\(\int \frac {x^4}{a x+b x^3+c x^5} \, dx\) [82]

Optimal result

Integrand size = 20, antiderivative size = 63 \[ \int \frac {x^4}{a x+b x^3+c x^5} \, dx=\frac {b \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c \sqrt {b^2-4 a c}}+\frac {\log \left (a+b x^2+c x^4\right )}{4 c} \]

[Out]

1/4*ln(c*x^4+b*x^2+a)/c+1/2*b*arctanh((2*c*x^2+b)/(-4*a*c+b^2)^(1/2))/c/(-4*a*c+b^2)^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1599, 1128, 648, 632, 212, 642} \[ \int \frac {x^4}{a x+b x^3+c x^5} \, dx=\frac {b \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c \sqrt {b^2-4 a c}}+\frac {\log \left (a+b x^2+c x^4\right )}{4 c} \]

[In]

Int[x^4/(a*x + b*x^3 + c*x^5),x]

[Out]

(b*ArcTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(2*c*Sqrt[b^2 - 4*a*c]) + Log[a + b*x^2 + c*x^4]/(4*c)

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[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]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 1128

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]

Rule 1599

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(m +
n*p)*(a + b*x^(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, m, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] &
& PosQ[r - p]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^3}{a+b x^2+c x^4} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {x}{a+b x+c x^2} \, dx,x,x^2\right ) \\ & = \frac {\text {Subst}\left (\int \frac {b+2 c x}{a+b x+c x^2} \, dx,x,x^2\right )}{4 c}-\frac {b \text {Subst}\left (\int \frac {1}{a+b x+c x^2} \, dx,x,x^2\right )}{4 c} \\ & = \frac {\log \left (a+b x^2+c x^4\right )}{4 c}+\frac {b \text {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x^2\right )}{2 c} \\ & = \frac {b \tanh ^{-1}\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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.98 \[ \int \frac {x^4}{a x+b x^3+c x^5} \, dx=\frac {-\frac {2 b \arctan \left (\frac {b+2 c x^2}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}+\log \left (a+b x^2+c x^4\right )}{4 c} \]

[In]

Integrate[x^4/(a*x + b*x^3 + c*x^5),x]

[Out]

((-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)

Maple [A] (verified)

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	\(\frac {\ln \left (\left (-4 a b c +b^{3}+\sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, b \right ) x^{2}+2 \sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, a \right ) a}{4 a c -b^{2}}-\frac {\ln \left (\left (-4 a b c +b^{3}+\sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, b \right ) x^{2}+2 \sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, a \right ) b^{2}}{4 c \left (4 a c -b^{2}\right )}+\frac {\ln \left (\left (-4 a b c +b^{3}+\sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, b \right ) x^{2}+2 \sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, a \right ) \sqrt {-b^{2} \left (4 a c -b^{2}\right )}}{4 c \left (4 a c -b^{2}\right )}+\frac {\ln \left (\left (-4 a b c +b^{3}-\sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, b \right ) x^{2}-2 \sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, a \right ) a}{4 a c -b^{2}}-\frac {\ln \left (\left (-4 a b c +b^{3}-\sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, b \right ) x^{2}-2 \sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, a \right ) b^{2}}{4 c \left (4 a c -b^{2}\right )}-\frac {\ln \left (\left (-4 a b c +b^{3}-\sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, b \right ) x^{2}-2 \sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, a \right ) \sqrt {-b^{2} \left (4 a c -b^{2}\right )}}{4 c \left (4 a c -b^{2}\right )}\)	\(465\)

[In]

int(x^4/(c*x^5+b*x^3+a*x),x,method=_RETURNVERBOSE)

[Out]

