\(\int \frac {1}{x (a x^2+b x^3+c x^4)^2} \, dx\) [42]

Optimal result

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

-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)*arcta 
nh((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
 

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 272, normalized size of antiderivative = 0.86 \[ \int \frac {1}{x \left (a x^2+b x^3+c x^4\right )^2} \, dx=\frac {-\frac {3 a^4}{x^4}+\frac {8 a^3 b}{x^3}+\frac {6 a^2 \left (-3 b^2+2 a c\right )}{x^2}-\frac {24 a b \left (-2 b^2+3 a c\right )}{x}-\frac {12 a \left (-b^6+6 a b^4 c-9 a^2 b^2 c^2+2 a^3 c^3-b^5 c x+5 a b^3 c^2 x-5 a^2 b c^3 x\right )}{\left (b^2-4 a c\right ) (a+x (b+c x))}+\frac {12 b \left (5 b^6-42 a b^4 c+105 a^2 b^2 c^2-70 a^3 c^3\right ) \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\left (-b^2+4 a c\right )^{3/2}}+12 \left (5 b^4-12 a b^2 c+3 a^2 c^2\right ) \log (x)-6 \left (5 b^4-12 a b^2 c+3 a^2 c^2\right ) \log (a+x (b+c x))}{12 a^6} \] Input:

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

Output:

((-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)
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 309, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {9, 1165, 25, 1200, 2009}

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.

\(\displaystyle \int \frac {1}{x \left (a x^2+b x^3+c x^4\right )^2} \, dx\)

\(\Big \downarrow \) 9

\(\displaystyle \int \frac {1}{x^5 \left (a+b x+c x^2\right )^2}dx\)

\(\Big \downarrow \) 1165

\(\displaystyle \frac {-2 a c+b^2+b c x}{a x^4 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {\int -\frac {5 b^2+5 c x b-12 a c}{x^5 \left (c x^2+b x+a\right )}dx}{a \left (b^2-4 a c\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {5 b^2+5 c x b-12 a c}{x^5 \left (c x^2+b x+a\right )}dx}{a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c x}{a x^4 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\)

\(\Big \downarrow \) 1200

\(\displaystyle \frac {\int \left (\frac {5 b^2-12 a c}{a x^5}+\frac {\left (b^2-4 a c\right ) \left (5 b^4-12 a c b^2+3 a^2 c^2\right )}{a^5 x}+\frac {-b \left (5 b^6-37 a c b^4+78 a^2 c^2 b^2-41 a^3 c^3\right )-c \left (b^2-4 a c\right ) \left (5 b^4-12 a c b^2+3 a^2 c^2\right ) x}{a^5 \left (c x^2+b x+a\right )}+\frac {-5 b^5+27 a c b^3-29 a^2 c^2 b}{a^4 x^2}+\frac {5 b^4-22 a c b^2+12 a^2 c^2}{a^3 x^3}+\frac {b \left (17 a c-5 b^2\right )}{a^2 x^4}\right )dx}{a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c x}{a x^4 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {b \left (5 b^2-17 a c\right )}{3 a^2 x^3}-\frac {\left (b^2-4 a c\right ) \left (3 a^2 c^2-12 a b^2 c+5 b^4\right ) \log \left (a+b x+c x^2\right )}{2 a^5}+\frac {\log (x) \left (b^2-4 a c\right ) \left (3 a^2 c^2-12 a b^2 c+5 b^4\right )}{a^5}+\frac {b \left (29 a^2 c^2-27 a b^2 c+5 b^4\right )}{a^4 x}-\frac {12 a^2 c^2-22 a b^2 c+5 b^4}{2 a^3 x^2}+\frac {b \left (-70 a^3 c^3+105 a^2 b^2 c^2-42 a b^4 c+5 b^6\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a^5 \sqrt {b^2-4 a c}}-\frac {5 b^2-12 a c}{4 a x^4}}{a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c x}{a x^4 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 9

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]
 

