### 3.2188 $$\int \frac{1}{(d+e x)^2 (a+b x+c x^2)} \, dx$$

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

[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)

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Rubi [A]  time = 0.285323, antiderivative size = 186, 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 = {709, 800, 634, 618, 206, 628} $-\frac{\left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\sqrt{b^2-4 a c} \left (a e^2-b d e+c d^2\right )^2}-\frac{e (2 c d-b e) \log \left (a+b x+c x^2\right )}{2 \left (a e^2-b d e+c d^2\right )^2}-\frac{e}{(d+e x) \left (a e^2-b d e+c d^2\right )}+\frac{e (2 c d-b e) \log (d+e x)}{\left (a e^2-b d e+c d^2\right )^2}$

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 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 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)^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*}

Mathematica [A]  time = 0.224112, size = 151, normalized size = 0.81 $\frac{\frac{2 \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )}{\sqrt{4 a c-b^2}}-\frac{2 e \left (e (a e-b d)+c d^2\right )}{d+e x}+e (b e-2 c d) \log (a+x (b+c x))-2 e (b e-2 c d) \log (d+e x)}{2 \left (e (a e-b d)+c d^2\right )^2}$

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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Maple [B]  time = 0.171, size = 386, normalized size = 2.1 \begin{align*}{\frac{\ln \left ( c{x}^{2}+bx+a \right ) b{e}^{2}}{2\, \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}}}-{\frac{c\ln \left ( c{x}^{2}+bx+a \right ) de}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}}}-2\,{\frac{ac{e}^{2}}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+{\frac{{b}^{2}{e}^{2}}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}}\arctan \left ({(2\,cx+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}-2\,{\frac{bcde}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+2\,{\frac{{c}^{2}{d}^{2}}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-{\frac{e}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) \left ( ex+d \right ) }}-{\frac{{e}^{2}\ln \left ( ex+d \right ) b}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}}}+2\,{\frac{e\ln \left ( ex+d \right ) cd}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}}} \end{align*}

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]

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-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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 19.6378, size = 2295, normalized size = 12.34 \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)^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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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)**2/(c*x**2+b*x+a),x)

[Out]

Timed out

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Giac [A]  time = 1.11342, size = 447, normalized size = 2.4 \begin{align*} -\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{align*}

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))