\(\int \frac {x}{(a x^2+b x^3+c x^4)^2} \, dx\) [26]

Optimal result

Integrand size = 20, antiderivative size = 202 \[ \int \frac {x}{\left (a x^2+b x^3+c x^4\right )^2} \, dx=-\frac {3 b^2-8 a c}{2 a^2 \left (b^2-4 a c\right ) x^2}+\frac {b \left (3 b^2-11 a c\right )}{a^3 \left (b^2-4 a c\right ) x}+\frac {b^2-2 a c+b c x}{a \left (b^2-4 a c\right ) x^2 \left (a+b x+c x^2\right )}+\frac {b \left (3 b^4-20 a b^2 c+30 a^2 c^2\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a^4 \left (b^2-4 a c\right )^{3/2}}+\frac {\left (3 b^2-2 a c\right ) \log (x)}{a^4}-\frac {\left (3 b^2-2 a c\right ) \log \left (a+b x+c x^2\right )}{2 a^4} \]

[Out]

1/2*(8*a*c-3*b^2)/a^2/(-4*a*c+b^2)/x^2+b*(-11*a*c+3*b^2)/a^3/(-4*a*c+b^2)/x+(b*c*x-2*a*c+b^2)/a/(-4*a*c+b^2)/x
^2/(c*x^2+b*x+a)+b*(30*a^2*c^2-20*a*b^2*c+3*b^4)*arctanh((2*c*x+b)/(-4*a*c+b^2)^(1/2))/a^4/(-4*a*c+b^2)^(3/2)+
(-2*a*c+3*b^2)*ln(x)/a^4-1/2*(-2*a*c+3*b^2)*ln(c*x^2+b*x+a)/a^4

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1599, 754, 814, 648, 632, 212, 642} \[ \int \frac {x}{\left (a x^2+b x^3+c x^4\right )^2} \, dx=-\frac {\left (3 b^2-2 a c\right ) \log \left (a+b x+c x^2\right )}{2 a^4}+\frac {\log (x) \left (3 b^2-2 a c\right )}{a^4}+\frac {b \left (3 b^2-11 a c\right )}{a^3 x \left (b^2-4 a c\right )}-\frac {3 b^2-8 a c}{2 a^2 x^2 \left (b^2-4 a c\right )}+\frac {b \left (30 a^2 c^2-20 a b^2 c+3 b^4\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a^4 \left (b^2-4 a c\right )^{3/2}}+\frac {-2 a c+b^2+b c x}{a x^2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )} \]

[In]

Int[x/(a*x^2 + b*x^3 + c*x^4)^2,x]

[Out]

-1/2*(3*b^2 - 8*a*c)/(a^2*(b^2 - 4*a*c)*x^2) + (b*(3*b^2 - 11*a*c))/(a^3*(b^2 - 4*a*c)*x) + (b^2 - 2*a*c + b*c
*x)/(a*(b^2 - 4*a*c)*x^2*(a + b*x + c*x^2)) + (b*(3*b^4 - 20*a*b^2*c + 30*a^2*c^2)*ArcTanh[(b + 2*c*x)/Sqrt[b^
2 - 4*a*c]])/(a^4*(b^2 - 4*a*c)^(3/2)) + ((3*b^2 - 2*a*c)*Log[x])/a^4 - ((3*b^2 - 2*a*c)*Log[a + b*x + c*x^2])
/(2*a^4)

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 754

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

Rule 814

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[(d + e*x)^m*((f + 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] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

