### 3.1512 $$\int \frac{1}{(d+e x)^2 (a^2+2 a b x+b^2 x^2)} \, dx$$

Optimal. Leaf size=81 $-\frac{b}{(a+b x) (b d-a e)^2}-\frac{e}{(d+e x) (b d-a e)^2}-\frac{2 b e \log (a+b x)}{(b d-a e)^3}+\frac{2 b e \log (d+e x)}{(b d-a e)^3}$

[Out]

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

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Rubi [A]  time = 0.0533884, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 26, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.077, Rules used = {27, 44} $-\frac{b}{(a+b x) (b d-a e)^2}-\frac{e}{(d+e x) (b d-a e)^2}-\frac{2 b e \log (a+b x)}{(b d-a e)^3}+\frac{2 b e \log (d+e x)}{(b d-a e)^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0681853, size = 66, normalized size = 0.81 $\frac{\frac{b (a e-b d)}{a+b x}+\frac{e (a e-b d)}{d+e x}-2 b e \log (a+b x)+2 b e \log (d+e x)}{(b d-a e)^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.052, size = 82, normalized size = 1. \begin{align*} -{\frac{e}{ \left ( ae-bd \right ) ^{2} \left ( ex+d \right ) }}-2\,{\frac{be\ln \left ( ex+d \right ) }{ \left ( ae-bd \right ) ^{3}}}-{\frac{b}{ \left ( ae-bd \right ) ^{2} \left ( bx+a \right ) }}+2\,{\frac{be\ln \left ( bx+a \right ) }{ \left ( ae-bd \right ) ^{3}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-e/(a*e-b*d)^2/(e*x+d)-2*e/(a*e-b*d)^3*b*ln(e*x+d)-b/(a*e-b*d)^2/(b*x+a)+2*e/(a*e-b*d)^3*b*ln(b*x+a)

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Maxima [B]  time = 1.02128, size = 281, normalized size = 3.47 \begin{align*} -\frac{2 \, b e \log \left (b x + a\right )}{b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}} + \frac{2 \, b e \log \left (e x + d\right )}{b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}} - \frac{2 \, b e x + b d + a e}{a b^{2} d^{3} - 2 \, a^{2} b d^{2} e + a^{3} d e^{2} +{\left (b^{3} d^{2} e - 2 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x^{2} +{\left (b^{3} d^{3} - a b^{2} d^{2} e - a^{2} b d e^{2} + a^{3} e^{3}\right )} x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*b*e*log(b*x + a)/(b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3) + 2*b*e*log(e*x + d)/(b^3*d^3 - 3*a*b^
2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3) - (2*b*e*x + b*d + a*e)/(a*b^2*d^3 - 2*a^2*b*d^2*e + a^3*d*e^2 + (b^3*d^2*e
- 2*a*b^2*d*e^2 + a^2*b*e^3)*x^2 + (b^3*d^3 - a*b^2*d^2*e - a^2*b*d*e^2 + a^3*e^3)*x)

