Integrand size = 30, antiderivative size = 118 \[ \int \frac {d+e x^2+f x^4}{x^3 \left (a+b x^2+c x^4\right )} \, dx=-\frac {d}{2 a x^2}-\frac {\left (b^2 d-a b e-2 a (c d-a f)\right ) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 a^2 \sqrt {b^2-4 a c}}-\frac {(b d-a e) \log (x)}{a^2}+\frac {(b d-a e) \log \left (a+b x^2+c x^4\right )}{4 a^2} \]
-1/2*d/a/x^2-(-a*e+b*d)*ln(x)/a^2+1/4*(-a*e+b*d)*ln(c*x^4+b*x^2+a)/a^2-1/2 *(b^2*d-a*b*e-2*a*(-a*f+c*d))*arctanh((2*c*x^2+b)/(-4*a*c+b^2)^(1/2))/a^2/ (-4*a*c+b^2)^(1/2)
Time = 0.09 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.72 \[ \int \frac {d+e x^2+f x^4}{x^3 \left (a+b x^2+c x^4\right )} \, dx=\frac {-\frac {2 a d}{x^2}+4 (-b d+a e) \log (x)+\frac {\left (b^2 d+b \left (\sqrt {b^2-4 a c} d-a e\right )+a \left (-2 c d-\sqrt {b^2-4 a c} e+2 a f\right )\right ) \log \left (b-\sqrt {b^2-4 a c}+2 c x^2\right )}{\sqrt {b^2-4 a c}}+\frac {\left (-b^2 d+b \left (\sqrt {b^2-4 a c} d+a e\right )-a \left (-2 c d+\sqrt {b^2-4 a c} e+2 a f\right )\right ) \log \left (b+\sqrt {b^2-4 a c}+2 c x^2\right )}{\sqrt {b^2-4 a c}}}{4 a^2} \]
((-2*a*d)/x^2 + 4*(-(b*d) + a*e)*Log[x] + ((b^2*d + b*(Sqrt[b^2 - 4*a*c]*d - a*e) + a*(-2*c*d - Sqrt[b^2 - 4*a*c]*e + 2*a*f))*Log[b - Sqrt[b^2 - 4*a *c] + 2*c*x^2])/Sqrt[b^2 - 4*a*c] + ((-(b^2*d) + b*(Sqrt[b^2 - 4*a*c]*d + a*e) - a*(-2*c*d + Sqrt[b^2 - 4*a*c]*e + 2*a*f))*Log[b + Sqrt[b^2 - 4*a*c] + 2*c*x^2])/Sqrt[b^2 - 4*a*c])/(4*a^2)
Time = 0.44 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.02, 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.
\(\displaystyle \int \frac {d+e x^2+f x^4}{x^3 \left (a+b x^2+c x^4\right )} \, dx\) |
\(\Big \downarrow \) 2194 |
\(\displaystyle \frac {1}{2} \int \frac {f x^4+e x^2+d}{x^4 \left (c x^4+b x^2+a\right )}dx^2\) |
\(\Big \downarrow \) 2159 |
\(\displaystyle \frac {1}{2} \int \left (\frac {d}{a x^4}+\frac {d b^2-a e b+c (b d-a e) x^2-a (c d-a f)}{a^2 \left (c x^4+b x^2+a\right )}+\frac {a e-b d}{a^2 x^2}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {\text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right ) \left (-a b e-2 a (c d-a f)+b^2 d\right )}{a^2 \sqrt {b^2-4 a c}}+\frac {(b d-a e) \log \left (a+b x^2+c x^4\right )}{2 a^2}-\frac {\log \left (x^2\right ) (b d-a e)}{a^2}-\frac {d}{a x^2}\right )\) |
(-(d/(a*x^2)) - ((b^2*d - a*b*e - 2*a*(c*d - a*f))*ArcTanh[(b + 2*c*x^2)/S qrt[b^2 - 4*a*c]])/(a^2*Sqrt[b^2 - 4*a*c]) - ((b*d - a*e)*Log[x^2])/a^2 + ((b*d - a*e)*Log[a + b*x^2 + c*x^4])/(2*a^2))/2
3.1.52.3.1 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 = 0.10 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.12
method | result | size |
default | \(-\frac {d}{2 a \,x^{2}}+\frac {\left (a e -b d \right ) \ln \left (x \right )}{a^{2}}+\frac {\frac {\left (-a c e +b c d \right ) \ln \left (c \,x^{4}+b \,x^{2}+a \right )}{2 c}+\frac {2 \left (f \,a^{2}-a b e -a c d +b^{2} d -\frac {\left (-a c e +b c d \right ) b}{2 c}\right ) \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{2 a^{2}}\) | \(132\) |