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 {1}{x^3 \left (a+b x+c x^2\right )^2} \, dx \\ & = \frac {b^2-2 a c+b c x}{a \left (b^2-4 a c\right ) x^2 \left (a+b x+c x^2\right )}-\frac {\int \frac {-3 b^2+8 a c-3 b c x}{x^3 \left (a+b x+c x^2\right )} \, dx}{a \left (b^2-4 a c\right )} \\ & = \frac {b^2-2 a c+b c x}{a \left (b^2-4 a c\right ) x^2 \left (a+b x+c x^2\right )}-\frac {\int \left (\frac {-3 b^2+8 a c}{a x^3}+\frac {3 b^3-11 a b c}{a^2 x^2}+\frac {\left (b^2-4 a c\right ) \left (-3 b^2+2 a c\right )}{a^3 x}+\frac {b \left (3 b^4-17 a b^2 c+19 a^2 c^2\right )+c \left (b^2-4 a c\right ) \left (3 b^2-2 a c\right ) x}{a^3 \left (a+b x+c x^2\right )}\right ) \, dx}{a \left (b^2-4 a c\right )} \\ & = -\frac {3 b^2-8 a c}{2 a^2 \left (b^2-4 a c\right ) x^2}+\frac {b \left (3 b^2-11 a c\right )}{a^3 \left (b^2-4 a c\right ) x}+\frac {b^2-2 a c+b c x}{a \left (b^2-4 a c\right ) x^2 \left (a+b x+c x^2\right )}+\frac {\left (3 b^2-2 a c\right ) \log (x)}{a^4}-\frac {\int \frac {b \left (3 b^4-17 a b^2 c+19 a^2 c^2\right )+c \left (b^2-4 a c\right ) \left (3 b^2-2 a c\right ) x}{a+b x+c x^2} \, dx}{a^4 \left (b^2-4 a c\right )} \\ & = -\frac {3 b^2-8 a c}{2 a^2 \left (b^2-4 a c\right ) x^2}+\frac {b \left (3 b^2-11 a c\right )}{a^3 \left (b^2-4 a c\right ) x}+\frac {b^2-2 a c+b c x}{a \left (b^2-4 a c\right ) x^2 \left (a+b x+c x^2\right )}+\frac {\left (3 b^2-2 a c\right ) \log (x)}{a^4}-\frac {\left (3 b^2-2 a c\right ) \int \frac {b+2 c x}{a+b x+c x^2} \, dx}{2 a^4}-\frac {\left (b \left (3 b^4-20 a b^2 c+30 a^2 c^2\right )\right ) \int \frac {1}{a+b x+c x^2} \, dx}{2 a^4 \left (b^2-4 a c\right )} \\ & = -\frac {3 b^2-8 a c}{2 a^2 \left (b^2-4 a c\right ) x^2}+\frac {b \left (3 b^2-11 a c\right )}{a^3 \left (b^2-4 a c\right ) x}+\frac {b^2-2 a c+b c x}{a \left (b^2-4 a c\right ) x^2 \left (a+b x+c x^2\right )}+\frac {\left (3 b^2-2 a c\right ) \log (x)}{a^4}-\frac {\left (3 b^2-2 a c\right ) \log \left (a+b x+c x^2\right )}{2 a^4}+\frac {\left (b \left (3 b^4-20 a b^2 c+30 a^2 c^2\right )\right ) \text {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{a^4 \left (b^2-4 a c\right )} \\ & = -\frac {3 b^2-8 a c}{2 a^2 \left (b^2-4 a c\right ) x^2}+\frac {b \left (3 b^2-11 a c\right )}{a^3 \left (b^2-4 a c\right ) x}+\frac {b^2-2 a c+b c x}{a \left (b^2-4 a c\right ) x^2 \left (a+b x+c x^2\right )}+\frac {b \left (3 b^4-20 a b^2 c+30 a^2 c^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a^4 \left (b^2-4 a c\right )^{3/2}}+\frac {\left (3 b^2-2 a c\right ) \log (x)}{a^4}-\frac {\left (3 b^2-2 a c\right ) \log \left (a+b x+c x^2\right )}{2 a^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.87 \[ \int \frac {x}{\left (a x^2+b x^3+c x^4\right )^2} \, dx=\frac {-\frac {a^2}{x^2}+\frac {4 a b}{x}+\frac {2 a \left (b^4-4 a b^2 c+2 a^2 c^2+b^3 c x-3 a b c^2 x\right )}{\left (b^2-4 a c\right ) (a+x (b+c x))}+\frac {2 b \left (3 b^4-20 a b^2 c+30 a^2 c^2\right ) \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\left (-b^2+4 a c\right )^{3/2}}+2 \left (3 b^2-2 a c\right ) \log (x)+\left (-3 b^2+2 a c\right ) \log (a+x (b+c x))}{2 a^4} \]

[In]

Integrate[x/(a*x^2 + b*x^3 + c*x^4)^2,x]

[Out]

(-(a^2/x^2) + (4*a*b)/x + (2*a*(b^4 - 4*a*b^2*c + 2*a^2*c^2 + b^3*c*x - 3*a*b*c^2*x))/((b^2 - 4*a*c)*(a + x*(b
+ c*x))) + (2*b*(3*b^4 - 20*a*b^2*c + 30*a^2*c^2)*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/(-b^2 + 4*a*c)^(3/2
) + 2*(3*b^2 - 2*a*c)*Log[x] + (-3*b^2 + 2*a*c)*Log[a + x*(b + c*x)])/(2*a^4)

