Integrand size = 20, antiderivative size = 63 \[ \int \frac {x^4}{a x+b x^3+c x^5} \, dx=\frac {b \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c \sqrt {b^2-4 a c}}+\frac {\log \left (a+b x^2+c x^4\right )}{4 c} \]
Time = 0.02 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.98 \[ \int \frac {x^4}{a x+b x^3+c x^5} \, dx=\frac {-\frac {2 b \arctan \left (\frac {b+2 c x^2}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}+\log \left (a+b x^2+c x^4\right )}{4 c} \]
((-2*b*ArcTan[(b + 2*c*x^2)/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c] + Log[ a + b*x^2 + c*x^4])/(4*c)
Time = 0.23 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {9, 1434, 1142, 1083, 219, 1103}
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.
\(\displaystyle \int \frac {x^4}{a x+b x^3+c x^5} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {x^3}{a+b x^2+c x^4}dx\) |
\(\Big \downarrow \) 1434 |
\(\displaystyle \frac {1}{2} \int \frac {x^2}{c x^4+b x^2+a}dx^2\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {2 c x^2+b}{c x^4+b x^2+a}dx^2}{2 c}-\frac {b \int \frac {1}{c x^4+b x^2+a}dx^2}{2 c}\right )\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {1}{2} \left (\frac {b \int \frac {1}{-x^4+b^2-4 a c}d\left (2 c x^2+b\right )}{c}+\frac {\int \frac {2 c x^2+b}{c x^4+b x^2+a}dx^2}{2 c}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {2 c x^2+b}{c x^4+b x^2+a}dx^2}{2 c}+\frac {b \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{c \sqrt {b^2-4 a c}}\right )\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {1}{2} \left (\frac {b \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{c \sqrt {b^2-4 a c}}+\frac {\log \left (a+b x^2+c x^4\right )}{2 c}\right )\) |
((b*ArcTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(c*Sqrt[b^2 - 4*a*c]) + Log[ a + b*x^2 + c*x^4]/(2*c))/2
3.1.82.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp [1/2 Subst[Int[x^((m - 1)/2)*(a + b*x + c*x^2)^p, x], x, x^2], x] /; Free Q[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]
Time = 0.05 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.95
method | result | size |
default | \(\frac {\ln \left (c \,x^{4}+b \,x^{2}+a \right )}{4 c}-\frac {b \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{2 c \sqrt {4 a c -b^{2}}}\) | \(60\) |
risch | \(\frac {\ln \left (\left (-4 a b c +b^{3}+\sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, b \right ) x^{2}+2 \sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, a \right ) a}{4 a c -b^{2}}-\frac {\ln \left (\left (-4 a b c +b^{3}+\sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, b \right ) x^{2}+2 \sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, a \right ) b^{2}}{4 c \left (4 a c -b^{2}\right )}+\frac {\ln \left (\left (-4 a b c +b^{3}+\sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, b \right ) x^{2}+2 \sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, a \right ) \sqrt {-b^{2} \left (4 a c -b^{2}\right )}}{4 c \left (4 a c -b^{2}\right )}+\frac {\ln \left (\left (-4 a b c +b^{3}-\sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, b \right ) x^{2}-2 \sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, a \right ) a}{4 a c -b^{2}}-\frac {\ln \left (\left (-4 a b c +b^{3}-\sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, b \right ) x^{2}-2 \sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, a \right ) b^{2}}{4 c \left (4 a c -b^{2}\right )}-\frac {\ln \left (\left (-4 a b c +b^{3}-\sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, b \right ) x^{2}-2 \sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, a \right ) \sqrt {-b^{2} \left (4 a c -b^{2}\right )}}{4 c \left (4 a c -b^{2}\right )}\) | \(465\) |
Time = 0.27 (sec) , antiderivative size = 197, normalized size of antiderivative = 3.13 \[ \int \frac {x^4}{a x+b x^3+c x^5} \, dx=\left [\frac {\sqrt {b^{2} - 4 \, a c} b \log \left (\frac {2 \, c^{2} x^{4} + 2 \, b c x^{2} + b^{2} - 2 \, a c + {\left (2 \, c x^{2} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{4} + b x^{2} + a}\right ) + {\left (b^{2} - 4 \, a c\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, {\left (b^{2} c - 4 \, a c^{2}\right )}}, \frac {2 \, \sqrt {-b^{2} + 4 \, a c} b \arctan \left (-\frac {{\left (2 \, c x^{2} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) + {\left (b^{2} - 4 \, a c\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, {\left (b^{2} c - 4 \, a c^{2}\right )}}\right ] \]
[1/4*(sqrt(b^2 - 4*a*c)*b*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^2 - 4*a*c)*log(c*x^4 + b*x^2 + a))/(b^2*c - 4*a*c^2), 1/4*(2*sqrt(-b^2 + 4*a*c)*b*arctan(-(2*c *x^2 + b)*sqrt(-b^2 + 4*a*c)/(b^2 - 4*a*c)) + (b^2 - 4*a*c)*log(c*x^4 + b* x^2 + a))/(b^2*c - 4*a*c^2)]
Leaf count of result is larger than twice the leaf count of optimal. 223 vs. \(2 (54) = 108\).
