### 3.2189 $$\int \frac{1}{(d+e x)^3 (a+b x+c x^2)} \, dx$$

Optimal. Leaf size=272 $-\frac{e \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right ) \log \left (a+b x+c x^2\right )}{2 \left (a e^2-b d e+c d^2\right )^3}+\frac{e \log (d+e x) \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right )}{\left (a e^2-b d e+c d^2\right )^3}-\frac{(2 c d-b e) \left (-c e (3 a e+b d)+b^2 e^2+c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\sqrt{b^2-4 a c} \left (a e^2-b d e+c d^2\right )^3}-\frac{e (2 c d-b e)}{(d+e x) \left (a e^2-b d e+c d^2\right )^2}-\frac{e}{2 (d+e x)^2 \left (a e^2-b d e+c d^2\right )}$

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.425177, antiderivative size = 272, 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} $-\frac{e \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right ) \log \left (a+b x+c x^2\right )}{2 \left (a e^2-b d e+c d^2\right )^3}+\frac{e \log (d+e x) \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right )}{\left (a e^2-b d e+c d^2\right )^3}-\frac{(2 c d-b e) \left (-c e (3 a e+b d)+b^2 e^2+c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\sqrt{b^2-4 a c} \left (a e^2-b d e+c d^2\right )^3}-\frac{e (2 c d-b e)}{(d+e x) \left (a e^2-b d e+c d^2\right )^2}-\frac{e}{2 (d+e x)^2 \left (a e^2-b d e+c d^2\right )}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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)^3 \left (a+b x+c x^2\right )} \, dx &=-\frac{e}{2 \left (c d^2-b d e+a e^2\right ) (d+e x)^2}+\frac{\int \frac{c d-b e-c e x}{(d+e x)^2 \left (a+b x+c x^2\right )} \, dx}{c d^2-b d e+a e^2}\\ &=-\frac{e}{2 \left (c d^2-b d e+a e^2\right ) (d+e x)^2}+\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)^2}+\frac{e^2 \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )}{\left (c d^2-b d e+a e^2\right )^2 (d+e x)}+\frac{c^3 d^3-b^3 e^3-3 c^2 d e (b d+a e)+b c e^2 (3 b d+2 a e)-c e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right ) x}{\left (c d^2-b d e+a e^2\right )^2 \left (a+b x+c x^2\right )}\right ) \, dx}{c d^2-b d e+a e^2}\\ &=-\frac{e}{2 \left (c d^2-b d e+a e^2\right ) (d+e x)^2}-\frac{e (2 c d-b e)}{\left (c d^2-b d e+a e^2\right )^2 (d+e x)}+\frac{e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right ) \log (d+e x)}{\left (c d^2-b d e+a e^2\right )^3}+\frac{\int \frac{c^3 d^3-b^3 e^3-3 c^2 d e (b d+a e)+b c e^2 (3 b d+2 a e)-c e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right ) x}{a+b x+c x^2} \, dx}{\left (c d^2-b d e+a e^2\right )^3}\\ &=-\frac{e}{2 \left (c d^2-b d e+a e^2\right ) (d+e x)^2}-\frac{e (2 c d-b e)}{\left (c d^2-b d e+a e^2\right )^2 (d+e x)}+\frac{e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right ) \log (d+e x)}{\left (c d^2-b d e+a e^2\right )^3}-\frac{\left (e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )\right ) \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 )^3}+\frac{\left ((2 c d-b e) \left (c^2 d^2+b^2 e^2-c e (b d+3 a e)\right )\right ) \int \frac{1}{a+b x+c x^2} \, dx}{2 \left (c d^2-b d e+a e^2\right )^3}\\ &=-\frac{e}{2 \left (c d^2-b d e+a e^2\right ) (d+e x)^2}-\frac{e (2 c d-b e)}{\left (c d^2-b d e+a e^2\right )^2 (d+e x)}+\frac{e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right ) \log (d+e x)}{\left (c d^2-b d e+a e^2\right )^3}-\frac{e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right ) \log \left (a+b x+c x^2\right )}{2 \left (c d^2-b d e+a e^2\right )^3}-\frac{\left ((2 c d-b e) \left (c^2 d^2+b^2 e^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 (c d^2-b d e+a e^2\right )^3}\\ &=-\frac{e}{2 \left (c d^2-b d e+a e^2\right ) (d+e x)^2}-\frac{e (2 c d-b e)}{\left (c d^2-b d e+a e^2\right )^2 (d+e x)}-\frac{(2 c d-b e) \left (c^2 d^2+b^2 e^2-c e (b d+3 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 )^3}+\frac{e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right ) \log (d+e x)}{\left (c d^2-b d e+a e^2\right )^3}-\frac{e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right ) \log \left (a+b x+c x^2\right )}{2 \left (c d^2-b d e+a e^2\right )^3}\\ \end{align*}

Mathematica [A]  time = 0.346421, size = 272, normalized size = 1. $\frac{e \log (d+e x) \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right )}{\left (e (a e-b d)+c d^2\right )^3}+\frac{e \left (c e (a e+3 b d)-b^2 e^2-3 c^2 d^2\right ) \log (a+x (b+c x))}{2 \left (e (a e-b d)+c d^2\right )^3}+\frac{(b e-2 c d) \left (-c e (3 a e+b d)+b^2 e^2+c^2 d^2\right ) \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )}{\sqrt{4 a c-b^2} \left (e (b d-a e)-c d^2\right )^3}+\frac{e (b e-2 c d)}{(d+e x) \left (e (a e-b d)+c d^2\right )^2}-\frac{e}{2 (d+e x)^2 \left (e (a e-b d)+c d^2\right )}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.162, size = 719, normalized size = 2.6 \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)^3/(c*x^2+b*x+a),x)

