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

Optimal. Leaf size=194 \[ -\frac {\left (-2 c \left (a d^2-c e^2\right )+b^2 d^2-2 b c d e\right ) \tanh ^{-1}\left (\frac {2 a x+b}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )^2}-\frac {d^2}{e (d+e x) \left (a d^2-b d e+c e^2\right )}-\frac {d (b d-2 c e) \log \left (a x^2+b x+c\right )}{2 \left (a d^2-e (b d-c e)\right )^2}+\frac {d (b d-2 c e) \log (d+e x)}{\left (a d^2-e (b d-c e)\right )^2} \]

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Rubi [A]  time = 0.31, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1445, 1628, 634, 618, 206, 628} \begin {gather*} -\frac {\left (-2 c \left (a d^2-c e^2\right )+b^2 d^2-2 b c d e\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 {d^2}{e (d+e x) \left (a d^2-b d e+c e^2\right )}-\frac {d (b d-2 c e) \log \left (a x^2+b x+c\right )}{2 \left (a d^2-e (b d-c e)\right )^2}+\frac {d (b d-2 c 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)*(d + e*x)^2),x]

[Out]

-(d^2/(e*(a*d^2 - b*d*e + c*e^2)*(d + e*x))) - ((b^2*d^2 - 2*b*c*d*e - 2*c*(a*d^2 - c*e^2))*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) + (d*(b*d - 2*c*e)*Log[d + e*x])/(a*d^2 -
e*(b*d - c*e))^2 - (d*(b*d - 2*c*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 1445

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

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin {align*} \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) (d+e x)^2} \, dx &=\int \frac {x^2}{(d+e x)^2 \left (c+b x+a x^2\right )} \, dx\\ &=\int \left (\frac {d^2}{\left (a d^2-e (b d-c e)\right ) (d+e x)^2}+\frac {d e (b d-2 c e)}{\left (a d^2-e (b d-c e)\right )^2 (d+e x)}+\frac {-c \left (a d^2-c e^2\right )-a d (b d-2 c e) x}{\left (a d^2-e (b d-c e)\right )^2 \left (c+b x+a x^2\right )}\right ) \, dx\\ &=-\frac {d^2}{e \left (a d^2-b d e+c e^2\right ) (d+e x)}+\frac {d (b d-2 c e) \log (d+e x)}{\left (a d^2-e (b d-c e)\right )^2}+\frac {\int \frac {-c \left (a d^2-c e^2\right )-a d (b d-2 c e) x}{c+b x+a x^2} \, dx}{\left (a d^2-e (b d-c e)\right )^2}\\ &=-\frac {d^2}{e \left (a d^2-b d e+c e^2\right ) (d+e x)}+\frac {d (b d-2 c e) \log (d+e x)}{\left (a d^2-e (b d-c e)\right )^2}-\frac {(d (b d-2 c 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 (b^2 d^2-2 b c d e-2 c \left (a d^2-c e^2\right )\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^2}{e \left (a d^2-b d e+c e^2\right ) (d+e x)}+\frac {d (b d-2 c e) \log (d+e x)}{\left (a d^2-e (b d-c e)\right )^2}-\frac {d (b d-2 c e) \log \left (c+b x+a x^2\right )}{2 \left (a d^2-e (b d-c e)\right )^2}-\frac {\left (b^2 d^2-2 b c d e-2 c \left (a d^2-c e^2\right )\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^2}{e \left (a d^2-b d e+c e^2\right ) (d+e x)}-\frac {\left (b^2 d^2-2 b c d e-2 c \left (a d^2-c e^2\right )\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 {d (b d-2 c e) \log (d+e x)}{\left (a d^2-e (b d-c e)\right )^2}-\frac {d (b d-2 c 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 = 0.23, size = 159, normalized size = 0.82 \begin {gather*} \frac {\frac {2 \left (2 c \left (c e^2-a d^2\right )+b^2 d^2-2 b c d e\right ) \tan ^{-1}\left (\frac {2 a x+b}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}-\frac {2 d^2 \left (a d^2+e (c e-b d)\right )}{e (d+e x)}-d (b d-2 c e) \log (x (a x+b)+c)+2 d (b d-2 c e) \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)*(d + e*x)^2),x]

[Out]

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 19.72, size = 1120, normalized size = 5.77

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.35, size = 331, normalized size = 1.71 \begin {gather*} \frac {{\left (b^{2} d^{2} e^{2} - 2 \, a c d^{2} e^{2} - 2 \, b c d e^{3} + 2 \, c^{2} 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 {d^{2} e}{{\left (a d^{2} e^{2} - b d e^{3} + c e^{4}\right )} {\left (x e + d\right )}} - \frac {{\left (b d^{2} - 2 \, c d e\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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 389, normalized size = 2.01 \begin {gather*} -\frac {2 a c \,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 {b^{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 b 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 {2 c^{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 {b \,d^{2} \ln \left (e x +d \right )}{\left (a \,d^{2}-d e b +c \,e^{2}\right )^{2}}-\frac {b \,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 {2 c d e \ln \left (e x +d \right )}{\left (a \,d^{2}-d e b +c \,e^{2}\right )^{2}}+\frac {c d e \ln \left (a \,x^{2}+b x +c \right )}{\left (a \,d^{2}-d e b +c \,e^{2}\right )^{2}}-\frac {d^{2}}{\left (a \,d^{2}-d e b +c \,e^{2}\right ) \left (e x +d \right ) e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)/(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 = 6.09, size = 1585, normalized size = 8.17

