### 3.1511 $$\int \frac{1}{(d+e x) (a^2+2 a b x+b^2 x^2)} \, dx$$

Optimal. Leaf size=57 $-\frac{1}{(a+b x) (b d-a e)}-\frac{e \log (a+b x)}{(b d-a e)^2}+\frac{e \log (d+e x)}{(b d-a e)^2}$

[Out]

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

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Rubi [A]  time = 0.0351234, antiderivative size = 57, 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} $-\frac{1}{(a+b x) (b d-a e)}-\frac{e \log (a+b x)}{(b d-a e)^2}+\frac{e \log (d+e x)}{(b d-a e)^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0249867, size = 53, normalized size = 0.93 $\frac{e (a+b x) \log (d+e x)-e (a+b x) \log (a+b x)+a e-b d}{(a+b x) (b d-a e)^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.057, size = 57, normalized size = 1. \begin{align*}{\frac{e\ln \left ( ex+d \right ) }{ \left ( ae-bd \right ) ^{2}}}+{\frac{1}{ \left ( ae-bd \right ) \left ( bx+a \right ) }}-{\frac{e\ln \left ( bx+a \right ) }{ \left ( ae-bd \right ) ^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.11861, size = 124, normalized size = 2.18 \begin{align*} -\frac{e \log \left (b x + a\right )}{b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}} + \frac{e \log \left (e x + d\right )}{b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}} - \frac{1}{a b d - a^{2} e +{\left (b^{2} d - a b e\right )} x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.76115, size = 200, normalized size = 3.51 \begin{align*} -\frac{b d - a e +{\left (b e x + a e\right )} \log \left (b x + a\right ) -{\left (b e x + a e\right )} \log \left (e x + d\right )}{a b^{2} d^{2} - 2 \, a^{2} b d e + a^{3} e^{2} +{\left (b^{3} d^{2} - 2 \, a b^{2} d e + a^{2} b e^{2}\right )} x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [B]  time = 0.829572, size = 233, normalized size = 4.09 \begin{align*} \frac{e \log{\left (x + \frac{- \frac{a^{3} e^{4}}{\left (a e - b d\right )^{2}} + \frac{3 a^{2} b d e^{3}}{\left (a e - b d\right )^{2}} - \frac{3 a b^{2} d^{2} e^{2}}{\left (a e - b d\right )^{2}} + a e^{2} + \frac{b^{3} d^{3} e}{\left (a e - b d\right )^{2}} + b d e}{2 b e^{2}} \right )}}{\left (a e - b d\right )^{2}} - \frac{e \log{\left (x + \frac{\frac{a^{3} e^{4}}{\left (a e - b d\right )^{2}} - \frac{3 a^{2} b d e^{3}}{\left (a e - b d\right )^{2}} + \frac{3 a b^{2} d^{2} e^{2}}{\left (a e - b d\right )^{2}} + a e^{2} - \frac{b^{3} d^{3} e}{\left (a e - b d\right )^{2}} + b d e}{2 b e^{2}} \right )}}{\left (a e - b d\right )^{2}} + \frac{1}{a^{2} e - a b d + x \left (a b e - b^{2} d\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14648, size = 128, normalized size = 2.25 \begin{align*} -\frac{b e \log \left ({\left | b x + a \right |}\right )}{b^{3} d^{2} - 2 \, a b^{2} d e + a^{2} b e^{2}} + \frac{e^{2} \log \left ({\left | x e + d \right |}\right )}{b^{2} d^{2} e - 2 \, a b d e^{2} + a^{2} e^{3}} - \frac{1}{{\left (b d - a e\right )}{\left (b x + a\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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