### 3.1457 $$\int \frac{a^2+2 a b x+b^2 x^2}{(d+e x)^2} \, dx$$

Optimal. Leaf size=51 $-\frac{(b d-a e)^2}{e^3 (d+e x)}-\frac{2 b (b d-a e) \log (d+e x)}{e^3}+\frac{b^2 x}{e^2}$

[Out]

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

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Rubi [A]  time = 0.0477378, antiderivative size = 51, 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} $-\frac{(b d-a e)^2}{e^3 (d+e x)}-\frac{2 b (b d-a e) \log (d+e x)}{e^3}+\frac{b^2 x}{e^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0369414, size = 47, normalized size = 0.92 $\frac{-\frac{(b d-a e)^2}{d+e x}+2 b (a e-b d) \log (d+e x)+b^2 e x}{e^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.046, size = 86, normalized size = 1.7 \begin{align*}{\frac{{b}^{2}x}{{e}^{2}}}+2\,{\frac{b\ln \left ( ex+d \right ) a}{{e}^{2}}}-2\,{\frac{{b}^{2}\ln \left ( ex+d \right ) d}{{e}^{3}}}-{\frac{{a}^{2}}{e \left ( ex+d \right ) }}+2\,{\frac{abd}{{e}^{2} \left ( ex+d \right ) }}-{\frac{{b}^{2}{d}^{2}}{{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)^2,x)

[Out]

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

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Maxima [A]  time = 1.18164, size = 90, normalized size = 1.76 \begin{align*} \frac{b^{2} x}{e^{2}} - \frac{b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}}{e^{4} x + d e^{3}} - \frac{2 \,{\left (b^{2} d - a b e\right )} \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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.74924, size = 184, normalized size = 3.61 \begin{align*} \frac{b^{2} e^{2} x^{2} + b^{2} d e x - b^{2} d^{2} + 2 \, a b d e - a^{2} e^{2} - 2 \,{\left (b^{2} d^{2} - a b d e +{\left (b^{2} d e - a b e^{2}\right )} x\right )} \log \left (e x + d\right )}{e^{4} x + d 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)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.553262, size = 60, normalized size = 1.18 \begin{align*} \frac{b^{2} x}{e^{2}} + \frac{2 b \left (a e - b d\right ) \log{\left (d + e x \right )}}{e^{3}} - \frac{a^{2} e^{2} - 2 a b d e + b^{2} d^{2}}{d e^{3} + e^{4} x} \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)**2,x)

[Out]

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

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Giac [B]  time = 1.14451, size = 150, normalized size = 2.94 \begin{align*} -2 \,{\left (e^{\left (-1\right )} \log \left (\frac{{\left | x e + d \right |} e^{\left (-1\right )}}{{\left (x e + d\right )}^{2}}\right ) - \frac{d e^{\left (-1\right )}}{x e + d}\right )} a b e^{\left (-1\right )} +{\left (2 \, d e^{\left (-3\right )} \log \left (\frac{{\left | x e + d \right |} e^{\left (-1\right )}}{{\left (x e + d\right )}^{2}}\right ) +{\left (x e + d\right )} e^{\left (-3\right )} - \frac{d^{2} e^{\left (-3\right )}}{x e + d}\right )} b^{2} - \frac{a^{2} e^{\left (-1\right )}}{x e + d} \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)^2,x, algorithm="giac")

[Out]

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