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Fricas [B]  time = 1.8012, size = 486, normalized size = 6. \begin{align*} -\frac{b^{2} d^{2} - a^{2} e^{2} + 2 \,{\left (b^{2} d e - a b e^{2}\right )} x + 2 \,{\left (b^{2} e^{2} x^{2} + a b d e +{\left (b^{2} d e + a b e^{2}\right )} x\right )} \log \left (b x + a\right ) - 2 \,{\left (b^{2} e^{2} x^{2} + a b d e +{\left (b^{2} d e + a b e^{2}\right )} x\right )} \log \left (e x + d\right )}{a b^{3} d^{4} - 3 \, a^{2} b^{2} d^{3} e + 3 \, a^{3} b d^{2} e^{2} - a^{4} d e^{3} +{\left (b^{4} d^{3} e - 3 \, a b^{3} d^{2} e^{2} + 3 \, a^{2} b^{2} d e^{3} - a^{3} b e^{4}\right )} x^{2} +{\left (b^{4} d^{4} - 2 \, a b^{3} d^{3} e + 2 \, a^{3} b d e^{3} - a^{4} e^{4}\right )} x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 1.32834, size = 405, normalized size = 5. \begin{align*} - \frac{2 b e \log{\left (x + \frac{- \frac{2 a^{4} b e^{5}}{\left (a e - b d\right )^{3}} + \frac{8 a^{3} b^{2} d e^{4}}{\left (a e - b d\right )^{3}} - \frac{12 a^{2} b^{3} d^{2} e^{3}}{\left (a e - b d\right )^{3}} + \frac{8 a b^{4} d^{3} e^{2}}{\left (a e - b d\right )^{3}} + 2 a b e^{2} - \frac{2 b^{5} d^{4} e}{\left (a e - b d\right )^{3}} + 2 b^{2} d e}{4 b^{2} e^{2}} \right )}}{\left (a e - b d\right )^{3}} + \frac{2 b e \log{\left (x + \frac{\frac{2 a^{4} b e^{5}}{\left (a e - b d\right )^{3}} - \frac{8 a^{3} b^{2} d e^{4}}{\left (a e - b d\right )^{3}} + \frac{12 a^{2} b^{3} d^{2} e^{3}}{\left (a e - b d\right )^{3}} - \frac{8 a b^{4} d^{3} e^{2}}{\left (a e - b d\right )^{3}} + 2 a b e^{2} + \frac{2 b^{5} d^{4} e}{\left (a e - b d\right )^{3}} + 2 b^{2} d e}{4 b^{2} e^{2}} \right )}}{\left (a e - b d\right )^{3}} - \frac{a e + b d + 2 b e x}{a^{3} d e^{2} - 2 a^{2} b d^{2} e + a b^{2} d^{3} + x^{2} \left (a^{2} b e^{3} - 2 a b^{2} d e^{2} + b^{3} d^{2} e\right ) + x \left (a^{3} e^{3} - a^{2} b d e^{2} - a b^{2} d^{2} e + b^{3} d^{3}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*b*e*log(x + (-2*a**4*b*e**5/(a*e - b*d)**3 + 8*a**3*b**2*d*e**4/(a*e - b*d)**3 - 12*a**2*b**3*d**2*e**3/(a*
e - b*d)**3 + 8*a*b**4*d**3*e**2/(a*e - b*d)**3 + 2*a*b*e**2 - 2*b**5*d**4*e/(a*e - b*d)**3 + 2*b**2*d*e)/(4*b
**2*e**2))/(a*e - b*d)**3 + 2*b*e*log(x + (2*a**4*b*e**5/(a*e - b*d)**3 - 8*a**3*b**2*d*e**4/(a*e - b*d)**3 +
12*a**2*b**3*d**2*e**3/(a*e - b*d)**3 - 8*a*b**4*d**3*e**2/(a*e - b*d)**3 + 2*a*b*e**2 + 2*b**5*d**4*e/(a*e -
b*d)**3 + 2*b**2*d*e)/(4*b**2*e**2))/(a*e - b*d)**3 - (a*e + b*d + 2*b*e*x)/(a**3*d*e**2 - 2*a**2*b*d**2*e + a
*b**2*d**3 + x**2*(a**2*b*e**3 - 2*a*b**2*d*e**2 + b**3*d**2*e) + x*(a**3*e**3 - a**2*b*d*e**2 - a*b**2*d**2*e
+ b**3*d**3))

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Giac [A]  time = 1.21836, size = 211, normalized size = 2.6 \begin{align*} -\frac{2 \, b e^{2} \log \left ({\left | b - \frac{b d}{x e + d} + \frac{a e}{x e + d} \right |}\right )}{b^{3} d^{3} e - 3 \, a b^{2} d^{2} e^{2} + 3 \, a^{2} b d e^{3} - a^{3} e^{4}} - \frac{e^{3}}{{\left (b^{2} d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )}{\left (x e + d\right )}} - \frac{b^{2} e}{{\left (b d - a e\right )}^{3}{\left (b - \frac{b d}{x e + d} + \frac{a e}{x e + d}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*b*e^2*log(abs(b - b*d/(x*e + d) + a*e/(x*e + d)))/(b^3*d^3*e - 3*a*b^2*d^2*e^2 + 3*a^2*b*d*e^3 - a^3*e^4) -
e^3/((b^2*d^2*e^2 - 2*a*b*d*e^3 + a^2*e^4)*(x*e + d)) - b^2*e/((b*d - a*e)^3*(b - b*d/(x*e + d) + a*e/(x*e +
d)))