3.5.36 \(\int \frac {1}{(c+\frac {a}{x^2}+\frac {b}{x})^3 x^5} \, dx\) [436]

Optimal. Leaf size=103 \[ \frac {2 a+b x}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac {3 b (b+2 c x)}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}+\frac {6 b c \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}} \]

[Out]

1/2*(b*x+2*a)/(-4*a*c+b^2)/(c*x^2+b*x+a)^2-3/2*b*(2*c*x+b)/(-4*a*c+b^2)^2/(c*x^2+b*x+a)+6*b*c*arctanh((2*c*x+b
)/(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(5/2)

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Rubi [A]
time = 0.03, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {1368, 652, 628, 632, 212} \begin {gather*} \frac {2 a+b x}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac {3 b (b+2 c x)}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}+\frac {6 b c \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((c + a/x^2 + b/x)^3*x^5),x]

[Out]

(2*a + b*x)/(2*(b^2 - 4*a*c)*(a + b*x + c*x^2)^2) - (3*b*(b + 2*c*x))/(2*(b^2 - 4*a*c)^2*(a + b*x + c*x^2)) +
(6*b*c*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(b^2 - 4*a*c)^(5/2)

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 628

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1
)*(b^2 - 4*a*c))), x] - Dist[2*c*((2*p + 3)/((p + 1)*(b^2 - 4*a*c))), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

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 652

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)/((p + 1)*(b^2 - 4*a*c)))*(a + b*x + c*x^2)^(p + 1), x] - Dist[(2*p + 3)*((2*c*d - b*e)/((p + 1)*(b^2 -
4*a*c))), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 1368

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 )^3 x^5} \, dx &=\int \frac {x}{\left (a+b x+c x^2\right )^3} \, dx\\ &=\frac {2 a+b x}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac {(3 b) \int \frac {1}{\left (a+b x+c x^2\right )^2} \, dx}{2 \left (b^2-4 a c\right )}\\ &=\frac {2 a+b x}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac {3 b (b+2 c x)}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac {(3 b c) \int \frac {1}{a+b x+c x^2} \, dx}{\left (b^2-4 a c\right )^2}\\ &=\frac {2 a+b x}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac {3 b (b+2 c x)}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}+\frac {(6 b c) \text {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{\left (b^2-4 a c\right )^2}\\ &=\frac {2 a+b x}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac {3 b (b+2 c x)}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}+\frac {6 b c \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 102, normalized size = 0.99 \begin {gather*} \frac {\frac {\left (b^2-4 a c\right ) (2 a+b x)}{(a+x (b+c x))^2}-\frac {3 b (b+2 c x)}{a+x (b+c x)}-\frac {12 b c \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}}{2 \left (b^2-4 a c\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((c + a/x^2 + b/x)^3*x^5),x]

[Out]

(((b^2 - 4*a*c)*(2*a + b*x))/(a + x*(b + c*x))^2 - (3*b*(b + 2*c*x))/(a + x*(b + c*x)) - (12*b*c*ArcTan[(b + 2
*c*x)/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c])/(2*(b^2 - 4*a*c)^2)

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Maple [A]
time = 0.05, size = 118, normalized size = 1.15