[Out]

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

________________________________________________________________________________________

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)^3/(c*x^2+b*x+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [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)^3/(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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)**3/(c*x**2+b*x+a),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.11838, size = 801, normalized size = 2.94 \begin{align*} -\frac{{\left (3 \, c^{2} d^{2} e - 3 \, b c d e^{2} + b^{2} e^{3} - a c e^{3}\right )} \log \left (c x^{2} + b x + a\right )}{2 \,{\left (c^{3} d^{6} - 3 \, b c^{2} d^{5} e + 3 \, b^{2} c d^{4} e^{2} + 3 \, a c^{2} d^{4} e^{2} - b^{3} d^{3} e^{3} - 6 \, a b c d^{3} e^{3} + 3 \, a b^{2} d^{2} e^{4} + 3 \, a^{2} c d^{2} e^{4} - 3 \, a^{2} b d e^{5} + a^{3} e^{6}\right )}} + \frac{{\left (3 \, c^{2} d^{2} e^{2} - 3 \, b c d e^{3} + b^{2} e^{4} - a c e^{4}\right )} \log \left ({\left | x e + d \right |}\right )}{c^{3} d^{6} e - 3 \, b c^{2} d^{5} e^{2} + 3 \, b^{2} c d^{4} e^{3} + 3 \, a c^{2} d^{4} e^{3} - b^{3} d^{3} e^{4} - 6 \, a b c d^{3} e^{4} + 3 \, a b^{2} d^{2} e^{5} + 3 \, a^{2} c d^{2} e^{5} - 3 \, a^{2} b d e^{6} + a^{3} e^{7}} + \frac{{\left (2 \, c^{3} d^{3} - 3 \, b c^{2} d^{2} e + 3 \, b^{2} c d e^{2} - 6 \, a c^{2} d e^{2} - b^{3} e^{3} + 3 \, a b c e^{3}\right )} \arctan \left (\frac{2 \, c x + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{{\left (c^{3} d^{6} - 3 \, b c^{2} d^{5} e + 3 \, b^{2} c d^{4} e^{2} + 3 \, a c^{2} d^{4} e^{2} - b^{3} d^{3} e^{3} - 6 \, a b c d^{3} e^{3} + 3 \, a b^{2} d^{2} e^{4} + 3 \, a^{2} c d^{2} e^{4} - 3 \, a^{2} b d e^{5} + a^{3} e^{6}\right )} \sqrt{-b^{2} + 4 \, a c}} - \frac{5 \, c^{2} d^{4} e - 8 \, b c d^{3} e^{2} + 3 \, b^{2} d^{2} e^{3} + 6 \, a c d^{2} e^{3} - 4 \, a b d e^{4} + a^{2} e^{5} + 2 \,{\left (2 \, c^{2} d^{3} e^{2} - 3 \, b c d^{2} e^{3} + b^{2} d e^{4} + 2 \, a c d e^{4} - a b e^{5}\right )} x}{2 \,{\left (c d^{2} - b d e + a e^{2}\right )}^{3}{\left (x e + d\right )}^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^3/(c*x^2+b*x+a),x, algorithm="giac")

[Out]

-1/2*(3*c^2*d^2*e - 3*b*c*d*e^2 + b^2*e^3 - a*c*e^3)*log(c*x^2 + b*x + a)/(c^3*d^6 - 3*b*c^2*d^5*e + 3*b^2*c*d
^4*e^2 + 3*a*c^2*d^4*e^2 - b^3*d^3*e^3 - 6*a*b*c*d^3*e^3 + 3*a*b^2*d^2*e^4 + 3*a^2*c*d^2*e^4 - 3*a^2*b*d*e^5 +
a^3*e^6) + (3*c^2*d^2*e^2 - 3*b*c*d*e^3 + b^2*e^4 - a*c*e^4)*log(abs(x*e + d))/(c^3*d^6*e - 3*b*c^2*d^5*e^2 +
3*b^2*c*d^4*e^3 + 3*a*c^2*d^4*e^3 - b^3*d^3*e^4 - 6*a*b*c*d^3*e^4 + 3*a*b^2*d^2*e^5 + 3*a^2*c*d^2*e^5 - 3*a^2
*b*d*e^6 + a^3*e^7) + (2*c^3*d^3 - 3*b*c^2*d^2*e + 3*b^2*c*d*e^2 - 6*a*c^2*d*e^2 - b^3*e^3 + 3*a*b*c*e^3)*arct
an((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((c^3*d^6 - 3*b*c^2*d^5*e + 3*b^2*c*d^4*e^2 + 3*a*c^2*d^4*e^2 - b^3*d^3*e^3
- 6*a*b*c*d^3*e^3 + 3*a*b^2*d^2*e^4 + 3*a^2*c*d^2*e^4 - 3*a^2*b*d*e^5 + a^3*e^6)*sqrt(-b^2 + 4*a*c)) - 1/2*(5
*c^2*d^4*e - 8*b*c*d^3*e^2 + 3*b^2*d^2*e^3 + 6*a*c*d^2*e^3 - 4*a*b*d*e^4 + a^2*e^5 + 2*(2*c^2*d^3*e^2 - 3*b*c*
d^2*e^3 + b^2*d*e^4 + 2*a*c*d*e^4 - a*b*e^5)*x)/((c*d^2 - b*d*e + a*e^2)^3*(x*e + d)^2)