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.26

method	result	size
default	\(-\frac {1}{2 a^{2} x^{2}}+\frac {\left (-2 a c +3 b^{2}\right ) \ln \left (x \right )}{a^{4}}+\frac {2 b}{a^{3} x}+\frac {\frac {\frac {a c b \left (3 a c -b^{2}\right ) x}{4 a c -b^{2}}-\frac {a \left (2 a^{2} c^{2}-4 a \,b^{2} c +b^{4}\right )}{4 a c -b^{2}}}{c \,x^{2}+b x +a}+\frac {\frac {\left (8 a^{2} c^{3}-14 b^{2} a \,c^{2}+3 b^{4} c \right ) \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {2 \left (19 a^{2} b \,c^{2}-17 a \,b^{3} c +3 b^{5}-\frac {\left (8 a^{2} c^{3}-14 b^{2} a \,c^{2}+3 b^{4} c \right ) b}{2 c}\right ) \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{4 a c -b^{2}}}{a^{4}}\)	\(255\)
risch	\(\frac {\frac {b c \left (11 a c -3 b^{2}\right ) x^{3}}{a^{3} \left (4 a c -b^{2}\right )}-\frac {\left (8 a^{2} c^{2}-25 a \,b^{2} c +6 b^{4}\right ) x^{2}}{2 a^{3} \left (4 a c -b^{2}\right )}+\frac {3 b x}{2 a^{2}}-\frac {1}{2 a}}{\left (c \,x^{2}+b x +a \right ) x^{2}}-\frac {2 \ln \left (x \right ) c}{a^{3}}+\frac {3 b^{2} \ln \left (x \right )}{a^{4}}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (64 a^{7} c^{3}-48 a^{6} b^{2} c^{2}+12 a^{5} b^{4} c -a^{4} b^{6}\right ) \textit {\_Z}^{2}+\left (-128 a^{4} c^{4}+288 a^{3} b^{2} c^{3}-168 a^{2} b^{4} c^{2}+38 a \,b^{6} c -3 b^{8}\right ) \textit {\_Z} +64 a \,c^{5}-15 b^{2} c^{4}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (96 a^{9} c^{3}-80 a^{8} b^{2} c^{2}+22 a^{7} b^{4} c -2 a^{6} b^{6}\right ) \textit {\_R}^{2}+\left (-96 a^{6} c^{4}+148 a^{5} b^{2} c^{3}-55 a^{4} b^{4} c^{2}+6 a^{3} b^{6} c \right ) \textit {\_R} +121 a^{2} b^{2} c^{4}-66 a \,b^{4} c^{3}+9 c^{2} b^{6}\right ) x +\left (-16 a^{9} b \,c^{2}+8 a^{8} b^{3} c -a^{7} b^{5}\right ) \textit {\_R}^{2}+\left (-76 a^{6} b \,c^{3}+87 a^{5} b^{3} c^{2}-29 a^{4} b^{5} c +3 a^{3} b^{7}\right ) \textit {\_R} -88 a^{3} b \,c^{4}+178 a^{2} b^{3} c^{3}-75 a \,b^{5} c^{2}+9 b^{7} c \right )\right )\)	\(450\)

[In]

int(x/(c*x^4+b*x^3+a*x^2)^2,x,method=_RETURNVERBOSE)

[Out]

-1/2/a^2/x^2+(-2*a*c+3*b^2)*ln(x)/a^4+2/a^3*b/x+1/a^4*((a*c*b*(3*a*c-b^2)/(4*a*c-b^2)*x-a*(2*a^2*c^2-4*a*b^2*c
+b^4)/(4*a*c-b^2))/(c*x^2+b*x+a)+1/(4*a*c-b^2)*(1/2*(8*a^2*c^3-14*a*b^2*c^2+3*b^4*c)/c*ln(c*x^2+b*x+a)+2*(19*a
^2*b*c^2-17*a*b^3*c+3*b^5-1/2*(8*a^2*c^3-14*a*b^2*c^2+3*b^4*c)*b/c)/(4*a*c-b^2)^(1/2)*arctan((2*c*x+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. 603 vs. \(2 (194) = 388\).