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(2*a*b^3*d^4 + b*c^3*e^4 - c^3*e^4*(b^2 - 4*a*c)^(1/2) + 16*a^2*c^2*d^3*e + 2*b^2*c^2*d*e^3 - b^3*c*d^2*e^
2 + a^2*b^2*d^4*x + b^2*c^2*e^4*x - b^4*d^2*e^2*x - 7*a^2*b*c*d^4 - 16*a*c^3*d*e^3 - 2*a^3*c*d^4*x - 2*a*c^3*e
^4*x + 2*a*b^2*d^4*(b^2 - 4*a*c)^(1/2) - a^2*c*d^4*(b^2 - 4*a*c)^(1/2) - 6*a*b^2*c*d^3*e + 2*a*b^3*d^3*e*x + 2
*b^3*c*d*e^3*x - 2*b*c^2*d*e^3*(b^2 - 4*a*c)^(1/2) + 3*a^2*b*d^4*x*(b^2 - 4*a*c)^(1/2) - b*c^2*e^4*x*(b^2 - 4*
a*c)^(1/2) + 10*a*b*c^2*d^2*e^2 + 14*a*c^2*d^2*e^2*(b^2 - 4*a*c)^(1/2) + b^2*c*d^2*e^2*(b^2 - 4*a*c)^(1/2) + b
^3*d^2*e^2*x*(b^2 - 4*a*c)^(1/2) + 28*a^2*c^2*d^2*e^2*x - 10*a*b*c*d^3*e*(b^2 - 4*a*c)^(1/2) - 12*a*b*c^2*d*e^
3*x - 12*a^2*b*c*d^3*e*x - 2*a*b^2*d^3*e*x*(b^2 - 4*a*c)^(1/2) + 8*a*c^2*d*e^3*x*(b^2 - 4*a*c)^(1/2) - 8*a^2*c
*d^3*e*x*(b^2 - 4*a*c)^(1/2) - 2*b^2*c*d*e^3*x*(b^2 - 4*a*c)^(1/2) + 2*a*b*c*d^2*e^2*x*(b^2 - 4*a*c)^(1/2))*(d
^2*(b^3/2 + (b^2*(b^2 - 4*a*c)^(1/2))/2) - c*(d^2*(2*a*b + a*(b^2 - 4*a*c)^(1/2)) + d*(b^2*e + b*e*(b^2 - 4*a*
c)^(1/2))) + c^2*(e^2*(b^2 - 4*a*c)^(1/2) + 4*a*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(2*a*b^3*d^4 + b*c^3*e^4 + c^3*e^4*(b^2 - 4*a*c)^(1/2) + 16*a^2*c^2*d^3*e + 2*b^2*c^2*d*e^3
- b^3*c*d^2*e^2 + a^2*b^2*d^4*x + b^2*c^2*e^4*x - b^4*d^2*e^2*x - 7*a^2*b*c*d^4 - 16*a*c^3*d*e^3 - 2*a^3*c*d^
4*x - 2*a*c^3*e^4*x - 2*a*b^2*d^4*(b^2 - 4*a*c)^(1/2) + a^2*c*d^4*(b^2 - 4*a*c)^(1/2) - 6*a*b^2*c*d^3*e + 2*a*
b^3*d^3*e*x + 2*b^3*c*d*e^3*x + 2*b*c^2*d*e^3*(b^2 - 4*a*c)^(1/2) - 3*a^2*b*d^4*x*(b^2 - 4*a*c)^(1/2) + b*c^2*
e^4*x*(b^2 - 4*a*c)^(1/2) + 10*a*b*c^2*d^2*e^2 - 14*a*c^2*d^2*e^2*(b^2 - 4*a*c)^(1/2) - b^2*c*d^2*e^2*(b^2 - 4
*a*c)^(1/2) - b^3*d^2*e^2*x*(b^2 - 4*a*c)^(1/2) + 28*a^2*c^2*d^2*e^2*x + 10*a*b*c*d^3*e*(b^2 - 4*a*c)^(1/2) -
12*a*b*c^2*d*e^3*x - 12*a^2*b*c*d^3*e*x + 2*a*b^2*d^3*e*x*(b^2 - 4*a*c)^(1/2) - 8*a*c^2*d*e^3*x*(b^2 - 4*a*c)^
(1/2) + 8*a^2*c*d^3*e*x*(b^2 - 4*a*c)^(1/2) + 2*b^2*c*d*e^3*x*(b^2 - 4*a*c)^(1/2) - 2*a*b*c*d^2*e^2*x*(b^2 - 4
*a*c)^(1/2))*(c*(d^2*(2*a*b - a*(b^2 - 4*a*c)^(1/2)) + d*(b^2*e - b*e*(b^2 - 4*a*c)^(1/2))) - d^2*(b^3/2 - (b^
2*(b^2 - 4*a*c)^(1/2))/2) + c^2*(e^2*(b^2 - 4*a*c)^(1/2) - 4*a*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*d^2 - 2*c*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) - d^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)/(e*x+d)**2,x)

[Out]

Timed out

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