\(\int \frac {1}{x (a x^2+b x^3+c x^4)^2} \, dx\) [28]

Optimal result

Integrand size = 22, antiderivative size = 318 \[ \int \frac {1}{x \left (a x^2+b x^3+c x^4\right )^2} \, dx=-\frac {5 b^2-12 a c}{4 a^2 \left (b^2-4 a c\right ) x^4}+\frac {b \left (5 b^2-17 a c\right )}{3 a^3 \left (b^2-4 a c\right ) x^3}-\frac {5 b^4-22 a b^2 c+12 a^2 c^2}{2 a^4 \left (b^2-4 a c\right ) x^2}+\frac {b \left (5 b^4-27 a b^2 c+29 a^2 c^2\right )}{a^5 \left (b^2-4 a c\right ) x}+\frac {b^2-2 a c+b c x}{a \left (b^2-4 a c\right ) x^4 \left (a+b x+c x^2\right )}+\frac {b \left (5 b^6-42 a b^4 c+105 a^2 b^2 c^2-70 a^3 c^3\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a^6 \left (b^2-4 a c\right )^{3/2}}+\frac {\left (5 b^4-12 a b^2 c+3 a^2 c^2\right ) \log (x)}{a^6}-\frac {\left (5 b^4-12 a b^2 c+3 a^2 c^2\right ) \log \left (a+b x+c x^2\right )}{2 a^6} \]

[Out]

1/4*(12*a*c-5*b^2)/a^2/(-4*a*c+b^2)/x^4+1/3*b*(-17*a*c+5*b^2)/a^3/(-4*a*c+b^2)/x^3+1/2*(-12*a^2*c^2+22*a*b^2*c
-5*b^4)/a^4/(-4*a*c+b^2)/x^2+b*(29*a^2*c^2-27*a*b^2*c+5*b^4)/a^5/(-4*a*c+b^2)/x+(b*c*x-2*a*c+b^2)/a/(-4*a*c+b^
2)/x^4/(c*x^2+b*x+a)+b*(-70*a^3*c^3+105*a^2*b^2*c^2-42*a*b^4*c+5*b^6)*arctanh((2*c*x+b)/(-4*a*c+b^2)^(1/2))/a^
6/(-4*a*c+b^2)^(3/2)+(3*a^2*c^2-12*a*b^2*c+5*b^4)*ln(x)/a^6-1/2*(3*a^2*c^2-12*a*b^2*c+5*b^4)*ln(c*x^2+b*x+a)/a
^6

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {1599, 754, 814, 648, 632, 212, 642} \[ \int \frac {1}{x \left (a x^2+b x^3+c x^4\right )^2} \, dx=\frac {b \left (5 b^2-17 a c\right )}{3 a^3 x^3 \left (b^2-4 a c\right )}-\frac {5 b^2-12 a c}{4 a^2 x^4 \left (b^2-4 a c\right )}-\frac {\left (3 a^2 c^2-12 a b^2 c+5 b^4\right ) \log \left (a+b x+c x^2\right )}{2 a^6}+\frac {\log (x) \left (3 a^2 c^2-12 a b^2 c+5 b^4\right )}{a^6}+\frac {b \left (29 a^2 c^2-27 a b^2 c+5 b^4\right )}{a^5 x \left (b^2-4 a c\right )}-\frac {12 a^2 c^2-22 a b^2 c+5 b^4}{2 a^4 x^2 \left (b^2-4 a c\right )}+\frac {b \left (-70 a^3 c^3+105 a^2 b^2 c^2-42 a b^4 c+5 b^6\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a^6 \left (b^2-4 a c\right )^{3/2}}+\frac {-2 a c+b^2+b c x}{a x^4 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )} \]

[In]

Int[1/(x*(a*x^2 + b*x^3 + c*x^4)^2),x]

[Out]