risch | \(-\frac {d}{2 a \,x^{2}}+\frac {\ln \left (x \right ) e}{a}-\frac {\ln \left (x \right ) b d}{a^{2}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (4 a^{3} c -a^{2} b^{2}\right ) \textit {\_Z}^{2}+\left (4 a^{2} c e -a \,b^{2} e -4 a b c d +b^{3} d \right ) \textit {\_Z} +a^{2} f^{2}-a b e f -2 a c d f +e^{2} a c +b^{2} d f -b c d e +c^{2} d^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (10 a^{3} c -3 a^{2} b^{2}\right ) \textit {\_R}^{2}+\left (-a^{2} b f +5 a^{2} c e -4 a b c d \right ) \textit {\_R} +2 a^{2} f^{2}-4 a c d f +2 c^{2} d^{2}\right ) x^{2}-a^{3} b \,\textit {\_R}^{2}+\left (-a^{3} f +2 a^{2} b e +a^{2} c d -2 a \,b^{2} d \right ) \textit {\_R} +2 a^{2} e f -2 a b d f -2 a c d e +2 b c \,d^{2}\right )\right )}{2}\) | \(262\) |
-1/2*d/a/x^2+(a*e-b*d)/a^2*ln(x)+1/2/a^2*(1/2*(-a*c*e+b*c*d)/c*ln(c*x^4+b* x^2+a)+2*(f*a^2-a*b*e-a*c*d+b^2*d-1/2*(-a*c*e+b*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 = 0.55 (sec) , antiderivative size = 399, normalized size of antiderivative = 3.38 \[ \int \frac {d+e x^2+f x^4}{x^3 \left (a+b x^2+c x^4\right )} \, dx=\left [-\frac {{\left (a b e - 2 \, a^{2} f - {\left (b^{2} - 2 \, a c\right )} d\right )} \sqrt {b^{2} - 4 \, a c} x^{2} \log \left (\frac {2 \, c^{2} x^{4} + 2 \, b c x^{2} + b^{2} - 2 \, a c - {\left (2 \, c x^{2} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{4} + b x^{2} + a}\right ) - {\left ({\left (b^{3} - 4 \, a b c\right )} d - {\left (a b^{2} - 4 \, a^{2} c\right )} e\right )} x^{2} \log \left (c x^{4} + b x^{2} + a\right ) + 4 \, {\left ({\left (b^{3} - 4 \, a b c\right )} d - {\left (a b^{2} - 4 \, a^{2} c\right )} e\right )} x^{2} \log \left (x\right ) + 2 \, {\left (a b^{2} - 4 \, a^{2} c\right )} d}{4 \, {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} x^{2}}, \frac {2 \, {\left (a b e - 2 \, a^{2} f - {\left (b^{2} - 2 \, a c\right )} d\right )} \sqrt {-b^{2} + 4 \, a c} x^{2} \arctan \left (-\frac {{\left (2 \, c x^{2} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) + {\left ({\left (b^{3} - 4 \, a b c\right )} d - {\left (a b^{2} - 4 \, a^{2} c\right )} e\right )} x^{2} \log \left (c x^{4} + b x^{2} + a\right ) - 4 \, {\left ({\left (b^{3} - 4 \, a b c\right )} d - {\left (a b^{2} - 4 \, a^{2} c\right )} e\right )} x^{2} \log \left (x\right ) - 2 \, {\left (a b^{2} - 4 \, a^{2} c\right )} d}{4 \, {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} x^{2}}\right ] \]
[-1/4*((a*b*e - 2*a^2*f - (b^2 - 2*a*c)*d)*sqrt(b^2 - 4*a*c)*x^2*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^3 - 4*a*b*c)*d - (a*b^2 - 4*a^2*c)*e)*x^2*log(c*x^4 + b*x^2 + a) + 4*((b^3 - 4*a*b*c)*d - (a*b^2 - 4*a^2*c)*e)*x^2*log(x) + 2*(a *b^2 - 4*a^2*c)*d)/((a^2*b^2 - 4*a^3*c)*x^2), 1/4*(2*(a*b*e - 2*a^2*f - (b ^2 - 2*a*c)*d)*sqrt(-b^2 + 4*a*c)*x^2*arctan(-(2*c*x^2 + b)*sqrt(-b^2 + 4* a*c)/(b^2 - 4*a*c)) + ((b^3 - 4*a*b*c)*d - (a*b^2 - 4*a^2*c)*e)*x^2*log(c* x^4 + b*x^2 + a) - 4*((b^3 - 4*a*b*c)*d - (a*b^2 - 4*a^2*c)*e)*x^2*log(x) - 2*(a*b^2 - 4*a^2*c)*d)/((a^2*b^2 - 4*a^3*c)*x^2)]
