3.2196 \(\int \frac{1}{(d+e x) (a+b x+c x^2)^2} \, dx\)

Optimal. Leaf size=224 \[ \frac{(2 c d-b e) \left (-2 c e (b d-3 a e)-b^2 e^2+2 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2} \left (a e^2-b d e+c d^2\right )^2}-\frac{2 a c e+b^2 (-e)+c x (2 c d-b e)+b c d}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \left (a e^2-b d e+c d^2\right )}-\frac{e^3 \log \left (a+b x+c x^2\right )}{2 \left (a e^2-b d e+c d^2\right )^2}+\frac{e^3 \log (d+e x)}{\left (a e^2-b d e+c d^2\right )^2} \]

[Out]

-((b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x)/((b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)*(a + b*x + c*x^2))) + (
(2*c*d - b*e)*(2*c^2*d^2 - b^2*e^2 - 2*c*e*(b*d - 3*a*e))*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/((b^2 - 4*a*
c)^(3/2)*(c*d^2 - b*d*e + a*e^2)^2) + (e^3*Log[d + e*x])/(c*d^2 - b*d*e + a*e^2)^2 - (e^3*Log[a + b*x + c*x^2]
)/(2*(c*d^2 - b*d*e + a*e^2)^2)

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Rubi [A]  time = 0.38123, antiderivative size = 224, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {740, 800, 634, 618, 206, 628} \[ \frac{(2 c d-b e) \left (-2 c e (b d-3 a e)-b^2 e^2+2 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2} \left (a e^2-b d e+c d^2\right )^2}-\frac{2 a c e+b^2 (-e)+c x (2 c d-b e)+b c d}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \left (a e^2-b d e+c d^2\right )}-\frac{e^3 \log \left (a+b x+c x^2\right )}{2 \left (a e^2-b d e+c d^2\right )^2}+\frac{e^3 \log (d+e x)}{\left (a e^2-b d e+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x)/((b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)*(a + b*x + c*x^2))) + (
(2*c*d - b*e)*(2*c^2*d^2 - b^2*e^2 - 2*c*e*(b*d - 3*a*e))*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/((b^2 - 4*a*
c)^(3/2)*(c*d^2 - b*d*e + a*e^2)^2) + (e^3*Log[d + e*x])/(c*d^2 - b*d*e + a*e^2)^2 - (e^3*Log[a + b*x + c*x^2]
)/(2*(c*d^2 - b*d*e + a*e^2)^2)

Rule 740

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e
+ a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*Simp[b*c*d*e*(2*p - m
+ 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x
, x]*(a + b*x + c*x^2)^(p + 1), 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[p, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]

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 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 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 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 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]

Rubi steps

\begin{align*} \int \frac{1}{(d+e x) \left (a+b x+c x^2\right )^2} \, dx &=-\frac{b c d-b^2 e+2 a c e+c (2 c d-b e) x}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )}-\frac{\int \frac{2 c^2 d^2-b^2 e^2-c e (b d-4 a e)+c e (2 c d-b e) x}{(d+e x) \left (a+b x+c x^2\right )} \, dx}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )}\\ &=-\frac{b c d-b^2 e+2 a c e+c (2 c d-b e) x}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )}-\frac{\int \left (-\frac{\left (b^2-4 a c\right ) e^4}{\left (c d^2-b d e+a e^2\right ) (d+e x)}+\frac{2 c^3 d^3+b^3 e^3-5 a b c e^3-3 c^2 d e (b d-2 a e)+c \left (b^2-4 a c\right ) e^3 x}{\left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )}\right ) \, dx}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )}\\ &=-\frac{b c d-b^2 e+2 a c e+c (2 c d-b e) x}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )}+\frac{e^3 \log (d+e x)}{\left (c d^2-b d e+a e^2\right )^2}-\frac{\int \frac{2 c^3 d^3+b^3 e^3-5 a b c e^3-3 c^2 d e (b d-2 a e)+c \left (b^2-4 a c\right ) e^3 x}{a+b x+c x^2} \, dx}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )^2}\\ &=-\frac{b c d-b^2 e+2 a c e+c (2 c d-b e) x}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )}+\frac{e^3 \log (d+e x)}{\left (c d^2-b d e+a e^2\right )^2}-\frac{e^3 \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 d-b e) \left (2 c^2 d^2-b^2 e^2-2 c e (b d-3 a e)\right )\right ) \int \frac{1}{a+b x+c x^2} \, dx}{2 \left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )^2}\\ &=-\frac{b c d-b^2 e+2 a c e+c (2 c d-b e) x}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )}+\frac{e^3 \log (d+e x)}{\left (c d^2-b d e+a e^2\right )^2}-\frac{e^3 \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 d-b e) \left (2 c^2 d^2-b^2 e^2-2 c e (b d-3 a e)\right )\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )^2}\\ &=-\frac{b c d-b^2 e+2 a c e+c (2 c d-b e) x}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )}+\frac{(2 c d-b e) \left (2 c^2 d^2-b^2 e^2-2 c e (b d-3 a e)\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2} \left (c d^2-b d e+a e^2\right )^2}+\frac{e^3 \log (d+e x)}{\left (c d^2-b d e+a e^2\right )^2}-\frac{e^3 \log \left (a+b x+c x^2\right )}{2 \left (c d^2-b d e+a e^2\right )^2}\\ \end{align*}

