\(\int \frac {x^9}{(a x+b x^3+c x^5)^2} \, dx\) [91]

Optimal result

Integrand size = 20, antiderivative size = 132 \[ \int \frac {x^9}{\left (a x+b x^3+c x^5\right )^2} \, dx=-\frac {b x^2}{2 c \left (b^2-4 a c\right )}+\frac {x^4 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac {b \left (b^2-6 a c\right ) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c^2 \left (b^2-4 a c\right )^{3/2}}+\frac {\log \left (a+b x^2+c x^4\right )}{4 c^2} \]

[Out]

-1/2*b*x^2/c/(-4*a*c+b^2)+1/2*x^4*(b*x^2+2*a)/(-4*a*c+b^2)/(c*x^4+b*x^2+a)+1/2*b*(-6*a*c+b^2)*arctanh((2*c*x^2
+b)/(-4*a*c+b^2)^(1/2))/c^2/(-4*a*c+b^2)^(3/2)+1/4*ln(c*x^4+b*x^2+a)/c^2

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {1599, 1128, 752, 787, 648, 632, 212, 642} \[ \int \frac {x^9}{\left (a x+b x^3+c x^5\right )^2} \, dx=\frac {b \left (b^2-6 a c\right ) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c^2 \left (b^2-4 a c\right )^{3/2}}-\frac {b x^2}{2 c \left (b^2-4 a c\right )}+\frac {x^4 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac {\log \left (a+b x^2+c x^4\right )}{4 c^2} \]

[In]

Int[x^9/(a*x + b*x^3 + c*x^5)^2,x]

[Out]

-1/2*(b*x^2)/(c*(b^2 - 4*a*c)) + (x^4*(2*a + b*x^2))/(2*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) + (b*(b^2 - 6*a*c)*
ArcTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(2*c^2*(b^2 - 4*a*c)^(3/2)) + Log[a + b*x^2 + c*x^4]/(4*c^2)

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 752

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m - 1)*(d
*b - 2*a*e + (2*c*d - b*e)*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Dist[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] &
& NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, b, c, d,
e, m, p, x]

Rule 787

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

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^7}{\left (a+b x^2+c x^4\right )^2} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {x^3}{\left (a+b x+c x^2\right )^2} \, dx,x,x^2\right ) \\ & = \frac {x^4 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\text {Subst}\left (\int \frac {x (4 a+b x)}{a+b x+c x^2} \, dx,x,x^2\right )}{2 \left (b^2-4 a c\right )} \\ & = -\frac {b x^2}{2 c \left (b^2-4 a c\right )}+\frac {x^4 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\text {Subst}\left (\int \frac {-a b+\left (-b^2+4 a c\right ) x}{a+b x+c x^2} \, dx,x,x^2\right )}{2 c \left (b^2-4 a c\right )} \\ & = -\frac {b x^2}{2 c \left (b^2-4 a c\right )}+\frac {x^4 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac {\text {Subst}\left (\int \frac {b+2 c x}{a+b x+c x^2} \, dx,x,x^2\right )}{4 c^2}-\frac {\left (b \left (b^2-6 a c\right )\right ) \text {Subst}\left (\int \frac {1}{a+b x+c x^2} \, dx,x,x^2\right )}{4 c^2 \left (b^2-4 a c\right )} \\ & = -\frac {b x^2}{2 c \left (b^2-4 a c\right )}+\frac {x^4 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac {\log \left (a+b x^2+c x^4\right )}{4 c^2}+\frac {\left (b \left (b^2-6 a c\right )\right ) \text {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x^2\right )}{2 c^2 \left (b^2-4 a c\right )} \\ & = -\frac {b x^2}{2 c \left (b^2-4 a c\right )}+\frac {x^4 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac {b \left (b^2-6 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c^2 \left (b^2-4 a c\right )^{3/2}}+\frac {\log \left (a+b x^2+c x^4\right )}{4 c^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.92 \[ \int \frac {x^9}{\left (a x+b x^3+c x^5\right )^2} \, dx=\frac {\frac {2 \left (-2 a^2 c+b^3 x^2+a b \left (b-3 c x^2\right )\right )}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac {2 b \left (b^2-6 a c\right ) \arctan \left (\frac {b+2 c x^2}{\sqrt {-b^2+4 a c}}\right )}{\left (-b^2+4 a c\right )^{3/2}}+\log \left (a+b x^2+c x^4\right )}{4 c^2} \]