Time = 0.45 (sec) , antiderivative size = 1226, normalized size of antiderivative = 6.07 \[ \int \frac {x}{\left (a x^2+b x^3+c x^4\right )^2} \, dx=\text {Too large to display} \]

[In]

integrate(x/(c*x^4+b*x^3+a*x^2)^2,x, algorithm="fricas")

[Out]

[-1/2*(a^3*b^4 - 8*a^4*b^2*c + 16*a^5*c^2 - 2*(3*a*b^5*c - 23*a^2*b^3*c^2 + 44*a^3*b*c^3)*x^3 - (6*a*b^6 - 49*
a^2*b^4*c + 108*a^3*b^2*c^2 - 32*a^4*c^3)*x^2 + ((3*b^5*c - 20*a*b^3*c^2 + 30*a^2*b*c^3)*x^4 + (3*b^6 - 20*a*b
^4*c + 30*a^2*b^2*c^2)*x^3 + (3*a*b^5 - 20*a^2*b^3*c + 30*a^3*b*c^2)*x^2)*sqrt(b^2 - 4*a*c)*log((2*c^2*x^2 + 2
*b*c*x + b^2 - 2*a*c - sqrt(b^2 - 4*a*c)*(2*c*x + b))/(c*x^2 + b*x + a)) - 3*(a^2*b^5 - 8*a^3*b^3*c + 16*a^4*b
*c^2)*x + ((3*b^6*c - 26*a*b^4*c^2 + 64*a^2*b^2*c^3 - 32*a^3*c^4)*x^4 + (3*b^7 - 26*a*b^5*c + 64*a^2*b^3*c^2 -
32*a^3*b*c^3)*x^3 + (3*a*b^6 - 26*a^2*b^4*c + 64*a^3*b^2*c^2 - 32*a^4*c^3)*x^2)*log(c*x^2 + b*x + a) - 2*((3*
b^6*c - 26*a*b^4*c^2 + 64*a^2*b^2*c^3 - 32*a^3*c^4)*x^4 + (3*b^7 - 26*a*b^5*c + 64*a^2*b^3*c^2 - 32*a^3*b*c^3)
*x^3 + (3*a*b^6 - 26*a^2*b^4*c + 64*a^3*b^2*c^2 - 32*a^4*c^3)*x^2)*log(x))/((a^4*b^4*c - 8*a^5*b^2*c^2 + 16*a^
6*c^3)*x^4 + (a^4*b^5 - 8*a^5*b^3*c + 16*a^6*b*c^2)*x^3 + (a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)*x^2), -1/2*(a^3
*b^4 - 8*a^4*b^2*c + 16*a^5*c^2 - 2*(3*a*b^5*c - 23*a^2*b^3*c^2 + 44*a^3*b*c^3)*x^3 - (6*a*b^6 - 49*a^2*b^4*c
+ 108*a^3*b^2*c^2 - 32*a^4*c^3)*x^2 - 2*((3*b^5*c - 20*a*b^3*c^2 + 30*a^2*b*c^3)*x^4 + (3*b^6 - 20*a*b^4*c + 3
0*a^2*b^2*c^2)*x^3 + (3*a*b^5 - 20*a^2*b^3*c + 30*a^3*b*c^2)*x^2)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c
)*(2*c*x + b)/(b^2 - 4*a*c)) - 3*(a^2*b^5 - 8*a^3*b^3*c + 16*a^4*b*c^2)*x + ((3*b^6*c - 26*a*b^4*c^2 + 64*a^2*
b^2*c^3 - 32*a^3*c^4)*x^4 + (3*b^7 - 26*a*b^5*c + 64*a^2*b^3*c^2 - 32*a^3*b*c^3)*x^3 + (3*a*b^6 - 26*a^2*b^4*c
+ 64*a^3*b^2*c^2 - 32*a^4*c^3)*x^2)*log(c*x^2 + b*x + a) - 2*((3*b^6*c - 26*a*b^4*c^2 + 64*a^2*b^2*c^3 - 32*a
^3*c^4)*x^4 + (3*b^7 - 26*a*b^5*c + 64*a^2*b^3*c^2 - 32*a^3*b*c^3)*x^3 + (3*a*b^6 - 26*a^2*b^4*c + 64*a^3*b^2*
c^2 - 32*a^4*c^3)*x^2)*log(x))/((a^4*b^4*c - 8*a^5*b^2*c^2 + 16*a^6*c^3)*x^4 + (a^4*b^5 - 8*a^5*b^3*c + 16*a^6
*b*c^2)*x^3 + (a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)*x^2)]

