\(\int \frac {1}{(c+\frac {a}{x^2}+\frac {b}{x}) x^6} \, dx\) [18]

Optimal result

Integrand size = 18, antiderivative size = 137 \[ \int \frac {1}{\left (c+\frac {a}{x^2}+\frac {b}{x}\right ) x^6} \, dx=-\frac {1}{3 a x^3}+\frac {b}{2 a^2 x^2}-\frac {b^2-a c}{a^3 x}-\frac {\left (b^4-4 a b^2 c+2 a^2 c^2\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a^4 \sqrt {b^2-4 a c}}-\frac {b \left (b^2-2 a c\right ) \log (x)}{a^4}+\frac {b \left (b^2-2 a c\right ) \log \left (a+b x+c x^2\right )}{2 a^4} \] Output:

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\left (c+\frac {a}{x^2}+\frac {b}{x}\right ) x^6} \, dx=\frac {-\frac {2 a^3}{x^3}+\frac {3 a^2 b}{x^2}+\frac {6 a \left (-b^2+a c\right )}{x}+\frac {6 \left (b^4-4 a b^2 c+2 a^2 c^2\right ) \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}-6 \left (b^3-2 a b c\right ) \log (x)+3 \left (b^3-2 a b c\right ) \log (a+x (b+c x))}{6 a^4} \] Input:

Integrate[1/((c + a/x^2 + b/x)*x^6),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {1692, 1145, 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.

\(\displaystyle \int \frac {1}{x^6 \left (\frac {a}{x^2}+\frac {b}{x}+c\right )} \, dx\)

\(\Big \downarrow \) 1692

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

\(\Big \downarrow \) 1145

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1200

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {-\frac {b \left (b^2-2 a c\right ) \log \left (a+b x+c x^2\right )}{2 a^3}+\frac {b \log (x) \left (b^2-2 a c\right )}{a^3}+\frac {b^2-a c}{a^2 x}+\frac {\left (2 a^2 c^2-4 a b^2 c+b^4\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a^3 \sqrt {b^2-4 a c}}-\frac {b}{2 a x^2}}{a}-\frac {1}{3 a x^3}\)

Input:

Int[1/((c + a/x^2 + b/x)*x^6),x]
 

Output:

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

Defintions of rubi rules used

rule 25

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

rule 1145

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

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 1692

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] 
 :> Int[x^(m + 2*n*p)*(c + b/x^n + a/x^(2*n))^p, x] /; FreeQ[{a, b, c, m, n 
}, x] && EqQ[n2, 2*n] && ILtQ[p, 0] && NegQ[n]
 

rule 2009

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

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.15

method	result	size
default	\(\frac {\frac {\left (-2 a b \,c^{2}+b^{3} c \right ) \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {2 \left (a^{2} c^{2}-3 a \,b^{2} c +b^{4}-\frac {\left (-2 a b \,c^{2}+b^{3} c \right ) b}{2 c}\right ) \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{a^{4}}-\frac {1}{3 a \,x^{3}}-\frac {-a c +b^{2}}{a^{3} x}+\frac {b \left (2 a c -b^{2}\right ) \ln \left (x \right )}{a^{4}}+\frac {b}{2 a^{2} x^{2}}\)	\(157\)
risch	\(\text {Expression too large to display}\)	\(3142\)

Input:

int(1/(c+a/x^2+b/x)/x^6,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 445, normalized size of antiderivative = 3.25 \[ \int \frac {1}{\left (c+\frac {a}{x^2}+\frac {b}{x}\right ) x^6} \, dx=\left [\frac {3 \, {\left (b^{4} - 4 \, a b^{2} c + 2 \, a^{2} c^{2}\right )} \sqrt {b^{2} - 4 \, a c} x^{3} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c - \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) - 2 \, a^{3} b^{2} + 8 \, a^{4} c + 3 \, {\left (b^{5} - 6 \, a b^{3} c + 8 \, a^{2} b c^{2}\right )} x^{3} \log \left (c x^{2} + b x + a\right ) - 6 \, {\left (b^{5} - 6 \, a b^{3} c + 8 \, a^{2} b c^{2}\right )} x^{3} \log \left (x\right ) - 6 \, {\left (a b^{4} - 5 \, a^{2} b^{2} c + 4 \, a^{3} c^{2}\right )} x^{2} + 3 \, {\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x}{6 \, {\left (a^{4} b^{2} - 4 \, a^{5} c\right )} x^{3}}, -\frac {6 \, {\left (b^{4} - 4 \, a b^{2} c + 2 \, a^{2} c^{2}\right )} \sqrt {-b^{2} + 4 \, a c} x^{3} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + 2 \, a^{3} b^{2} - 8 \, a^{4} c - 3 \, {\left (b^{5} - 6 \, a b^{3} c + 8 \, a^{2} b c^{2}\right )} x^{3} \log \left (c x^{2} + b x + a\right ) + 6 \, {\left (b^{5} - 6 \, a b^{3} c + 8 \, a^{2} b c^{2}\right )} x^{3} \log \left (x\right ) + 6 \, {\left (a b^{4} - 5 \, a^{2} b^{2} c + 4 \, a^{3} c^{2}\right )} x^{2} - 3 \, {\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x}{6 \, {\left (a^{4} b^{2} - 4 \, a^{5} c\right )} x^{3}}\right ] \] Input:

integrate(1/(c+a/x^2+b/x)/x^6,x, algorithm="fricas")
 