-1/4*(5*b^2 - 12*a*c)/(a^2*(b^2 - 4*a*c)*x^4) + (b*(5*b^2 - 17*a*c))/(3*a^3*(b^2 - 4*a*c)*x^3) - (5*b^4 - 22*a
*b^2*c + 12*a^2*c^2)/(2*a^4*(b^2 - 4*a*c)*x^2) + (b*(5*b^4 - 27*a*b^2*c + 29*a^2*c^2))/(a^5*(b^2 - 4*a*c)*x) +
(b^2 - 2*a*c + b*c*x)/(a*(b^2 - 4*a*c)*x^4*(a + b*x + c*x^2)) + (b*(5*b^6 - 42*a*b^4*c + 105*a^2*b^2*c^2 - 70
*a^3*c^3)*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(a^6*(b^2 - 4*a*c)^(3/2)) + ((5*b^4 - 12*a*b^2*c + 3*a^2*c^2
)*Log[x])/a^6 - ((5*b^4 - 12*a*b^2*c + 3*a^2*c^2)*Log[a + b*x + c*x^2])/(2*a^6)

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^5 \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^4 \left (a+b x+c x^2\right )}-\frac {\int \frac {-5 b^2+12 a c-5 b c x}{x^5 \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^4 \left (a+b x+c x^2\right )}-\frac {\int \left (\frac {-5 b^2+12 a c}{a x^5}+\frac {5 b^3-17 a b c}{a^2 x^4}+\frac {-5 b^4+22 a b^2 c-12 a^2 c^2}{a^3 x^3}+\frac {5 b^5-27 a b^3 c+29 a^2 b c^2}{a^4 x^2}+\frac {\left (b^2-4 a c\right ) \left (-5 b^4+12 a b^2 c-3 a^2 c^2\right )}{a^5 x}+\frac {b \left (5 b^6-37 a b^4 c+78 a^2 b^2 c^2-41 a^3 c^3\right )+c \left (5 b^6-32 a b^4 c+51 a^2 b^2 c^2-12 a^3 c^3\right ) x}{a^5 \left (a+b x+c x^2\right )}\right ) \, dx}{a \left (b^2-4 a c\right )} \\ & = -\frac {5 b^2-12 a c}{4 a^2 \left (b^2-4 a c\right ) x^4}+\frac {b \left (5 b^2-17 a c\right )}{3 a^3 \left (b^2-4 a c\right ) x^3}-\frac {5 b^4-22 a b^2 c+12 a^2 c^2}{2 a^4 \left (b^2-4 a c\right ) x^2}+\frac {b \left (5 b^4-27 a b^2 c+29 a^2 c^2\right )}{a^5 \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^4 \left (a+b x+c x^2\right )}+\frac {\left (5 b^4-12 a b^2 c+3 a^2 c^2\right ) \log (x)}{a^6}-\frac {\int \frac {b \left (5 b^6-37 a b^4 c+78 a^2 b^2 c^2-41 a^3 c^3\right )+c \left (5 b^6-32 a b^4 c+51 a^2 b^2 c^2-12 a^3 c^3\right ) x}{a+b x+c x^2} \, dx}{a^6 \left (b^2-4 a c\right )} \\ & = -\frac {5 b^2-12 a c}{4 a^2 \left (b^2-4 a c\right ) x^4}+\frac {b \left (5 b^2-17 a c\right )}{3 a^3 \left (b^2-4 a c\right ) x^3}-\frac {5 b^4-22 a b^2 c+12 a^2 c^2}{2 a^4 \left (b^2-4 a c\right ) x^2}+\frac {b \left (5 b^4-27 a b^2 c+29 a^2 c^2\right )}{a^5 \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^4 \left (a+b x+c x^2\right )}+\frac {\left (5 b^4-12 a b^2 c+3 a^2 c^2\right ) \log (x)}{a^6}-\frac {\left (5 b^4-12 a b^2 c+3 a^2 c^2\right ) \int \frac {b+2 c x}{a+b x+c x^2} \, dx}{2 a^6}-\frac {\left (b \left (5 b^6-42 a b^4 c+105 a^2 b^2 c^2-70 a^3 c^3\right )\right ) \int \frac {1}{a+b x+c x^2} \, dx}{2 a^6 \left (b^2-4 a c\right )} \\ & = -\frac {5 b^2-12 a c}{4 a^2 \left (b^2-4 a c\right ) x^4}+\frac {b \left (5 b^2-17 a c\right )}{3 a^3 \left (b^2-4 a c\right ) x^3}-\frac {5 b^4-22 a b^2 c+12 a^2 c^2}{2 a^4 \left (b^2-4 a c\right ) x^2}+\frac {b \left (5 b^4-27 a b^2 c+29 a^2 c^2\right )}{a^5 \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^4 \left (a+b x+c x^2\right )}+\frac {\left (5 b^4-12 a b^2 c+3 a^2 c^2\right ) \log (x)}{a^6}-\frac {\left (5 b^4-12 a b^2 c+3 a^2 c^2\right ) \log \left (a+b x+c x^2\right )}{2 a^6}+\frac {\left (b \left (5 b^6-42 a b^4 c+105 a^2 b^2 c^2-70 a^3 c^3\right )\right ) \text {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{a^6 \left (b^2-4 a c\right )} \\ & = -\frac {5 b^2-12 a c}{4 a^2 \left (b^2-4 a c\right ) x^4}+\frac {b \left (5 b^2-17 a c\right )}{3 a^3 \left (b^2-4 a c\right ) x^3}-\frac {5 b^4-22 a b^2 c+12 a^2 c^2}{2 a^4 \left (b^2-4 a c\right ) x^2}+\frac {b \left (5 b^4-27 a b^2 c+29 a^2 c^2\right )}{a^5 \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^4 \left (a+b x+c x^2\right )}+\frac {b \left (5 b^6-42 a b^4 c+105 a^2 b^2 c^2-70 a^3 c^3\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a^6 \left (b^2-4 a c\right )^{3/2}}+\frac {\left (5 b^4-12 a b^2 c+3 a^2 c^2\right ) \log (x)}{a^6}-\frac {\left (5 b^4-12 a b^2 c+3 a^2 c^2\right ) \log \left (a+b x+c x^2\right )}{2 a^6} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 272, normalized size of antiderivative = 0.86 \[ \int \frac {1}{x \left (a x^2+b x^3+c x^4\right )^2} \, dx=\frac {-\frac {3 a^4}{x^4}+\frac {8 a^3 b}{x^3}+\frac {6 a^2 \left (-3 b^2+2 a c\right )}{x^2}-\frac {24 a b \left (-2 b^2+3 a c\right )}{x}-\frac {12 a \left (-b^6+6 a b^4 c-9 a^2 b^2 c^2+2 a^3 c^3-b^5 c x+5 a b^3 c^2 x-5 a^2 b c^3 x\right )}{\left (b^2-4 a c\right ) (a+x (b+c x))}+\frac {12 b \left (5 b^6-42 a b^4 c+105 a^2 b^2 c^2-70 a^3 c^3\right ) \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\left (-b^2+4 a c\right )^{3/2}}+12 \left (5 b^4-12 a b^2 c+3 a^2 c^2\right ) \log (x)-6 \left (5 b^4-12 a b^2 c+3 a^2 c^2\right ) \log (a+x (b+c x))}{12 a^6} \]

