3.412 \(\int \frac {x^3}{c+\frac {a}{x^2}+\frac {b}{x}} \, dx\)

Optimal. Leaf size=147 \[ \frac {\left (a^2 c^2-3 a b^2 c+b^4\right ) \log \left (a+b x+c x^2\right )}{2 c^5}+\frac {b \left (5 a^2 c^2-5 a b^2 c+b^4\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^5 \sqrt {b^2-4 a c}}-\frac {b x \left (b^2-2 a c\right )}{c^4}+\frac {x^2 \left (b^2-a c\right )}{2 c^3}-\frac {b x^3}{3 c^2}+\frac {x^4}{4 c} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1354, 701, 634, 618, 206, 628} \[ \frac {\left (a^2 c^2-3 a b^2 c+b^4\right ) \log \left (a+b x+c x^2\right )}{2 c^5}+\frac {b \left (5 a^2 c^2-5 a b^2 c+b^4\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^5 \sqrt {b^2-4 a c}}+\frac {x^2 \left (b^2-a c\right )}{2 c^3}-\frac {b x \left (b^2-2 a c\right )}{c^4}-\frac {b x^3}{3 c^2}+\frac {x^4}{4 c} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(c + a/x^2 + b/x),x]

[Out]

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

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 701

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[(d + e*x)
^m, a + b*x + c*x^2, x], x] /; FreeQ[{a, b, c, d, e}, 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] && IGtQ[m, 1] && (NeQ[d, 0] || GtQ[m, 2])

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

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Mathematica [A]  time = 0.12, size = 140, normalized size = 0.95 \[ \frac {6 \left (a^2 c^2-3 a b^2 c+b^4\right ) \log (a+x (b+c x))-\frac {12 b \left (5 a^2 c^2-5 a b^2 c+b^4\right ) \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}+c x \left (-4 b c \left (c x^2-6 a\right )+3 c^2 x \left (c x^2-2 a\right )-12 b^3+6 b^2 c x\right )}{12 c^5} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/(c + a/x^2 + b/x),x]

[Out]

(c*x*(-12*b^3 + 6*b^2*c*x - 4*b*c*(-6*a + c*x^2) + 3*c^2*x*(-2*a + c*x^2)) - (12*b*(b^4 - 5*a*b^2*c + 5*a^2*c^
2)*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c] + 6*(b^4 - 3*a*b^2*c + a^2*c^2)*Log[a + x*(b + c
*x)])/(12*c^5)

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fricas [A]  time = 0.90, size = 466, normalized size = 3.17 \[ \left [\frac {3 \, {\left (b^{2} c^{4} - 4 \, a c^{5}\right )} x^{4} - 4 \, {\left (b^{3} c^{3} - 4 \, a b c^{4}\right )} x^{3} + 6 \, {\left (b^{4} c^{2} - 5 \, a b^{2} c^{3} + 4 \, a^{2} c^{4}\right )} x^{2} + 6 \, {\left (b^{5} - 5 \, a b^{3} c + 5 \, a^{2} b c^{2}\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) - 12 \, {\left (b^{5} c - 6 \, a b^{3} c^{2} + 8 \, a^{2} b c^{3}\right )} x + 6 \, {\left (b^{6} - 7 \, a b^{4} c + 13 \, a^{2} b^{2} c^{2} - 4 \, a^{3} c^{3}\right )} \log \left (c x^{2} + b x + a\right )}{12 \, {\left (b^{2} c^{5} - 4 \, a c^{6}\right )}}, \frac {3 \, {\left (b^{2} c^{4} - 4 \, a c^{5}\right )} x^{4} - 4 \, {\left (b^{3} c^{3} - 4 \, a b c^{4}\right )} x^{3} + 6 \, {\left (b^{4} c^{2} - 5 \, a b^{2} c^{3} + 4 \, a^{2} c^{4}\right )} x^{2} + 12 \, {\left (b^{5} - 5 \, a b^{3} c + 5 \, a^{2} b c^{2}\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) - 12 \, {\left (b^{5} c - 6 \, a b^{3} c^{2} + 8 \, a^{2} b c^{3}\right )} x + 6 \, {\left (b^{6} - 7 \, a b^{4} c + 13 \, a^{2} b^{2} c^{2} - 4 \, a^{3} c^{3}\right )} \log \left (c x^{2} + b x + a\right )}{12 \, {\left (b^{2} c^{5} - 4 \, a c^{6}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(3*(b^2*c^4 - 4*a*c^5)*x^4 - 4*(b^3*c^3 - 4*a*b*c^4)*x^3 + 6*(b^4*c^2 - 5*a*b^2*c^3 + 4*a^2*c^4)*x^2 + 6
*(b^5 - 5*a*b^3*c + 5*a^2*b*c^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)) - 12*(b^5*c - 6*a*b^3*c^2 + 8*a^2*b*c^3)*x + 6*(b^6 - 7*a*b^4*c + 13*a^2*b^2*c
^2 - 4*a^3*c^3)*log(c*x^2 + b*x + a))/(b^2*c^5 - 4*a*c^6), 1/12*(3*(b^2*c^4 - 4*a*c^5)*x^4 - 4*(b^3*c^3 - 4*a*
b*c^4)*x^3 + 6*(b^4*c^2 - 5*a*b^2*c^3 + 4*a^2*c^4)*x^2 + 12*(b^5 - 5*a*b^3*c + 5*a^2*b*c^2)*sqrt(-b^2 + 4*a*c)
*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) - 12*(b^5*c - 6*a*b^3*c^2 + 8*a^2*b*c^3)*x + 6*(b^6 - 7
*a*b^4*c + 13*a^2*b^2*c^2 - 4*a^3*c^3)*log(c*x^2 + b*x + a))/(b^2*c^5 - 4*a*c^6)]

