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

Optimal. Leaf size=183 \[ \frac {\left (a d (b d-4 c e)+b c e^2\right ) \tanh ^{-1}\left (\frac {2 a x+b}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )^2}+\frac {\left (a d^2-c e^2\right ) \log \left (a x^2+b x+c\right )}{2 \left (a d^2-e (b d-c e)\right )^2}+\frac {d}{(d+e x) \left (a d^2-b d e+c e^2\right )}-\frac {\left (a d^2-c e^2\right ) \log (d+e x)}{\left (a d^2-e (b d-c e)\right )^2} \]

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Rubi [A]  time = 0.24, antiderivative size = 183, 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, 800, 634, 618, 206, 628} \begin {gather*} \frac {\left (a d (b d-4 c e)+b c e^2\right ) \tanh ^{-1}\left (\frac {2 a x+b}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )^2}+\frac {\left (a d^2-c e^2\right ) \log \left (a x^2+b x+c\right )}{2 \left (a d^2-e (b d-c e)\right )^2}+\frac {d}{(d+e x) \left (a d^2-b d e+c e^2\right )}-\frac {\left (a d^2-c e^2\right ) \log (d+e x)}{\left (a d^2-e (b d-c e)\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.24, size = 148, normalized size = 0.81 \begin {gather*} \frac {-\frac {2 \left (a d (b d-4 c e)+b c e^2\right ) \tan ^{-1}\left (\frac {2 a x+b}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}+\left (a d^2-c e^2\right ) \log (x (a x+b)+c)+\frac {2 d \left (a d^2+e (c e-b d)\right )}{d+e x}+\left (2 c e^2-2 a d^2\right ) \log (d+e x)}{2 \left (a d^2+e (c e-b d)\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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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 (d+e x)^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 16.70, size = 1059, normalized size = 5.79

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*(a*b^2 - 4*a^2*c)*d^3 - 2*(b^3 - 4*a*b*c)*d^2*e + 2*(b^2*c - 4*a*c^2)*d*e^2 + (a*b*d^3 - 4*a*c*d^2*e +
b*c*d*e^2 + (a*b*d^2*e - 4*a*c*d*e^2 + b*c*e^3)*x)*sqrt(b^2 - 4*a*c)*log((2*a^2*x^2 + 2*a*b*x + b^2 - 2*a*c +
sqrt(b^2 - 4*a*c)*(2*a*x + b))/(a*x^2 + b*x + c)) + ((a*b^2 - 4*a^2*c)*d^3 - (b^2*c - 4*a*c^2)*d*e^2 + ((a*b^
2 - 4*a^2*c)*d^2*e - (b^2*c - 4*a*c^2)*e^3)*x)*log(a*x^2 + b*x + c) - 2*((a*b^2 - 4*a^2*c)*d^3 - (b^2*c - 4*a*
c^2)*d*e^2 + ((a*b^2 - 4*a^2*c)*d^2*e - (b^2*c - 4*a*c^2)*e^3)*x)*log(e*x + d))/((a^2*b^2 - 4*a^3*c)*d^5 - 2*(
a*b^3 - 4*a^2*b*c)*d^4*e + (b^4 - 2*a*b^2*c - 8*a^2*c^2)*d^3*e^2 - 2*(b^3*c - 4*a*b*c^2)*d^2*e^3 + (b^2*c^2 -
4*a*c^3)*d*e^4 + ((a^2*b^2 - 4*a^3*c)*d^4*e - 2*(a*b^3 - 4*a^2*b*c)*d^3*e^2 + (b^4 - 2*a*b^2*c - 8*a^2*c^2)*d^
2*e^3 - 2*(b^3*c - 4*a*b*c^2)*d*e^4 + (b^2*c^2 - 4*a*c^3)*e^5)*x), 1/2*(2*(a*b^2 - 4*a^2*c)*d^3 - 2*(b^3 - 4*a
*b*c)*d^2*e + 2*(b^2*c - 4*a*c^2)*d*e^2 + 2*(a*b*d^3 - 4*a*c*d^2*e + b*c*d*e^2 + (a*b*d^2*e - 4*a*c*d*e^2 + b*
c*e^3)*x)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*a*x + b)/(b^2 - 4*a*c)) + ((a*b^2 - 4*a^2*c)*d^3 -
(b^2*c - 4*a*c^2)*d*e^2 + ((a*b^2 - 4*a^2*c)*d^2*e - (b^2*c - 4*a*c^2)*e^3)*x)*log(a*x^2 + b*x + c) - 2*((a*b^
2 - 4*a^2*c)*d^3 - (b^2*c - 4*a*c^2)*d*e^2 + ((a*b^2 - 4*a^2*c)*d^2*e - (b^2*c - 4*a*c^2)*e^3)*x)*log(e*x + d)
)/((a^2*b^2 - 4*a^3*c)*d^5 - 2*(a*b^3 - 4*a^2*b*c)*d^4*e + (b^4 - 2*a*b^2*c - 8*a^2*c^2)*d^3*e^2 - 2*(b^3*c -
4*a*b*c^2)*d^2*e^3 + (b^2*c^2 - 4*a*c^3)*d*e^4 + ((a^2*b^2 - 4*a^3*c)*d^4*e - 2*(a*b^3 - 4*a^2*b*c)*d^3*e^2 +
(b^4 - 2*a*b^2*c - 8*a^2*c^2)*d^2*e^3 - 2*(b^3*c - 4*a*b*c^2)*d*e^4 + (b^2*c^2 - 4*a*c^3)*e^5)*x)]

