3.854 \(\int \frac {1}{x^5 (a+b x^2+c x^4)} \, dx\)

Optimal. Leaf size=114 \[ \frac {b \left (b^2-3 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 a^3 \sqrt {b^2-4 a c}}-\frac {\left (b^2-a c\right ) \log \left (a+b x^2+c x^4\right )}{4 a^3}+\frac {\log (x) \left (b^2-a c\right )}{a^3}+\frac {b}{2 a^2 x^2}-\frac {1}{4 a x^4} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 114, 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 = {1114, 709, 800, 634, 618, 206, 628} \[ -\frac {\left (b^2-a c\right ) \log \left (a+b x^2+c x^4\right )}{4 a^3}+\frac {b \left (b^2-3 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 a^3 \sqrt {b^2-4 a c}}+\frac {\log (x) \left (b^2-a c\right )}{a^3}+\frac {b}{2 a^2 x^2}-\frac {1}{4 a x^4} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^5*(a + b*x^2 + c*x^4)),x]

[Out]

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

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 1114

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

Rubi steps

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

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Mathematica [A]  time = 0.23, size = 188, normalized size = 1.65 \[ \frac {-\frac {a^2}{x^4}+4 \log (x) \left (b^2-a c\right )-\frac {\left (b^2 \sqrt {b^2-4 a c}-a c \sqrt {b^2-4 a c}-3 a b c+b^3\right ) \log \left (-\sqrt {b^2-4 a c}+b+2 c x^2\right )}{\sqrt {b^2-4 a c}}+\frac {\left (-b^2 \sqrt {b^2-4 a c}+a c \sqrt {b^2-4 a c}-3 a b c+b^3\right ) \log \left (\sqrt {b^2-4 a c}+b+2 c x^2\right )}{\sqrt {b^2-4 a c}}+\frac {2 a b}{x^2}}{4 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*(a + b*x^2 + c*x^4)),x]

[Out]

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

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fricas [A]  time = 0.78, size = 374, normalized size = 3.28 \[ \left [-\frac {{\left (b^{3} - 3 \, a b c\right )} \sqrt {b^{2} - 4 \, a c} x^{4} \log \left (\frac {2 \, c^{2} x^{4} + 2 \, b c x^{2} + b^{2} - 2 \, a c - {\left (2 \, c x^{2} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{4} + b x^{2} + a}\right ) + {\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} x^{4} \log \left (c x^{4} + b x^{2} + a\right ) - 4 \, {\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} x^{4} \log \relax (x) + a^{2} b^{2} - 4 \, a^{3} c - 2 \, {\left (a b^{3} - 4 \, a^{2} b c\right )} x^{2}}{4 \, {\left (a^{3} b^{2} - 4 \, a^{4} c\right )} x^{4}}, \frac {2 \, {\left (b^{3} - 3 \, a b c\right )} \sqrt {-b^{2} + 4 \, a c} x^{4} \arctan \left (-\frac {{\left (2 \, c x^{2} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) - {\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} x^{4} \log \left (c x^{4} + b x^{2} + a\right ) + 4 \, {\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} x^{4} \log \relax (x) - a^{2} b^{2} + 4 \, a^{3} c + 2 \, {\left (a b^{3} - 4 \, a^{2} b c\right )} x^{2}}{4 \, {\left (a^{3} b^{2} - 4 \, a^{4} c\right )} x^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(c*x^4+b*x^2+a),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.55, size = 126, normalized size = 1.11 \[ -\frac {{\left (b^{2} - a c\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, a^{3}} + \frac {{\left (b^{2} - a c\right )} \log \left (x^{2}\right )}{2 \, a^{3}} - \frac {{\left (b^{3} - 3 \, a b c\right )} \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{2 \, \sqrt {-b^{2} + 4 \, a c} a^{3}} - \frac {3 \, b^{2} x^{4} - 3 \, a c x^{4} - 2 \, a b x^{2} + a^{2}}{4 \, a^{3} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(c*x^4+b*x^2+a),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 159, normalized size = 1.39 \[ \frac {3 b c \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{2 \sqrt {4 a c -b^{2}}\, a^{2}}-\frac {b^{3} \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{2 \sqrt {4 a c -b^{2}}\, a^{3}}-\frac {c \ln \relax (x )}{a^{2}}+\frac {c \ln \left (c \,x^{4}+b \,x^{2}+a \right )}{4 a^{2}}+\frac {b^{2} \ln \relax (x )}{a^{3}}-\frac {b^{2} \ln \left (c \,x^{4}+b \,x^{2}+a \right )}{4 a^{3}}+\frac {b}{2 a^{2} x^{2}}-\frac {1}{4 a \,x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(c*x^4+b*x^2+a),x)

