3.1.60 \(\int \frac {1}{(a+\frac {c}{x^2}+\frac {b}{x}) x^4 (d+e x)} \, dx\)

Optimal. Leaf size=193 \[ \frac {\left (2 a^2 c d-a b (b d+3 c e)+b^3 e\right ) \tanh ^{-1}\left (\frac {2 a x+b}{\sqrt {b^2-4 a c}}\right )}{c^2 \sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )}+\frac {\left (a b d+a c e+b^2 (-e)\right ) \log \left (a x^2+b x+c\right )}{2 c^2 \left (a d^2-e (b d-c e)\right )}+\frac {e^3 \log (d+e x)}{d^2 \left (a d^2-e (b d-c e)\right )}-\frac {\log (x) (b d+c e)}{c^2 d^2}-\frac {1}{c d x} \]

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Rubi [A]  time = 0.34, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {1569, 893, 634, 618, 206, 628} \begin {gather*} \frac {\left (2 a^2 c d-a b (b d+3 c e)+b^3 e\right ) \tanh ^{-1}\left (\frac {2 a x+b}{\sqrt {b^2-4 a c}}\right )}{c^2 \sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )}+\frac {\left (a b d+a c e+b^2 (-e)\right ) \log \left (a x^2+b x+c\right )}{2 c^2 \left (a d^2-e (b d-c e)\right )}+\frac {e^3 \log (d+e x)}{d^2 \left (a d^2-e (b d-c e)\right )}-\frac {\log (x) (b d+c e)}{c^2 d^2}-\frac {1}{c d x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + c/x^2 + b/x)*x^4*(d + e*x)),x]

[Out]

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

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 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 1569

Int[(x_)^(m_.)*((a_.) + (b_.)*(x_)^(mn_.) + (c_.)*(x_)^(mn2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_.))^(q_.), x_Symbo
l] :> Int[x^(m - 2*n*p)*(d + e*x^n)^q*(c + b*x^n + a*x^(2*n))^p, x] /; FreeQ[{a, b, c, d, e, m, n, q}, x] && E
qQ[mn, -n] && EqQ[mn2, 2*mn] && IntegerQ[p]

Rubi steps

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

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Mathematica [A]  time = 0.17, size = 194, normalized size = 1.01 \begin {gather*} \frac {\left (2 a^2 c d-a b (b d+3 c e)+b^3 e\right ) \tan ^{-1}\left (\frac {2 a x+b}{\sqrt {4 a c-b^2}}\right )}{c^2 \sqrt {4 a c-b^2} \left (e (b d-c e)-a d^2\right )}+\frac {\left (a b d+a c e+b^2 (-e)\right ) \log (x (a x+b)+c)}{2 c^2 \left (a d^2+e (c e-b d)\right )}+\frac {e^3 \log (d+e x)}{a d^4+d^2 e (c e-b d)}-\frac {\log (x) (b d+c e)}{c^2 d^2}-\frac {1}{c d x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^4 (d+e x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[1/((a + c/x^2 + b/x)*x^4*(d + e*x)),x]

[Out]

IntegrateAlgebraic[1/((a + c/x^2 + b/x)*x^4*(d + e*x)), x]

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [A]  time = 0.34, size = 210, normalized size = 1.09 \begin {gather*} \frac {{\left (a b d - b^{2} e + a c e\right )} \log \left (a x^{2} + b x + c\right )}{2 \, {\left (a c^{2} d^{2} - b c^{2} d e + c^{3} e^{2}\right )}} + \frac {e^{4} \log \left ({\left | x e + d \right |}\right )}{a d^{4} e - b d^{3} e^{2} + c d^{2} e^{3}} + \frac {{\left (a b^{2} d - 2 \, a^{2} c d - b^{3} e + 3 \, a b c e\right )} \arctan \left (\frac {2 \, a x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (a c^{2} d^{2} - b c^{2} d e + c^{3} e^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {{\left (b d + c e\right )} \log \left ({\left | x \right |}\right )}{c^{2} d^{2}} - \frac {1}{c d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 412, normalized size = 2.13 \begin {gather*} -\frac {2 a^{2} d \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,d^{2}-d e b +c \,e^{2}\right ) \sqrt {4 a c -b^{2}}\, c}+\frac {a \,b^{2} d \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,d^{2}-d e b +c \,e^{2}\right ) \sqrt {4 a c -b^{2}}\, c^{2}}+\frac {3 a b e \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,d^{2}-d e b +c \,e^{2}\right ) \sqrt {4 a c -b^{2}}\, c}-\frac {b^{3} e \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,d^{2}-d e b +c \,e^{2}\right ) \sqrt {4 a c -b^{2}}\, c^{2}}+\frac {a b d \ln \left (a \,x^{2}+b x +c \right )}{2 \left (a \,d^{2}-d e b +c \,e^{2}\right ) c^{2}}+\frac {a e \ln \left (a \,x^{2}+b x +c \right )}{2 \left (a \,d^{2}-d e b +c \,e^{2}\right ) c}-\frac {b^{2} e \ln \left (a \,x^{2}+b x +c \right )}{2 \left (a \,d^{2}-d e b +c \,e^{2}\right ) c^{2}}+\frac {e^{3} \ln \left (e x +d \right )}{\left (a \,d^{2}-d e b +c \,e^{2}\right ) d^{2}}-\frac {b \ln \relax (x )}{c^{2} d}-\frac {e \ln \relax (x )}{c \,d^{2}}-\frac {1}{c d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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/(a+c/x^2+b/x)/x^4/(e*x+d),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 = 20.39, size = 2388, normalized size = 12.37

