### 3.1458 $$\int \frac{a^2+2 a b x+b^2 x^2}{(d+e x)^3} \, dx$$

Optimal. Leaf size=59 $\frac{2 b (b d-a e)}{e^3 (d+e x)}-\frac{(b d-a e)^2}{2 e^3 (d+e x)^2}+\frac{b^2 \log (d+e x)}{e^3}$

[Out]

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

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Rubi [A]  time = 0.0379732, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.083, Rules used = {27, 43} $\frac{2 b (b d-a e)}{e^3 (d+e x)}-\frac{(b d-a e)^2}{2 e^3 (d+e x)^2}+\frac{b^2 \log (d+e x)}{e^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(b*d - a*e)^2/(2*e^3*(d + e*x)^2) + (2*b*(b*d - a*e))/(e^3*(d + e*x)) + (b^2*Log[d + e*x])/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 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0239186, size = 48, normalized size = 0.81 $\frac{\frac{(b d-a e) (a e+3 b d+4 b e x)}{(d+e x)^2}+2 b^2 \log (d+e x)}{2 e^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.046, size = 92, normalized size = 1.6 \begin{align*} -{\frac{{a}^{2}}{2\,e \left ( ex+d \right ) ^{2}}}+{\frac{abd}{{e}^{2} \left ( ex+d \right ) ^{2}}}-{\frac{{b}^{2}{d}^{2}}{2\,{e}^{3} \left ( ex+d \right ) ^{2}}}+{\frac{{b}^{2}\ln \left ( ex+d \right ) }{{e}^{3}}}-2\,{\frac{ab}{{e}^{2} \left ( ex+d \right ) }}+2\,{\frac{{b}^{2}d}{{e}^{3} \left ( ex+d \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^2+2*a*b*x+a^2)/(e*x+d)^3,x)

[Out]

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

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Maxima [A]  time = 1.14444, size = 108, normalized size = 1.83 \begin{align*} \frac{3 \, b^{2} d^{2} - 2 \, a b d e - a^{2} e^{2} + 4 \,{\left (b^{2} d e - a b e^{2}\right )} x}{2 \,{\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} + \frac{b^{2} \log \left (e x + d\right )}{e^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(3*b^2*d^2 - 2*a*b*d*e - a^2*e^2 + 4*(b^2*d*e - a*b*e^2)*x)/(e^5*x^2 + 2*d*e^4*x + d^2*e^3) + b^2*log(e*x
+ d)/e^3

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Fricas [A]  time = 1.6027, size = 205, normalized size = 3.47 \begin{align*} \frac{3 \, b^{2} d^{2} - 2 \, a b d e - a^{2} e^{2} + 4 \,{\left (b^{2} d e - a b e^{2}\right )} x + 2 \,{\left (b^{2} e^{2} x^{2} + 2 \, b^{2} d e x + b^{2} d^{2}\right )} \log \left (e x + d\right )}{2 \,{\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(3*b^2*d^2 - 2*a*b*d*e - a^2*e^2 + 4*(b^2*d*e - a*b*e^2)*x + 2*(b^2*e^2*x^2 + 2*b^2*d*e*x + b^2*d^2)*log(e
*x + d))/(e^5*x^2 + 2*d*e^4*x + d^2*e^3)

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Sympy [A]  time = 0.795259, size = 80, normalized size = 1.36 \begin{align*} \frac{b^{2} \log{\left (d + e x \right )}}{e^{3}} - \frac{a^{2} e^{2} + 2 a b d e - 3 b^{2} d^{2} + x \left (4 a b e^{2} - 4 b^{2} d e\right )}{2 d^{2} e^{3} + 4 d e^{4} x + 2 e^{5} x^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**2+2*a*b*x+a**2)/(e*x+d)**3,x)

[Out]

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

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Giac [A]  time = 1.19965, size = 93, normalized size = 1.58 \begin{align*} b^{2} e^{\left (-3\right )} \log \left ({\left | x e + d \right |}\right ) + \frac{{\left (4 \,{\left (b^{2} d - a b e\right )} x +{\left (3 \, b^{2} d^{2} - 2 \, a b d e - a^{2} e^{2}\right )} e^{\left (-1\right )}\right )} e^{\left (-2\right )}}{2 \,{\left (x e + d\right )}^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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