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giac [A]  time = 0.29, size = 145, normalized size = 0.99 \[ \frac {3 \, c^{3} x^{4} - 4 \, b c^{2} x^{3} + 6 \, b^{2} c x^{2} - 6 \, a c^{2} x^{2} - 12 \, b^{3} x + 24 \, a b c x}{12 \, c^{4}} + \frac {{\left (b^{4} - 3 \, a b^{2} c + a^{2} c^{2}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, c^{5}} - \frac {{\left (b^{5} - 5 \, a b^{3} c + 5 \, a^{2} b c^{2}\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c} c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*c^3*x^4 - 4*b*c^2*x^3 + 6*b^2*c*x^2 - 6*a*c^2*x^2 - 12*b^3*x + 24*a*b*c*x)/c^4 + 1/2*(b^4 - 3*a*b^2*c
+ a^2*c^2)*log(c*x^2 + b*x + a)/c^5 - (b^5 - 5*a*b^3*c + 5*a^2*b*c^2)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/(
sqrt(-b^2 + 4*a*c)*c^5)

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maple [A]  time = 0.01, size = 236, normalized size = 1.61 \[ \frac {x^{4}}{4 c}-\frac {b \,x^{3}}{3 c^{2}}-\frac {5 a^{2} b \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c^{3}}+\frac {5 a \,b^{3} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c^{4}}-\frac {a \,x^{2}}{2 c^{2}}-\frac {b^{5} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c^{5}}+\frac {b^{2} x^{2}}{2 c^{3}}+\frac {a^{2} \ln \left (c \,x^{2}+b x +a \right )}{2 c^{3}}-\frac {3 a \,b^{2} \ln \left (c \,x^{2}+b x +a \right )}{2 c^{4}}+\frac {2 a b x}{c^{3}}+\frac {b^{4} \ln \left (c \,x^{2}+b x +a \right )}{2 c^{5}}-\frac {b^{3} x}{c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(x^3/(c+a/x^2+b/x),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 = 0.16, size = 183, normalized size = 1.24 \[ x\,\left (\frac {b\,\left (\frac {a}{c^2}-\frac {b^2}{c^3}\right )}{c}+\frac {a\,b}{c^3}\right )+\frac {x^4}{4\,c}-x^2\,\left (\frac {a}{2\,c^2}-\frac {b^2}{2\,c^3}\right )-\frac {\ln \left (c\,x^2+b\,x+a\right )\,\left (-4\,a^3\,c^3+13\,a^2\,b^2\,c^2-7\,a\,b^4\,c+b^6\right )}{2\,\left (4\,a\,c^6-b^2\,c^5\right )}-\frac {b\,x^3}{3\,c^2}-\frac {b\,\mathrm {atan}\left (\frac {b+2\,c\,x}{\sqrt {4\,a\,c-b^2}}\right )\,\left (5\,a^2\,c^2-5\,a\,b^2\,c+b^4\right )}{c^5\,\sqrt {4\,a\,c-b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(c + a/x^2 + b/x),x)

[Out]

