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

Optimal. Leaf size=149 \[ -\frac {\left (-2 a c d+b^2 d-b c e\right ) \tanh ^{-1}\left (\frac {2 a x+b}{\sqrt {b^2-4 a c}}\right )}{a \sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )}+\frac {d^2 \log (d+e x)}{e \left (a d^2-b d e+c e^2\right )}-\frac {(b d-c e) \log \left (a x^2+b x+c\right )}{2 a \left (a d^2-e (b d-c e)\right )} \]

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Rubi [A]  time = 0.21, antiderivative size = 149, 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 a c d+b^2 d-b c e\right ) \tanh ^{-1}\left (\frac {2 a x+b}{\sqrt {b^2-4 a c}}\right )}{a \sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )}+\frac {d^2 \log (d+e x)}{e \left (a d^2-b d e+c e^2\right )}-\frac {(b d-c e) \log \left (a x^2+b x+c\right )}{2 a \left (a d^2-e (b d-c e)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.12, size = 132, normalized size = 0.89 \begin {gather*} -\frac {\sqrt {4 a c-b^2} \left (e (b d-c e) \log (x (a x+b)+c)-2 a d^2 \log (d+e x)\right )+2 e \left (2 a c d+b^2 (-d)+b c e\right ) \tan ^{-1}\left (\frac {2 a x+b}{\sqrt {4 a c-b^2}}\right )}{2 a e \sqrt {4 a c-b^2} \left (a d^2+e (c e-b d)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(2*e*(-(b^2*d) + 2*a*c*d + b*c*e)*ArcTan[(b + 2*a*x)/Sqrt[-b^2 + 4*a*c]] + Sqrt[-b^2 + 4*a*c]*(-2*a*d^2*L
og[d + e*x] + e*(b*d - c*e)*Log[c + x*(b + a*x)]))/(a*Sqrt[-b^2 + 4*a*c]*e*(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 ) (d+e x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 5.76, size = 405, normalized size = 2.72 \begin {gather*} \left [\frac {2 \, {\left (a b^{2} - 4 \, a^{2} c\right )} d^{2} \log \left (e x + d\right ) + {\left (b c e^{2} - {\left (b^{2} - 2 \, a c\right )} d e\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, a^{2} x^{2} + 2 \, a b x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, a x + b\right )}}{a x^{2} + b x + c}\right ) - {\left ({\left (b^{3} - 4 \, a b c\right )} d e - {\left (b^{2} c - 4 \, a c^{2}\right )} e^{2}\right )} \log \left (a x^{2} + b x + c\right )}{2 \, {\left ({\left (a^{2} b^{2} - 4 \, a^{3} c\right )} d^{2} e - {\left (a b^{3} - 4 \, a^{2} b c\right )} d e^{2} + {\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} e^{3}\right )}}, \frac {2 \, {\left (a b^{2} - 4 \, a^{2} c\right )} d^{2} \log \left (e x + d\right ) + 2 \, {\left (b c e^{2} - {\left (b^{2} - 2 \, a c\right )} d e\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, a x + b\right )}}{b^{2} - 4 \, a c}\right ) - {\left ({\left (b^{3} - 4 \, a b c\right )} d e - {\left (b^{2} c - 4 \, a c^{2}\right )} e^{2}\right )} \log \left (a x^{2} + b x + c\right )}{2 \, {\left ({\left (a^{2} b^{2} - 4 \, a^{3} c\right )} d^{2} e - {\left (a b^{3} - 4 \, a^{2} b c\right )} d e^{2} + {\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} e^{3}\right )}}\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),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.37, size = 149, normalized size = 1.00 \begin {gather*} \frac {d^{2} \log \left ({\left | x e + d \right |}\right )}{a d^{2} e - b d e^{2} + c e^{3}} - \frac {{\left (b d - c e\right )} \log \left (a x^{2} + b x + c\right )}{2 \, {\left (a^{2} d^{2} - a b d e + a c e^{2}\right )}} + \frac {{\left (b^{2} d - 2 \, a c d - b c e\right )} \arctan \left (\frac {2 \, a x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (a^{2} d^{2} - a b d e + a c e^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 275, normalized size = 1.85 \begin {gather*} \frac {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}}\, a}-\frac {b c 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}}\, a}-\frac {2 c 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}}}-\frac {b d \ln \left (a \,x^{2}+b x +c \right )}{2 \left (a \,d^{2}-d e b +c \,e^{2}\right ) a}+\frac {c e \ln \left (a \,x^{2}+b x +c \right )}{2 \left (a \,d^{2}-d e b +c \,e^{2}\right ) a}+\frac {d^{2} \ln \left (e x +d \right )}{\left (a \,d^{2}-d e b +c \,e^{2}\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),x)