Timed out. \[ \int \frac {d+e x^2+f x^4}{x^3 \left (a+b x^2+c x^4\right )} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {d+e x^2+f x^4}{x^3 \left (a+b x^2+c x^4\right )} \, dx=\text {Exception raised: ValueError} \]
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 = 0.63 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.11 \[ \int \frac {d+e x^2+f x^4}{x^3 \left (a+b x^2+c x^4\right )} \, dx=\frac {{\left (b d - a e\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, a^{2}} - \frac {{\left (b d - a e\right )} \log \left (x^{2}\right )}{2 \, a^{2}} + \frac {{\left (b^{2} d - 2 \, a c d - a b e + 2 \, a^{2} f\right )} \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{2 \, \sqrt {-b^{2} + 4 \, a c} a^{2}} + \frac {b d x^{2} - a e x^{2} - a d}{2 \, a^{2} x^{2}} \]
1/4*(b*d - a*e)*log(c*x^4 + b*x^2 + a)/a^2 - 1/2*(b*d - a*e)*log(x^2)/a^2 + 1/2*(b^2*d - 2*a*c*d - a*b*e + 2*a^2*f)*arctan((2*c*x^2 + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4*a*c)*a^2) + 1/2*(b*d*x^2 - a*e*x^2 - a*d)/(a^2*x^2 )
Time = 12.76 (sec) , antiderivative size = 4437, normalized size of antiderivative = 37.60 \[ \int \frac {d+e x^2+f x^4}{x^3 \left (a+b x^2+c x^4\right )} \, dx=\text {Too large to display} \]
(log(x)*(a*e - b*d))/a^2 - d/(2*a*x^2) - (log(((c^2*(a*e - b*d)*(a*f - c*d )^2)/a^3 - ((b*d - a*e + a^2*(-(b^2*d + 2*a^2*f - a*b*e - 2*a*c*d)^2/(a^4* (4*a*c - b^2)))^(1/2))*(((b*d - a*e + a^2*(-(b^2*d + 2*a^2*f - a*b*e - 2*a *c*d)^2/(a^4*(4*a*c - b^2)))^(1/2))*((2*c^2*x^2*(10*a*c^2*d + 4*a*b^2*f + b^2*c*d - 10*a^2*c*f - 5*a*b*c*e))/a + (4*b*c^2*(b^2*d + a^2*f - a*b*e - a *c*d))/a + (b*c^2*(b*d - a*e + a^2*(-(b^2*d + 2*a^2*f - a*b*e - 2*a*c*d)^2 /(a^4*(4*a*c - b^2)))^(1/2))*(a*b + 3*b^2*x^2 - 10*a*c*x^2))/a^2))/(4*a^2) + (c^2*(a*f - c*d)*(4*b^2*d + a^2*f - 4*a*b*e - a*c*d))/a^2 - (c^2*x^2*(a *f - c*d)*(a*b*f + 5*a*c*e - 6*b*c*d))/a^2))/(4*a^2) + (c^2*x^2*(a*f - c*d )^3)/a^3)*((c^2*(a*e - b*d)*(a*f - c*d)^2)/a^3 - ((a*e - b*d + a^2*(-(b^2* d + 2*a^2*f - a*b*e - 2*a*c*d)^2/(a^4*(4*a*c - b^2)))^(1/2))*(((a*e - b*d + a^2*(-(b^2*d + 2*a^2*f - a*b*e - 2*a*c*d)^2/(a^4*(4*a*c - b^2)))^(1/2))* ((2*c^2*x^2*(10*a*c^2*d + 4*a*b^2*f + b^2*c*d - 10*a^2*c*f - 5*a*b*c*e))/a + (4*b*c^2*(b^2*d + a^2*f - a*b*e - a*c*d))/a - (b*c^2*(a*e - b*d + a^2*( -(b^2*d + 2*a^2*f - a*b*e - 2*a*c*d)^2/(a^4*(4*a*c - b^2)))^(1/2))*(a*b + 3*b^2*x^2 - 10*a*c*x^2))/a^2))/(4*a^2) - (c^2*(a*f - c*d)*(4*b^2*d + a^2*f - 4*a*b*e - a*c*d))/a^2 + (c^2*x^2*(a*f - c*d)*(a*b*f + 5*a*c*e - 6*b*c*d ))/a^2))/(4*a^2) + (c^2*x^2*(a*f - c*d)^3)/a^3))*(2*b^3*d - 2*a*b^2*e + 8* a^2*c*e - 8*a*b*c*d))/(2*(16*a^3*c - 4*a^2*b^2)) - (atan((16*a^6*(4*a*c - b^2)^(3/2)*(x^2*((((c^5*d^3 - a^3*c^2*f^3 + 3*a^2*c^3*d*f^2 - 3*a*c^4*d...