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

Optimal. Leaf size=189 \[ -\frac {\left (2 a^2 d^2-2 a e (b d+c e)+b^2 e^2\right ) \tanh ^{-1}\left (\frac {2 a x+b}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )^2}-\frac {e}{(d+e x) \left (a d^2-b d e+c e^2\right )}-\frac {e (2 a d-b e) \log \left (a x^2+b x+c\right )}{2 \left (a d^2-e (b d-c e)\right )^2}+\frac {e (2 a d-b e) \log (d+e x)}{\left (a d^2-e (b d-c e)\right )^2} \]

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Rubi [A]  time = 0.31, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {1569, 709, 800, 634, 618, 206, 628} \begin {gather*} -\frac {\left (2 a^2 d^2-2 a e (b d+c e)+b^2 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 {e}{(d+e x) \left (a d^2-b d e+c e^2\right )}-\frac {e (2 a d-b e) \log \left (a x^2+b x+c\right )}{2 \left (a d^2-e (b d-c e)\right )^2}+\frac {e (2 a d-b e) \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^2*(d + e*x)^2),x]

[Out]

-(e/((a*d^2 - b*d*e + c*e^2)*(d + e*x))) - ((2*a^2*d^2 + b^2*e^2 - 2*a*e*(b*d + 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) + (e*(2*a*d - b*e)*Log[d + e*x])/(a*d^2 - e*(b*d
- c*e))^2 - (e*(2*a*d - b*e)*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 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 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^2 (d+e x)^2} \, dx &=\int \frac {1}{(d+e x)^2 \left (c+b x+a x^2\right )} \, dx\\ &=-\frac {e}{\left (a d^2-b d e+c e^2\right ) (d+e x)}+\frac {\int \frac {a d-b e-a e x}{(d+e x) \left (c+b x+a x^2\right )} \, dx}{a d^2-b d e+c e^2}\\ &=-\frac {e}{\left (a d^2-b d e+c e^2\right ) (d+e x)}+\frac {\int \left (\frac {e^2 (2 a d-b e)}{\left (a d^2-e (b d-c e)\right ) (d+e x)}+\frac {a^2 d^2+b^2 e^2-a e (2 b d+c e)-a e (2 a d-b e) x}{\left (a d^2-e (b d-c e)\right ) \left (c+b x+a x^2\right )}\right ) \, dx}{a d^2-b d e+c e^2}\\ &=-\frac {e}{\left (a d^2-b d e+c e^2\right ) (d+e x)}+\frac {e (2 a d-b e) \log (d+e x)}{\left (a d^2-e (b d-c e)\right )^2}+\frac {\int \frac {a^2 d^2+b^2 e^2-a e (2 b d+c e)-a e (2 a d-b e) x}{c+b x+a x^2} \, dx}{\left (a d^2-e (b d-c e)\right )^2}\\ &=-\frac {e}{\left (a d^2-b d e+c e^2\right ) (d+e x)}+\frac {e (2 a d-b e) \log (d+e x)}{\left (a d^2-e (b d-c e)\right )^2}-\frac {(e (2 a d-b e)) \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 (2 a^2 d^2+b^2 e^2-2 a e (b d+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 {e}{\left (a d^2-b d e+c e^2\right ) (d+e x)}+\frac {e (2 a d-b e) \log (d+e x)}{\left (a d^2-e (b d-c e)\right )^2}-\frac {e (2 a d-b e) \log \left (c+b x+a x^2\right )}{2 \left (a d^2-e (b d-c e)\right )^2}-\frac {\left (2 a^2 d^2+b^2 e^2-2 a e (b d+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 {e}{\left (a d^2-b d e+c e^2\right ) (d+e x)}-\frac {\left (2 a^2 d^2+b^2 e^2-2 a e (b d+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 {e (2 a d-b e) \log (d+e x)}{\left (a d^2-e (b d-c e)\right )^2}-\frac {e (2 a d-b e) \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 = 0.21, size = 151, normalized size = 0.80 \begin {gather*} \frac {\frac {2 \left (2 a^2 d^2-2 a e (b d+c e)+b^2 e^2\right ) \tan ^{-1}\left (\frac {2 a x+b}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}-\frac {2 e \left (a d^2+e (c e-b d)\right )}{d+e x}+e (b e-2 a d) \log (x (a x+b)+c)-2 e (b e-2 a d) \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^2*(d + e*x)^2),x]

[Out]

((-2*e*(a*d^2 + e*(-(b*d) + c*e)))/(d + e*x) + (2*(2*a^2*d^2 + b^2*e^2 - 2*a*e*(b*d + c*e))*ArcTan[(b + 2*a*x)
/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c] - 2*e*(-2*a*d + b*e)*Log[d + e*x] + e*(-2*a*d + b*e)*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^2 (d+e x)^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 9.28, size = 1079, normalized size = 5.71