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giac [A]  time = 0.37, size = 323, normalized size = 1.77 \begin {gather*} -\frac {1}{2} \, {\left (\frac {2 \, {\left (a b d^{2} e - 4 \, a c d e^{2} + b c e^{3}\right )} \arctan \left (\frac {{\left (2 \, a d - \frac {2 \, a d^{2}}{x e + d} - b e + \frac {2 \, b d e}{x e + d} - \frac {2 \, c e^{2}}{x e + d}\right )} e^{\left (-1\right )}}{\sqrt {-b^{2} + 4 \, a c}}\right ) e^{\left (-2\right )}}{{\left (a^{2} d^{4} - 2 \, a b d^{3} e + b^{2} d^{2} e^{2} + 2 \, a c d^{2} e^{2} - 2 \, b c d e^{3} + c^{2} e^{4}\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {{\left (a d^{2} - c e^{2}\right )} \log \left (a - \frac {2 \, a d}{x e + d} + \frac {a d^{2}}{{\left (x e + d\right )}^{2}} + \frac {b e}{x e + d} - \frac {b d e}{{\left (x e + d\right )}^{2}} + \frac {c e^{2}}{{\left (x e + d\right )}^{2}}\right )}{a^{2} d^{4} e - 2 \, a b d^{3} e^{2} + b^{2} d^{2} e^{3} + 2 \, a c d^{2} e^{3} - 2 \, b c d e^{4} + c^{2} e^{5}} - \frac {2 \, d e}{{\left (a d^{2} e^{2} - b d e^{3} + c e^{4}\right )} {\left (x e + d\right )}}\right )} e \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(a*b*d^2*e - 4*a*c*d*e^2 + b*c*e^3)*arctan((2*a*d - 2*a*d^2/(x*e + d) - b*e + 2*b*d*e/(x*e + d) - 2*c*
e^2/(x*e + d))*e^(-1)/sqrt(-b^2 + 4*a*c))*e^(-2)/((a^2*d^4 - 2*a*b*d^3*e + b^2*d^2*e^2 + 2*a*c*d^2*e^2 - 2*b*c
*d*e^3 + c^2*e^4)*sqrt(-b^2 + 4*a*c)) - (a*d^2 - c*e^2)*log(a - 2*a*d/(x*e + d) + a*d^2/(x*e + d)^2 + b*e/(x*e
+ d) - b*d*e/(x*e + d)^2 + c*e^2/(x*e + d)^2)/(a^2*d^4*e - 2*a*b*d^3*e^2 + b^2*d^2*e^3 + 2*a*c*d^2*e^3 - 2*b*
c*d*e^4 + c^2*e^5) - 2*d*e/((a*d^2*e^2 - b*d*e^3 + c*e^4)*(x*e + d)))*e