Sympy [F(-1)]

Timed out. \[ \int \frac {x}{\left (a x^2+b x^3+c x^4\right )^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {x}{\left (a x^2+b x^3+c x^4\right )^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(x/(c*x^4+b*x^3+a*x^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.13 \[ \int \frac {x}{\left (a x^2+b x^3+c x^4\right )^2} \, dx=-\frac {{\left (3 \, b^{5} - 20 \, a b^{3} c + 30 \, a^{2} b c^{2}\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (a^{4} b^{2} - 4 \, a^{5} c\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {{\left (3 \, b^{2} - 2 \, a c\right )} \log \left (c x^{2} + b x + a\right )}{2 \, a^{4}} + \frac {{\left (3 \, b^{2} - 2 \, a c\right )} \log \left ({\left | x \right |}\right )}{a^{4}} - \frac {a^{3} b^{2} - 4 \, a^{4} c - 2 \, {\left (3 \, a b^{3} c - 11 \, a^{2} b c^{2}\right )} x^{3} - {\left (6 \, a b^{4} - 25 \, a^{2} b^{2} c + 8 \, a^{3} c^{2}\right )} x^{2} - 3 \, {\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x}{2 \, {\left (c x^{2} + b x + a\right )} {\left (b^{2} - 4 \, a c\right )} a^{4} x^{2}} \]

[In]

integrate(x/(c*x^4+b*x^3+a*x^2)^2,x, algorithm="giac")

[Out]

-(3*b^5 - 20*a*b^3*c + 30*a^2*b*c^2)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((a^4*b^2 - 4*a^5*c)*sqrt(-b^2 + 4
*a*c)) - 1/2*(3*b^2 - 2*a*c)*log(c*x^2 + b*x + a)/a^4 + (3*b^2 - 2*a*c)*log(abs(x))/a^4 - 1/2*(a^3*b^2 - 4*a^4
*c - 2*(3*a*b^3*c - 11*a^2*b*c^2)*x^3 - (6*a*b^4 - 25*a^2*b^2*c + 8*a^3*c^2)*x^2 - 3*(a^2*b^3 - 4*a^3*b*c)*x)/
((c*x^2 + b*x + a)*(b^2 - 4*a*c)*a^4*x^2)

Mupad [B] (verification not implemented)

Time = 9.21 (sec) , antiderivative size = 914, normalized size of antiderivative = 4.52 \[ \int \frac {x}{\left (a x^2+b x^3+c x^4\right )^2} \, dx=\frac {\ln \left (6\,a\,b^8+6\,b^9\,x+192\,a^5\,c^4-6\,a\,b^5\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-73\,a^2\,b^6\,c-6\,b^6\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+307\,a^3\,b^4\,c^2-492\,a^4\,b^2\,c^3+31\,a^2\,b^3\,c\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-27\,a^3\,b\,c^2\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+339\,a^2\,b^5\,c^2\,x-602\,a^3\,b^3\,c^3\,x+24\,a^3\,c^3\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-76\,a\,b^7\,c\,x+312\,a^4\,b\,c^4\,x+40\,a\,b^4\,c\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-69\,a^2\,b^2\,c^2\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}\right )\,\left (3\,b^8+128\,a^4\,c^4-3\,b^5\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+168\,a^2\,b^4\,c^2-288\,a^3\,b^2\,c^3-38\,a\,b^6\,c-30\,a^2\,b\,c^2\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+20\,a\,b^3\,c\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}\right )}{2\,a^4\,{\left (4\,a\,c-b^2\right )}^3}-\frac {\ln \left (x\right )\,\left (2\,a\,c-3\,b^2\right )}{a^4}-\frac {\frac {1}{2\,a}-\frac {3\,b\,x}{2\,a^2}+\frac {x^2\,\left (8\,a^2\,c^2-25\,a\,b^2\,c+6\,b^4\right )}{2\,a^3\,\left (4\,a\,c-b^2\right )}-\frac {b\,c\,x^3\,\left (11\,a\,c-3\,b^2\right )}{a^3\,\left (4\,a\,c-b^2\right )}}{c\,x^4+b\,x^3+a\,x^2}+\frac {\ln \left (6\,a\,b^8+6\,b^9\,x+192\,a^5\,c^4+6\,a\,b^5\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-73\,a^2\,b^6\,c+6\,b^6\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+307\,a^3\,b^4\,c^2-492\,a^4\,b^2\,c^3-31\,a^2\,b^3\,c\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+27\,a^3\,b\,c^2\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+339\,a^2\,b^5\,c^2\,x-602\,a^3\,b^3\,c^3\,x-24\,a^3\,c^3\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-76\,a\,b^7\,c\,x+312\,a^4\,b\,c^4\,x-40\,a\,b^4\,c\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+69\,a^2\,b^2\,c^2\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}\right )\,\left (3\,b^8+128\,a^4\,c^4+3\,b^5\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+168\,a^2\,b^4\,c^2-288\,a^3\,b^2\,c^3-38\,a\,b^6\,c+30\,a^2\,b\,c^2\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-20\,a\,b^3\,c\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}\right )}{2\,a^4\,{\left (4\,a\,c-b^2\right )}^3} \]