[Out]

-1/2/(a*d^2-b*d*e+c*e^2)/a*ln(a*x^2+b*x+c)*b*d+1/2/(a*d^2-b*d*e+c*e^2)/a*ln(a*x^2+b*x+c)*c*e-2/(a*d^2-b*d*e+c*
e^2)/(4*a*c-b^2)^(1/2)*arctan((2*a*x+b)/(4*a*c-b^2)^(1/2))*c*d+1/(a*d^2-b*d*e+c*e^2)/(4*a*c-b^2)^(1/2)*arctan(
(2*a*x+b)/(4*a*c-b^2)^(1/2))/a*b^2*d-1/(a*d^2-b*d*e+c*e^2)/(4*a*c-b^2)^(1/2)*arctan((2*a*x+b)/(4*a*c-b^2)^(1/2
))/a*b*c*e+d^2*ln(e*x+d)/e/(a*d^2-b*d*e+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)/(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 = 3.67, size = 966, normalized size = 6.48 \begin {gather*} \frac {d^2\,\ln \left (d+e\,x\right )}{a\,d^2\,e-b\,d\,e^2+c\,e^3}-\frac {\ln \left (a\,b^2\,d^4-2\,c^3\,e^4-4\,a^2\,c\,d^4+b^3\,d^3\,e+c^2\,e^4\,x\,\sqrt {b^2-4\,a\,c}+10\,a\,c^2\,d^2\,e^2-4\,b^2\,c\,d^2\,e^2-b^3\,d^2\,e^2\,x+a\,b\,d^4\,\sqrt {b^2-4\,a\,c}+3\,b\,c^2\,d\,e^3-b\,c^2\,e^4\,x+b^2\,d^3\,e\,\sqrt {b^2-4\,a\,c}+3\,c^2\,d\,e^3\,\sqrt {b^2-4\,a\,c}+2\,a^2\,d^4\,x\,\sqrt {b^2-4\,a\,c}+3\,a\,b^2\,d^3\,e\,x+6\,a\,c^2\,d\,e^3\,x-10\,a^2\,c\,d^3\,e\,x-2\,b\,c\,d^2\,e^2\,\sqrt {b^2-4\,a\,c}-3\,a\,b\,c\,d^3\,e+b^2\,d^2\,e^2\,x\,\sqrt {b^2-4\,a\,c}-5\,a\,c\,d^3\,e\,\sqrt {b^2-4\,a\,c}-a\,b\,d^3\,e\,x\,\sqrt {b^2-4\,a\,c}+a\,b\,c\,d^2\,e^2\,x-5\,a\,c\,d^2\,e^2\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (e\,\left (\frac {b^2\,c}{2}-2\,a\,c^2+\frac {b\,c\,\sqrt {b^2-4\,a\,c}}{2}\right )-\frac {b^3\,d}{2}-\frac {b^2\,d\,\sqrt {b^2-4\,a\,c}}{2}+a\,c\,d\,\sqrt {b^2-4\,a\,c}+2\,a\,b\,c\,d\right )}{4\,a^3\,c\,d^2-a^2\,b^2\,d^2-4\,a^2\,b\,c\,d\,e+4\,a^2\,c^2\,e^2+a\,b^3\,d\,e-a\,b^2\,c\,e^2}+\frac {\ln \left (2\,c^3\,e^4-a\,b^2\,d^4+4\,a^2\,c\,d^4-b^3\,d^3\,e+c^2\,e^4\,x\,\sqrt {b^2-4\,a\,c}-10\,a\,c^2\,d^2\,e^2+4\,b^2\,c\,d^2\,e^2+b^3\,d^2\,e^2\,x+a\,b\,d^4\,\sqrt {b^2-4\,a\,c}-3\,b\,c^2\,d\,e^3+b\,c^2\,e^4\,x+b^2\,d^3\,e\,\sqrt {b^2-4\,a\,c}+3\,c^2\,d\,e^3\,\sqrt {b^2-4\,a\,c}+2\,a^2\,d^4\,x\,\sqrt {b^2-4\,a\,c}-3\,a\,b^2\,d^3\,e\,x-6\,a\,c^2\,d\,e^3\,x+10\,a^2\,c\,d^3\,e\,x-2\,b\,c\,d^2\,e^2\,\sqrt {b^2-4\,a\,c}+3\,a\,b\,c\,d^3\,e+b^2\,d^2\,e^2\,x\,\sqrt {b^2-4\,a\,c}-5\,a\,c\,d^3\,e\,\sqrt {b^2-4\,a\,c}-a\,b\,d^3\,e\,x\,\sqrt {b^2-4\,a\,c}-a\,b\,c\,d^2\,e^2\,x-5\,a\,c\,d^2\,e^2\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (\frac {b^3\,d}{2}+e\,\left (2\,a\,c^2-\frac {b^2\,c}{2}+\frac {b\,c\,\sqrt {b^2-4\,a\,c}}{2}\right )-\frac {b^2\,d\,\sqrt {b^2-4\,a\,c}}{2}+a\,c\,d\,\sqrt {b^2-4\,a\,c}-2\,a\,b\,c\,d\right )}{4\,a^3\,c\,d^2-a^2\,b^2\,d^2-4\,a^2\,b\,c\,d\,e+4\,a^2\,c^2\,e^2+a\,b^3\,d\,e-a\,b^2\,c\,e^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d^2*log(d + e*x))/(c*e^3 + a*d^2*e - b*d*e^2) - (log(a*b^2*d^4 - 2*c^3*e^4 - 4*a^2*c*d^4 + b^3*d^3*e + c^2*e^
4*x*(b^2 - 4*a*c)^(1/2) + 10*a*c^2*d^2*e^2 - 4*b^2*c*d^2*e^2 - b^3*d^2*e^2*x + a*b*d^4*(b^2 - 4*a*c)^(1/2) + 3
*b*c^2*d*e^3 - b*c^2*e^4*x + b^2*d^3*e*(b^2 - 4*a*c)^(1/2) + 3*c^2*d*e^3*(b^2 - 4*a*c)^(1/2) + 2*a^2*d^4*x*(b^
2 - 4*a*c)^(1/2) + 3*a*b^2*d^3*e*x + 6*a*c^2*d*e^3*x - 10*a^2*c*d^3*e*x - 2*b*c*d^2*e^2*(b^2 - 4*a*c)^(1/2) -
3*a*b*c*d^3*e + b^2*d^2*e^2*x*(b^2 - 4*a*c)^(1/2) - 5*a*c*d^3*e*(b^2 - 4*a*c)^(1/2) - a*b*d^3*e*x*(b^2 - 4*a*c
)^(1/2) + a*b*c*d^2*e^2*x - 5*a*c*d^2*e^2*x*(b^2 - 4*a*c)^(1/2))*(e*((b^2*c)/2 - 2*a*c^2 + (b*c*(b^2 - 4*a*c)^
(1/2))/2) - (b^3*d)/2 - (b^2*d*(b^2 - 4*a*c)^(1/2))/2 + a*c*d*(b^2 - 4*a*c)^(1/2) + 2*a*b*c*d))/(4*a^3*c*d^2 -
a^2*b^2*d^2 + 4*a^2*c^2*e^2 + a*b^3*d*e - a*b^2*c*e^2 - 4*a^2*b*c*d*e) + (log(2*c^3*e^4 - a*b^2*d^4 + 4*a^2*c
*d^4 - b^3*d^3*e + c^2*e^4*x*(b^2 - 4*a*c)^(1/2) - 10*a*c^2*d^2*e^2 + 4*b^2*c*d^2*e^2 + b^3*d^2*e^2*x + a*b*d^
4*(b^2 - 4*a*c)^(1/2) - 3*b*c^2*d*e^3 + b*c^2*e^4*x + b^2*d^3*e*(b^2 - 4*a*c)^(1/2) + 3*c^2*d*e^3*(b^2 - 4*a*c
)^(1/2) + 2*a^2*d^4*x*(b^2 - 4*a*c)^(1/2) - 3*a*b^2*d^3*e*x - 6*a*c^2*d*e^3*x + 10*a^2*c*d^3*e*x - 2*b*c*d^2*e
^2*(b^2 - 4*a*c)^(1/2) + 3*a*b*c*d^3*e + b^2*d^2*e^2*x*(b^2 - 4*a*c)^(1/2) - 5*a*c*d^3*e*(b^2 - 4*a*c)^(1/2) -
a*b*d^3*e*x*(b^2 - 4*a*c)^(1/2) - a*b*c*d^2*e^2*x - 5*a*c*d^2*e^2*x*(b^2 - 4*a*c)^(1/2))*((b^3*d)/2 + e*(2*a*
c^2 - (b^2*c)/2 + (b*c*(b^2 - 4*a*c)^(1/2))/2) - (b^2*d*(b^2 - 4*a*c)^(1/2))/2 + a*c*d*(b^2 - 4*a*c)^(1/2) - 2
*a*b*c*d))/(4*a^3*c*d^2 - a^2*b^2*d^2 + 4*a^2*c^2*e^2 + a*b^3*d*e - a*b^2*c*e^2 - 4*a^2*b*c*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),x)

[Out]

Timed out

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