Output:

[1/6*(3*(b^4 - 4*a*b^2*c + 2*a^2*c^2)*sqrt(b^2 - 4*a*c)*x^3*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 
)) - 2*a^3*b^2 + 8*a^4*c + 3*(b^5 - 6*a*b^3*c + 8*a^2*b*c^2)*x^3*log(c*x^2 
 + b*x + a) - 6*(b^5 - 6*a*b^3*c + 8*a^2*b*c^2)*x^3*log(x) - 6*(a*b^4 - 5* 
a^2*b^2*c + 4*a^3*c^2)*x^2 + 3*(a^2*b^3 - 4*a^3*b*c)*x)/((a^4*b^2 - 4*a^5* 
c)*x^3), -1/6*(6*(b^4 - 4*a*b^2*c + 2*a^2*c^2)*sqrt(-b^2 + 4*a*c)*x^3*arct 
an(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) + 2*a^3*b^2 - 8*a^4*c - 
3*(b^5 - 6*a*b^3*c + 8*a^2*b*c^2)*x^3*log(c*x^2 + b*x + a) + 6*(b^5 - 6*a* 
b^3*c + 8*a^2*b*c^2)*x^3*log(x) + 6*(a*b^4 - 5*a^2*b^2*c + 4*a^3*c^2)*x^2 
- 3*(a^2*b^3 - 4*a^3*b*c)*x)/((a^4*b^2 - 4*a^5*c)*x^3)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\left (c+\frac {a}{x^2}+\frac {b}{x}\right ) x^6} \, dx=\text {Timed out} \] Input:

integrate(1/(c+a/x**2+b/x)/x**6,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\left (c+\frac {a}{x^2}+\frac {b}{x}\right ) x^6} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/(c+a/x^2+b/x)/x^6,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 = 0.13 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.99 \[ \int \frac {1}{\left (c+\frac {a}{x^2}+\frac {b}{x}\right ) x^6} \, dx=\frac {{\left (b^{3} - 2 \, a b c\right )} \log \left (c x^{2} + b x + a\right )}{2 \, a^{4}} - \frac {{\left (b^{3} - 2 \, a b c\right )} \log \left ({\left | x \right |}\right )}{a^{4}} + \frac {{\left (b^{4} - 4 \, a b^{2} c + 2 \, a^{2} c^{2}\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c} a^{4}} + \frac {3 \, a^{2} b x - 2 \, a^{3} - 6 \, {\left (a b^{2} - a^{2} c\right )} x^{2}}{6 \, a^{4} x^{3}} \] Input:

integrate(1/(c+a/x^2+b/x)/x^6,x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 20.56 (sec) , antiderivative size = 524, normalized size of antiderivative = 3.82 \[ \int \frac {1}{\left (c+\frac {a}{x^2}+\frac {b}{x}\right ) x^6} \, dx=\ln \left (2\,a\,b^4\,\sqrt {b^2-4\,a\,c}-2\,b^6\,x-2\,a\,b^5+2\,b^5\,x\,\sqrt {b^2-4\,a\,c}+11\,a^2\,b^3\,c-13\,a^3\,b\,c^2+2\,a^3\,c^3\,x+a^3\,c^2\,\sqrt {b^2-4\,a\,c}-17\,a^2\,b^2\,c^2\,x+12\,a\,b^4\,c\,x-5\,a^2\,b^2\,c\,\sqrt {b^2-4\,a\,c}-8\,a\,b^3\,c\,x\,\sqrt {b^2-4\,a\,c}+7\,a^2\,b\,c^2\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (\frac {b^3}{2\,a^4}-\frac {b^2\,\sqrt {b^2-4\,a\,c}}{2\,a^4}-\frac {b\,c}{a^3}+\frac {a^2\,c^2\,\sqrt {b^2-4\,a\,c}}{4\,a^5\,c-a^4\,b^2}\right )+\ln \left (2\,a\,b^5+2\,b^6\,x+2\,a\,b^4\,\sqrt {b^2-4\,a\,c}+2\,b^5\,x\,\sqrt {b^2-4\,a\,c}-11\,a^2\,b^3\,c+13\,a^3\,b\,c^2-2\,a^3\,c^3\,x+a^3\,c^2\,\sqrt {b^2-4\,a\,c}+17\,a^2\,b^2\,c^2\,x-12\,a\,b^4\,c\,x-5\,a^2\,b^2\,c\,\sqrt {b^2-4\,a\,c}-8\,a\,b^3\,c\,x\,\sqrt {b^2-4\,a\,c}+7\,a^2\,b\,c^2\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (\frac {b^3}{2\,a^4}+\frac {b^2\,\sqrt {b^2-4\,a\,c}}{2\,a^4}-\frac {b\,c}{a^3}-\frac {a^2\,c^2\,\sqrt {b^2-4\,a\,c}}{4\,a^5\,c-a^4\,b^2}\right )+\frac {\frac {x^2\,\left (a\,c-b^2\right )}{a^3}-\frac {1}{3\,a}+\frac {b\,x}{2\,a^2}}{x^3}+\frac {b\,\ln \left (x\right )\,\left (2\,a\,c-b^2\right )}{a^4} \] Input:

int(1/(x^6*(c + a/x^2 + b/x)),x)
 