[In]

int(x/(a*x^2 + b*x^3 + c*x^4)^2,x)

[Out]

(log(6*a*b^8 + 6*b^9*x + 192*a^5*c^4 - 6*a*b^5*(-(4*a*c - b^2)^3)^(1/2) - 73*a^2*b^6*c - 6*b^6*x*(-(4*a*c - b^
2)^3)^(1/2) + 307*a^3*b^4*c^2 - 492*a^4*b^2*c^3 + 31*a^2*b^3*c*(-(4*a*c - b^2)^3)^(1/2) - 27*a^3*b*c^2*(-(4*a*
c - b^2)^3)^(1/2) + 339*a^2*b^5*c^2*x - 602*a^3*b^3*c^3*x + 24*a^3*c^3*x*(-(4*a*c - b^2)^3)^(1/2) - 76*a*b^7*c
*x + 312*a^4*b*c^4*x + 40*a*b^4*c*x*(-(4*a*c - b^2)^3)^(1/2) - 69*a^2*b^2*c^2*x*(-(4*a*c - b^2)^3)^(1/2))*(3*b
^8 + 128*a^4*c^4 - 3*b^5*(-(4*a*c - b^2)^3)^(1/2) + 168*a^2*b^4*c^2 - 288*a^3*b^2*c^3 - 38*a*b^6*c - 30*a^2*b*
c^2*(-(4*a*c - b^2)^3)^(1/2) + 20*a*b^3*c*(-(4*a*c - b^2)^3)^(1/2)))/(2*a^4*(4*a*c - b^2)^3) - (log(x)*(2*a*c
- 3*b^2))/a^4 - (1/(2*a) - (3*b*x)/(2*a^2) + (x^2*(6*b^4 + 8*a^2*c^2 - 25*a*b^2*c))/(2*a^3*(4*a*c - b^2)) - (b
*c*x^3*(11*a*c - 3*b^2))/(a^3*(4*a*c - b^2)))/(a*x^2 + b*x^3 + c*x^4) + (log(6*a*b^8 + 6*b^9*x + 192*a^5*c^4 +
6*a*b^5*(-(4*a*c - b^2)^3)^(1/2) - 73*a^2*b^6*c + 6*b^6*x*(-(4*a*c - b^2)^3)^(1/2) + 307*a^3*b^4*c^2 - 492*a^
4*b^2*c^3 - 31*a^2*b^3*c*(-(4*a*c - b^2)^3)^(1/2) + 27*a^3*b*c^2*(-(4*a*c - b^2)^3)^(1/2) + 339*a^2*b^5*c^2*x
- 602*a^3*b^3*c^3*x - 24*a^3*c^3*x*(-(4*a*c - b^2)^3)^(1/2) - 76*a*b^7*c*x + 312*a^4*b*c^4*x - 40*a*b^4*c*x*(-
(4*a*c - b^2)^3)^(1/2) + 69*a^2*b^2*c^2*x*(-(4*a*c - b^2)^3)^(1/2))*(3*b^8 + 128*a^4*c^4 + 3*b^5*(-(4*a*c - b^
2)^3)^(1/2) + 168*a^2*b^4*c^2 - 288*a^3*b^2*c^3 - 38*a*b^6*c + 30*a^2*b*c^2*(-(4*a*c - b^2)^3)^(1/2) - 20*a*b^
3*c*(-(4*a*c - b^2)^3)^(1/2)))/(2*a^4*(4*a*c - b^2)^3)