[In]

Integrate[x^9/(a*x + b*x^3 + c*x^5)^2,x]

[Out]

((2*(-2*a^2*c + b^3*x^2 + a*b*(b - 3*c*x^2)))/((b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) + (2*b*(b^2 - 6*a*c)*ArcTan[
(b + 2*c*x^2)/Sqrt[-b^2 + 4*a*c]])/(-b^2 + 4*a*c)^(3/2) + Log[a + b*x^2 + c*x^4])/(4*c^2)

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.36

method	result	size
default	\(\frac {\frac {b \left (3 a c -b^{2}\right ) x^{2}}{c^{2} \left (4 a c -b^{2}\right )}+\frac {a \left (2 a c -b^{2}\right )}{\left (4 a c -b^{2}\right ) c^{2}}}{2 c \,x^{4}+2 b \,x^{2}+2 a}+\frac {\frac {\left (4 a c -b^{2}\right ) \ln \left (c \,x^{4}+b \,x^{2}+a \right )}{2 c}+\frac {2 \left (-a b -\frac {\left (4 a c -b^{2}\right ) b}{2 c}\right ) \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{2 c \left (4 a c -b^{2}\right )}\)	\(179\)
risch	\(\text {Expression too large to display}\)	\(1017\)

[In]

int(x^9/(c*x^5+b*x^3+a*x)^2,x,method=_RETURNVERBOSE)

[Out]

1/2*(b*(3*a*c-b^2)/c^2/(4*a*c-b^2)*x^2+a*(2*a*c-b^2)/(4*a*c-b^2)/c^2)/(c*x^4+b*x^2+a)+1/2/c/(4*a*c-b^2)*(1/2*(
4*a*c-b^2)/c*ln(c*x^4+b*x^2+a)+2*(-a*b-1/2*(4*a*c-b^2)*b/c)/(4*a*c-b^2)^(1/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(
1/2)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 320 vs. \(2 (120) = 240\).

Time = 0.30 (sec) , antiderivative size = 663, normalized size of antiderivative = 5.02 \[ \int \frac {x^9}{\left (a x+b x^3+c x^5\right )^2} \, dx=\left [\frac {2 \, a b^{4} - 12 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + 2 \, {\left (b^{5} - 7 \, a b^{3} c + 12 \, a^{2} b c^{2}\right )} x^{2} + {\left ({\left (b^{3} c - 6 \, a b c^{2}\right )} x^{4} + a b^{3} - 6 \, a^{2} b c + {\left (b^{4} - 6 \, a b^{2} c\right )} x^{2}\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{4} + 2 \, b c x^{2} + b^{2} - 2 \, a c + {\left (2 \, c x^{2} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{4} + b x^{2} + a}\right ) + {\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{4} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x^{2}\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, {\left (a b^{4} c^{2} - 8 \, a^{2} b^{2} c^{3} + 16 \, a^{3} c^{4} + {\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} x^{4} + {\left (b^{5} c^{2} - 8 \, a b^{3} c^{3} + 16 \, a^{2} b c^{4}\right )} x^{2}\right )}}, \frac {2 \, a b^{4} - 12 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + 2 \, {\left (b^{5} - 7 \, a b^{3} c + 12 \, a^{2} b c^{2}\right )} x^{2} + 2 \, {\left ({\left (b^{3} c - 6 \, a b c^{2}\right )} x^{4} + a b^{3} - 6 \, a^{2} b c + {\left (b^{4} - 6 \, a b^{2} c\right )} x^{2}\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {{\left (2 \, c x^{2} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) + {\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{4} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x^{2}\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, {\left (a b^{4} c^{2} - 8 \, a^{2} b^{2} c^{3} + 16 \, a^{3} c^{4} + {\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} x^{4} + {\left (b^{5} c^{2} - 8 \, a b^{3} c^{3} + 16 \, a^{2} b c^{4}\right )} x^{2}\right )}}\right ] \]