Output:

log(2*a*b^4*(b^2 - 4*a*c)^(1/2) - 2*b^6*x - 2*a*b^5 + 2*b^5*x*(b^2 - 4*a*c 
)^(1/2) + 11*a^2*b^3*c - 13*a^3*b*c^2 + 2*a^3*c^3*x + a^3*c^2*(b^2 - 4*a*c 
)^(1/2) - 17*a^2*b^2*c^2*x + 12*a*b^4*c*x - 5*a^2*b^2*c*(b^2 - 4*a*c)^(1/2 
) - 8*a*b^3*c*x*(b^2 - 4*a*c)^(1/2) + 7*a^2*b*c^2*x*(b^2 - 4*a*c)^(1/2))*( 
b^3/(2*a^4) - (b^2*(b^2 - 4*a*c)^(1/2))/(2*a^4) - (b*c)/a^3 + (a^2*c^2*(b^ 
2 - 4*a*c)^(1/2))/(4*a^5*c - a^4*b^2)) + log(2*a*b^5 + 2*b^6*x + 2*a*b^4*( 
b^2 - 4*a*c)^(1/2) + 2*b^5*x*(b^2 - 4*a*c)^(1/2) - 11*a^2*b^3*c + 13*a^3*b 
*c^2 - 2*a^3*c^3*x + a^3*c^2*(b^2 - 4*a*c)^(1/2) + 17*a^2*b^2*c^2*x - 12*a 
*b^4*c*x - 5*a^2*b^2*c*(b^2 - 4*a*c)^(1/2) - 8*a*b^3*c*x*(b^2 - 4*a*c)^(1/ 
2) + 7*a^2*b*c^2*x*(b^2 - 4*a*c)^(1/2))*(b^3/(2*a^4) + (b^2*(b^2 - 4*a*c)^ 
(1/2))/(2*a^4) - (b*c)/a^3 - (a^2*c^2*(b^2 - 4*a*c)^(1/2))/(4*a^5*c - a^4* 
b^2)) + ((x^2*(a*c - b^2))/a^3 - 1/(3*a) + (b*x)/(2*a^2))/x^3 + (b*log(x)* 
(2*a*c - b^2))/a^4
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 308, normalized size of antiderivative = 2.25 \[ \int \frac {1}{\left (c+\frac {a}{x^2}+\frac {b}{x}\right ) x^6} \, dx=\frac {12 \sqrt {4 a c -b^{2}}\, \mathit {atan} \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right ) a^{2} c^{2} x^{3}-24 \sqrt {4 a c -b^{2}}\, \mathit {atan} \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right ) a \,b^{2} c \,x^{3}+6 \sqrt {4 a c -b^{2}}\, \mathit {atan} \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right ) b^{4} x^{3}-24 \,\mathrm {log}\left (c \,x^{2}+b x +a \right ) a^{2} b \,c^{2} x^{3}+18 \,\mathrm {log}\left (c \,x^{2}+b x +a \right ) a \,b^{3} c \,x^{3}-3 \,\mathrm {log}\left (c \,x^{2}+b x +a \right ) b^{5} x^{3}+48 \,\mathrm {log}\left (x \right ) a^{2} b \,c^{2} x^{3}-36 \,\mathrm {log}\left (x \right ) a \,b^{3} c \,x^{3}+6 \,\mathrm {log}\left (x \right ) b^{5} x^{3}-8 a^{4} c +2 a^{3} b^{2}+12 a^{3} b c x +24 a^{3} c^{2} x^{2}-3 a^{2} b^{3} x -30 a^{2} b^{2} c \,x^{2}+6 a \,b^{4} x^{2}}{6 a^{4} x^{3} \left (4 a c -b^{2}\right )} \] Input:

int(1/(c+a/x^2+b/x)/x^6,x)
 

Output:

(12*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**2*c**2*x**3 
 - 24*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a*b**2*c*x** 
3 + 6*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b**4*x**3 - 
24*log(a + b*x + c*x**2)*a**2*b*c**2*x**3 + 18*log(a + b*x + c*x**2)*a*b** 
3*c*x**3 - 3*log(a + b*x + c*x**2)*b**5*x**3 + 48*log(x)*a**2*b*c**2*x**3 
- 36*log(x)*a*b**3*c*x**3 + 6*log(x)*b**5*x**3 - 8*a**4*c + 2*a**3*b**2 + 
12*a**3*b*c*x + 24*a**3*c**2*x**2 - 3*a**2*b**3*x - 30*a**2*b**2*c*x**2 + 
6*a*b**4*x**2)/(6*a**4*x**3*(4*a*c - b**2))