[In]

Integrate[1/(x*(a*x^2 + b*x^3 + c*x^4)^2),x]

[Out]

((-3*a^4)/x^4 + (8*a^3*b)/x^3 + (6*a^2*(-3*b^2 + 2*a*c))/x^2 - (24*a*b*(-2*b^2 + 3*a*c))/x - (12*a*(-b^6 + 6*a
*b^4*c - 9*a^2*b^2*c^2 + 2*a^3*c^3 - b^5*c*x + 5*a*b^3*c^2*x - 5*a^2*b*c^3*x))/((b^2 - 4*a*c)*(a + x*(b + c*x)
)) + (12*b*(5*b^6 - 42*a*b^4*c + 105*a^2*b^2*c^2 - 70*a^3*c^3)*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/(-b^2 +
4*a*c)^(3/2) + 12*(5*b^4 - 12*a*b^2*c + 3*a^2*c^2)*Log[x] - 6*(5*b^4 - 12*a*b^2*c + 3*a^2*c^2)*Log[a + x*(b +
c*x)])/(12*a^6)

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 359, normalized size of antiderivative = 1.13

method	result	size
default	\(-\frac {1}{4 a^{2} x^{4}}-\frac {-2 a c +3 b^{2}}{2 x^{2} a^{4}}+\frac {\left (3 a^{2} c^{2}-12 a \,b^{2} c +5 b^{4}\right ) \ln \left (x \right )}{a^{6}}+\frac {2 b}{3 a^{3} x^{3}}-\frac {2 b \left (3 a c -2 b^{2}\right )}{a^{5} x}-\frac {\frac {\frac {a c b \left (5 a^{2} c^{2}-5 a \,b^{2} c +b^{4}\right ) x}{4 a c -b^{2}}-\frac {a \left (2 c^{3} a^{3}-9 a^{2} b^{2} c^{2}+6 a \,b^{4} c -b^{6}\right )}{4 a c -b^{2}}}{c \,x^{2}+b x +a}+\frac {\frac {\left (12 a^{3} c^{4}-51 a^{2} b^{2} c^{3}+32 a \,b^{4} c^{2}-5 b^{6} c \right ) \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {2 \left (41 b \,c^{3} a^{3}-78 b^{3} c^{2} a^{2}+37 b^{5} c a -5 b^{7}-\frac {\left (12 a^{3} c^{4}-51 a^{2} b^{2} c^{3}+32 a \,b^{4} c^{2}-5 b^{6} c \right ) b}{2 c}\right ) \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{4 a c -b^{2}}}{a^{6}}\)	\(359\)
risch	\(\text {Expression too large to display}\)	\(8920\)

[In]

int(1/x/(c*x^4+b*x^3+a*x^2)^2,x,method=_RETURNVERBOSE)

[Out]

-1/4/a^2/x^4-1/2*(-2*a*c+3*b^2)/x^2/a^4+(3*a^2*c^2-12*a*b^2*c+5*b^4)*ln(x)/a^6+2/3/a^3*b/x^3-2*b*(3*a*c-2*b^2)
/a^5/x-1/a^6*((a*c*b*(5*a^2*c^2-5*a*b^2*c+b^4)/(4*a*c-b^2)*x-a*(2*a^3*c^3-9*a^2*b^2*c^2+6*a*b^4*c-b^6)/(4*a*c-
b^2))/(c*x^2+b*x+a)+1/(4*a*c-b^2)*(1/2*(12*a^3*c^4-51*a^2*b^2*c^3+32*a*b^4*c^2-5*b^6*c)/c*ln(c*x^2+b*x+a)+2*(4
1*b*c^3*a^3-78*b^3*c^2*a^2+37*b^5*c*a-5*b^7-1/2*(12*a^3*c^4-51*a^2*b^2*c^3+32*a*b^4*c^2-5*b^6*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. 810 vs. \(2 (306) = 612\).

Time = 0.73 (sec) , antiderivative size = 1640, normalized size of antiderivative = 5.16 \[ \int \frac {1}{x \left (a x^2+b x^3+c x^4\right )^2} \, dx=\text {Too large to display} \]

[In]

integrate(1/x/(c*x^4+b*x^3+a*x^2)^2,x, algorithm="fricas")

[Out]