Mathematica [A]  time = 0.3572, size = 223, normalized size = 1. \[ \frac{(b e-2 c d) \left (2 c e (b d-3 a e)+b^2 e^2-2 c^2 d^2\right ) \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{3/2} \left (e (a e-b d)+c d^2\right )^2}+\frac{2 c (a e+c d x)+b^2 (-e)+b c (d-e x)}{\left (b^2-4 a c\right ) (a+x (b+c x)) \left (e (b d-a e)-c d^2\right )}+\frac{e^3 \log (d+e x)}{\left (e (a e-b d)+c d^2\right )^2}-\frac{e^3 \log (a+x (b+c x))}{2 \left (e (a e-b d)+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(b^2*e) + 2*c*(a*e + c*d*x) + b*c*(d - e*x))/((b^2 - 4*a*c)*(-(c*d^2) + e*(b*d - a*e))*(a + x*(b + c*x))) +
((-2*c*d + b*e)*(-2*c^2*d^2 + b^2*e^2 + 2*c*e*(b*d - 3*a*e))*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/((-b^2 +
4*a*c)^(3/2)*(c*d^2 + e*(-(b*d) + a*e))^2) + (e^3*Log[d + e*x])/(c*d^2 + e*(-(b*d) + a*e))^2 - (e^3*Log[a + x*
(b + c*x)])/(2*(c*d^2 + e*(-(b*d) + a*e))^2)