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e^3*log(d + e*x))/(a*d^4 + c*d^2*e^2 - b*d^3*e) + (log((a^4*e^4*x)/(c^2*d^2) - (((a*e*x*(a^4*d^4 + b^4*e^4 +
2*a^2*c^2*e^4 + 2*a^3*c*d^2*e^2 - 4*a*b^2*c*e^4 + 2*a^2*b*c*d*e^3))/(c^2*d^2) - (((a*e*(a^2*b^2*d^4 - 4*a*c^3*
e^4 - a^3*c*d^4 + b^2*c^2*e^4 + b^4*d^2*e^2 + 4*a^2*c^2*d^2*e^2 - 2*a*b^3*d^3*e + b^3*c*d*e^3 - 4*a*b*c^2*d*e^
3 + 5*a^2*b*c*d^3*e - 5*a*b^2*c*d^2*e^2))/(c*d) + (a*e*x*(2*a^3*b*d^4 + 2*b^3*c*e^4 + 2*b^4*d*e^3 - 2*a*b^3*d^
2*e^2 - 2*a^2*b^2*d^3*e + 12*a^2*c^2*d*e^3 - 8*a*b*c^2*e^4 + a^3*c*d^3*e - 11*a*b^2*c*d*e^3 + 8*a^2*b*c*d^2*e^
2))/(c*d) + (a*e*(b^4*e + b^3*e*(b^2 - 4*a*c)^(1/2) + 4*a^2*c^2*e - a*b^3*d + 4*a^2*b*c*d - 5*a*b^2*c*e - a*b^
2*d*(b^2 - 4*a*c)^(1/2) + 2*a^2*c*d*(b^2 - 4*a*c)^(1/2) - 3*a*b*c*e*(b^2 - 4*a*c)^(1/2))*(4*a^2*c^2*d^3*e + b^
2*c^2*d*e^3 + b^3*c*d^2*e^2 + 2*a^2*b^2*d^4*x + 2*b^2*c^2*e^4*x + 2*b^4*d^2*e^2*x + a^2*b*c*d^4 - 4*a*c^3*d*e^
3 - 6*a^3*c*d^4*x - 8*a*c^3*e^4*x - 2*a*b^2*c*d^3*e - 4*a*b^3*d^3*e*x - 2*b^3*c*d*e^3*x - 3*a*b*c^2*d^2*e^2 -
6*a^2*c^2*d^2*e^2*x + 8*a*b*c^2*d*e^3*x + 14*a^2*b*c*d^3*e*x - 6*a*b^2*c*d^2*e^2*x))/(2*c^2*(4*a*c - b^2)*(a*d
^2 + c*e^2 - b*d*e)))*(b^4*e + b^3*e*(b^2 - 4*a*c)^(1/2) + 4*a^2*c^2*e - a*b^3*d + 4*a^2*b*c*d - 5*a*b^2*c*e -
a*b^2*d*(b^2 - 4*a*c)^(1/2) + 2*a^2*c*d*(b^2 - 4*a*c)^(1/2) - 3*a*b*c*e*(b^2 - 4*a*c)^(1/2)))/(2*c^2*(4*a*c -
b^2)*(a*d^2 + c*e^2 - b*d*e)) + (a*e*(b*d + c*e)*(a^3*d^3 + b^3*e^3 - 3*a*b*c*e^3))/(c^2*d^2))*(b^4*e + b^3*e
*(b^2 - 4*a*c)^(1/2) + 4*a^2*c^2*e - a*b^3*d + 4*a^2*b*c*d - 5*a*b^2*c*e - a*b^2*d*(b^2 - 4*a*c)^(1/2) + 2*a^2
*c*d*(b^2 - 4*a*c)^(1/2) - 3*a*b*c*e*(b^2 - 4*a*c)^(1/2)))/(2*c^2*(4*a*c - b^2)*(a*d^2 + c*e^2 - b*d*e)))*(b^4
*e + b^3*e*(b^2 - 4*a*c)^(1/2) + 4*a^2*c^2*e - a*b^3*d + 4*a^2*b*c*d - 5*a*b^2*c*e - a*b^2*d*(b^2 - 4*a*c)^(1/
2) + 2*a^2*c*d*(b^2 - 4*a*c)^(1/2) - 3*a*b*c*e*(b^2 - 4*a*c)^(1/2)))/(2*(4*a*c^4*e^2 + 4*a^2*c^3*d^2 - b^2*c^3
*e^2 - a*b^2*c^2*d^2 + b^3*c^2*d*e - 4*a*b*c^3*d*e)) + (log((a^4*e^4*x)/(c^2*d^2) - (((a*e*x*(a^4*d^4 + b^4*e^