Time = 0.50 (sec) , antiderivative size = 223, normalized size of antiderivative = 3.54 \[ \int \frac {x^4}{a x+b x^3+c x^5} \, dx=\left (- \frac {b \sqrt {- 4 a c + b^{2}}}{4 c \left (4 a c - b^{2}\right )} + \frac {1}{4 c}\right ) \log {\left (x^{2} + \frac {- 8 a c \left (- \frac {b \sqrt {- 4 a c + b^{2}}}{4 c \left (4 a c - b^{2}\right )} + \frac {1}{4 c}\right ) + 2 a + 2 b^{2} \left (- \frac {b \sqrt {- 4 a c + b^{2}}}{4 c \left (4 a c - b^{2}\right )} + \frac {1}{4 c}\right )}{b} \right )} + \left (\frac {b \sqrt {- 4 a c + b^{2}}}{4 c \left (4 a c - b^{2}\right )} + \frac {1}{4 c}\right ) \log {\left (x^{2} + \frac {- 8 a c \left (\frac {b \sqrt {- 4 a c + b^{2}}}{4 c \left (4 a c - b^{2}\right )} + \frac {1}{4 c}\right ) + 2 a + 2 b^{2} \left (\frac {b \sqrt {- 4 a c + b^{2}}}{4 c \left (4 a c - b^{2}\right )} + \frac {1}{4 c}\right )}{b} \right )} \]
(-b*sqrt(-4*a*c + b**2)/(4*c*(4*a*c - b**2)) + 1/(4*c))*log(x**2 + (-8*a*c *(-b*sqrt(-4*a*c + b**2)/(4*c*(4*a*c - b**2)) + 1/(4*c)) + 2*a + 2*b**2*(- b*sqrt(-4*a*c + b**2)/(4*c*(4*a*c - b**2)) + 1/(4*c)))/b) + (b*sqrt(-4*a*c + b**2)/(4*c*(4*a*c - b**2)) + 1/(4*c))*log(x**2 + (-8*a*c*(b*sqrt(-4*a*c + b**2)/(4*c*(4*a*c - b**2)) + 1/(4*c)) + 2*a + 2*b**2*(b*sqrt(-4*a*c + b **2)/(4*c*(4*a*c - b**2)) + 1/(4*c)))/b)
\[ \int \frac {x^4}{a x+b x^3+c x^5} \, dx=\int { \frac {x^{4}}{c x^{5} + b x^{3} + a x} \,d x } \]
Time = 0.28 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.94 \[ \int \frac {x^4}{a x+b x^3+c x^5} \, dx=-\frac {b \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{2 \, \sqrt {-b^{2} + 4 \, a c} c} + \frac {\log \left (c x^{4} + b x^{2} + a\right )}{4 \, c} \]
-1/2*b*arctan((2*c*x^2 + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4*a*c)*c) + 1 /4*log(c*x^4 + b*x^2 + a)/c
Time = 0.09 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.87 \[ \int \frac {x^4}{a x+b x^3+c x^5} \, dx=\frac {4\,a\,c\,\ln \left (c\,x^4+b\,x^2+a\right )}{16\,a\,c^2-4\,b^2\,c}-\frac {b^2\,\ln \left (c\,x^4+b\,x^2+a\right )}{16\,a\,c^2-4\,b^2\,c}-\frac {b\,\mathrm {atan}\left (\frac {b}{\sqrt {4\,a\,c-b^2}}+\frac {2\,c\,x^2}{\sqrt {4\,a\,c-b^2}}\right )}{2\,c\,\sqrt {4\,a\,c-b^2}} \]