### 3.272 $$\int \frac{1}{(b x+c x^2)^2} \, dx$$

Optimal. Leaf size=43 $-\frac{b+2 c x}{b^2 \left (b x+c x^2\right )}-\frac{2 c \log (x)}{b^3}+\frac{2 c \log (b+c x)}{b^3}$

[Out]

-((b + 2*c*x)/(b^2*(b*x + c*x^2))) - (2*c*Log[x])/b^3 + (2*c*Log[b + c*x])/b^3

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Rubi [A]  time = 0.009857, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 11, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.182, Rules used = {614, 615} $-\frac{b+2 c x}{b^2 \left (b x+c x^2\right )}-\frac{2 c \log (x)}{b^3}+\frac{2 c \log (b+c x)}{b^3}$

Antiderivative was successfully veriﬁed.

[In]

Int[(b*x + c*x^2)^(-2),x]

[Out]

-((b + 2*c*x)/(b^2*(b*x + c*x^2))) - (2*c*Log[x])/b^3 + (2*c*Log[b + c*x])/b^3

Rule 614

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^(p + 1))/((p +
1)*(b^2 - 4*a*c)), x] - Dist[(2*c*(2*p + 3))/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

Rule 615

Int[((b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[Log[x]/b, x] - Simp[Log[RemoveContent[b + c*x, x]]/b,
x] /; FreeQ[{b, c}, x]

Rubi steps

\begin{align*} \int \frac{1}{\left (b x+c x^2\right )^2} \, dx &=-\frac{b+2 c x}{b^2 \left (b x+c x^2\right )}-\frac{(2 c) \int \frac{1}{b x+c x^2} \, dx}{b^2}\\ &=-\frac{b+2 c x}{b^2 \left (b x+c x^2\right )}-\frac{2 c \log (x)}{b^3}+\frac{2 c \log (b+c x)}{b^3}\\ \end{align*}

Mathematica [A]  time = 0.0381159, size = 35, normalized size = 0.81 $-\frac{b \left (\frac{c}{b+c x}+\frac{1}{x}\right )-2 c \log (b+c x)+2 c \log (x)}{b^3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(b*x + c*x^2)^(-2),x]

[Out]

-((b*(x^(-1) + c/(b + c*x)) + 2*c*Log[x] - 2*c*Log[b + c*x])/b^3)

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Maple [A]  time = 0.056, size = 43, normalized size = 1. \begin{align*} -{\frac{1}{{b}^{2}x}}-2\,{\frac{c\ln \left ( x \right ) }{{b}^{3}}}-{\frac{c}{{b}^{2} \left ( cx+b \right ) }}+2\,{\frac{c\ln \left ( cx+b \right ) }{{b}^{3}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x^2+b*x)^2,x)

[Out]

-1/b^2/x-2*c*ln(x)/b^3-c/b^2/(c*x+b)+2*c*ln(c*x+b)/b^3

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Maxima [A]  time = 1.13675, size = 61, normalized size = 1.42 \begin{align*} -\frac{2 \, c x + b}{b^{2} c x^{2} + b^{3} x} + \frac{2 \, c \log \left (c x + b\right )}{b^{3}} - \frac{2 \, c \log \left (x\right )}{b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*c*x + b)/(b^2*c*x^2 + b^3*x) + 2*c*log(c*x + b)/b^3 - 2*c*log(x)/b^3

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Fricas [A]  time = 1.71884, size = 138, normalized size = 3.21 \begin{align*} -\frac{2 \, b c x + b^{2} - 2 \,{\left (c^{2} x^{2} + b c x\right )} \log \left (c x + b\right ) + 2 \,{\left (c^{2} x^{2} + b c x\right )} \log \left (x\right )}{b^{3} c x^{2} + b^{4} x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^2+b*x)^2,x, algorithm="fricas")

[Out]

-(2*b*c*x + b^2 - 2*(c^2*x^2 + b*c*x)*log(c*x + b) + 2*(c^2*x^2 + b*c*x)*log(x))/(b^3*c*x^2 + b^4*x)

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Sympy [A]  time = 1.22442, size = 36, normalized size = 0.84 \begin{align*} - \frac{b + 2 c x}{b^{3} x + b^{2} c x^{2}} + \frac{2 c \left (- \log{\left (x \right )} + \log{\left (\frac{b}{c} + x \right )}\right )}{b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x**2+b*x)**2,x)

[Out]

-(b + 2*c*x)/(b**3*x + b**2*c*x**2) + 2*c*(-log(x) + log(b/c + x))/b**3

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Giac [A]  time = 1.3025, size = 61, normalized size = 1.42 \begin{align*} \frac{2 \, c \log \left ({\left | c x + b \right |}\right )}{b^{3}} - \frac{2 \, c \log \left ({\left | x \right |}\right )}{b^{3}} - \frac{2 \, c x + b}{{\left (c x^{2} + b x\right )} b^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*c*log(abs(c*x + b))/b^3 - 2*c*log(abs(x))/b^3 - (2*c*x + b)/((c*x^2 + b*x)*b^2)