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maple [A]  time = 0.01, size = 328, normalized size = 1.79 \begin {gather*} -\frac {a b \,d^{2} \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 )^{2} \sqrt {4 a c -b^{2}}}+\frac {4 a c d 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 )^{2} \sqrt {4 a c -b^{2}}}-\frac {b c \,e^{2} \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 )^{2} \sqrt {4 a c -b^{2}}}-\frac {a \,d^{2} \ln \left (e x +d \right )}{\left (a \,d^{2}-d e b +c \,e^{2}\right )^{2}}+\frac {a \,d^{2} \ln \left (a \,x^{2}+b x +c \right )}{2 \left (a \,d^{2}-d e b +c \,e^{2}\right )^{2}}+\frac {c \,e^{2} \ln \left (e x +d \right )}{\left (a \,d^{2}-d e b +c \,e^{2}\right )^{2}}-\frac {c \,e^{2} \ln \left (a \,x^{2}+b x +c \right )}{2 \left (a \,d^{2}-d e b +c \,e^{2}\right )^{2}}+\frac {d}{\left (a \,d^{2}-d e b +c \,e^{2}\right ) \left (e x +d \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/(a*d^2-b*d*e+c*e^2)^2*a*ln(a*x^2+b*x+c)*d^2-1/2/(a*d^2-b*d*e+c*e^2)^2*ln(a*x^2+b*x+c)*c*e^2-1/(a*d^2-b*d*e
+c*e^2)^2/(4*a*c-b^2)^(1/2)*arctan((2*a*x+b)/(4*a*c-b^2)^(1/2))*a*b*d^2+4/(a*d^2-b*d*e+c*e^2)^2/(4*a*c-b^2)^(1
/2)*arctan((2*a*x+b)/(4*a*c-b^2)^(1/2))*a*c*d*e-1/(a*d^2-b*d*e+c*e^2)^2/(4*a*c-b^2)^(1/2)*arctan((2*a*x+b)/(4*
a*c-b^2)^(1/2))*b*c*e^2+d/(a*d^2-b*d*e+c*e^2)/(e*x+d)-1/(a*d^2-b*d*e+c*e^2)^2*ln(e*x+d)*a*d^2+1/(a*d^2-b*d*e+c
*e^2)^2*ln(e*x+d)*c*e^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/(a+c/x^2+b/x)/x/(e*x+d)^2,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 = 8.07, size = 1768, normalized size = 9.66