method	result	size
default	\(\frac {-b x -2 a}{2 \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{2}}-\frac {3 b \left (\frac {2 c x +b}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )}+\frac {4 c \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {3}{2}}}\right )}{2 \left (4 a c -b^{2}\right )}\)	\(118\)
risch	\(\frac {-\frac {3 b \,c^{2} x^{3}}{16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}}-\frac {9 b^{2} c \,x^{2}}{2 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}-\frac {\left (5 a c +b^{2}\right ) b x}{16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}}-\frac {a \left (8 a c +b^{2}\right )}{2 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}}{\left (c \,x^{2}+b x +a \right )^{2}}-\frac {3 c b \ln \left (\left (32 a^{2} c^{3}-16 a \,b^{2} c^{2}+2 b^{4} c \right ) x -\left (-4 a c +b^{2}\right )^{\frac {5}{2}}+16 a^{2} b \,c^{2}-8 a \,b^{3} c +b^{5}\right )}{\left (-4 a c +b^{2}\right )^{\frac {5}{2}}}+\frac {3 c b \ln \left (\left (-32 a^{2} c^{3}+16 a \,b^{2} c^{2}-2 b^{4} c \right ) x -\left (-4 a c +b^{2}\right )^{\frac {5}{2}}-16 a^{2} b \,c^{2}+8 a \,b^{3} c -b^{5}\right )}{\left (-4 a c +b^{2}\right )^{\frac {5}{2}}}\)	\(289\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c+a/x^2+b/x)^3/x^5,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+a/x^2+b/x)^3/x^5,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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 384 vs. \(2 (95) = 190\).
time = 0.35, size = 788, normalized size = 7.65 \begin {gather*} \left [-\frac {a b^{4} + 4 \, a^{2} b^{2} c - 32 \, a^{3} c^{2} + 6 \, {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{3} + 9 \, {\left (b^{4} c - 4 \, a b^{2} c^{2}\right )} x^{2} - 6 \, {\left (b c^{3} x^{4} + 2 \, b^{2} c^{2} x^{3} + 2 \, a b^{2} c x + a^{2} b c + {\left (b^{3} c + 2 \, a b c^{2}\right )} x^{2}\right )} \sqrt {b^{2} - 4 \, a c} \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 \, {\left (b^{5} + a b^{3} c - 20 \, a^{2} b c^{2}\right )} x}{2 \, {\left (a^{2} b^{6} - 12 \, a^{3} b^{4} c + 48 \, a^{4} b^{2} c^{2} - 64 \, a^{5} c^{3} + {\left (b^{6} c^{2} - 12 \, a b^{4} c^{3} + 48 \, a^{2} b^{2} c^{4} - 64 \, a^{3} c^{5}\right )} x^{4} + 2 \, {\left (b^{7} c - 12 \, a b^{5} c^{2} + 48 \, a^{2} b^{3} c^{3} - 64 \, a^{3} b c^{4}\right )} x^{3} + {\left (b^{8} - 10 \, a b^{6} c + 24 \, a^{2} b^{4} c^{2} + 32 \, a^{3} b^{2} c^{3} - 128 \, a^{4} c^{4}\right )} x^{2} + 2 \, {\left (a b^{7} - 12 \, a^{2} b^{5} c + 48 \, a^{3} b^{3} c^{2} - 64 \, a^{4} b c^{3}\right )} x\right )}}, -\frac {a b^{4} + 4 \, a^{2} b^{2} c - 32 \, a^{3} c^{2} + 6 \, {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{3} + 9 \, {\left (b^{4} c - 4 \, a b^{2} c^{2}\right )} x^{2} - 12 \, {\left (b c^{3} x^{4} + 2 \, b^{2} c^{2} x^{3} + 2 \, a b^{2} c x + a^{2} b c + {\left (b^{3} c + 2 \, a b c^{2}\right )} x^{2}\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + 2 \, {\left (b^{5} + a b^{3} c - 20 \, a^{2} b c^{2}\right )} x}{2 \, {\left (a^{2} b^{6} - 12 \, a^{3} b^{4} c + 48 \, a^{4} b^{2} c^{2} - 64 \, a^{5} c^{3} + {\left (b^{6} c^{2} - 12 \, a b^{4} c^{3} + 48 \, a^{2} b^{2} c^{4} - 64 \, a^{3} c^{5}\right )} x^{4} + 2 \, {\left (b^{7} c - 12 \, a b^{5} c^{2} + 48 \, a^{2} b^{3} c^{3} - 64 \, a^{3} b c^{4}\right )} x^{3} + {\left (b^{8} - 10 \, a b^{6} c + 24 \, a^{2} b^{4} c^{2} + 32 \, a^{3} b^{2} c^{3} - 128 \, a^{4} c^{4}\right )} x^{2} + 2 \, {\left (a b^{7} - 12 \, a^{2} b^{5} c + 48 \, a^{3} b^{3} c^{2} - 64 \, a^{4} b c^{3}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+a/x^2+b/x)^3/x^5,x, algorithm="fricas")