[In]

integrate(x^9/(c*x^5+b*x^3+a*x)^2,x, algorithm="fricas")

[Out]

[1/4*(2*a*b^4 - 12*a^2*b^2*c + 16*a^3*c^2 + 2*(b^5 - 7*a*b^3*c + 12*a^2*b*c^2)*x^2 + ((b^3*c - 6*a*b*c^2)*x^4
+ a*b^3 - 6*a^2*b*c + (b^4 - 6*a*b^2*c)*x^2)*sqrt(b^2 - 4*a*c)*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)) + (a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2 + (b^4*c - 8*a*b^2*c^2 +
16*a^2*c^3)*x^4 + (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*x^2)*log(c*x^4 + b*x^2 + a))/(a*b^4*c^2 - 8*a^2*b^2*c^3 +
16*a^3*c^4 + (b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*x^4 + (b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*x^2), 1/4*(2*a*
b^4 - 12*a^2*b^2*c + 16*a^3*c^2 + 2*(b^5 - 7*a*b^3*c + 12*a^2*b*c^2)*x^2 + 2*((b^3*c - 6*a*b*c^2)*x^4 + a*b^3
- 6*a^2*b*c + (b^4 - 6*a*b^2*c)*x^2)*sqrt(-b^2 + 4*a*c)*arctan(-(2*c*x^2 + b)*sqrt(-b^2 + 4*a*c)/(b^2 - 4*a*c)
) + (a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2 + (b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*x^4 + (b^5 - 8*a*b^3*c + 16*a^2*b*c
^2)*x^2)*log(c*x^4 + b*x^2 + a))/(a*b^4*c^2 - 8*a^2*b^2*c^3 + 16*a^3*c^4 + (b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5
)*x^4 + (b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*x^2)]

Sympy [F(-1)]

Timed out. \[ \int \frac {x^9}{\left (a x+b x^3+c x^5\right )^2} \, dx=\text {Timed out} \]

[In]

integrate(x**9/(c*x**5+b*x**3+a*x)**2,x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {x^9}{\left (a x+b x^3+c x^5\right )^2} \, dx=\int { \frac {x^{9}}{{\left (c x^{5} + b x^{3} + a x\right )}^{2}} \,d x } \]

[In]

integrate(x^9/(c*x^5+b*x^3+a*x)^2,x, algorithm="maxima")

[Out]

1/2*(a*b^2 - 2*a^2*c + (b^3 - 3*a*b*c)*x^2)/(a*b^2*c^2 - 4*a^2*c^3 + (b^2*c^3 - 4*a*c^4)*x^4 + (b^3*c^2 - 4*a*
b*c^3)*x^2) - integrate(-((b^2 - 4*a*c)*x^3 + a*b*x)/(c*x^4 + b*x^2 + a), x)/(b^2*c - 4*a*c^2)

Giac [A] (verification not implemented)

none

Time = 0.69 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.15 \[ \int \frac {x^9}{\left (a x+b x^3+c x^5\right )^2} \, dx=-\frac {{\left (b^{3} - 6 \, a b c\right )} \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{2 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {b^{2} c x^{4} - 4 \, a c^{2} x^{4} - b^{3} x^{2} + 2 \, a b c x^{2} - a b^{2}}{4 \, {\left (c x^{4} + b x^{2} + a\right )} {\left (b^{2} c^{2} - 4 \, a c^{3}\right )}} + \frac {\log \left (c x^{4} + b x^{2} + a\right )}{4 \, c^{2}} \]