4 + 2*a^2*c^2*e^4 + 2*a^3*c*d^2*e^2 - 4*a*b^2*c*e^4 + 2*a^2*b*c*d*e^3))/(c^2*d^2) - (((a*e*(a^2*b^2*d^4 - 4*a*
c^3*e^4 - a^3*c*d^4 + b^2*c^2*e^4 + b^4*d^2*e^2 + 4*a^2*c^2*d^2*e^2 - 2*a*b^3*d^3*e + b^3*c*d*e^3 - 4*a*b*c^2*
d*e^3 + 5*a^2*b*c*d^3*e - 5*a*b^2*c*d^2*e^2))/(c*d) + (a*e*x*(2*a^3*b*d^4 + 2*b^3*c*e^4 + 2*b^4*d*e^3 - 2*a*b^
3*d^2*e^2 - 2*a^2*b^2*d^3*e + 12*a^2*c^2*d*e^3 - 8*a*b*c^2*e^4 + a^3*c*d^3*e - 11*a*b^2*c*d*e^3 + 8*a^2*b*c*d^
2*e^2))/(c*d) + (a*e*(b^4*e - b^3*e*(b^2 - 4*a*c)^(1/2) + 4*a^2*c^2*e - a*b^3*d + 4*a^2*b*c*d - 5*a*b^2*c*e +
a*b^2*d*(b^2 - 4*a*c)^(1/2) - 2*a^2*c*d*(b^2 - 4*a*c)^(1/2) + 3*a*b*c*e*(b^2 - 4*a*c)^(1/2))*(4*a^2*c^2*d^3*e
+ b^2*c^2*d*e^3 + b^3*c*d^2*e^2 + 2*a^2*b^2*d^4*x + 2*b^2*c^2*e^4*x + 2*b^4*d^2*e^2*x + a^2*b*c*d^4 - 4*a*c^3*
d*e^3 - 6*a^3*c*d^4*x - 8*a*c^3*e^4*x - 2*a*b^2*c*d^3*e - 4*a*b^3*d^3*e*x - 2*b^3*c*d*e^3*x - 3*a*b*c^2*d^2*e^
2 - 6*a^2*c^2*d^2*e^2*x + 8*a*b*c^2*d*e^3*x + 14*a^2*b*c*d^3*e*x - 6*a*b^2*c*d^2*e^2*x))/(2*c^2*(4*a*c - b^2)*
(a*d^2 + c*e^2 - b*d*e)))*(b^4*e - b^3*e*(b^2 - 4*a*c)^(1/2) + 4*a^2*c^2*e - a*b^3*d + 4*a^2*b*c*d - 5*a*b^2*c
*e + a*b^2*d*(b^2 - 4*a*c)^(1/2) - 2*a^2*c*d*(b^2 - 4*a*c)^(1/2) + 3*a*b*c*e*(b^2 - 4*a*c)^(1/2)))/(2*c^2*(4*a
*c - b^2)*(a*d^2 + c*e^2 - b*d*e)) + (a*e*(b*d + c*e)*(a^3*d^3 + b^3*e^3 - 3*a*b*c*e^3))/(c^2*d^2))*(b^4*e - b
^3*e*(b^2 - 4*a*c)^(1/2) + 4*a^2*c^2*e - a*b^3*d + 4*a^2*b*c*d - 5*a*b^2*c*e + a*b^2*d*(b^2 - 4*a*c)^(1/2) - 2
*a^2*c*d*(b^2 - 4*a*c)^(1/2) + 3*a*b*c*e*(b^2 - 4*a*c)^(1/2)))/(2*c^2*(4*a*c - b^2)*(a*d^2 + c*e^2 - b*d*e)))*
(b^4*e - b^3*e*(b^2 - 4*a*c)^(1/2) + 4*a^2*c^2*e - a*b^3*d + 4*a^2*b*c*d - 5*a*b^2*c*e + a*b^2*d*(b^2 - 4*a*c)
^(1/2) - 2*a^2*c*d*(b^2 - 4*a*c)^(1/2) + 3*a*b*c*e*(b^2 - 4*a*c)^(1/2)))/(2*(4*a*c^4*e^2 + 4*a^2*c^3*d^2 - b^2
*c^3*e^2 - a*b^2*c^2*d^2 + b^3*c^2*d*e - 4*a*b*c^3*d*e)) - 1/(c*d*x) - (log(x)*(b*d + c*e))/(c^2*d^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+c/x**2+b/x)/x**4/(e*x+d),x)

[Out]

Timed out

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