[-1/12*(3*a^5*b^4 - 24*a^6*b^2*c + 48*a^7*c^2 - 12*(5*a*b^7*c - 47*a^2*b^5*c^2 + 137*a^3*b^3*c^3 - 116*a^4*b*c
^4)*x^5 - 6*(10*a*b^8 - 99*a^2*b^6*c + 316*a^3*b^4*c^2 - 332*a^4*b^2*c^3 + 48*a^5*c^4)*x^4 - 2*(15*a^2*b^7 - 1
46*a^3*b^5*c + 448*a^4*b^3*c^2 - 416*a^5*b*c^3)*x^3 + (10*a^3*b^6 - 89*a^4*b^4*c + 232*a^5*b^2*c^2 - 144*a^6*c
^3)*x^2 - 6*((5*b^7*c - 42*a*b^5*c^2 + 105*a^2*b^3*c^3 - 70*a^3*b*c^4)*x^6 + (5*b^8 - 42*a*b^6*c + 105*a^2*b^4
*c^2 - 70*a^3*b^2*c^3)*x^5 + (5*a*b^7 - 42*a^2*b^5*c + 105*a^3*b^3*c^2 - 70*a^4*b*c^3)*x^4)*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)) - 5*(a^4*b^5 - 8*a^
5*b^3*c + 16*a^6*b*c^2)*x + 6*((5*b^8*c - 52*a*b^6*c^2 + 179*a^2*b^4*c^3 - 216*a^3*b^2*c^4 + 48*a^4*c^5)*x^6 +
(5*b^9 - 52*a*b^7*c + 179*a^2*b^5*c^2 - 216*a^3*b^3*c^3 + 48*a^4*b*c^4)*x^5 + (5*a*b^8 - 52*a^2*b^6*c + 179*a
^3*b^4*c^2 - 216*a^4*b^2*c^3 + 48*a^5*c^4)*x^4)*log(c*x^2 + b*x + a) - 12*((5*b^8*c - 52*a*b^6*c^2 + 179*a^2*b
^4*c^3 - 216*a^3*b^2*c^4 + 48*a^4*c^5)*x^6 + (5*b^9 - 52*a*b^7*c + 179*a^2*b^5*c^2 - 216*a^3*b^3*c^3 + 48*a^4*
b*c^4)*x^5 + (5*a*b^8 - 52*a^2*b^6*c + 179*a^3*b^4*c^2 - 216*a^4*b^2*c^3 + 48*a^5*c^4)*x^4)*log(x))/((a^6*b^4*
c - 8*a^7*b^2*c^2 + 16*a^8*c^3)*x^6 + (a^6*b^5 - 8*a^7*b^3*c + 16*a^8*b*c^2)*x^5 + (a^7*b^4 - 8*a^8*b^2*c + 16
*a^9*c^2)*x^4), -1/12*(3*a^5*b^4 - 24*a^6*b^2*c + 48*a^7*c^2 - 12*(5*a*b^7*c - 47*a^2*b^5*c^2 + 137*a^3*b^3*c^
3 - 116*a^4*b*c^4)*x^5 - 6*(10*a*b^8 - 99*a^2*b^6*c + 316*a^3*b^4*c^2 - 332*a^4*b^2*c^3 + 48*a^5*c^4)*x^4 - 2*
(15*a^2*b^7 - 146*a^3*b^5*c + 448*a^4*b^3*c^2 - 416*a^5*b*c^3)*x^3 + (10*a^3*b^6 - 89*a^4*b^4*c + 232*a^5*b^2*
c^2 - 144*a^6*c^3)*x^2 - 12*((5*b^7*c - 42*a*b^5*c^2 + 105*a^2*b^3*c^3 - 70*a^3*b*c^4)*x^6 + (5*b^8 - 42*a*b^6
*c + 105*a^2*b^4*c^2 - 70*a^3*b^2*c^3)*x^5 + (5*a*b^7 - 42*a^2*b^5*c + 105*a^3*b^3*c^2 - 70*a^4*b*c^3)*x^4)*sq
rt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) - 5*(a^4*b^5 - 8*a^5*b^3*c + 16*a^6*b*c
^2)*x + 6*((5*b^8*c - 52*a*b^6*c^2 + 179*a^2*b^4*c^3 - 216*a^3*b^2*c^4 + 48*a^4*c^5)*x^6 + (5*b^9 - 52*a*b^7*c
+ 179*a^2*b^5*c^2 - 216*a^3*b^3*c^3 + 48*a^4*b*c^4)*x^5 + (5*a*b^8 - 52*a^2*b^6*c + 179*a^3*b^4*c^2 - 216*a^4
*b^2*c^3 + 48*a^5*c^4)*x^4)*log(c*x^2 + b*x + a) - 12*((5*b^8*c - 52*a*b^6*c^2 + 179*a^2*b^4*c^3 - 216*a^3*b^2
*c^4 + 48*a^4*c^5)*x^6 + (5*b^9 - 52*a*b^7*c + 179*a^2*b^5*c^2 - 216*a^3*b^3*c^3 + 48*a^4*b*c^4)*x^5 + (5*a*b^
8 - 52*a^2*b^6*c + 179*a^3*b^4*c^2 - 216*a^4*b^2*c^3 + 48*a^5*c^4)*x^4)*log(x))/((a^6*b^4*c - 8*a^7*b^2*c^2 +
16*a^8*c^3)*x^6 + (a^6*b^5 - 8*a^7*b^3*c + 16*a^8*b*c^2)*x^5 + (a^7*b^4 - 8*a^8*b^2*c + 16*a^9*c^2)*x^4)]