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 1165

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> 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] + Simp[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] && LtQ[p, -1] 
 && IntQuadraticQ[a, b, c, d, e, m, p, x]
 

rule 1200

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* 
(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g* 
x)^n/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && In 
tegersQ[n]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 359, normalized size of antiderivative = 1.13

method	result	size
default	\(-\frac {\frac {\frac {a c b \left (5 a^{2} c^{2}-5 a \,b^{2} c +b^{4}\right ) x}{4 a c -b^{2}}-\frac {a \left (2 a^{3} c^{3}-9 a^{2} b^{2} c^{2}+6 a \,b^{4} c -b^{6}\right )}{4 a c -b^{2}}}{c \,x^{2}+b x +a}+\frac {\frac {\left (12 a^{3} c^{4}-51 a^{2} c^{3} b^{2}+32 a \,b^{4} c^{2}-5 b^{6} c \right ) \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {2 \left (41 a^{3} b \,c^{3}-78 a^{2} b^{3} c^{2}+37 a \,b^{5} c -5 b^{7}-\frac {\left (12 a^{3} c^{4}-51 a^{2} c^{3} b^{2}+32 a \,b^{4} c^{2}-5 b^{6} c \right ) b}{2 c}\right ) \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{4 a c -b^{2}}}{a^{6}}-\frac {1}{4 a^{2} x^{4}}-\frac {-2 a c +3 b^{2}}{2 x^{2} a^{4}}+\frac {\left (3 a^{2} c^{2}-12 a \,b^{2} c +5 b^{4}\right ) \ln \left (x \right )}{a^{6}}+\frac {2 b}{3 a^{3} x^{3}}-\frac {2 b \left (3 a c -2 b^{2}\right )}{a^{5} x}\)	\(359\)
risch	\(\frac {-\frac {b c \left (29 a^{2} c^{2}-27 a \,b^{2} c +5 b^{4}\right ) x^{5}}{a^{5} \left (4 a c -b^{2}\right )}+\frac {\left (12 a^{3} c^{3}-80 a^{2} b^{2} c^{2}+59 a \,b^{4} c -10 b^{6}\right ) x^{4}}{2 a^{5} \left (4 a c -b^{2}\right )}-\frac {b \left (26 a c -15 b^{2}\right ) x^{3}}{6 a^{4}}+\frac {\left (9 a c -10 b^{2}\right ) x^{2}}{12 a^{3}}+\frac {5 b x}{12 a^{2}}-\frac {1}{4 a}}{x^{4} \left (c \,x^{2}+b x +a \right )}+\frac {3 \ln \left (x \right ) c^{2}}{a^{4}}-\frac {12 \ln \left (x \right ) b^{2} c}{a^{5}}+\frac {5 \ln \left (x \right ) b^{4}}{a^{6}}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (64 a^{9} c^{3}-48 a^{8} b^{2} c^{2}+12 a^{7} b^{4} c -a^{6} b^{6}\right ) \textit {\_Z}^{2}+\left (192 a^{5} c^{5}-912 a^{4} b^{2} c^{4}+932 a^{3} b^{4} c^{3}-387 a^{2} b^{6} c^{2}+72 c a \,b^{8}-5 b^{10}\right ) \textit {\_Z} +144 a \,c^{7}-35 b^{2} c^{6}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (96 a^{13} c^{3}-80 a^{12} b^{2} c^{2}+22 a^{11} b^{4} c -2 a^{10} b^{6}\right ) \textit {\_R}^{2}+\left (144 a^{9} c^{5}-532 a^{8} b^{2} c^{4}+400 a^{7} b^{4} c^{3}-109 a^{6} b^{6} c^{2}+10 a^{5} b^{8} c \right ) \textit {\_R} +841 a^{4} b^{2} c^{6}-1566 a^{3} b^{4} c^{5}+1019 a^{2} b^{6} c^{4}-270 a \,b^{8} c^{3}+25 b^{10} c^{2}\right ) x +\left (-16 a^{13} b \,c^{2}+8 a^{12} b^{3} c -a^{11} b^{5}\right ) \textit {\_R}^{2}+\left (164 a^{9} b \,c^{4}-353 a^{8} b^{3} c^{3}+226 a^{7} b^{5} c^{2}-57 a^{6} b^{7} c +5 a^{5} b^{9}\right ) \textit {\_R} -348 a^{5} b \,c^{6}+1803 a^{4} b^{3} c^{5}-2365 a^{3} b^{5} c^{4}+1264 a^{2} b^{7} c^{3}-295 b^{9} c^{2} a +25 b^{11} c \right )\right )\)	\(600\)

Input:

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

Output:

-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*(41*a^3*b* 
c^3-78*a^2*b^3*c^2+37*a*b^5*c-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))) 
)-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
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 810 vs. \(2 (306) = 612\).