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

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sympy [B]  time = 1.25, size = 605, normalized size = 4.12 \[ - \frac {b x^{3}}{3 c^{2}} + x^{2} \left (- \frac {a}{2 c^{2}} + \frac {b^{2}}{2 c^{3}}\right ) + x \left (\frac {2 a b}{c^{3}} - \frac {b^{3}}{c^{4}}\right ) + \left (- \frac {b \sqrt {- 4 a c + b^{2}} \left (5 a^{2} c^{2} - 5 a b^{2} c + b^{4}\right )}{2 c^{5} \left (4 a c - b^{2}\right )} + \frac {a^{2} c^{2} - 3 a b^{2} c + b^{4}}{2 c^{5}}\right ) \log {\left (x + \frac {2 a^{3} c^{2} - 4 a^{2} b^{2} c + a b^{4} - 4 a c^{5} \left (- \frac {b \sqrt {- 4 a c + b^{2}} \left (5 a^{2} c^{2} - 5 a b^{2} c + b^{4}\right )}{2 c^{5} \left (4 a c - b^{2}\right )} + \frac {a^{2} c^{2} - 3 a b^{2} c + b^{4}}{2 c^{5}}\right ) + b^{2} c^{4} \left (- \frac {b \sqrt {- 4 a c + b^{2}} \left (5 a^{2} c^{2} - 5 a b^{2} c + b^{4}\right )}{2 c^{5} \left (4 a c - b^{2}\right )} + \frac {a^{2} c^{2} - 3 a b^{2} c + b^{4}}{2 c^{5}}\right )}{5 a^{2} b c^{2} - 5 a b^{3} c + b^{5}} \right )} + \left (\frac {b \sqrt {- 4 a c + b^{2}} \left (5 a^{2} c^{2} - 5 a b^{2} c + b^{4}\right )}{2 c^{5} \left (4 a c - b^{2}\right )} + \frac {a^{2} c^{2} - 3 a b^{2} c + b^{4}}{2 c^{5}}\right ) \log {\left (x + \frac {2 a^{3} c^{2} - 4 a^{2} b^{2} c + a b^{4} - 4 a c^{5} \left (\frac {b \sqrt {- 4 a c + b^{2}} \left (5 a^{2} c^{2} - 5 a b^{2} c + b^{4}\right )}{2 c^{5} \left (4 a c - b^{2}\right )} + \frac {a^{2} c^{2} - 3 a b^{2} c + b^{4}}{2 c^{5}}\right ) + b^{2} c^{4} \left (\frac {b \sqrt {- 4 a c + b^{2}} \left (5 a^{2} c^{2} - 5 a b^{2} c + b^{4}\right )}{2 c^{5} \left (4 a c - b^{2}\right )} + \frac {a^{2} c^{2} - 3 a b^{2} c + b^{4}}{2 c^{5}}\right )}{5 a^{2} b c^{2} - 5 a b^{3} c + b^{5}} \right )} + \frac {x^{4}}{4 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*x**3/(3*c**2) + x**2*(-a/(2*c**2) + b**2/(2*c**3)) + x*(2*a*b/c**3 - b**3/c**4) + (-b*sqrt(-4*a*c + b**2)*(
5*a**2*c**2 - 5*a*b**2*c + b**4)/(2*c**5*(4*a*c - b**2)) + (a**2*c**2 - 3*a*b**2*c + b**4)/(2*c**5))*log(x + (
2*a**3*c**2 - 4*a**2*b**2*c + a*b**4 - 4*a*c**5*(-b*sqrt(-4*a*c + b**2)*(5*a**2*c**2 - 5*a*b**2*c + b**4)/(2*c
**5*(4*a*c - b**2)) + (a**2*c**2 - 3*a*b**2*c + b**4)/(2*c**5)) + b**2*c**4*(-b*sqrt(-4*a*c + b**2)*(5*a**2*c*
*2 - 5*a*b**2*c + b**4)/(2*c**5*(4*a*c - b**2)) + (a**2*c**2 - 3*a*b**2*c + b**4)/(2*c**5)))/(5*a**2*b*c**2 -
5*a*b**3*c + b**5)) + (b*sqrt(-4*a*c + b**2)*(5*a**2*c**2 - 5*a*b**2*c + b**4)/(2*c**5*(4*a*c - b**2)) + (a**2
*c**2 - 3*a*b**2*c + b**4)/(2*c**5))*log(x + (2*a**3*c**2 - 4*a**2*b**2*c + a*b**4 - 4*a*c**5*(b*sqrt(-4*a*c +
b**2)*(5*a**2*c**2 - 5*a*b**2*c + b**4)/(2*c**5*(4*a*c - b**2)) + (a**2*c**2 - 3*a*b**2*c + b**4)/(2*c**5)) +
b**2*c**4*(b*sqrt(-4*a*c + b**2)*(5*a**2*c**2 - 5*a*b**2*c + b**4)/(2*c**5*(4*a*c - b**2)) + (a**2*c**2 - 3*a
*b**2*c + b**4)/(2*c**5)))/(5*a**2*b*c**2 - 5*a*b**3*c + b**5)) + x**4/(4*c)

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