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^2/(e*x+d)^2,x, algorithm="fricas")

[Out]

[-1/2*(2*(a*b^2 - 4*a^2*c)*d^2*e - 2*(b^3 - 4*a*b*c)*d*e^2 + 2*(b^2*c - 4*a*c^2)*e^3 + (2*a^2*d^3 - 2*a*b*d^2*
e + (b^2 - 2*a*c)*d*e^2 + (2*a^2*d^2*e - 2*a*b*d*e^2 + (b^2 - 2*a*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)) + (2*(a*b^2 - 4*a^2*c)*d^2*e - (b^
3 - 4*a*b*c)*d*e^2 + (2*(a*b^2 - 4*a^2*c)*d*e^2 - (b^3 - 4*a*b*c)*e^3)*x)*log(a*x^2 + b*x + c) - 2*(2*(a*b^2 -
4*a^2*c)*d^2*e - (b^3 - 4*a*b*c)*d*e^2 + (2*(a*b^2 - 4*a^2*c)*d*e^2 - (b^3 - 4*a*b*c)*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^2*e - 2*(b^3 - 4*a*b*c)*d*e^2 + 2*(b^2*c - 4*a*c^2)*e^3 + 2*(2*a^2*d^3 - 2*a*b*d^2*e + (b^2 - 2*
a*c)*d*e^2 + (2*a^2*d^2*e - 2*a*b*d*e^2 + (b^2 - 2*a*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)) + (2*(a*b^2 - 4*a^2*c)*d^2*e - (b^3 - 4*a*b*c)*d*e^2 + (2*(a*b^2 - 4*a^2*c)*d*e^2 -
(b^3 - 4*a*b*c)*e^3)*x)*log(a*x^2 + b*x + c) - 2*(2*(a*b^2 - 4*a^2*c)*d^2*e - (b^3 - 4*a*b*c)*d*e^2 + (2*(a*b
^2 - 4*a^2*c)*d*e^2 - (b^3 - 4*a*b*c)*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 = 0.35, size = 331, normalized size = 1.75 \begin {gather*} -\frac {{\left (2 \, a^{2} d^{2} e^{2} - 2 \, a b d e^{3} + b^{2} e^{4} - 2 \, a c e^{4}\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 (2 \, a d e - b 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 )}{2 \, {\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 )}} - \frac {e^{3}}{{\left (a d^{2} e^{2} - b d e^{3} + c e^{4}\right )} {\left (x e + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 386, normalized size = 2.04 \begin {gather*} \frac {2 a^{2} 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 {2 a b 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 {2 a 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 {b^{2} 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 {2 a d e \ln \left (e x +d \right )}{\left (a \,d^{2}-d e b +c \,e^{2}\right )^{2}}-\frac {a d e \ln \left (a \,x^{2}+b x +c \right )}{\left (a \,d^{2}-d e b +c \,e^{2}\right )^{2}}-\frac {b \,e^{2} \ln \left (e x +d \right )}{\left (a \,d^{2}-d e b +c \,e^{2}\right )^{2}}+\frac {b \,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 {e}{\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^2/(e*x+d)^2,x)

[Out]

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

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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^2/(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 = 8.11, size = 1782, normalized size = 9.43