Time = 0.55 (sec) , antiderivative size = 1640, normalized size of antiderivative = 5.16 \[ \int \frac {1}{x \left (a x^2+b x^3+c x^4\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

[-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 - 6*((5*b^7*c - 42*a*b^5*c^2 + 1 
05*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 + s 
qrt(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...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.09 \[ \int \frac {1}{x \left (a x^2+b x^3+c x^4\right )^2} \, dx=-\frac {{\left (5 \, b^{7} - 42 \, a b^{5} c + 105 \, a^{2} b^{3} c^{2} - 70 \, a^{3} b c^{3}\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (a^{6} b^{2} - 4 \, a^{7} c\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {{\left (5 \, b^{4} - 12 \, a b^{2} c + 3 \, a^{2} c^{2}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, a^{6}} + \frac {{\left (5 \, b^{4} - 12 \, a b^{2} c + 3 \, a^{2} c^{2}\right )} \log \left ({\left | x \right |}\right )}{a^{6}} - \frac {3 \, a^{5} b^{2} - 12 \, a^{6} c - 12 \, {\left (5 \, a b^{5} c - 27 \, a^{2} b^{3} c^{2} + 29 \, a^{3} b c^{3}\right )} x^{5} - 6 \, {\left (10 \, a b^{6} - 59 \, a^{2} b^{4} c + 80 \, a^{3} b^{2} c^{2} - 12 \, a^{4} c^{3}\right )} x^{4} - 2 \, {\left (15 \, a^{2} b^{5} - 86 \, a^{3} b^{3} c + 104 \, a^{4} b c^{2}\right )} x^{3} + {\left (10 \, a^{3} b^{4} - 49 \, a^{4} b^{2} c + 36 \, a^{5} c^{2}\right )} x^{2} - 5 \, {\left (a^{4} b^{3} - 4 \, a^{5} b c\right )} x}{12 \, {\left (c x^{2} + b x + a\right )} {\left (b^{2} - 4 \, a c\right )} a^{6} x^{4}} \] Input:

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

Output:

-(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 = 22.78 (sec) , antiderivative size = 1260, normalized size of antiderivative = 3.96 \[ \int \frac {1}{x \left (a x^2+b x^3+c x^4\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(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 - 1 
0*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) + 7...
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 1425, normalized size of antiderivative = 4.48 \[ \int \frac {1}{x \left (a x^2+b x^3+c x^4\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 840*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**4*b*c** 
3*x**4 + 1260*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**3 
*b**3*c**2*x**4 - 840*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b** 
2))*a**3*b**2*c**3*x**5 - 840*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a 
*c - b**2))*a**3*b*c**4*x**6 - 504*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqr 
t(4*a*c - b**2))*a**2*b**5*c*x**4 + 1260*sqrt(4*a*c - b**2)*atan((b + 2*c* 
x)/sqrt(4*a*c - b**2))*a**2*b**4*c**2*x**5 + 1260*sqrt(4*a*c - b**2)*atan( 
(b + 2*c*x)/sqrt(4*a*c - b**2))*a**2*b**3*c**3*x**6 + 60*sqrt(4*a*c - b**2 
)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a*b**7*x**4 - 504*sqrt(4*a*c - b**2 
)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a*b**6*c*x**5 - 504*sqrt(4*a*c - b* 
*2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a*b**5*c**2*x**6 + 60*sqrt(4*a*c 
- b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b**8*x**5 + 60*sqrt(4*a*c - b 
**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b**7*c*x**6 - 288*log(a + b*x + 
c*x**2)*a**5*c**4*x**4 + 1296*log(a + b*x + c*x**2)*a**4*b**2*c**3*x**4 - 
288*log(a + b*x + c*x**2)*a**4*b*c**4*x**5 - 288*log(a + b*x + c*x**2)*a** 
4*c**5*x**6 - 1074*log(a + b*x + c*x**2)*a**3*b**4*c**2*x**4 + 1296*log(a 
+ b*x + c*x**2)*a**3*b**3*c**3*x**5 + 1296*log(a + b*x + c*x**2)*a**3*b**2 
*c**4*x**6 + 312*log(a + b*x + c*x**2)*a**2*b**6*c*x**4 - 1074*log(a + b*x 
 + c*x**2)*a**2*b**5*c**2*x**5 - 1074*log(a + b*x + c*x**2)*a**2*b**4*c**3 
*x**6 - 30*log(a + b*x + c*x**2)*a*b**8*x**4 + 312*log(a + b*x + c*x**2...