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{x \left (a x^2+b x^3+c x^4\right )^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{x \left (a x^2+b x^3+c x^4\right )^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/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 = 0.31 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.09 \[ \int \frac {1}{x \left (a x^2+b x^3+c x^4\right )^2} \, dx=-\frac {{\left (5 \, b^{7} - 42 \, a b^{5} c + 105 \, a^{2} b^{3} c^{2} - 70 \, a^{3} b c^{3}\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (a^{6} b^{2} - 4 \, a^{7} c\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {{\left (5 \, b^{4} - 12 \, a b^{2} c + 3 \, a^{2} c^{2}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, a^{6}} + \frac {{\left (5 \, b^{4} - 12 \, a b^{2} c + 3 \, a^{2} c^{2}\right )} \log \left ({\left | x \right |}\right )}{a^{6}} - \frac {3 \, a^{5} b^{2} - 12 \, a^{6} c - 12 \, {\left (5 \, a b^{5} c - 27 \, a^{2} b^{3} c^{2} + 29 \, a^{3} b c^{3}\right )} x^{5} - 6 \, {\left (10 \, a b^{6} - 59 \, a^{2} b^{4} c + 80 \, a^{3} b^{2} c^{2} - 12 \, a^{4} c^{3}\right )} x^{4} - 2 \, {\left (15 \, a^{2} b^{5} - 86 \, a^{3} b^{3} c + 104 \, a^{4} b c^{2}\right )} x^{3} + {\left (10 \, a^{3} b^{4} - 49 \, a^{4} b^{2} c + 36 \, a^{5} c^{2}\right )} x^{2} - 5 \, {\left (a^{4} b^{3} - 4 \, a^{5} b c\right )} x}{12 \, {\left (c x^{2} + b x + a\right )} {\left (b^{2} - 4 \, a c\right )} a^{6} x^{4}} \]

[In]

integrate(1/x/(c*x^4+b*x^3+a*x^2)^2,x, algorithm="giac")

[Out]

-(5*b^7 - 42*a*b^5*c + 105*a^2*b^3*c^2 - 70*a^3*b*c^3)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((a^6*b^2 - 4*a^
7*c)*sqrt(-b^2 + 4*a*c)) - 1/2*(5*b^4 - 12*a*b^2*c + 3*a^2*c^2)*log(c*x^2 + b*x + a)/a^6 + (5*b^4 - 12*a*b^2*c
+ 3*a^2*c^2)*log(abs(x))/a^6 - 1/12*(3*a^5*b^2 - 12*a^6*c - 12*(5*a*b^5*c - 27*a^2*b^3*c^2 + 29*a^3*b*c^3)*x^
5 - 6*(10*a*b^6 - 59*a^2*b^4*c + 80*a^3*b^2*c^2 - 12*a^4*c^3)*x^4 - 2*(15*a^2*b^5 - 86*a^3*b^3*c + 104*a^4*b*c
^2)*x^3 + (10*a^3*b^4 - 49*a^4*b^2*c + 36*a^5*c^2)*x^2 - 5*(a^4*b^3 - 4*a^5*b*c)*x)/((c*x^2 + b*x + a)*(b^2 -
4*a*c)*a^6*x^4)

Mupad [B] (verification not implemented)

