 a^{3} b^{2} e -3 a^{3} b c d +2 a^{2} b^{3} d \right ) \textit {\_R} -2 a^{3} c e f +2 a^{2} b c d f +2 a^{2} b c \,e^{2}+2 a^{2} c^{2} d e -4 b^{2} c d e a -2 b \,c^{2} d^{2} a +2 b^{3} c \,d^{2}\right )\right )}{2}\) | \(411\) |
-1/4*d/a/x^4-1/2*(a*e-b*d)/a^2/x^2+(a^2*f-a*b*e-a*c*d+b^2*d)/a^3*ln(x)-1/2 /a^3*(1/2*(a^2*c*f-a*b*c*e-a*c^2*d+b^2*c*d)/c*ln(c*x^4+b*x^2+a)+2*(a^2*b*f +a^2*c*e-a*b^2*e-2*a*b*c*d+b^3*d-1/2*(a^2*c*f-a*b*c*e-a*c^2*d+b^2*c*d)*b/c )/(4*a*c-b^2)^(1/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2)))
Time = 0.81 (sec) , antiderivative size = 609, normalized size of antiderivative = 3.50 \[ \int \frac {d+e x^2+f x^4}{x^5 \left (a+b x^2+c x^4\right )} \, dx=\left [\frac {{\left (a^{2} b f + {\left (b^{3} - 3 \, a b c\right )} d - {\left (a b^{2} - 2 \, a^{2} c\right )} e\right )} \sqrt {b^{2} - 4 \, a c} x^{4} \log \left (\frac {2 \, c^{2} x^{4} + 2 \, b c x^{2} + b^{2} - 2 \, a c + {\left (2 \, c x^{2} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{4} + b x^{2} + a}\right ) - {\left ({\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} d - {\left (a b^{3} - 4 \, a^{2} b c\right )} e + {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} f\right )} x^{4} \log \left (c x^{4} + b x^{2} + a\right ) + 4 \, {\left ({\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} d - {\left (a b^{3} - 4 \, a^{2} b c\right )} e + {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} f\right )} x^{4} \log \left (x\right ) + 2 \, {\left ({\left (a b^{3} - 4 \, a^{2} b c\right )} d - {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} e\right )} x^{2} - {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} d}{4 \, {\left (a^{3} b^{2} - 4 \, a^{4} c\right )} x^{4}}, \frac {2 \, {\left (a^{2} b f + {\left (b^{3} - 3 \, a b c\right )} d - {\left (a b^{2} - 2 \, a^{2} c\right )} e\right )} \sqrt {-b^{2} + 4 \, a c} x^{4} \arctan \left (-\frac {{\left (2 \, c x^{2} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) - {\left ({\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} d - {\left (a b^{3} - 4 \, a^{2} b c\right )} e + {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} f\right )} x^{4} \log \left (c x^{4} + b x^{2} + a\right ) + 4 \, {\left ({\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} d - {\left (a b^{3} - 4 \, a^{2} b c\right )} e + {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} f\right )} x^{4} \log \left (x\right ) + 2 \, {\left ({\left (a b^{3} - 4 \, a^{2} b c\right )} d - {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} e\right )} x^{2} - {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} d}{4 \, {\left (a^{3} b^{2} - 4 \, a^{4} c\right )} x^{4}}\right ] \]
[1/4*((a^2*b*f + (b^3 - 3*a*b*c)*d - (a*b^2 - 2*a^2*c)*e)*sqrt(b^2 - 4*a*c )*x^4*log((2*c^2*x^4 + 2*b*c*x^2 + b^2 - 2*a*c + (2*c*x^2 + b)*sqrt(b^2 - 4*a*c))/(c*x^4 + b*x^2 + a)) - ((b^4 - 5*a*b^2*c + 4*a^2*c^2)*d - (a*b^3 - 4*a^2*b*c)*e + (a^2*b^2 - 4*a^3*c)*f)*x^4*log(c*x^4 + b*x^2 + a) + 4*((b^ 4 - 5*a*b^2*c + 4*a^2*c^2)*d - (a*b^3 - 4*a^2*b*c)*e + (a^2*b^2 - 4*a^3*c) *f)*x^4*log(x) + 2*((a*b^3 - 4*a^2*b*c)*d - (a^2*b^2 - 4*a^3*c)*e)*x^2 - ( a^2*b^2 - 4*a^3*c)*d)/((a^3*b^2 - 4*a^4*c)*x^4), 1/4*(2*(a^2*b*f + (b^3 - 3*a*b*c)*d - (a*b^2 - 2*a^2*c)*e)*sqrt(-b^2 + 4*a*c)*x^4*arctan(-(2*c*x^2 + b)*sqrt(-b^2 + 4*a*c)/(b^2 - 4*a*c)) - ((b^4 - 5*a*b^2*c + 4*a^2*c^2)*d - (a*b^3 - 4*a^2*b*c)*e + (a^2*b^2 - 4*a^3*c)*f)*x^4*log(c*x^4 + b*x^2 + a ) + 4*((b^4 - 5*a*b^2*c + 4*a^2*c^2)*d - (a*b^3 - 4*a^2*b*c)*e + (a^2*b^2 - 4*a^3*c)*f)*x^4*log(x) + 2*((a*b^3 - 4*a^2*b*c)*d - (a^2*b^2 - 4*a^3*c)* e)*x^2 - (a^2*b^2 - 4*a^3*c)*d)/((a^3*b^2 - 4*a^4*c)*x^4)]