[Out]

[-1/2*(a*b^4 + 4*a^2*b^2*c - 32*a^3*c^2 + 6*(b^3*c^2 - 4*a*b*c^3)*x^3 + 9*(b^4*c - 4*a*b^2*c^2)*x^2 - 6*(b*c^3
*x^4 + 2*b^2*c^2*x^3 + 2*a*b^2*c*x + a^2*b*c + (b^3*c + 2*a*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)) + 2*(b^5 + a*b^3*c - 20*a^2*b*c^2)*x)/(
a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 64*a^5*c^3 + (b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)*
x^4 + 2*(b^7*c - 12*a*b^5*c^2 + 48*a^2*b^3*c^3 - 64*a^3*b*c^4)*x^3 + (b^8 - 10*a*b^6*c + 24*a^2*b^4*c^2 + 32*a
^3*b^2*c^3 - 128*a^4*c^4)*x^2 + 2*(a*b^7 - 12*a^2*b^5*c + 48*a^3*b^3*c^2 - 64*a^4*b*c^3)*x), -1/2*(a*b^4 + 4*a
^2*b^2*c - 32*a^3*c^2 + 6*(b^3*c^2 - 4*a*b*c^3)*x^3 + 9*(b^4*c - 4*a*b^2*c^2)*x^2 - 12*(b*c^3*x^4 + 2*b^2*c^2*
x^3 + 2*a*b^2*c*x + a^2*b*c + (b^3*c + 2*a*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)) + 2*(b^5 + a*b^3*c - 20*a^2*b*c^2)*x)/(a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 64*a^5*c^3
+ (b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)*x^4 + 2*(b^7*c - 12*a*b^5*c^2 + 48*a^2*b^3*c^3 - 64*a
^3*b*c^4)*x^3 + (b^8 - 10*a*b^6*c + 24*a^2*b^4*c^2 + 32*a^3*b^2*c^3 - 128*a^4*c^4)*x^2 + 2*(a*b^7 - 12*a^2*b^5
*c + 48*a^3*b^3*c^2 - 64*a^4*b*c^3)*x)]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 481 vs. \(2 (95) = 190\).
time = 0.62, size = 481, normalized size = 4.67 \begin {gather*} 3 b c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \log {\left (x + \frac {- 192 a^{3} b c^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 144 a^{2} b^{3} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} - 36 a b^{5} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 3 b^{7} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 3 b^{2} c}{6 b c^{2}} \right )} - 3 b c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \log {\left (x + \frac {192 a^{3} b c^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} - 144 a^{2} b^{3} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 36 a b^{5} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} - 3 b^{7} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 3 b^{2} c}{6 b c^{2}} \right )} + \frac {- 8 a^{2} c - a b^{2} - 9 b^{2} c x^{2} - 6 b c^{2} x^{3} + x \left (- 10 a b c - 2 b^{3}\right )}{32 a^{4} c^{2} - 16 a^{3} b^{2} c + 2 a^{2} b^{4} + x^{4} \cdot \left (32 a^{2} c^{4} - 16 a b^{2} c^{3} + 2 b^{4} c^{2}\right ) + x^{3} \cdot \left (64 a^{2} b c^{3} - 32 a b^{3} c^{2} + 4 b^{5} c\right ) + x^{2} \cdot \left (64 a^{3} c^{3} - 12 a b^{4} c + 2 b^{6}\right ) + x \left (64 a^{3} b c^{2} - 32 a^{2} b^{3} c + 4 a b^{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+a/x**2+b/x)**3/x**5,x)

[Out]