Time = 9.35 (sec) , antiderivative size = 1260, normalized size of antiderivative = 3.96 \[ \int \frac {1}{x \left (a x^2+b x^3+c x^4\right )^2} \, dx=\frac {\ln \left (x\right )\,\left (3\,a^2\,c^2-12\,a\,b^2\,c+5\,b^4\right )}{a^6}-\frac {\frac {1}{4\,a}-\frac {x^2\,\left (9\,a\,c-10\,b^2\right )}{12\,a^3}-\frac {5\,b\,x}{12\,a^2}+\frac {x^4\,\left (-12\,a^3\,c^3+80\,a^2\,b^2\,c^2-59\,a\,b^4\,c+10\,b^6\right )}{2\,a^5\,\left (4\,a\,c-b^2\right )}+\frac {b\,x^3\,\left (26\,a\,c-15\,b^2\right )}{6\,a^4}+\frac {b\,c\,x^5\,\left (29\,a^2\,c^2-27\,a\,b^2\,c+5\,b^4\right )}{a^5\,\left (4\,a\,c-b^2\right )}}{c\,x^6+b\,x^5+a\,x^4}+\frac {\ln \left (288\,a^6\,c^5-10\,b^{11}\,x-10\,a\,b^{10}+10\,a\,b^7\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+139\,a^2\,b^8\,c+10\,b^8\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-717\,a^3\,b^6\,c^2+1643\,a^4\,b^4\,c^3-1508\,a^5\,b^2\,c^4-69\,a^2\,b^5\,c\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-53\,a^4\,b\,c^3\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-779\,a^2\,b^7\,c^2\,x+1916\,a^3\,b^5\,c^3\,x-1998\,a^4\,b^3\,c^4\,x+36\,a^4\,c^4\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+144\,a\,b^9\,c\,x+129\,a^3\,b^3\,c^2\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+568\,a^5\,b\,c^5\,x-84\,a\,b^6\,c\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+225\,a^2\,b^4\,c^2\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-206\,a^3\,b^2\,c^3\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}\right )\,\left (a^3\,\left (466\,b^4\,c^3-35\,b\,c^3\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}\right )-a^2\,\left (\frac {387\,b^6\,c^2}{2}-\frac {105\,b^3\,c^2\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}}{2}\right )-\frac {5\,b^{10}}{2}+96\,a^5\,c^5+\frac {5\,b^7\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}}{2}+a\,\left (36\,b^8\,c-21\,b^5\,c\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}\right )-456\,a^4\,b^2\,c^4\right )}{-64\,a^9\,c^3+48\,a^8\,b^2\,c^2-12\,a^7\,b^4\,c+a^6\,b^6}-\frac {\ln \left (10\,a\,b^{10}+10\,b^{11}\,x-288\,a^6\,c^5+10\,a\,b^7\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-139\,a^2\,b^8\,c+10\,b^8\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+717\,a^3\,b^6\,c^2-1643\,a^4\,b^4\,c^3+1508\,a^5\,b^2\,c^4-69\,a^2\,b^5\,c\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-53\,a^4\,b\,c^3\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+779\,a^2\,b^7\,c^2\,x-1916\,a^3\,b^5\,c^3\,x+1998\,a^4\,b^3\,c^4\,x+36\,a^4\,c^4\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-144\,a\,b^9\,c\,x+129\,a^3\,b^3\,c^2\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-568\,a^5\,b\,c^5\,x-84\,a\,b^6\,c\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+225\,a^2\,b^4\,c^2\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-206\,a^3\,b^2\,c^3\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}\right )\,\left (a^2\,\left (\frac {387\,b^6\,c^2}{2}+\frac {105\,b^3\,c^2\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}}{2}\right )-a^3\,\left (466\,b^4\,c^3+35\,b\,c^3\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}\right )+\frac {5\,b^{10}}{2}-96\,a^5\,c^5+\frac {5\,b^7\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}}{2}-a\,\left (36\,b^8\,c+21\,b^5\,c\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}\right )+456\,a^4\,b^2\,c^4\right )}{-64\,a^9\,c^3+48\,a^8\,b^2\,c^2-12\,a^7\,b^4\,c+a^6\,b^6} \]

[In]

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

[Out]