Timed out. \[ \int \frac {d+e x^2+f x^4}{x^5 \left (a+b x^2+c x^4\right )} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {d+e x^2+f x^4}{x^5 \left (a+b x^2+c x^4\right )} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for more deta
Time = 0.57 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.18 \[ \int \frac {d+e x^2+f x^4}{x^5 \left (a+b x^2+c x^4\right )} \, dx=-\frac {{\left (b^{2} d - a c d - a b e + a^{2} f\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, a^{3}} + \frac {{\left (b^{2} d - a c d - a b e + a^{2} f\right )} \log \left (x^{2}\right )}{2 \, a^{3}} - \frac {{\left (b^{3} d - 3 \, a b c d - a b^{2} e + 2 \, a^{2} c e + a^{2} b f\right )} \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{2 \, \sqrt {-b^{2} + 4 \, a c} a^{3}} - \frac {3 \, b^{2} d x^{4} - 3 \, a c d x^{4} - 3 \, a b e x^{4} + 3 \, a^{2} f x^{4} - 2 \, a b d x^{2} + 2 \, a^{2} e x^{2} + a^{2} d}{4 \, a^{3} x^{4}} \]
-1/4*(b^2*d - a*c*d - a*b*e + a^2*f)*log(c*x^4 + b*x^2 + a)/a^3 + 1/2*(b^2 *d - a*c*d - a*b*e + a^2*f)*log(x^2)/a^3 - 1/2*(b^3*d - 3*a*b*c*d - a*b^2* e + 2*a^2*c*e + a^2*b*f)*arctan((2*c*x^2 + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b ^2 + 4*a*c)*a^3) - 1/4*(3*b^2*d*x^4 - 3*a*c*d*x^4 - 3*a*b*e*x^4 + 3*a^2*f* x^4 - 2*a*b*d*x^2 + 2*a^2*e*x^2 + a^2*d)/(a^3*x^4)
Time = 14.76 (sec) , antiderivative size = 6187, normalized size of antiderivative = 35.56 \[ \int \frac {d+e x^2+f x^4}{x^5 \left (a+b x^2+c x^4\right )} \, dx=\text {Too large to display} \]
(log(x)*(b^2*d + a^2*f - a*b*e - a*c*d))/a^3 - (d/(4*a) + (x^2*(a*e - b*d) )/(2*a^2))/x^4 + (log(((((((2*c^3*x^2*(b^3*d - a*b^2*e + 5*a^2*b*f - 10*a^ 2*c*e + 5*a*b*c*d))/a^2 + (4*b*c^2*(b^3*d - a*b^2*e + a^2*b*f + a^2*c*e - 2*a*b*c*d))/a^2 + (b*c^2*(a*b + 3*b^2*x^2 - 10*a*c*x^2)*(b^2*d + a^2*f + a ^3*(-(b^3*d - a*b^2*e + a^2*b*f + 2*a^2*c*e - 3*a*b*c*d)^2/(a^6*(4*a*c - b ^2)))^(1/2) - a*b*e - a*c*d))/a^3)*(b^2*d + a^2*f + a^3*(-(b^3*d - a*b^2*e + a^2*b*f + 2*a^2*c*e - 3*a*b*c*d)^2/(a^6*(4*a*c - b^2)))^(1/2) - a*b*e - a*c*d))/(4*a^3) + (c^3*(a*e - b*d)*(4*b^3*d - 4*a*b^2*e + 4*a^2*b*f + a^2 *c*e - 5*a*b*c*d))/a^4 + (c^4*x^2*(a*e - b*d)*(6*b^2*d + 5*a^2*f - 6*a*b*e - 5*a*c*d))/a^4)*(b^2*d + a^2*f + a^3*(-(b^3*d - a*b^2*e + a^2*b*f + 2*a^ 2*c*e - 3*a*b*c*d)^2/(a^6*(4*a*c - b^2)))^(1/2) - a*b*e - a*c*d))/(4*a^3) + (c^4*(a*e - b*d)^2*(b^2*d + a^2*f - a*b*e - a*c*d))/a^6 - (c^5*x^2*(a*e - b*d)^3)/a^6)*((((c^3*(a*e - b*d)*(4*b^3*d - 4*a*b^2*e + 4*a^2*b*f + a^2* c*e - 5*a*b*c*d))/a^4 - (((2*c^3*x^2*(b^3*d - a*b^2*e + 5*a^2*b*f - 10*a^2 *c*e + 5*a*b*c*d))/a^2 + (4*b*c^2*(b^3*d - a*b^2*e + a^2*b*f + a^2*c*e - 2 *a*b*c*d))/a^2 - (b*c^2*(a*b + 3*b^2*x^2 - 10*a*c*x^2)*(a^3*(-(b^3*d - a*b ^2*e + a^2*b*f + 2*a^2*c*e - 3*a*b*c*d)^2/(a^6*(4*a*c - b^2)))^(1/2) - a^2 *f - b^2*d + a*b*e + a*c*d))/a^3)*(a^3*(-(b^3*d - a*b^2*e + a^2*b*f + 2*a^ 2*c*e - 3*a*b*c*d)^2/(a^6*(4*a*c - b^2)))^(1/2) - a^2*f - b^2*d + a*b*e + a*c*d))/(4*a^3) + (c^4*x^2*(a*e - b*d)*(6*b^2*d + 5*a^2*f - 6*a*b*e - 5...