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

Optimal. Leaf size=137 \[ \frac {b \left (b^2-2 a c\right ) \log \left (a+b x+c x^2\right )}{2 a^4}-\frac {b \log (x) \left (b^2-2 a c\right )}{a^4}-\frac {b^2-a c}{a^3 x}+\frac {b}{2 a^2 x^2}-\frac {\left (2 a^2 c^2-4 a b^2 c+b^4\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a^4 \sqrt {b^2-4 a c}}-\frac {1}{3 a x^3} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {1354, 709, 800, 634, 618, 206, 628} \[ -\frac {\left (2 a^2 c^2-4 a b^2 c+b^4\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a^4 \sqrt {b^2-4 a c}}+\frac {b \left (b^2-2 a c\right ) \log \left (a+b x+c x^2\right )}{2 a^4}-\frac {b^2-a c}{a^3 x}-\frac {b \log (x) \left (b^2-2 a c\right )}{a^4}+\frac {b}{2 a^2 x^2}-\frac {1}{3 a x^3} \]

Antiderivative was successfully verified.

[In]

Int[1/((c + a/x^2 + b/x)*x^6),x]

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 618

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 628

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 634

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 709

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] + Dist[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, 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[m, -1]

Rule 800

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 1354

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]

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 131, normalized size = 0.96 \[ \frac {-\frac {2 a^3}{x^3}+\frac {6 \left (2 a^2 c^2-4 a b^2 c+b^4\right ) \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}+\frac {3 a^2 b}{x^2}-6 \log (x) \left (b^3-2 a b c\right )+3 \left (b^3-2 a b c\right ) \log (a+x (b+c x))+\frac {6 a \left (a c-b^2\right )}{x}}{6 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-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)/Sqr
t[-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)

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fricas [A]  time = 1.04, size = 445, normalized size = 3.25 \[ \left [\frac {3 \, {\left (b^{4} - 4 \, a b^{2} c + 2 \, a^{2} c^{2}\right )} \sqrt {b^{2} - 4 \, a c} x^{3} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c - \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) - 2 \, a^{3} b^{2} + 8 \, a^{4} c + 3 \, {\left (b^{5} - 6 \, a b^{3} c + 8 \, a^{2} b c^{2}\right )} x^{3} \log \left (c x^{2} + b x + a\right ) - 6 \, {\left (b^{5} - 6 \, a b^{3} c + 8 \, a^{2} b c^{2}\right )} x^{3} \log \relax (x) - 6 \, {\left (a b^{4} - 5 \, a^{2} b^{2} c + 4 \, a^{3} c^{2}\right )} x^{2} + 3 \, {\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x}{6 \, {\left (a^{4} b^{2} - 4 \, a^{5} c\right )} x^{3}}, -\frac {6 \, {\left (b^{4} - 4 \, a b^{2} c + 2 \, a^{2} c^{2}\right )} \sqrt {-b^{2} + 4 \, a c} x^{3} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + 2 \, a^{3} b^{2} - 8 \, a^{4} c - 3 \, {\left (b^{5} - 6 \, a b^{3} c + 8 \, a^{2} b c^{2}\right )} x^{3} \log \left (c x^{2} + b x + a\right ) + 6 \, {\left (b^{5} - 6 \, a b^{3} c + 8 \, a^{2} b c^{2}\right )} x^{3} \log \relax (x) + 6 \, {\left (a b^{4} - 5 \, a^{2} b^{2} c + 4 \, a^{3} c^{2}\right )} x^{2} - 3 \, {\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x}{6 \, {\left (a^{4} b^{2} - 4 \, a^{5} c\right )} x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[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*
arctan(-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)]

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giac [A]  time = 0.38, size = 136, normalized size = 0.99 \[ \frac {{\left (b^{3} - 2 \, a b c\right )} \log \left (c x^{2} + b x + a\right )}{2 \, a^{4}} - \frac {{\left (b^{3} - 2 \, a b c\right )} \log \left ({\left | x \right |}\right )}{a^{4}} + \frac {{\left (b^{4} - 4 \, a b^{2} c + 2 \, a^{2} c^{2}\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c} a^{4}} + \frac {3 \, a^{2} b x - 2 \, a^{3} - 6 \, {\left (a b^{2} - a^{2} c\right )} x^{2}}{6 \, a^{4} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)

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maple [A]  time = 0.01, size = 214, normalized size = 1.56 \[ \frac {2 c^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, a^{2}}-\frac {4 b^{2} c \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, a^{3}}+\frac {b^{4} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, a^{4}}+\frac {2 b c \ln \relax (x )}{a^{3}}-\frac {b c \ln \left (c \,x^{2}+b x +a \right )}{a^{3}}-\frac {b^{3} \ln \relax (x )}{a^{4}}+\frac {b^{3} \ln \left (c \,x^{2}+b x +a \right )}{2 a^{4}}+\frac {c}{a^{2} x}-\frac {b^{2}}{a^{3} x}+\frac {b}{2 a^{2} x^{2}}-\frac {1}{3 a \,x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+a/x^2+b/x)/x^6,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 details)Is 4*a*c-b^2 positive or negative?

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mupad [B]  time = 1.92, size = 524, normalized size = 3.82 \[ \ln \left (2\,a\,b^4\,\sqrt {b^2-4\,a\,c}-2\,b^6\,x-2\,a\,b^5+2\,b^5\,x\,\sqrt {b^2-4\,a\,c}+11\,a^2\,b^3\,c-13\,a^3\,b\,c^2+2\,a^3\,c^3\,x+a^3\,c^2\,\sqrt {b^2-4\,a\,c}-17\,a^2\,b^2\,c^2\,x+12\,a\,b^4\,c\,x-5\,a^2\,b^2\,c\,\sqrt {b^2-4\,a\,c}-8\,a\,b^3\,c\,x\,\sqrt {b^2-4\,a\,c}+7\,a^2\,b\,c^2\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (\frac {b^3}{2\,a^4}-\frac {b^2\,\sqrt {b^2-4\,a\,c}}{2\,a^4}-\frac {b\,c}{a^3}+\frac {a^2\,c^2\,\sqrt {b^2-4\,a\,c}}{4\,a^5\,c-a^4\,b^2}\right )+\ln \left (2\,a\,b^5+2\,b^6\,x+2\,a\,b^4\,\sqrt {b^2-4\,a\,c}+2\,b^5\,x\,\sqrt {b^2-4\,a\,c}-11\,a^2\,b^3\,c+13\,a^3\,b\,c^2-2\,a^3\,c^3\,x+a^3\,c^2\,\sqrt {b^2-4\,a\,c}+17\,a^2\,b^2\,c^2\,x-12\,a\,b^4\,c\,x-5\,a^2\,b^2\,c\,\sqrt {b^2-4\,a\,c}-8\,a\,b^3\,c\,x\,\sqrt {b^2-4\,a\,c}+7\,a^2\,b\,c^2\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (\frac {b^3}{2\,a^4}+\frac {b^2\,\sqrt {b^2-4\,a\,c}}{2\,a^4}-\frac {b\,c}{a^3}-\frac {a^2\,c^2\,\sqrt {b^2-4\,a\,c}}{4\,a^5\,c-a^4\,b^2}\right )+\frac {\frac {x^2\,\left (a\,c-b^2\right )}{a^3}-\frac {1}{3\,a}+\frac {b\,x}{2\,a^2}}{x^3}+\frac {b\,\ln \relax (x)\,\left (2\,a\,c-b^2\right )}{a^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^6*(c + a/x^2 + b/x)),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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