(log(x)*(5*b^4 + 3*a^2*c^2 - 12*a*b^2*c))/a^6 - (1/(4*a) - (x^2*(9*a*c - 10*b^2))/(12*a^3) - (5*b*x)/(12*a^2)
+ (x^4*(10*b^6 - 12*a^3*c^3 + 80*a^2*b^2*c^2 - 59*a*b^4*c))/(2*a^5*(4*a*c - b^2)) + (b*x^3*(26*a*c - 15*b^2))/
(6*a^4) + (b*c*x^5*(5*b^4 + 29*a^2*c^2 - 27*a*b^2*c))/(a^5*(4*a*c - b^2)))/(a*x^4 + b*x^5 + c*x^6) + (log(288*
a^6*c^5 - 10*b^11*x - 10*a*b^10 + 10*a*b^7*(-(4*a*c - b^2)^3)^(1/2) + 139*a^2*b^8*c + 10*b^8*x*(-(4*a*c - b^2)
^3)^(1/2) - 717*a^3*b^6*c^2 + 1643*a^4*b^4*c^3 - 1508*a^5*b^2*c^4 - 69*a^2*b^5*c*(-(4*a*c - b^2)^3)^(1/2) - 53
*a^4*b*c^3*(-(4*a*c - b^2)^3)^(1/2) - 779*a^2*b^7*c^2*x + 1916*a^3*b^5*c^3*x - 1998*a^4*b^3*c^4*x + 36*a^4*c^4
*x*(-(4*a*c - b^2)^3)^(1/2) + 144*a*b^9*c*x + 129*a^3*b^3*c^2*(-(4*a*c - b^2)^3)^(1/2) + 568*a^5*b*c^5*x - 84*
a*b^6*c*x*(-(4*a*c - b^2)^3)^(1/2) + 225*a^2*b^4*c^2*x*(-(4*a*c - b^2)^3)^(1/2) - 206*a^3*b^2*c^3*x*(-(4*a*c -
b^2)^3)^(1/2))*(a^3*(466*b^4*c^3 - 35*b*c^3*(-(4*a*c - b^2)^3)^(1/2)) - a^2*((387*b^6*c^2)/2 - (105*b^3*c^2*(
-(4*a*c - b^2)^3)^(1/2))/2) - (5*b^10)/2 + 96*a^5*c^5 + (5*b^7*(-(4*a*c - b^2)^3)^(1/2))/2 + a*(36*b^8*c - 21*
b^5*c*(-(4*a*c - b^2)^3)^(1/2)) - 456*a^4*b^2*c^4))/(a^6*b^6 - 64*a^9*c^3 - 12*a^7*b^4*c + 48*a^8*b^2*c^2) - (
log(10*a*b^10 + 10*b^11*x - 288*a^6*c^5 + 10*a*b^7*(-(4*a*c - b^2)^3)^(1/2) - 139*a^2*b^8*c + 10*b^8*x*(-(4*a*
c - b^2)^3)^(1/2) + 717*a^3*b^6*c^2 - 1643*a^4*b^4*c^3 + 1508*a^5*b^2*c^4 - 69*a^2*b^5*c*(-(4*a*c - b^2)^3)^(1
/2) - 53*a^4*b*c^3*(-(4*a*c - b^2)^3)^(1/2) + 779*a^2*b^7*c^2*x - 1916*a^3*b^5*c^3*x + 1998*a^4*b^3*c^4*x + 36
*a^4*c^4*x*(-(4*a*c - b^2)^3)^(1/2) - 144*a*b^9*c*x + 129*a^3*b^3*c^2*(-(4*a*c - b^2)^3)^(1/2) - 568*a^5*b*c^5
*x - 84*a*b^6*c*x*(-(4*a*c - b^2)^3)^(1/2) + 225*a^2*b^4*c^2*x*(-(4*a*c - b^2)^3)^(1/2) - 206*a^3*b^2*c^3*x*(-
(4*a*c - b^2)^3)^(1/2))*(a^2*((387*b^6*c^2)/2 + (105*b^3*c^2*(-(4*a*c - b^2)^3)^(1/2))/2) - a^3*(466*b^4*c^3 +
35*b*c^3*(-(4*a*c - b^2)^3)^(1/2)) + (5*b^10)/2 - 96*a^5*c^5 + (5*b^7*(-(4*a*c - b^2)^3)^(1/2))/2 - a*(36*b^8
*c + 21*b^5*c*(-(4*a*c - b^2)^3)^(1/2)) + 456*a^4*b^2*c^4))/(a^6*b^6 - 64*a^9*c^3 - 12*a^7*b^4*c + 48*a^8*b^2*
c^2)