[Out]

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

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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/x^5/(c*x^4+b*x^2+a),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 = 6.37, size = 2451, normalized size = 21.50 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^5*(a + b*x^2 + c*x^4)),x)

[Out]

(log(a + b*x^2 + c*x^4)*(2*b^4 + 8*a^2*c^2 - 10*a*b^2*c))/(2*(16*a^4*c - 4*a^3*b^2)) - (1/(4*a) - (b*x^2)/(2*a
^2))/x^4 - (log(x)*(a*c - b^2))/a^3 + (b*atan((2*a^6*(4*a*c - b^2)*((((b*(3*a*c - b^2)*((4*a^4*b^4*c^2 - 8*a^5
*b^2*c^3)/a^6 - (2*a*b^2*c^2*(2*b^4 + 8*a^2*c^2 - 10*a*b^2*c))/(16*a^4*c - 4*a^3*b^2)))/(4*a^3*(4*a*c - b^2)^(
1/2)) - (b^3*c^2*(3*a*c - b^2)*(2*b^4 + 8*a^2*c^2 - 10*a*b^2*c))/(2*a^2*(4*a*c - b^2)^(1/2)*(16*a^4*c - 4*a^3*
b^2)))*(2*b^4 + 8*a^2*c^2 - 10*a*b^2*c))/(2*(16*a^4*c - 4*a^3*b^2)) + (b^5*c^2*(3*a*c - b^2)^3)/(16*a^8*(4*a*c
- b^2)^(3/2)) + (b*(3*a*c - b^2)*((4*a^2*b^4*c^3 - 5*a^3*b^2*c^4)/a^6 + (((4*a^4*b^4*c^2 - 8*a^5*b^2*c^3)/a^6
- (2*a*b^2*c^2*(2*b^4 + 8*a^2*c^2 - 10*a*b^2*c))/(16*a^4*c - 4*a^3*b^2))*(2*b^4 + 8*a^2*c^2 - 10*a*b^2*c))/(2
*(16*a^4*c - 4*a^3*b^2))))/(4*a^3*(4*a*c - b^2)^(1/2)))*(3*b^6 - 10*a^3*c^3 + 27*a^2*b^2*c^2 - 18*a*b^4*c))/(c
^2*(b^6*c^2 - 6*a*b^4*c^3 + 9*a^2*b^2*c^4)*(6*b^6 - 25*a^3*c^3 + 54*a^2*b^2*c^2 - 36*a*b^4*c)) - (16*a^9*x^2*(
(3*b*(b^4 + 3*a^2*c^2 - 4*a*b^2*c)*((((5*a^3*b*c^5 - 6*a^2*b^3*c^4)/a^6 - (((10*a^5*b*c^4 + 2*a^4*b^3*c^3)/a^6
+ ((40*a^7*b*c^3 - 12*a^6*b^3*c^2)*(2*b^4 + 8*a^2*c^2 - 10*a*b^2*c))/(2*a^6*(16*a^4*c - 4*a^3*b^2)))*(2*b^4 +
8*a^2*c^2 - 10*a*b^2*c))/(2*(16*a^4*c - 4*a^3*b^2)))*(2*b^4 + 8*a^2*c^2 - 10*a*b^2*c))/(2*(16*a^4*c - 4*a^3*b
^2)) - (b^3*c^5)/a^6 + (b*(3*a*c - b^2)*((b*((10*a^5*b*c^4 + 2*a^4*b^3*c^3)/a^6 + ((40*a^7*b*c^3 - 12*a^6*b^3*
c^2)*(2*b^4 + 8*a^2*c^2 - 10*a*b^2*c))/(2*a^6*(16*a^4*c - 4*a^3*b^2)))*(3*a*c - b^2))/(4*a^3*(4*a*c - b^2)^(1/
2)) + (b*(40*a^7*b*c^3 - 12*a^6*b^3*c^2)*(3*a*c - b^2)*(2*b^4 + 8*a^2*c^2 - 10*a*b^2*c))/(8*a^9*(4*a*c - b^2)^
(1/2)*(16*a^4*c - 4*a^3*b^2))))/(4*a^3*(4*a*c - b^2)^(1/2)) + (b^2*(40*a^7*b*c^3 - 12*a^6*b^3*c^2)*(3*a*c - b^