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d/((d + e*x)*(a*d^2 + c*e^2 - b*d*e)) - (log(56*a^3*b^2*c*d^4 - 96*a^4*c^2*d^4 - 96*a^2*c^4*e^4 - 8*b^4*c^2*e^
4 - 8*a^2*b^4*d^4 + 56*a*b^2*c^3*e^4 - 4*a^3*b^3*d^4*x + 320*a^3*c^3*d^2*e^2 + 8*a*d^3*e*(b^2 - 4*a*c)^(5/2) -
8*c*d*e^3*(b^2 - 4*a*c)^(5/2) - 3*c*e^4*x*(b^2 - 4*a*c)^(5/2) - 8*b^5*c*e^4*x + 8*a^2*b*d^4*(b^2 - 4*a*c)^(3/
2) - 8*b*c^2*e^4*(b^2 - 4*a*c)^(3/2) + 12*a^3*d^4*x*(b^2 - 4*a*c)^(3/2) - 6*b*d*e^3*x*(b^2 - 4*a*c)^(5/2) + 16
*a^4*b*c*d^4*x - 112*a^2*b^2*c^2*d^2*e^2 - 8*a*b^2*d^3*e*(b^2 - 4*a*c)^(3/2) + 8*b^2*c*d*e^3*(b^2 - 4*a*c)^(3/
2) + 10*a*d^2*e^2*x*(b^2 - 4*a*c)^(5/2) - 5*b^2*c*e^4*x*(b^2 - 4*a*c)^(3/2) + 6*b^3*d*e^3*x*(b^2 - 4*a*c)^(3/2
) + 16*a*b^3*c^2*d*e^3 + 8*a*b^4*c*d^2*e^2 - 64*a^2*b*c^3*d*e^3 + 16*a^2*b^3*c*d^3*e - 64*a^3*b*c^2*d^3*e + 60
*a*b^3*c^2*e^4*x - 112*a^2*b*c^3*e^4*x + 4*a*b^5*d^2*e^2*x - 8*a^2*b^4*d^3*e*x + 256*a^3*c^3*d*e^3*x - 256*a^4
*c^2*d^3*e*x - 6*a*b^2*d^2*e^2*x*(b^2 - 4*a*c)^(3/2) - 160*a^2*b^2*c^2*d*e^3*x - 56*a^2*b^3*c*d^2*e^2*x + 160*
a^3*b*c^2*d^2*e^2*x + 24*a*b^4*c*d*e^3*x - 8*a^2*b*d^3*e*x*(b^2 - 4*a*c)^(3/2) + 96*a^3*b^2*c*d^3*e*x)*(b^2*((
a*d^2)/2 - (c*e^2)/2) - b*((a*d^2*(b^2 - 4*a*c)^(1/2))/2 + (c*e^2*(b^2 - 4*a*c)^(1/2))/2) - 2*a^2*c*d^2 + 2*a*
c^2*e^2 + 2*a*c*d*e*(b^2 - 4*a*c)^(1/2)))/(4*a^3*c*d^4 + 4*a*c^3*e^4 - a^2*b^2*d^4 - b^2*c^2*e^4 - b^4*d^2*e^2
+ 8*a^2*c^2*d^2*e^2 + 2*a*b^3*d^3*e + 2*b^3*c*d*e^3 - 8*a*b*c^2*d*e^3 - 8*a^2*b*c*d^3*e + 2*a*b^2*c*d^2*e^2)
- (log(8*a^2*b^4*d^4 + 96*a^4*c^2*d^4 + 96*a^2*c^4*e^4 + 8*b^4*c^2*e^4 - 56*a^3*b^2*c*d^4 - 56*a*b^2*c^3*e^4 +
4*a^3*b^3*d^4*x - 320*a^3*c^3*d^2*e^2 + 8*a*d^3*e*(b^2 - 4*a*c)^(5/2) - 8*c*d*e^3*(b^2 - 4*a*c)^(5/2) - 3*c*e
^4*x*(b^2 - 4*a*c)^(5/2) + 8*b^5*c*e^4*x + 8*a^2*b*d^4*(b^2 - 4*a*c)^(3/2) - 8*b*c^2*e^4*(b^2 - 4*a*c)^(3/2) +
12*a^3*d^4*x*(b^2 - 4*a*c)^(3/2) - 6*b*d*e^3*x*(b^2 - 4*a*c)^(5/2) - 16*a^4*b*c*d^4*x + 112*a^2*b^2*c^2*d^2*e
^2 - 8*a*b^2*d^3*e*(b^2 - 4*a*c)^(3/2) + 8*b^2*c*d*e^3*(b^2 - 4*a*c)^(3/2) + 10*a*d^2*e^2*x*(b^2 - 4*a*c)^(5/2
) - 5*b^2*c*e^4*x*(b^2 - 4*a*c)^(3/2) + 6*b^3*d*e^3*x*(b^2 - 4*a*c)^(3/2) - 16*a*b^3*c^2*d*e^3 - 8*a*b^4*c*d^2
*e^2 + 64*a^2*b*c^3*d*e^3 - 16*a^2*b^3*c*d^3*e + 64*a^3*b*c^2*d^3*e - 60*a*b^3*c^2*e^4*x + 112*a^2*b*c^3*e^4*x
- 4*a*b^5*d^2*e^2*x + 8*a^2*b^4*d^3*e*x - 256*a^3*c^3*d*e^3*x + 256*a^4*c^2*d^3*e*x - 6*a*b^2*d^2*e^2*x*(b^2
- 4*a*c)^(3/2) + 160*a^2*b^2*c^2*d*e^3*x + 56*a^2*b^3*c*d^2*e^2*x - 160*a^3*b*c^2*d^2*e^2*x - 24*a*b^4*c*d*e^3
*x - 8*a^2*b*d^3*e*x*(b^2 - 4*a*c)^(3/2) - 96*a^3*b^2*c*d^3*e*x)*(b*((a*d^2*(b^2 - 4*a*c)^(1/2))/2 + (c*e^2*(b
^2 - 4*a*c)^(1/2))/2) + b^2*((a*d^2)/2 - (c*e^2)/2) - 2*a^2*c*d^2 + 2*a*c^2*e^2 - 2*a*c*d*e*(b^2 - 4*a*c)^(1/2
)))/(4*a^3*c*d^4 + 4*a*c^3*e^4 - a^2*b^2*d^4 - b^2*c^2*e^4 - b^4*d^2*e^2 + 8*a^2*c^2*d^2*e^2 + 2*a*b^3*d^3*e +
2*b^3*c*d*e^3 - 8*a*b*c^2*d*e^3 - 8*a^2*b*c*d^3*e + 2*a*b^2*c*d^2*e^2) - (log(d + e*x)*(a*d^2 - c*e^2))/(a^2*
d^4 + c^2*e^4 + b^2*d^2*e^2 - 2*a*b*d^3*e - 2*b*c*d*e^3 + 2*a*c*d^2*e^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/(e*x+d)**2,x)

[Out]

Timed out

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