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(c*e^4*(b^2 - 4*a*c)^(5/2) - 8*b^5*c*e^4 - 8*b^6*e^4*x - 4*a^3*d^4*(b^2 - 4*a*c)^(3/2) - 4*a^3*b^3*d^4 + 4
*b^3*e^4*x*(b^2 - 4*a*c)^(3/2) + 60*a*b^3*c^2*e^4 - 112*a^2*b*c^3*e^4 + 4*a*b^5*d^2*e^2 - 8*a^2*b^4*d^3*e + 25
6*a^3*c^3*d*e^3 - 256*a^4*c^2*d^3*e - 8*a^4*b^2*d^4*x + 32*a^3*c^3*e^4*x + 10*b*d*e^3*(b^2 - 4*a*c)^(5/2) + 4*
b*e^4*x*(b^2 - 4*a*c)^(5/2) + 16*a^4*b*c*d^4 + 32*a^5*c*d^4*x - 14*a*d^2*e^2*(b^2 - 4*a*c)^(5/2) + 7*b^2*c*e^4
*(b^2 - 4*a*c)^(3/2) - 10*b^3*d*e^3*(b^2 - 4*a*c)^(3/2) - 8*a*d*e^3*x*(b^2 - 4*a*c)^(5/2) + 24*a*b^4*c*d*e^3 +
64*a*b^4*c*e^4*x + 32*a*b^5*d*e^3*x - 8*a^2*b*d^3*e*(b^2 - 4*a*c)^(3/2) - 32*a^3*d^3*e*x*(b^2 - 4*a*c)^(3/2)
+ 96*a^3*b^2*c*d^3*e + 16*a^3*b^3*d^3*e*x + 18*a*b^2*d^2*e^2*(b^2 - 4*a*c)^(3/2) - 160*a^2*b^2*c^2*d*e^3 - 56*
a^2*b^3*c*d^2*e^2 + 160*a^3*b*c^2*d^2*e^2 - 136*a^2*b^2*c^2*e^4*x - 40*a^2*b^4*d^2*e^2*x - 448*a^4*c^2*d^2*e^2
*x + 48*a^2*b*d^2*e^2*x*(b^2 - 4*a*c)^(3/2) + 272*a^3*b^2*c*d^2*e^2*x - 64*a^4*b*c*d^3*e*x - 24*a*b^2*d*e^3*x*
(b^2 - 4*a*c)^(3/2) - 240*a^2*b^3*c*d*e^3*x + 448*a^3*b*c^2*d*e^3*x)*(a*(e^2*(2*b*c - c*(b^2 - 4*a*c)^(1/2)) +
e*(b^2*d - b*d*(b^2 - 4*a*c)^(1/2))) - e^2*(b^3/2 - (b^2*(b^2 - 4*a*c)^(1/2))/2) + a^2*(d^2*(b^2 - 4*a*c)^(1/
2) - 4*c*d*e)))/(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)*(b*e^2 - 2
*a*d*e))/(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) - (log(c*e^4*(b^2 - 4*a
*c)^(5/2) + 8*b^5*c*e^4 + 8*b^6*e^4*x - 4*a^3*d^4*(b^2 - 4*a*c)^(3/2) + 4*a^3*b^3*d^4 + 4*b^3*e^4*x*(b^2 - 4*a
*c)^(3/2) - 60*a*b^3*c^2*e^4 + 112*a^2*b*c^3*e^4 - 4*a*b^5*d^2*e^2 + 8*a^2*b^4*d^3*e - 256*a^3*c^3*d*e^3 + 256
*a^4*c^2*d^3*e + 8*a^4*b^2*d^4*x - 32*a^3*c^3*e^4*x + 10*b*d*e^3*(b^2 - 4*a*c)^(5/2) + 4*b*e^4*x*(b^2 - 4*a*c)
^(5/2) - 16*a^4*b*c*d^4 - 32*a^5*c*d^4*x - 14*a*d^2*e^2*(b^2 - 4*a*c)^(5/2) + 7*b^2*c*e^4*(b^2 - 4*a*c)^(3/2)
- 10*b^3*d*e^3*(b^2 - 4*a*c)^(3/2) - 8*a*d*e^3*x*(b^2 - 4*a*c)^(5/2) - 24*a*b^4*c*d*e^3 - 64*a*b^4*c*e^4*x - 3
2*a*b^5*d*e^3*x - 8*a^2*b*d^3*e*(b^2 - 4*a*c)^(3/2) - 32*a^3*d^3*e*x*(b^2 - 4*a*c)^(3/2) - 96*a^3*b^2*c*d^3*e
- 16*a^3*b^3*d^3*e*x + 18*a*b^2*d^2*e^2*(b^2 - 4*a*c)^(3/2) + 160*a^2*b^2*c^2*d*e^3 + 56*a^2*b^3*c*d^2*e^2 - 1
60*a^3*b*c^2*d^2*e^2 + 136*a^2*b^2*c^2*e^4*x + 40*a^2*b^4*d^2*e^2*x + 448*a^4*c^2*d^2*e^2*x + 48*a^2*b*d^2*e^2
*x*(b^2 - 4*a*c)^(3/2) - 272*a^3*b^2*c*d^2*e^2*x + 64*a^4*b*c*d^3*e*x - 24*a*b^2*d*e^3*x*(b^2 - 4*a*c)^(3/2) +
240*a^2*b^3*c*d*e^3*x - 448*a^3*b*c^2*d*e^3*x)*(e^2*(b^3/2 + (b^2*(b^2 - 4*a*c)^(1/2))/2) - a*(e^2*(2*b*c + c
*(b^2 - 4*a*c)^(1/2)) + e*(b^2*d + b*d*(b^2 - 4*a*c)^(1/2))) + a^2*(d^2*(b^2 - 4*a*c)^(1/2) + 4*c*d*e)))/(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) - e/((d + e*x)*(a*d^2 + c*e^2 - b*d*e))

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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**2/(e*x+d)**2,x)

[Out]

Timed out

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