### 3.1508 $$\int \frac{(d+e x)^2}{a^2+2 a b x+b^2 x^2} \, dx$$

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

[Out]

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

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Rubi [A]  time = 0.0393365, antiderivative size = 51, 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, 43} $-\frac{(b d-a e)^2}{b^3 (a+b x)}+\frac{2 e (b d-a e) \log (a+b x)}{b^3}+\frac{e^2 x}{b^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0371345, size = 47, normalized size = 0.92 $\frac{-\frac{(b d-a e)^2}{a+b x}+2 e (b d-a e) \log (a+b x)+b e^2 x}{b^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.047, size = 86, normalized size = 1.7 \begin{align*}{\frac{{e}^{2}x}{{b}^{2}}}-2\,{\frac{{e}^{2}\ln \left ( bx+a \right ) a}{{b}^{3}}}+2\,{\frac{e\ln \left ( bx+a \right ) d}{{b}^{2}}}-{\frac{{a}^{2}{e}^{2}}{{b}^{3} \left ( bx+a \right ) }}+2\,{\frac{ade}{{b}^{2} \left ( bx+a \right ) }}-{\frac{{d}^{2}}{b \left ( bx+a \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.13578, size = 90, normalized size = 1.76 \begin{align*} \frac{e^{2} x}{b^{2}} - \frac{b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}}{b^{4} x + a b^{3}} + \frac{2 \,{\left (b d e - a e^{2}\right )} \log \left (b x + a\right )}{b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.67671, size = 184, normalized size = 3.61 \begin{align*} \frac{b^{2} e^{2} x^{2} + a b e^{2} x - b^{2} d^{2} + 2 \, a b d e - a^{2} e^{2} + 2 \,{\left (a b d e - a^{2} e^{2} +{\left (b^{2} d e - a b e^{2}\right )} x\right )} \log \left (b x + a\right )}{b^{4} x + a b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.592445, size = 60, normalized size = 1.18 \begin{align*} - \frac{a^{2} e^{2} - 2 a b d e + b^{2} d^{2}}{a b^{3} + b^{4} x} + \frac{e^{2} x}{b^{2}} - \frac{2 e \left (a e - b d\right ) \log{\left (a + b x \right )}}{b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.24279, size = 86, normalized size = 1.69 \begin{align*} \frac{x e^{2}}{b^{2}} + \frac{2 \,{\left (b d e - a e^{2}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{3}} - \frac{b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}}{{\left (b x + a\right )} b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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