[In]

integrate(x^9/(c*x^5+b*x^3+a*x)^2,x, algorithm="giac")

[Out]

-1/2*(b^3 - 6*a*b*c)*arctan((2*c*x^2 + b)/sqrt(-b^2 + 4*a*c))/((b^2*c^2 - 4*a*c^3)*sqrt(-b^2 + 4*a*c)) - 1/4*(
b^2*c*x^4 - 4*a*c^2*x^4 - b^3*x^2 + 2*a*b*c*x^2 - a*b^2)/((c*x^4 + b*x^2 + a)*(b^2*c^2 - 4*a*c^3)) + 1/4*log(c
*x^4 + b*x^2 + a)/c^2

Mupad [B] (verification not implemented)

Time = 9.06 (sec) , antiderivative size = 1336, normalized size of antiderivative = 10.12 \[ \int \frac {x^9}{\left (a x+b x^3+c x^5\right )^2} \, dx=\frac {\frac {a\,\left (2\,a\,c-b^2\right )}{2\,c^2\,\left (4\,a\,c-b^2\right )}+\frac {b\,x^2\,\left (3\,a\,c-b^2\right )}{2\,c^2\,\left (4\,a\,c-b^2\right )}}{c\,x^4+b\,x^2+a}-\frac {\ln \left (c\,x^4+b\,x^2+a\right )\,\left (-128\,a^3\,c^3+96\,a^2\,b^2\,c^2-24\,a\,b^4\,c+2\,b^6\right )}{2\,\left (256\,a^3\,c^5-192\,a^2\,b^2\,c^4+48\,a\,b^4\,c^3-4\,b^6\,c^2\right )}+\frac {b\,\mathrm {atan}\left (\frac {\left (8\,a\,c^3\,{\left (4\,a\,c-b^2\right )}^3-2\,b^2\,c^2\,{\left (4\,a\,c-b^2\right )}^3\right )\,\left (x^2\,\left (\frac {\frac {b\,\left (\frac {6\,b^3\,c^2-28\,a\,b\,c^3}{4\,a\,c^3-b^2\,c^2}+\frac {\left (8\,b^3\,c^4-32\,a\,b\,c^5\right )\,\left (-128\,a^3\,c^3+96\,a^2\,b^2\,c^2-24\,a\,b^4\,c+2\,b^6\right )}{2\,\left (4\,a\,c^3-b^2\,c^2\right )\,\left (256\,a^3\,c^5-192\,a^2\,b^2\,c^4+48\,a\,b^4\,c^3-4\,b^6\,c^2\right )}\right )\,\left (6\,a\,c-b^2\right )}{8\,c^2\,{\left (4\,a\,c-b^2\right )}^{3/2}}+\frac {b\,\left (8\,b^3\,c^4-32\,a\,b\,c^5\right )\,\left (6\,a\,c-b^2\right )\,\left (-128\,a^3\,c^3+96\,a^2\,b^2\,c^2-24\,a\,b^4\,c+2\,b^6\right )}{16\,c^2\,{\left (4\,a\,c-b^2\right )}^{3/2}\,\left (4\,a\,c^3-b^2\,c^2\right )\,\left (256\,a^3\,c^5-192\,a^2\,b^2\,c^4+48\,a\,b^4\,c^3-4\,b^6\,c^2\right )}}{a\,\left (4\,a\,c-b^2\right )}-\frac {b\,\left (\frac {b^3-5\,a\,b\,c}{4\,a\,c^3-b^2\,c^2}+\frac {\left (\frac {6\,b^3\,c^2-28\,a\,b\,c^3}{4\,a\,c^3-b^2\,c^2}+\frac {\left (8\,b^3\,c^4-32\,a\,b\,c^5\right )\,\left (-128\,a^3\,c^3+96\,a^2\,b^2\,c^2-24\,a\,b^4\,c+2\,b^6\right )}{2\,\left (4\,a\,c^3-b^2\,c^2\right )\,\left (256\,a^3\,c^5-192\,a^2\,b^2\,c^4+48\,a\,b^4\,c^3-4\,b^6\,c^2\right )}\right )\,\left (-128\,a^3\,c^3+96\,a^2\,b^2\,c^2-24\,a\,b^4\,c+2\,b^6\right )}{2\,\left (256\,a^3\,c^5-192\,a^2\,b^2\,c^4+48\,a\,b^4\,c^3-4\,b^6\,c^2\right )}-\frac {b^2\,\left (\frac {b^3\,c^4}{2}-2\,a\,b\,c^5\right )\,{\left (6\,a\,c-b^2\right )}^2}{c^4\,{\left (4\,a\,c-b^2\right )}^3\,\left (4\,a\,c^3-b^2\,c^2\right )}\right )}{2\,a\,{\left (4\,a\,c-b^2\right )}^{3/2}}\right )-\frac {\frac {b\,\left (6\,a\,c-b^2\right )\,\left (8\,a+\frac {8\,a\,c^2\,\left (-128\,a^3\,c^3+96\,a^2\,b^2\,c^2-24\,a\,b^4\,c+2\,b^6\right )}{256\,a^3\,c^5-192\,a^2\,b^2\,c^4+48\,a\,b^4\,c^3-4\,b^6\,c^2}\right )}{8\,c^2\,{\left (4\,a\,c-b^2\right )}^{3/2}}+\frac {a\,b\,\left (6\,a\,c-b^2\right )\,\left (-128\,a^3\,c^3+96\,a^2\,b^2\,c^2-24\,a\,b^4\,c+2\,b^6\right )}{{\left (4\,a\,c-b^2\right )}^{3/2}\,\left (256\,a^3\,c^5-192\,a^2\,b^2\,c^4+48\,a\,b^4\,c^3-4\,b^6\,c^2\right )}}{a\,\left (4\,a\,c-b^2\right )}+\frac {b\,\left (\frac {a}{c^2}+\frac {\left (8\,a+\frac {8\,a\,c^2\,\left (-128\,a^3\,c^3+96\,a^2\,b^2\,c^2-24\,a\,b^4\,c+2\,b^6\right )}{256\,a^3\,c^5-192\,a^2\,b^2\,c^4+48\,a\,b^4\,c^3-4\,b^6\,c^2}\right )\,\left (-128\,a^3\,c^3+96\,a^2\,b^2\,c^2-24\,a\,b^4\,c+2\,b^6\right )}{2\,\left (256\,a^3\,c^5-192\,a^2\,b^2\,c^4+48\,a\,b^4\,c^3-4\,b^6\,c^2\right )}-\frac {a\,b^2\,{\left (6\,a\,c-b^2\right )}^2}{c^2\,{\left (4\,a\,c-b^2\right )}^3}\right )}{2\,a\,{\left (4\,a\,c-b^2\right )}^{3/2}}\right )}{36\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6}\right )\,\left (6\,a\,c-b^2\right )}{2\,c^2\,{\left (4\,a\,c-b^2\right )}^{3/2}} \]

