### 3.1509 $$\int \frac{d+e x}{a^2+2 a b x+b^2 x^2} \, dx$$

Optimal. Leaf size=32 $\frac{e \log (a+b x)}{b^2}-\frac{b d-a e}{b^2 (a+b x)}$

[Out]

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

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Rubi [A]  time = 0.0216175, antiderivative size = 32, 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} $\frac{e \log (a+b x)}{b^2}-\frac{b d-a e}{b^2 (a+b x)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0110032, size = 31, normalized size = 0.97 $\frac{a e-b d}{b^2 (a+b x)}+\frac{e \log (a+b x)}{b^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.043, size = 39, normalized size = 1.2 \begin{align*}{\frac{e\ln \left ( bx+a \right ) }{{b}^{2}}}+{\frac{ae}{{b}^{2} \left ( bx+a \right ) }}-{\frac{d}{b \left ( bx+a \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.15544, size = 47, normalized size = 1.47 \begin{align*} -\frac{b d - a e}{b^{3} x + a b^{2}} + \frac{e \log \left (b x + a\right )}{b^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.69713, size = 80, normalized size = 2.5 \begin{align*} -\frac{b d - a e -{\left (b e x + a e\right )} \log \left (b x + a\right )}{b^{3} x + a b^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.391281, size = 27, normalized size = 0.84 \begin{align*} \frac{a e - b d}{a b^{2} + b^{3} x} + \frac{e \log{\left (a + b x \right )}}{b^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.17031, size = 47, normalized size = 1.47 \begin{align*} \frac{e \log \left ({\left | b x + a \right |}\right )}{b^{2}} - \frac{b d - a e}{{\left (b x + a\right )} b^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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