3*b*c*sqrt(-1/(4*a*c - b**2)**5)*log(x + (-192*a**3*b*c**4*sqrt(-1/(4*a*c - b**2)**5) + 144*a**2*b**3*c**3*sqr
t(-1/(4*a*c - b**2)**5) - 36*a*b**5*c**2*sqrt(-1/(4*a*c - b**2)**5) + 3*b**7*c*sqrt(-1/(4*a*c - b**2)**5) + 3*
b**2*c)/(6*b*c**2)) - 3*b*c*sqrt(-1/(4*a*c - b**2)**5)*log(x + (192*a**3*b*c**4*sqrt(-1/(4*a*c - b**2)**5) - 1
44*a**2*b**3*c**3*sqrt(-1/(4*a*c - b**2)**5) + 36*a*b**5*c**2*sqrt(-1/(4*a*c - b**2)**5) - 3*b**7*c*sqrt(-1/(4
*a*c - b**2)**5) + 3*b**2*c)/(6*b*c**2)) + (-8*a**2*c - a*b**2 - 9*b**2*c*x**2 - 6*b*c**2*x**3 + x*(-10*a*b*c
- 2*b**3))/(32*a**4*c**2 - 16*a**3*b**2*c + 2*a**2*b**4 + x**4*(32*a**2*c**4 - 16*a*b**2*c**3 + 2*b**4*c**2) +
x**3*(64*a**2*b*c**3 - 32*a*b**3*c**2 + 4*b**5*c) + x**2*(64*a**3*c**3 - 12*a*b**4*c + 2*b**6) + x*(64*a**3*b
*c**2 - 32*a**2*b**3*c + 4*a*b**5))

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Giac [A]
time = 4.05, size = 135, normalized size = 1.31 \begin {gather*} -\frac {6 \, b c \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {6 \, b c^{2} x^{3} + 9 \, b^{2} c x^{2} + 2 \, b^{3} x + 10 \, a b c x + a b^{2} + 8 \, a^{2} c}{2 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} {\left (c x^{2} + b x + a\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+a/x^2+b/x)^3/x^5,x, algorithm="giac")

[Out]

-6*b*c*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(-b^2 + 4*a*c)) - 1/2*(6*b*c
^2*x^3 + 9*b^2*c*x^2 + 2*b^3*x + 10*a*b*c*x + a*b^2 + 8*a^2*c)/((b^4 - 8*a*b^2*c + 16*a^2*c^2)*(c*x^2 + b*x +
a)^2)

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Mupad [B]
time = 1.43, size = 253, normalized size = 2.46 \begin {gather*} -\frac {\frac {8\,c\,a^2+a\,b^2}{2\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {9\,b^2\,c\,x^2}{2\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {3\,b\,c^2\,x^3}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}+\frac {b\,x\,\left (b^2+5\,a\,c\right )}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}}{x^2\,\left (b^2+2\,a\,c\right )+a^2+c^2\,x^4+2\,a\,b\,x+2\,b\,c\,x^3}-\frac {6\,b\,c\,\mathrm {atan}\left (\frac {\left (\frac {3\,b^2\,c}{{\left (4\,a\,c-b^2\right )}^{5/2}}+\frac {6\,b\,c^2\,x}{{\left (4\,a\,c-b^2\right )}^{5/2}}\right )\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}{3\,b\,c}\right )}{{\left (4\,a\,c-b^2\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^5*(c + a/x^2 + b/x)^3),x)

[Out]

- ((a*b^2 + 8*a^2*c)/(2*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (9*b^2*c*x^2)/(2*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (
3*b*c^2*x^3)/(b^4 + 16*a^2*c^2 - 8*a*b^2*c) + (b*x*(5*a*c + b^2))/(b^4 + 16*a^2*c^2 - 8*a*b^2*c))/(x^2*(2*a*c
+ b^2) + a^2 + c^2*x^4 + 2*a*b*x + 2*b*c*x^3) - (6*b*c*atan((((3*b^2*c)/(4*a*c - b^2)^(5/2) + (6*b*c^2*x)/(4*a
*c - b^2)^(5/2))*(b^4 + 16*a^2*c^2 - 8*a*b^2*c))/(3*b*c)))/(4*a*c - b^2)^(5/2)

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