[In]

int(x^9/(a*x + b*x^3 + c*x^5)^2,x)

[Out]

((a*(2*a*c - b^2))/(2*c^2*(4*a*c - b^2)) + (b*x^2*(3*a*c - b^2))/(2*c^2*(4*a*c - b^2)))/(a + b*x^2 + c*x^4) -
(log(a + b*x^2 + c*x^4)*(2*b^6 - 128*a^3*c^3 + 96*a^2*b^2*c^2 - 24*a*b^4*c))/(2*(256*a^3*c^5 - 4*b^6*c^2 + 48*
a*b^4*c^3 - 192*a^2*b^2*c^4)) + (b*atan(((8*a*c^3*(4*a*c - b^2)^3 - 2*b^2*c^2*(4*a*c - b^2)^3)*(x^2*(((b*((6*b
^3*c^2 - 28*a*b*c^3)/(4*a*c^3 - b^2*c^2) + ((8*b^3*c^4 - 32*a*b*c^5)*(2*b^6 - 128*a^3*c^3 + 96*a^2*b^2*c^2 - 2
4*a*b^4*c))/(2*(4*a*c^3 - b^2*c^2)*(256*a^3*c^5 - 4*b^6*c^2 + 48*a*b^4*c^3 - 192*a^2*b^2*c^4)))*(6*a*c - b^2))
/(8*c^2*(4*a*c - b^2)^(3/2)) + (b*(8*b^3*c^4 - 32*a*b*c^5)*(6*a*c - b^2)*(2*b^6 - 128*a^3*c^3 + 96*a^2*b^2*c^2
- 24*a*b^4*c))/(16*c^2*(4*a*c - b^2)^(3/2)*(4*a*c^3 - b^2*c^2)*(256*a^3*c^5 - 4*b^6*c^2 + 48*a*b^4*c^3 - 192*
a^2*b^2*c^4)))/(a*(4*a*c - b^2)) - (b*((b^3 - 5*a*b*c)/(4*a*c^3 - b^2*c^2) + (((6*b^3*c^2 - 28*a*b*c^3)/(4*a*c
^3 - b^2*c^2) + ((8*b^3*c^4 - 32*a*b*c^5)*(2*b^6 - 128*a^3*c^3 + 96*a^2*b^2*c^2 - 24*a*b^4*c))/(2*(4*a*c^3 - b
^2*c^2)*(256*a^3*c^5 - 4*b^6*c^2 + 48*a*b^4*c^3 - 192*a^2*b^2*c^4)))*(2*b^6 - 128*a^3*c^3 + 96*a^2*b^2*c^2 - 2
4*a*b^4*c))/(2*(256*a^3*c^5 - 4*b^6*c^2 + 48*a*b^4*c^3 - 192*a^2*b^2*c^4)) - (b^2*((b^3*c^4)/2 - 2*a*b*c^5)*(6
*a*c - b^2)^2)/(c^4*(4*a*c - b^2)^3*(4*a*c^3 - b^2*c^2))))/(2*a*(4*a*c - b^2)^(3/2))) - ((b*(6*a*c - b^2)*(8*a
+ (8*a*c^2*(2*b^6 - 128*a^3*c^3 + 96*a^2*b^2*c^2 - 24*a*b^4*c))/(256*a^3*c^5 - 4*b^6*c^2 + 48*a*b^4*c^3 - 192
*a^2*b^2*c^4)))/(8*c^2*(4*a*c - b^2)^(3/2)) + (a*b*(6*a*c - b^2)*(2*b^6 - 128*a^3*c^3 + 96*a^2*b^2*c^2 - 24*a*
b^4*c))/((4*a*c - b^2)^(3/2)*(256*a^3*c^5 - 4*b^6*c^2 + 48*a*b^4*c^3 - 192*a^2*b^2*c^4)))/(a*(4*a*c - b^2)) +
(b*(a/c^2 + ((8*a + (8*a*c^2*(2*b^6 - 128*a^3*c^3 + 96*a^2*b^2*c^2 - 24*a*b^4*c))/(256*a^3*c^5 - 4*b^6*c^2 + 4
8*a*b^4*c^3 - 192*a^2*b^2*c^4))*(2*b^6 - 128*a^3*c^3 + 96*a^2*b^2*c^2 - 24*a*b^4*c))/(2*(256*a^3*c^5 - 4*b^6*c
^2 + 48*a*b^4*c^3 - 192*a^2*b^2*c^4)) - (a*b^2*(6*a*c - b^2)^2)/(c^2*(4*a*c - b^2)^3)))/(2*a*(4*a*c - b^2)^(3/
2))))/(b^6 + 36*a^2*b^2*c^2 - 12*a*b^4*c))*(6*a*c - b^2))/(2*c^2*(4*a*c - b^2)^(3/2))