2)^2*(2*b^4 + 8*a^2*c^2 - 10*a*b^2*c))/(32*a^12*(4*a*c - b^2)*(16*a^4*c - 4*a^3*b^2))))/(8*a^3*c^2*(6*b^6 - 25
*a^3*c^3 + 54*a^2*b^2*c^2 - 36*a*b^4*c)) + (((b^3*(40*a^7*b*c^3 - 12*a^6*b^3*c^2)*(3*a*c - b^2)^3)/(64*a^15*(4
*a*c - b^2)^(3/2)) - (((b*((10*a^5*b*c^4 + 2*a^4*b^3*c^3)/a^6 + ((40*a^7*b*c^3 - 12*a^6*b^3*c^2)*(2*b^4 + 8*a^
2*c^2 - 10*a*b^2*c))/(2*a^6*(16*a^4*c - 4*a^3*b^2)))*(3*a*c - b^2))/(4*a^3*(4*a*c - b^2)^(1/2)) + (b*(40*a^7*b
*c^3 - 12*a^6*b^3*c^2)*(3*a*c - b^2)*(2*b^4 + 8*a^2*c^2 - 10*a*b^2*c))/(8*a^9*(4*a*c - b^2)^(1/2)*(16*a^4*c -
4*a^3*b^2)))*(2*b^4 + 8*a^2*c^2 - 10*a*b^2*c))/(2*(16*a^4*c - 4*a^3*b^2)) + (b*((5*a^3*b*c^5 - 6*a^2*b^3*c^4)/
a^6 - (((10*a^5*b*c^4 + 2*a^4*b^3*c^3)/a^6 + ((40*a^7*b*c^3 - 12*a^6*b^3*c^2)*(2*b^4 + 8*a^2*c^2 - 10*a*b^2*c)
)/(2*a^6*(16*a^4*c - 4*a^3*b^2)))*(2*b^4 + 8*a^2*c^2 - 10*a*b^2*c))/(2*(16*a^4*c - 4*a^3*b^2)))*(3*a*c - b^2))
/(4*a^3*(4*a*c - b^2)^(1/2)))*(3*b^6 - 10*a^3*c^3 + 27*a^2*b^2*c^2 - 18*a*b^4*c))/(8*a^3*c^2*(4*a*c - b^2)^(1/
2)*(6*b^6 - 25*a^3*c^3 + 54*a^2*b^2*c^2 - 36*a*b^4*c)))*(4*a*c - b^2)^(3/2))/(b^6*c^2 - 6*a*b^4*c^3 + 9*a^2*b^
2*c^4) + (6*a^6*b*(4*a*c - b^2)^(3/2)*(b^4 + 3*a^2*c^2 - 4*a*b^2*c)*((b^4*c^4 - a*b^2*c^5)/a^6 + (((4*a^2*b^4*
c^3 - 5*a^3*b^2*c^4)/a^6 + (((4*a^4*b^4*c^2 - 8*a^5*b^2*c^3)/a^6 - (2*a*b^2*c^2*(2*b^4 + 8*a^2*c^2 - 10*a*b^2*
c))/(16*a^4*c - 4*a^3*b^2))*(2*b^4 + 8*a^2*c^2 - 10*a*b^2*c))/(2*(16*a^4*c - 4*a^3*b^2)))*(2*b^4 + 8*a^2*c^2 -
10*a*b^2*c))/(2*(16*a^4*c - 4*a^3*b^2)) - (b*((b*(3*a*c - b^2)*((4*a^4*b^4*c^2 - 8*a^5*b^2*c^3)/a^6 - (2*a*b^
2*c^2*(2*b^4 + 8*a^2*c^2 - 10*a*b^2*c))/(16*a^4*c - 4*a^3*b^2)))/(4*a^3*(4*a*c - b^2)^(1/2)) - (b^3*c^2*(3*a*c
- b^2)*(2*b^4 + 8*a^2*c^2 - 10*a*b^2*c))/(2*a^2*(4*a*c - b^2)^(1/2)*(16*a^4*c - 4*a^3*b^2)))*(3*a*c - b^2))/(
4*a^3*(4*a*c - b^2)^(1/2)) + (b^4*c^2*(3*a*c - b^2)^2*(2*b^4 + 8*a^2*c^2 - 10*a*b^2*c))/(8*a^5*(4*a*c - b^2)*(
16*a^4*c - 4*a^3*b^2))))/(c^2*(b^6*c^2 - 6*a*b^4*c^3 + 9*a^2*b^2*c^4)*(6*b^6 - 25*a^3*c^3 + 54*a^2*b^2*c^2 - 3
6*a*b^4*c)))*(3*a*c - b^2))/(2*a^3*(4*a*c - b^2)^(1/2))

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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/x**5/(c*x**4+b*x**2+a),x)

[Out]

Timed out

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