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

Optimal. Leaf size=186 \[ -\frac {\left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (a e^2-b d e+c d^2\right )^2}-\frac {e (2 c d-b e) \log \left (a+b x+c x^2\right )}{2 \left (a e^2-b d e+c d^2\right )^2}-\frac {e}{(d+e x) \left (a e^2-b d e+c d^2\right )}+\frac {e (2 c d-b e) \log (d+e x)}{\left (a e^2-b d e+c d^2\right )^2} \]

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Rubi [A]  time = 0.29, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {709, 800, 634, 618, 206, 628} \begin {gather*} -\frac {\left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (a e^2-b d e+c d^2\right )^2}-\frac {e (2 c d-b e) \log \left (a+b x+c x^2\right )}{2 \left (a e^2-b d e+c d^2\right )^2}-\frac {e}{(d+e x) \left (a e^2-b d e+c d^2\right )}+\frac {e (2 c d-b e) \log (d+e x)}{\left (a e^2-b d e+c d^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.20, size = 151, normalized size = 0.81 \begin {gather*} \frac {\frac {2 \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}-\frac {2 e \left (e (a e-b d)+c d^2\right )}{d+e x}+e (b e-2 c d) \log (a+x (b+c x))-2 e (b e-2 c d) \log (d+e x)}{2 \left (e (a e-b d)+c d^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 2.97, size = 1079, normalized size = 5.80

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 325, normalized size = 1.75 \begin {gather*} \frac {{\left (2 \, c^{2} d^{2} e^{2} - 2 \, b c d e^{3} + b^{2} e^{4} - 2 \, a c e^{4}\right )} \arctan \left (\frac {{\left (2 \, c d - \frac {2 \, c d^{2}}{x e + d} - b e + \frac {2 \, b d e}{x e + d} - \frac {2 \, a e^{2}}{x e + d}\right )} e^{\left (-1\right )}}{\sqrt {-b^{2} + 4 \, a c}}\right ) e^{\left (-2\right )}}{{\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2} + 2 \, a c d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {{\left (2 \, c d e - b e^{2}\right )} \log \left (c - \frac {2 \, c d}{x e + d} + \frac {c 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 {a e^{2}}{{\left (x e + d\right )}^{2}}\right )}{2 \, {\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2} + 2 \, a c d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )}} - \frac {e^{3}}{{\left (c d^{2} e^{2} - b d e^{3} + a e^{4}\right )} {\left (x e + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.05, size = 386, normalized size = 2.08 \begin {gather*} -\frac {2 a c \,e^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right )^{2} \sqrt {4 a c -b^{2}}}+\frac {b^{2} e^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right )^{2} \sqrt {4 a c -b^{2}}}-\frac {2 b c d e \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right )^{2} \sqrt {4 a c -b^{2}}}+\frac {2 c^{2} d^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right )^{2} \sqrt {4 a c -b^{2}}}-\frac {b \,e^{2} \ln \left (e x +d \right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right )^{2}}+\frac {b \,e^{2} \ln \left (c \,x^{2}+b x +a \right )}{2 \left (a \,e^{2}-b d e +c \,d^{2}\right )^{2}}+\frac {2 c d e \ln \left (e x +d \right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right )^{2}}-\frac {c d e \ln \left (c \,x^{2}+b x +a \right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right )^{2}}-\frac {e}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \left (e x +d \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-e/(a*e^2-b*d*e+c*d^2)/(e*x+d)-e^2/(a*e^2-b*d*e+c*d^2)^2*ln(e*x+d)*b+2*e/(a*e^2-b*d*e+c*d^2)^2*ln(e*x+d)*c*d+1
/2/(a*e^2-b*d*e+c*d^2)^2*ln(c*x^2+b*x+a)*b*e^2-1/(a*e^2-b*d*e+c*d^2)^2*c*ln(c*x^2+b*x+a)*d*e-2/(a*e^2-b*d*e+c*
d^2)^2/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*a*c*e^2+1/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)^(1/2)
*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*b^2*e^2-2/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c
-b^2)^(1/2))*b*c*d*e+2/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*c^2*d^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/(e*x+d)^2/(c*x^2+b*x+a),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 = 7.09, size = 1786, normalized size = 9.60

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

[Out]

Timed out

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