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Maple [B]  time = 0.193, size = 1037, normalized size = 4.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(a*e^2-b*d*e+c*d^2)^2/(c*x^2+b*x+a)*c/(4*a*c-b^2)*x*a*b*e^3+2/(a*e^2-b*d*e+c*d^2)^2/(c*x^2+b*x+a)*c^2/(4*a*
c-b^2)*x*a*d*e^2+1/(a*e^2-b*d*e+c*d^2)^2/(c*x^2+b*x+a)*c/(4*a*c-b^2)*x*b^2*d*e^2-3/(a*e^2-b*d*e+c*d^2)^2/(c*x^
2+b*x+a)*c^2/(4*a*c-b^2)*x*b*d^2*e+2/(a*e^2-b*d*e+c*d^2)^2/(c*x^2+b*x+a)*c^3/(4*a*c-b^2)*x*d^3+2/(a*e^2-b*d*e+
c*d^2)^2/(c*x^2+b*x+a)/(4*a*c-b^2)*a^2*c*e^3-1/(a*e^2-b*d*e+c*d^2)^2/(c*x^2+b*x+a)/(4*a*c-b^2)*a*b^2*e^3-1/(a*
e^2-b*d*e+c*d^2)^2/(c*x^2+b*x+a)/(4*a*c-b^2)*a*b*c*d*e^2+2/(a*e^2-b*d*e+c*d^2)^2/(c*x^2+b*x+a)/(4*a*c-b^2)*a*c
^2*d^2*e+1/(a*e^2-b*d*e+c*d^2)^2/(c*x^2+b*x+a)/(4*a*c-b^2)*b^3*d*e^2-2/(a*e^2-b*d*e+c*d^2)^2/(c*x^2+b*x+a)/(4*
a*c-b^2)*b^2*c*d^2*e+1/(a*e^2-b*d*e+c*d^2)^2/(c*x^2+b*x+a)/(4*a*c-b^2)*b*c^2*d^3-2/(a*e^2-b*d*e+c*d^2)^2/(4*a*
c-b^2)*c*ln(c*x^2+b*x+a)*a*e^3+1/2/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)*ln(c*x^2+b*x+a)*b^2*e^3-6/(a*e^2-b*d*e+c*
d^2)^2/(4*a*c-b^2)^(3/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*a*b*c*e^3+12/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)^(3
/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*c^2*a*d*e^2+1/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)^(3/2)*arctan((2*c*x+b)
/(4*a*c-b^2)^(1/2))*b^3*e^3-6/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)^(3/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*b*c^
2*d^2*e+4/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)^(3/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*c^3*d^3+e^3*ln(e*x+d)/(a
*e^2-b*d*e+c*d^2)^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 159.644, size = 4417, normalized size = 19.72 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(2*(b^3*c^2 - 4*a*b*c^3)*d^3 - 4*(b^4*c - 5*a*b^2*c^2 + 4*a^2*c^3)*d^2*e + 2*(b^5 - 5*a*b^3*c + 4*a^2*b*
c^2)*d*e^2 - 2*(a*b^4 - 6*a^2*b^2*c + 8*a^3*c^2)*e^3 - (4*a*c^3*d^3 - 6*a*b*c^2*d^2*e + 12*a^2*c^2*d*e^2 + (a*
b^3 - 6*a^2*b*c)*e^3 + (4*c^4*d^3 - 6*b*c^3*d^2*e + 12*a*c^3*d*e^2 + (b^3*c - 6*a*b*c^2)*e^3)*x^2 + (4*b*c^3*d
^3 - 6*b^2*c^2*d^2*e + 12*a*b*c^2*d*e^2 + (b^4 - 6*a*b^2*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*(2*(b^2*c^3 - 4*a*c^4)*d^3 - 3*(b^3*c^2
- 4*a*b*c^3)*d^2*e + (b^4*c - 2*a*b^2*c^2 - 8*a^2*c^3)*d*e^2 - (a*b^3*c - 4*a^2*b*c^2)*e^3)*x + ((b^4*c - 8*a
*b^2*c^2 + 16*a^2*c^3)*e^3*x^2 + (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*e^3*x + (a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2)*e
^3)*log(c*x^2 + b*x + a) - 2*((b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*e^3*x^2 + (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*e^
3*x + (a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2)*e^3)*log(e*x + d))/((a*b^4*c^2 - 8*a^2*b^2*c^3 + 16*a^3*c^4)*d^4 - 2*
(a*b^5*c - 8*a^2*b^3*c^2 + 16*a^3*b*c^3)*d^3*e + (a*b^6 - 6*a^2*b^4*c + 32*a^4*c^3)*d^2*e^2 - 2*(a^2*b^5 - 8*a
^3*b^3*c + 16*a^4*b*c^2)*d*e^3 + (a^3*b^4 - 8*a^4*b^2*c + 16*a^5*c^2)*e^4 + ((b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c
^5)*d^4 - 2*(b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^3*e + (b^6*c - 6*a*b^4*c^2 + 32*a^3*c^4)*d^2*e^2 - 2*(a*b
^5*c - 8*a^2*b^3*c^2 + 16*a^3*b*c^3)*d*e^3 + (a^2*b^4*c - 8*a^3*b^2*c^2 + 16*a^4*c^3)*e^4)*x^2 + ((b^5*c^2 - 8
*a*b^3*c^3 + 16*a^2*b*c^4)*d^4 - 2*(b^6*c - 8*a*b^4*c^2 + 16*a^2*b^2*c^3)*d^3*e + (b^7 - 6*a*b^5*c + 32*a^3*b*
c^3)*d^2*e^2 - 2*(a*b^6 - 8*a^2*b^4*c + 16*a^3*b^2*c^2)*d*e^3 + (a^2*b^5 - 8*a^3*b^3*c + 16*a^4*b*c^2)*e^4)*x)
, -1/2*(2*(b^3*c^2 - 4*a*b*c^3)*d^3 - 4*(b^4*c - 5*a*b^2*c^2 + 4*a^2*c^3)*d^2*e + 2*(b^5 - 5*a*b^3*c + 4*a^2*b