1/4*ln(c*x^4+b*x^2+a)/c-1/2*b/c/(4*a*c-b^2)^(1/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 197, normalized size of antiderivative = 3.13 \[ \int \frac {x^4}{a x+b x^3+c x^5} \, dx=\left [\frac {\sqrt {b^{2} - 4 \, a c} b \log \left (\frac {2 \, c^{2} x^{4} + 2 \, b c x^{2} + b^{2} - 2 \, a c + {\left (2 \, c x^{2} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{4} + b x^{2} + a}\right ) + {\left (b^{2} - 4 \, a c\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, {\left (b^{2} c - 4 \, a c^{2}\right )}}, \frac {2 \, \sqrt {-b^{2} + 4 \, a c} b \arctan \left (-\frac {{\left (2 \, c x^{2} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) + {\left (b^{2} - 4 \, a c\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, {\left (b^{2} c - 4 \, a c^{2}\right )}}\right ] \]

[In]

integrate(x^4/(c*x^5+b*x^3+a*x),x, algorithm="fricas")

[Out]

[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)]

Sympy [B] (verification not implemented)

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 \[ \int \frac {x^4}{a x+b x^3+c x^5} \, dx=\left (- \frac {b \sqrt {- 4 a c + b^{2}}}{4 c \left (4 a c - b^{2}\right )} + \frac {1}{4 c}\right ) \log {\left (x^{2} + \frac {- 8 a c \left (- \frac {b \sqrt {- 4 a c + b^{2}}}{4 c \left (4 a c - b^{2}\right )} + \frac {1}{4 c}\right ) + 2 a + 2 b^{2} \left (- \frac {b \sqrt {- 4 a c + b^{2}}}{4 c \left (4 a c - b^{2}\right )} + \frac {1}{4 c}\right )}{b} \right )} + \left (\frac {b \sqrt {- 4 a c + b^{2}}}{4 c \left (4 a c - b^{2}\right )} + \frac {1}{4 c}\right ) \log {\left (x^{2} + \frac {- 8 a c \left (\frac {b \sqrt {- 4 a c + b^{2}}}{4 c \left (4 a c - b^{2}\right )} + \frac {1}{4 c}\right ) + 2 a + 2 b^{2} \left (\frac {b \sqrt {- 4 a c + b^{2}}}{4 c \left (4 a c - b^{2}\right )} + \frac {1}{4 c}\right )}{b} \right )} \]

[In]

integrate(x**4/(c*x**5+b*x**3+a*x),x)

[Out]

(-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)

Maxima [F]

\[ \int \frac {x^4}{a x+b x^3+c x^5} \, dx=\int { \frac {x^{4}}{c x^{5} + b x^{3} + a x} \,d x } \]

[In]

integrate(x^4/(c*x^5+b*x^3+a*x),x, algorithm="maxima")

[Out]

integrate(x^4/(c*x^5 + b*x^3 + a*x), x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.94 \[ \int \frac {x^4}{a x+b x^3+c x^5} \, dx=-\frac {b \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{2 \, \sqrt {-b^{2} + 4 \, a c} c} + \frac {\log \left (c x^{4} + b x^{2} + a\right )}{4 \, c} \]

[In]

integrate(x^4/(c*x^5+b*x^3+a*x),x, algorithm="giac")

[Out]

-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

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.87 \[ \int \frac {x^4}{a x+b x^3+c x^5} \, dx=\frac {4\,a\,c\,\ln \left (c\,x^4+b\,x^2+a\right )}{16\,a\,c^2-4\,b^2\,c}-\frac {b^2\,\ln \left (c\,x^4+b\,x^2+a\right )}{16\,a\,c^2-4\,b^2\,c}-\frac {b\,\mathrm {atan}\left (\frac {b}{\sqrt {4\,a\,c-b^2}}+\frac {2\,c\,x^2}{\sqrt {4\,a\,c-b^2}}\right )}{2\,c\,\sqrt {4\,a\,c-b^2}} \]

[In]

int(x^4/(a*x + b*x^3 + c*x^5),x)

[Out]

(4*a*c*log(a + b*x^2 + c*x^4))/(16*a*c^2 - 4*b^2*c) - (b^2*log(a + b*x^2 + c*x^4))/(16*a*c^2 - 4*b^2*c) - (b*a
tan(b/(4*a*c - b^2)^(1/2) + (2*c*x^2)/(4*a*c - b^2)^(1/2)))/(2*c*(4*a*c - b^2)^(1/2))