*c^2)*d*e^2 - 2*(a*b^4 - 6*a^2*b^2*c + 8*a^3*c^2)*e^3 - 2*(4*a*c^3*d^3 - 6*a*b*c^2*d^2*e + 12*a^2*c^2*d*e^2 +
(a*b^3 - 6*a^2*b*c)*e^3 + (4*c^4*d^3 - 6*b*c^3*d^2*e + 12*a*c^3*d*e^2 + (b^3*c - 6*a*b*c^2)*e^3)*x^2 + (4*b*c^
3*d^3 - 6*b^2*c^2*d^2*e + 12*a*b*c^2*d*e^2 + (b^4 - 6*a*b^2*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*(2*(b^2*c^3 - 4*a*c^4)*d^3 - 3*(b^3*c^2 - 4*a*b*c^3)*d^2*e + (b^4*c - 2*
a*b^2*c^2 - 8*a^2*c^3)*d*e^2 - (a*b^3*c - 4*a^2*b*c^2)*e^3)*x + ((b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*e^3*x^2 +
(b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*e^3*x + (a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2)*e^3)*log(c*x^2 + b*x + a) - 2*((b^
4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*e^3*x^2 + (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*e^3*x + (a*b^4 - 8*a^2*b^2*c + 16*a
^3*c^2)*e^3)*log(e*x + d))/((a*b^4*c^2 - 8*a^2*b^2*c^3 + 16*a^3*c^4)*d^4 - 2*(a*b^5*c - 8*a^2*b^3*c^2 + 16*a^3
*b*c^3)*d^3*e + (a*b^6 - 6*a^2*b^4*c + 32*a^4*c^3)*d^2*e^2 - 2*(a^2*b^5 - 8*a^3*b^3*c + 16*a^4*b*c^2)*d*e^3 +
(a^3*b^4 - 8*a^4*b^2*c + 16*a^5*c^2)*e^4 + ((b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^4 - 2*(b^5*c^2 - 8*a*b^3*c^
3 + 16*a^2*b*c^4)*d^3*e + (b^6*c - 6*a*b^4*c^2 + 32*a^3*c^4)*d^2*e^2 - 2*(a*b^5*c - 8*a^2*b^3*c^2 + 16*a^3*b*c
^3)*d*e^3 + (a^2*b^4*c - 8*a^3*b^2*c^2 + 16*a^4*c^3)*e^4)*x^2 + ((b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^4 -
2*(b^6*c - 8*a*b^4*c^2 + 16*a^2*b^2*c^3)*d^3*e + (b^7 - 6*a*b^5*c + 32*a^3*b*c^3)*d^2*e^2 - 2*(a*b^6 - 8*a^2*b
^4*c + 16*a^3*b^2*c^2)*d*e^3 + (a^2*b^5 - 8*a^3*b^3*c + 16*a^4*b*c^2)*e^4)*x)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.10915, size = 635, normalized size = 2.83 \begin{align*} -\frac{e^{3} \log \left (c x^{2} + b x + a\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^{4} \log \left ({\left | x e + d \right |}\right )}{c^{2} d^{4} e - 2 \, b c d^{3} e^{2} + b^{2} d^{2} e^{3} + 2 \, a c d^{2} e^{3} - 2 \, a b d e^{4} + a^{2} e^{5}} - \frac{{\left (4 \, c^{3} d^{3} - 6 \, b c^{2} d^{2} e + 12 \, a c^{2} d e^{2} + b^{3} e^{3} - 6 \, a b c e^{3}\right )} \arctan \left (\frac{2 \, c x + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{{\left (b^{2} c^{2} d^{4} - 4 \, a c^{3} d^{4} - 2 \, b^{3} c d^{3} e + 8 \, a b c^{2} d^{3} e + b^{4} d^{2} e^{2} - 2 \, a b^{2} c d^{2} e^{2} - 8 \, a^{2} c^{2} d^{2} e^{2} - 2 \, a b^{3} d e^{3} + 8 \, a^{2} b c d e^{3} + a^{2} b^{2} e^{4} - 4 \, a^{3} c e^{4}\right )} \sqrt{-b^{2} + 4 \, a c}} - \frac{b c^{2} d^{3} - 2 \, b^{2} c d^{2} e + 2 \, a c^{2} d^{2} e + b^{3} d e^{2} - a b c d e^{2} - a b^{2} e^{3} + 2 \, a^{2} c e^{3} +{\left (2 \, c^{3} d^{3} - 3 \, b c^{2} d^{2} e + b^{2} c d e^{2} + 2 \, a c^{2} d e^{2} - a b c e^{3}\right )} x}{{\left (c d^{2} - b d e + a e^{2}\right )}^{2}{\left (c x^{2} + b x + a\right )}{\left (b^{2} - 4 \, a c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*e^3*log(c*x^2 + b*x + a)/(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^4*log(abs(x*e + d))/(c^2*d^4*e - 2*b*c*d^3*e^2 + b^2*d^2*e^3 + 2*a*c*d^2*e^3 - 2*a*b*d*e^4 + a^2*e^5) - (4*c
^3*d^3 - 6*b*c^2*d^2*e + 12*a*c^2*d*e^2 + b^3*e^3 - 6*a*b*c*e^3)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((b^2*
c^2*d^4 - 4*a*c^3*d^4 - 2*b^3*c*d^3*e + 8*a*b*c^2*d^3*e + b^4*d^2*e^2 - 2*a*b^2*c*d^2*e^2 - 8*a^2*c^2*d^2*e^2
- 2*a*b^3*d*e^3 + 8*a^2*b*c*d*e^3 + a^2*b^2*e^4 - 4*a^3*c*e^4)*sqrt(-b^2 + 4*a*c)) - (b*c^2*d^3 - 2*b^2*c*d^2*
e + 2*a*c^2*d^2*e + b^3*d*e^2 - a*b*c*d*e^2 - a*b^2*e^3 + 2*a^2*c*e^3 + (2*c^3*d^3 - 3*b*c^2*d^2*e + b^2*c*d*e
^2 + 2*a*c^2*d*e^2 - a*b*c*e^3)*x)/((c*d^2 - b*d*e + a*e^2)^2*(c*x^2 + b*x + a)*(b^2 - 4*a*c))